Questão 7 (Física)

**Analyzing physics equations** I need to scrutinize the physics surrounding the energy ratio and variations. P2's spring starts relaxed, but the gravitational force along BC is 2N, while F is 0.18N, indicating it's not in equilibrium initially. The stiff spring oscillates, and at its maximum, I estimate Fspring is equal to Felec plus mg*sin at that point. The oscillator's dynamics involve slow variation and oscillation that complicate force balance. When they expect equilibrium, the actual forces need careful evaluation, especially regarding gravity's influence on P2.**Evaluating expected ratios** I need to derive the expected values for the forces and energy ratios related to elastic and gravitational variations. The statement indicates that they're balancing forces on P2, which has a stiff spring constant, rather than focusing on dynamics. It emphasizes that the spring is nearly compressed, which implies an approximate equilibrium for P2. I also need to consider that for both particles, a maximum approach can lead to the same velocity around zero, and this might influence their behavior during oscillation.**Parsing figure details** I need to clarify the dimensions in figure 1, focusing on the length of BC. The orientation of the top dodecagon starts with AB being horizontal, then BC has a slope of 30 degrees. I noted that q2 is positive, specifically set at +1 µC to ensure validity. It might be helpful to mention this configuration, such as saying, "Taking \(q_2=+1\,\mu C\), as indicated." I just want to make sure I'm approaching the calculations accurately.

**Interpreting dodecagon geometry** I need to solve and interpret the geometry of a regular dodecagon, specifically the B-C side. I think P1 starts at B and P2 springs at C, both sliding along segment BC. The initial distance d0 is 1.0 m, and the side length of the dodecagon could also be d0. It seems like P1 moves toward P2, which is roughly fixed due to the high k spring. So, as P1 approaches P2, the spring compresses. I'm piecing this together!**Understanding particle positioning** In Figure 2, P1 starts at point B and P2 is near C. The spring is relaxed at length L0, meaning P2 is initially L0 away from C, while P1 is at B. Since the distance d0 between the particles is 1.0 m, it suggests that the length of segment BC is d0 plus L0, which totals 1.15 m. As P1 moves right, the repulsive force pushes P2 toward C, compressing the spring against the fixed stop. I've got to fully grasp this setup!**Analyzing maximum proximity and energy transfer** At maximum proximity, P1 is transferring kinetic energy into electric potential and spring elastic energy while P2 is mostly fixed. I need to solve for the maximum separation shrink, which can be expressed as x = d0 - r. It looks like we also have to consider the gravitational energy due to the slope along BC, given the dodecagon’s orientation, and then account for the electrostatic and elastic spring energies. Since P1 is sliding downhill along BC, this could impact the energy calculations.**Analyzing spring dynamics** I’m working through how an infinite stiff spring affects the P2 position, which has minimal movement. Even with high spring constant k and tiny displacement, the potential energy remains finite. I’m considering the energy balance at the turning point, combining gravitational, electric potential, and spring energy. If k is large, I wonder how the displacement relates to the maximum force applied. At the closest position, both P1 and P2 likely stop momentarily, leading to interesting dynamics!**Calculating spring constant** I need to figure out the spring constant k, which is described as "very high." To simplify, I should consider the geometric and energy aspects of the mechanics. So, P1 and P2 are sliding along BC; P1 starts at B, while P2 is anchored at a relaxed spring position. As P1 moves downhill due to gravity and repulsion, it compresses the spring when P2 moves towards C. Given the stiffness, P2's movement can be neglected for P1's calculations!**Analyzing forces and ratios** I'm breaking down the relationship between force, Coulomb's law, and gravitational components. I need to compare spring potential energy with gravitational energy variation, exploring the ratios involved. My calculations suggest a significant relationship based on certain assumptions about forces and angles. I recognize the importance of understanding the geometry and forces at play, especially when examining particle interactions after rotation. Understanding these details is crucial for determining the velocity at a specific point in the scenario.**Calculating velocity after rotation** I'm considering the scenario in Fig 2 after it rotates to Fig 3. It asks for the velocity of P2 when it reaches a vertex of the dodecagon. In this rotated state, P2 starts horizontally on segment BC and experiences an electric force from P1, pushing it to the right. With no friction or gravity acting on this horizontal movement, P2 moves towards vertex C. I need to calculate its speed at that point using conservation of momentum and potential energy, as both particles interact freely now.**Considering particle velocities at rotation** I'm thinking about the moment of rotation when the particles reach their maximum proximity. I presume their velocities are effectively zero at this point since they’ve just rotated into position. This makes sense because, as they come closer, their motion might slow down until the rotation stabilizes. It’s important to keep this in mind when calculating any further movements or interactions after this critical moment in the scenario.**Assessing distance and momentum** I'm calculating the distance between points B and C, figuring out where P1 and P2 are located. It looks like P1 moves slightly left while P2 nudges right, creating a separation of .45. After a rotation, P2 moves to point C, while P1 maintains a bit of distance from B. I’m also considering conservation of momentum with equal masses, but I’m unsure about the initial velocities as I analyze their approach. The details are quite intricate!**Analyzing forces on a spring** I'm considering the net force equation, which combines electrostatic forces and gravity along the incline to equal the spring force. It seems that there might be a general expectation for force balance here. However, I'm also pondering how a dynamic spring particle behaves under the influence of forces while moving rapidly. It's a bit complex, but I'm curious about how these different forces interact in this scenario.**Evaluating energy dynamics** I'm considering energy accounting at maximum approach between two points, where coordinates shift slightly. I wonder how the spring's energy relates to force. I need to make equilibrium assumptions; P2 may not be at static equilibrium but has a large spring constant, which tracks forces. The quasi-static approximation might work because the spring constant is high and displacement is small, but oscillation periods are short, meaning it’s not truly quasi-static. The gravitational force complicates things as well.**Analyzing spring energy and forces** I'm thinking about how to treat P2 as essentially fixed while considering its geometry and mass, which allows me to ignore its gravitational dynamics. However, this leads to a question: how can I derive the energy ratio due to gravitational variation from tiny displacement? I might use a balance of spring forces for this approximation. The ratio can come from energy transformed via forces, leading to a calculated ratio of around 1.1, factoring in spring energy considerations.**Analyzing momentum and energy** I’m thinking about how, for equal mass particles influenced only by internal forces, the momentum remains zero if they start at rest. In that case, their velocities go in opposite directions. The potential energy changes, and the distance increases based on the displacement of P2. When P2 reaches vertex C, the momentum and center of mass are conserved, maintaining the same magnitude. I'm noting that the final distance involves a slight adjustment due to the compression factor.**Exploring spring compression** I’m thinking about spring compression and how it's defined. It seems like when P2 moves towards C, the spring is initially relaxed at position C-L0. Then, as P2 compresses the spring, it shifts from that relaxed state. I need to clarify what happens during that compression phase. It's interesting to explore these dynamics and ensure I understand how the forces interact when the spring is being compressed. This could be a fun concept to dive deeper into!**Considering force balance** I'm thinking about how P2 might be quasi-equilibrated at maximum approach, even if P1 stops. The concept of equilibrium here is interesting because P2 seems to maintain a virtually unchanged position, which implies that the change in gravitational potential energy (ΔUg2) is nearly zero. I wonder if both energies being small allows for a finite ratio with force balance. It’s quite a puzzle!**Evaluating spring compression** I'm considering part c after rotation, focusing on P2's initial coordinates relative to C. There's spring compression represented as δ, meaning the distance to C is L0 - δ. With the spring removed, I have to think about the separation: final distance r_f is roughly r + 2(L0 - δ), which is around 0.75. But I need to remember that P2 had nonzero velocity when it was removed. It feels so tricky, assuming it's negligible for the quasi-static approximation!**Considering energy dynamics** I'm thinking about how at very high k during late maximum, the acceleration dynamics might not be negligible. Maybe the test expects a ratio that involves balance. I could set up an assumption for estimation. I might use Hooke's law and approximate balance here, which seems likely. For instance, I can calculate energy changes, looking at how potential energy is lost versus electric energy that rises and factoring in the coefficient.

I'm setting up the geometry: a regular dodecagon with AB horizontal, BC sloped at 30°, P1 starting at rest at B and sliding down to C, where a spring with relaxed length 0.15 m connects to P2 which can also slide along BC. For part (a), I use energy conservation between P1's initial position (1 m from P2) and closest approach distance d, balancing gravitational potential energy loss against electrostatic repulsion. Solving the equation gives d = 0.45 m, and I note I should also check whether P2's own motion (gravity plus repulsion against the stiff spring) needs separate consideration. For part (b), I'm comparing the spring's elastic potential energy to the change in P2's gravitational potential energy at maximum compression, computing the spring force from equilibrium between gravity and electrostatic repulsion, then taking the ratio of the two energies — getting approximately 1.61. Trying energy balance instead: if the spring stores (1/2)kx² equal to work done by gravity plus electric force, I get inconsistent results compared to the equilibrium approach — the discrepancy must come from P2's oscillation kinetic energy. So I reconsider: if the electric force builds up gradually, P2 adiabatically follows equilibrium, meaning the spring starts relaxed with P2 at rest before anything happens. Since the problem asks for an approximate ratio, I'll go with the quasi-static estimate: ratio = (mg sin30 + Fe)/(2 mg sin30) = 6.44/4 ≈ 1.61, which seems like the most defensible answer given the setup. Now moving to part (c): after rotating the system 30° counterclockwise so BC is horizontal and removing the spring, the particles P1 and P2 are at rest, 0.45 m apart along the side (BC = 1.15 m total), with P2 at 0.15 m from C and P1 at 0.55 m from B. Since the masses are equal and the system starts from rest, momentum conservation means they'll move with equal and opposite velocities, and I need to figure out which particle reaches its corner first — P2 is closer to C at 0.15 m, so it likely arrives there first. At that moment the separation between the two particles has grown to 0.75 m, so applying energy conservation with Coulomb's constant times the charges gives me the kinetic energy gained: working through the numbers, I get 0.8 J total, which splits evenly since the masses and speeds are equal, leading to v² = 2 and a clean final answer of v = √2 m/s for each particle.

I'm working out the dodecagon geometry: AB horizontal, BC angled 30° below due to the exterior angle, with P1 starting at B a distance 1.0 m from P2, and the spring (0.15 m) setting the side length at 1.15 m. Now I'm setting up the energy equation as P1 slides down toward P2 along BC. Computing kq1q2 = 0.9 N·m² and mg sin30 = 2 N, I solve the energy balance equation and get d = 0.45 m as the physical closest-approach distance (discarding the trivial root d=1). Then for part b, I find the Coulomb force at that separation (4.444 N), add the gravity component along the incline (2 N) to get total spring force 6.444 N, and I'm now checking the ratio between the spring's elastic energy and the change in P2's gravitational potential energy, getting roughly 1.61, though I'm second-guessing whether the spring starts relaxed or already under tension. After rotating BC counterclockwise 30°, it becomes horizontal, with B left, C right, P1 near B at 0.55 m and P2 near C at 0.15 m, spring removed. Since both particles repel with no friction and equal masses, momentum conservation means they move with equal and opposite velocities, covering equal distances — so P2 reaches C first after traveling only 0.15 m. Confirming P2 is indeed at rest at this moment, so the energy accounting holds.

I'm setting up the geometry of the regular dodecagon, noting the 150° interior angle and the 30° incline of side BC relative to AB, then working out the initial setup with P1 released from rest at B, a distance of 1.0 m from P2 near the relaxed spring end at C with natural length 0.15 m. Using energy conservation along the incline for P1, factoring gravity component (mg sin30° = 2 N) and the Coulomb constant kq1q2 = 0.9 N·m², I set up the quadratic 2d² - 2.9d + 0.9 = 0, solving to get d = 1 or d = 0.45 m, taking the physically meaningful closest approach at d = 0.45 m. For part b, at this closest approach both charges are momentarily at rest, so I balance forces on P2 along the incline: gravity (2 N) plus Coulomb repulsion (0.9/0.45² ≈ 4.444 N) pushing P2 down the incline must be countered by the spring force pushing up, giving a spring force of about 6.444 N, which I'll use to find the spring constant. With a very stiff spring, the compression x is tiny, so I compare the elastic energy (F²/2K) to the change in gravitational potential energy of P2 (2x), getting a ratio of F/4 ≈ 1.61 — confirming the elastic energy contribution is small but calculable. For part c, I'm now visualizing the setup after the system rotates 30° anticlockwise so BC lies horizontal, with the spring removed and the charges at rest 0.45 m apart on a 1.15 m segment, then tracking how P1 and P2 move apart due to repulsion from their horizontal starting positions. I use conservation of energy as P2 moves 0.15 m to reach point C and P1 moves an equal distance, so the separation grows from 0.45 m to 0.75 m. Working out the Coulomb potential energy difference and setting it equal to the total kinetic energy gained (since masses are equal, both move at the same speed), I solve for v² and get v = √2 ≈ 1.41 m/s, checking that P1 hasn't yet reached B before this occurs.

The problem is in Portuguese from an IME exam. A regular dodecagon has two positively charged particles sliding on segment BC. One sits at vertex B and the other sits at vertex C with a spring attached. They approach until maximum distance, then the system rotates 30° counterclockwise without adding energy. The spring is removed and the particles move freely. The particles have mass 0.4 kg each and charges +20 μC and +5 μC. The spring constant is very high, so P2's position can be treated as fixed during P1's motion. The initial distance is 1.0 m, the relaxed spring length is 0.15 m, and g = 10 m/s². The first question asks for the distance between the particles at maximum approach. The remaining questions ask for the ratio of elastic potential energy to gravitational potential energy variation at maximum approach, and for P2's velocity when it reaches a dodecagon vertex. The geometry is being mapped from the figures. The dodecagon's bottom side is horizontal. The external angle is 30°, so BC slopes downward from B toward C. The 30° counterclockwise rotation makes BC horizontal. This means BC was originally inclined at -30° to the horizontal. The original orientation of BC is being reconsidered. If the 30° rotation makes BC horizontal, then originally BC was rotated 30° clockwise from horizontal. The figure suggests A to B is nearly flat or slightly upward, then B to C slopes downward. The exact angle must be derived from the regular dodecagon geometry. The dodecagon is being mapped by tracing its sides. The bottom side is horizontal, and the top side is also horizontal opposite it. In Figure 2, AB appears horizontal while BC slopes downward. Both particles carry positive charges, so they repel each other. P1 starts at B and slides toward C, where P2 is held by a spring of relaxed length 0.15 m. The initial distance between them is 1.0 m, which implies the BC segment must be longer than that. The length of BC is 1.15 m. P2 sits at distance L0 = 0.15 m from C, so the total segment length is d0 + L0. P1 approaches P2 from B toward C. The repulsive force slows P1 and compresses the spring, pushing P2 toward C. The spring constant is very high, so the compression is tiny and P2 barely moves. P1 is repelled by P2 and also feels gravity. The spring compresses slightly, but because k is very high, the compression is tiny. P1 cannot actually pass through P2, so it stops at a small finite distance before they collide. The minimum distance is found by equating P1's initial kinetic energy to the sum of the electrostatic and gravitational potential energies at that point. Because P2 barely moves, its position can be treated as fixed. The spring only compresses if P2 moves. Because k is very high, P2 barely shifts, so the spring stays almost relaxed. The repulsive force from P1 pushes P2 slightly until the spring force balances it. This creates a brief oscillation or impulsive-like behavior. Treating P2 as fixed is an approximation for a very stiff spring. The dodecagon's internal angle is 150°, so each external turn is 30°. Traversing the boundary counterclockwise from the bottom side left to right, the next side turns downward at 30° below horizontal. After the rotation, the spring is removed at maximum approach. The particles then move freely on the horizontal segment BC with no gravity component along that direction. They repel each other. The elastic potential energy stored in the spring is released when it is removed. The spring is fully removed, so its stored energy is not returned to the particles. They are left at their velocities at maximum approach and then slide freely along the horizontal segment BC. They will eventually reach a vertex of the dodecagon. After the spring is removed, P1 and P2 repel each other along the horizontal segment BC. P1 moves toward B and P2 toward C. The question asks for P2's velocity when it reaches a vertex of the dodecagon. It is unclear whether the particles remain confined to BC or can continue onto adjacent sides. The particles are confined to segment BC. Because P2 was treated as fixed, both particles are momentarily at rest at the minimum distance. After the spring is removed, repulsion pushes them apart along the horizontal track. Equal masses mean they will gain equal speeds. The rotation occurs exactly when the particles reach maximum approach, and the spring is removed only after that rotation. This means the spring is fully compressed before it is detached. Part a is the distance at maximum approach before rotation. Part b requires estimating the ratio of elastic potential energy to the gravitational potential energy variation of P2. Because the spring constant is very high, the compression is small and P2 moves only slightly. The elastic energy is large but the height change is tiny. A proper approximation is needed. The two-body spring problem is being reconsidered. Because the spring constant is very large, P2 barely shifts and P1 almost stops relative to it. The minimum distance is therefore determined by energy conservation between the two particles, not just by treating P2 as fixed. At maximum approach, P1 stops and P2 remains fixed. Energy conservation then yields the minimum distance r_min. Because the spring is effectively inactive for P1 while P2 is held in place, only the electrostatic and gravitational potential energies need to be considered. A coordinate system is being set with P2 at the origin and P1 starting at x = 1.0 m. The height loss as P1 slides down the 30° incline is Δh = −s·sin30°. The initial electrostatic potential energy is Ke·q1·q2/1, and at minimum approach it becomes Ke·q1·q2/r_min. The energy conservation equation for part a is derived. The initial height of P1 is set to zero. The final height drop is (d0 − r_min) sinθ. The change in electrostatic energy is Ke·q1·q2(1/r_min − 1/d0). The change in gravitational energy is −m·g(d0 − r_min) sinθ. These balance because both particles are momentarily at rest. The product Ke·q1·q2 evaluates to 0.9. The energy equation is solved numerically. With m = 0.4 kg, g = 10 m/s², and θ = 30°, the minimum distance is r_min = 0.45 m. The algebra checks out: 0.9(1/0.45 − 1) = 2(1 − 0.45). The validity of the 30° angle is now being questioned. The exact value of 0.45 m is a direct consequence of the ratio Ke·q1·q2 divided by mg·sinθ. The cancellation of the (1−u) factor in the energy equation explains why this elegant result holds. The energy equation factors into (1−u)(0.9−2u)=0, so u=1 is an extraneous solution. The only physically relevant root is u=0.45, confirming the minimum distance. The effective potential energy is being examined to confirm whether the particles can actually reach the minimum distance. The total energy equals the initial value at r = 0.45, so the motion is energetically allowed. The question is whether a potential barrier prevents the approach or if the particles can simply slide down the incline. The effective potential energy is evaluated at r = 0.45 and r = 0.7. At r = 0.45 the total equals the initial value of 0.9, confirming the particles can reach that distance with zero velocity. At r = 0.7 the potential drops to about 0.6857, which is lower than the initial energy. This means the particles slide down the incline, pass through a minimum, and return to the same total energy at r = 0.45. The motion is therefore energetically allowed and the approach is physically valid. The net force at the minimum distance is repulsive. The electric force of about 4.44 N exceeds the 2 N component of gravity down the incline, so P1 turns back and stops. This confirms the approach is stable. The answer for part a is 0.45 m. Whether P2 can truly remain fixed is now being questioned. The 0.45 m result for part a is the requested approximation, since P2's slight displacement is ignored. For part b, P2 does move slightly during compression, so the ratio of elastic potential energy to P2's gravitational potential variation must be estimated with that motion included. Because the spring constant is very large, P2's displacement is tiny. The motion at maximum approach is being reconsidered: with an effectively infinite spring constant, P1 behaves as if P2 were fixed, yet the spring energy must still be estimated accurately. The relative velocity is zero at closest approach, so the spring compression and P2's position adjustment require a careful dynamic approximation. The spring constant is very large, so P2 oscillates rapidly with a tiny amplitude around its equilibrium position. Because the problem asks to ignore P2's positional variation, its motion is effectively frozen. The ratio of elastic potential energy to P2's gravitational potential change must therefore be estimated without assuming P2 is completely fixed. The high spring constant means the particles momentarily stop relative to each other at maximum compression. Because P2's positional variation is ignored, its motion is effectively frozen during this instant. The energy that would normally go into P2's gravitational potential and spring compression is therefore treated as negligible for the calculation. The forces at the minimum distance r = 0.45 m are being calculated to determine the spring compression and P2's small positional shift. The ratio of elastic potential energy to P2's gravitational potential change is derived using the spring compression δ = F/k. The result simplifies to F/(2mg sinθ). With the Coulomb force at minimum distance being 40/9 N and mg sinθ equal to 2 N, the ratio evaluates to approximately 1.11. The exact ratio is 10/9, not 40/36. The derivation is being questioned: it may describe the energies at the instant of maximum approach rather than the variation during the compression process. The problem asks for the ratio of elastic potential energy to P2's gravitational potential variation. The compression δ = F/k is valid under quasi-static equilibrium, where P2 balances spring, gravity, and Coulomb forces. The initial Coulomb force is 0.9 N. Gravity pulls P2 down the incline toward C, which may already compress the spring. The question is whether the spring is initially relaxed or already in equilibrium. P2 is initially at rest, but the net force down the incline is 2.9 N because the 0.9 N Coulomb repulsion and the 2 N gravity component both push it toward C. The spring is initially relaxed, so P2 would start moving and compressing. The high spring constant would then make it oscillate around a new equilibrium. The approximation of ignoring P2's positional variation means this initial motion is being set aside. The limit k→∞ means P2 is rigidly fixed in place. The spring force adjusts to whatever value is needed to keep P2 from moving. This makes the calculation for part a) exact in that limit. The minimum distance is exactly 0.45 m. The question now is how to estimate the elastic potential energy in this scenario. As the spring constant approaches infinity, the spring compression vanishes and both energies approach zero. The ratio of elastic potential energy to P2's gravitational potential variation therefore converges to a finite limit. This confirms the approximation is valid in the high-k limit. The spring constant is unknown, so the ratio can be found without it. A variable x is introduced for the spring compression, and the position of P1 is being derived from the total distance d0 minus the separation r plus the extra displacement x. The position of P1 is derived as s1 = −r + x, where x is the spring compression. The total descent of P1 down the incline is therefore d0 − r + x. Because x is small and r ≈ 0.45 m, this simplifies to approximately d0 − 0.45. It is also noted that at maximum approach the velocities may not both be zero; motion of the center of mass could still occur. The high spring constant causes rapid oscillation, so the maximum approach likely refers to the envelope of motion. The problem explicitly asks for an approximate estimate of the ratio. The ratio in part b is being estimated by considering the energy that would be stored if P2 were fixed, using only P1's potential and kinetic energies. The problem is from IME 2027, so it must be solved directly from the given information. The observations are being re-read carefully to ensure no detail is missed. At maximum approach, P2 is assumed to remain nearly in equilibrium because its velocity and acceleration are small. The spring force therefore balances the net forces acting on it. The equilibrium condition is kx = F_e + mg sinθ. The elastic potential energy is U_el = (1/2)(F_e + mg sinθ)·x, and P2's gravitational potential change is ΔU_grav = −mg·x·sinθ. The ratio of these two quantities is therefore (F_e + mg sinθ)/(2mg sinθ). The ratio is being reinterpreted as the elastic energy divided by the absolute value of P2's gravitational potential change. This yields (1/2)(F_e / mg sinθ + 1). The ratio evaluates to 29/18, or about 1.61. The equilibrium includes gravity. Because the spring constant is very large, P2 follows equilibrium closely and the compression x increases as the particles approach each other. The earlier estimate for part a) ignored this small displacement. The problem asks for an approximate ratio, so some simplification may be intended. The spring oscillates around its equilibrium compression x_eq(r). The extra oscillation energy is small, so the total elastic energy is roughly twice the static value. Because the problem asks for an approximate estimate, the ratio can be approximated to order of magnitude. The fixed spring end exerts an external force, so momentum is not conserved along the incline. Mechanical energy is conserved instead. The initial state has zero velocity and only the Coulomb potential energy. At the final configuration both particles are momentarily at rest; the energies include the Coulomb term, the spring compression, and the gravitational potential changes for both particles. The total gravitational potential change is −mg sinθ (d0 − r + 2x). The final spring energy is (1/2)kx². The assumption that both particles are at rest at the maximum approach may be incorrect, since the configuration could instead be a turning point in the relative motion. The coupled oscillation for finite k is too complex for an exam problem. The request for an approximate estimate implies a finite numerical answer. The charges and masses are identical. The ratio of elastic energy to P2's gravitational potential variation is finite and of order one, because both quantities scale with the same small compression x. The exact limit is (1/2)(F/(mg sinθ) + 1). The quasi-static approximation is being reconsidered for this high spring constant case. The elastic energy and P2's gravitational variation both scale as O(1/k), so they vanish in the limit k→∞. The minimum separation therefore remains 0.45 m, and the approximate ratio is simply the limiting value computed from the force at that distance. The next step is to formalize the quasi-static equilibrium. The quasi-static equilibrium is formalized. P2's compression is x_eq = (F_e + mg sinθ)/k, and the spring energy is U_el ≈ (1/2)kx_eq² = (F_e + mg sinθ)²/(2k). P2's gravitational variation is |ΔU_g| = mg sinθ·x_eq. The ratio therefore becomes ρ = (F_e + mg sinθ)/(2mg sinθ), or equivalently [F_e/(mg sinθ) + 1]/2. The ratio is confirmed as 29/18, or about 1.61. The author notes this is an unusual fraction for an IME problem and wonders if the absolute value was taken correctly. The author suspects the angle of inclination may be wrong. They are re-examining the dodecagon geometry to verify the correct angle for side BC. The dodecagon is regular with a flat bottom. The top side AB is horizontal. Side BC inclines downward at 30° to the horizontal. This confirms the angle of the incline. The numbers in part a) were deliberately chosen so that Ke·q1·q2 = 0.9 and mg sin30° = 2 yield exactly r = 0.45 m. The author now doubts whether the ratio in part b is actually small, since both energies are vanishingly small. The author is now looking at the problem from P2's perspective alone. Because the spring constant is very large, P2's displacement is tiny, so its motion can be analyzed separately to estimate the ratio of elastic energy to its own gravitational potential variation. The author derives an exact energy relation for P2. Because the particle starts and ends at rest, the work done by all forces is zero. The spring and gravity do negative work equal to their energy changes, while the Coulomb force does positive work. Therefore the elastic energy equals the work done by the electrostatic force minus the gravitational potential variation. The author notes that the electrostatic force is not constant during the approach, because P1 also moves and r changes. This invalidates the assumption that P2 could be treated as moving in a fixed field. The quasi-static condition with zero kinetic energy for P2 therefore requires a different formulation. The author is now questioning whether gravity should be included in the equilibrium force balance. They suspect the 30° anti-clockwise rotation may have reversed the original slope, and plan to check Figure 3 to verify the true incline direction. Figura 3 shows the side BC horizontal after the 30° anti-clockwise rotation. The original direction of BC is being reconstructed by reversing that rotation. The original BC direction was confirmed to be 30° below the horizontal, pointing downward from B to C. The author now turns to part c to clarify the maximum approach configuration and the associated velocities and energies. The instant rotation preserves the particles' velocities because no energy is added. Both particles remain at rest at the minimum separation of 0.45 m on the now horizontal side BC. The spring is fully removed after the rotation. The particles remain at rest at 0.45 m on the horizontal frictionless surface. They then repel each other freely under electrostatic force alone. Because both particles have equal mass, they will move apart symmetrically. The charges are q1 = 20 and q2 = 5, so they differ. Because the particles have equal mass, the electrostatic force produces equal-magnitude accelerations in opposite directions. They therefore move apart symmetrically with equal speeds, and the center of mass remains at rest. The author is now calculating the velocity of P2 when it reaches a vertex of the dodecagon. They note that P2 is closer to vertex C than to B, and that the side length BC is 1.15 m. P2 is 0.15 m from vertex C and P1 is 0.60 m from C after the rotation. P2 will move right to reach C, covering a distance of 0.15 m. P2 reaches vertex C first. The separation then becomes 0.75 m. The author applies energy conservation. The decrease in electrostatic potential energy equals the total kinetic energy. With equal masses and speeds, this yields v² = 2 and v = √2 m/s. The author wonders if the particles are constrained to segment BC or if they fly off at the vertex. The question asks for velocity exactly when P2 reaches a vertex, so this boundary behavior is uncertain. The author now questions whether removing the spring after rotation is physically consistent. In the limit of infinite spring constant, the elastic energy vanishes, so the spring would not store any energy to release when removed. The author now doubts the energy calculation. Removing the spring after rotation would abruptly unbalance P2, which was held in equilibrium by the spring force. The particle would then accelerate from that equilibrium state rather than starting from rest under pure electrostatic repulsion. The author realized the 30° rotation occurs while the spring is still compressed, changing the incline. After the rotation, the spring is removed in the new horizontal configuration. Because the rotation preserves velocities and adds no energy, both particles remain at rest at the maximum approach distance. After the 30° rotation, the side BC becomes horizontal. The component of gravity along BC therefore vanishes, so the equilibrium condition changes even though the spring remains compressed and the electrostatic force stays the same. The spring force kx is finite and non-zero after the rotation, while the component of gravity along BC is now zero. This means the net force on P2 is not zero, so the equilibrium condition changes even though the spring remains compressed. The author notes that the spring is then removed immediately in the new configuration. The spring is removed in the horizontal configuration. Because the spring constant is very large, the compression x is vanishingly small, so the particles remain essentially at rest with no energy input. After removal, only the electrostatic force acts on P2, and gravity is perpendicular to the motion. The earlier energy calculation therefore holds, giving a velocity of √2 m/s when P2 reaches the vertex. The author now realizes part b concerns the maximum approach condition before the rotation, while part c concerns the motion after. They are calculating the numerical value of the ratio 29/18 and questioning whether the problem expects an approximate or exact answer. The author is now exploring whether part b should be solved by neglecting gravity's role in compression, or by assuming the electrostatic force dominates. They also consider that the spring energy might be found through energy conservation in the limit of large k. The author now suspects the expected answers are exact fractions: 0,45 m for a), 29/18 for b), and √2 m/s for c). They also note that the variation in gravitational potential energy is negative because gravity does negative work during compression. The author now suspects the ratio in part b should be negative, since the change in gravitational potential energy is negative. They wonder whether the problem expects the absolute value or the signed ratio. The author suspects the ratio in part b is finite and not large, since both energies scale as 1/k. They are now looking at Figure 2, where small squares might indicate a horizontal spring or a fixed support. The author now interprets the "very high spring constant" as meaning the spring barely compresses, so P1 and P2 approach almost to zero separation. The small squares in Figure 2 are noted but not clearly identified. The author rejects the idea that the particles could approach to zero separation, since that would require infinite electrostatic energy. The calculation for part a) is confirmed sound, yielding a distance of 0.45 m. For part b), the ratio depends on two small quantities: the spring energy and the change in gravitational potential energy. The author now considers estimating this ratio directly from the derived expression. The author is now doubting the 30° angle assumption. They are reconsidering whether the dodecagon's orientation in Figure 1 places side AB at a different angle, which would change the equilibrium condition. The author now suspects the ratio in part b is very large, because the elastic energy dominates the small change in gravitational potential energy. They are calculating the force at closest approach to estimate this ratio numerically. The author models P2 as a driven oscillator with high spring constant. The total force is the electrostatic repulsion plus gravity. The particular solution is x ≈ F(t)/k. The homogeneous oscillation depends on initial conditions. The oscillator model is too complex for a rough estimate. The ratio is finite and of order one, not large. The author considers whether the ratio in part b justifies neglecting gravity's energy variation. They note that even with maximum spring compression, the elastic energy and gravitational variation remain comparable in magnitude, so this does not explain why one should be ignored. The author now suspects the expected ratio is approximately 10/9. They note that P2 initially compresses the spring to balance both the electrostatic force and gravity, so the total displacement is 3.544/k. The change in gravitational potential energy must therefore be calculated from this initial equilibrium position rather than from zero. The author realized the spring is initially relaxed, so its compression starts from zero rather than an equilibrium position. This invalidates the earlier assumption that P2 was precompressed to balance gravity and electrostatic forces. The spring therefore begins at zero compression, and the ratio in part b must be calculated from this initial condition. The spring starts from zero compression but immediately oscillates around the equilibrium position x_eq. Because the spring constant is very large, the amplitude of this oscillation is small and can be ignored. The elastic energy therefore averages to roughly (1/2)k x_eq². The author is now deriving the exact equation of motion for P2. Because P1's trajectory is known and its displacement is negligible, the electrostatic force F_e(t) can be treated as a prescribed function. The equation becomes m x'' = F_e(t) + mg sinθ − kx, with the spring constant k very large. The author writes the equation of motion as x'' + ω²x = [F_e(t) + mg sinθ]/m. The particular solution is x_particular ≈ [F_e + mg sinθ]/k. Because the spring constant is very large, the homogeneous oscillation has frequency ω and amplitude determined by initial conditions. At t=0 the spring is relaxed and F_e(0)=0.9, so the equilibrium position is x_eq(0) = 2.9/k. The author derives the full solution for P2's motion. Because the spring constant is very large, the oscillation amplitude stays approximately equal to the initial equilibrium displacement x_eq(0) = 2.9/k. The motion therefore remains confined to a small region around the slowly varying equilibrium position. The fast oscillation makes the elastic energy at maximum approach unpredictable. The ratio is therefore estimated as order one. The spring is attached to P2 and the fixed wall at C, not between the two charged particles. The particles interact through Coulomb force in vacuum, so P1 never touches the spring. This confirms that the minimum distance between P1 and P2 is 0.45 m. The author now realizes that with an infinite spring constant, the support behaves like a rigid rod and keeps P2 fixed in place. This would make the elastic energy zero at all times, so the ratio would be undefined. They note that the approximate finite ratio they calculated is only valid for a large but finite spring constant. The ratio is approximately 1.6, not large. The author suspects they may have mistakenly included gravity in the total force F, or that the spring support is horizontal rather than fixed. They are now reconsidering how to express the change in gravitational potential energy of P2. The author now suspects the ratio in part b is approximately 1.6, or possibly 29/18, because both energies scale as 1/k. They briefly reconsidered whether the angle θ might differ from 30°, but confirmed it is correct. The author suspects the 30° rotation might orient side BC differently. They note that parts a) and c) yielded clean numerical answers, so part b) should likely have a clean ratio too. They briefly reconsidered whether the high spring constant makes the ratio infinite, but rejected this idea. The author recalculates the ratio in part b) and finds exactly 29/18. They also note that the change in gravitational potential energy is negative, since P2 moves downward when compressed, so the ratio could be −29/18. The author now doubts the negative sign in the ratio. They suspect gravity might oppose the spring force, or that the slope orientation is reversed, which would change the equilibrium condition and the sign of the energy variation. The author confirmed the initial force direction is correct. Gravity exceeds the electrostatic repulsion, so P1 moves down the incline toward P2. The slope orientation is therefore not reversed. The author now accepts the negative sign in the ratio. Because the change in gravitational potential energy is negative, the approximate ratio is −29/18. The author now doubts the negative sign in the ratio. In Brazilian textbooks, "variação de energia potencial" is usually the signed difference Uf − Ui, but a negative ratio is uncommon. They suspect the problem may instead ask for the absolute ratio or how many times larger one energy is than the other. They also wonder whether the maximum approach of the particles corresponds to the spring's maximum compression. The author briefly considered calculating the electric energy from the charges, but realized quasi-static conditions are unnecessary for this ratio. They noted that with a very large spring constant, the particles interact only briefly when close, so P2 barely moves during that interval. The problem asks for an estimate, so they may simply approximate the ratio using the change in force. The author now estimates the maximum possible ratio by considering the spring's full oscillation amplitude. The maximum elastic energy is roughly 2F²/k and the maximum change in gravitational energy is about mg sinθ·2F/k, so the peak ratio is approximately F/(mg sinθ). This yields a value around 1.6, confirming that neither energy term can be neglected. The ratio is order-one and independent of the spring constant, so the answer should be 29/18 or approximately 1.6. The author now suspects the ratio should be calculated using a quasi-static approximation, ignoring the fast oscillation. This would yield exactly −29/18, but they are unsure why the cosine term disappears in that limit. The author now doubts that the oscillatory motion can be safely ignored. Because both the equilibrium displacement and the oscillation amplitude scale as 1/k, they are of the same order, so the ratio depends on the phase at the moment of maximum approach. They are now trying to determine exactly when this maximum approach occurs. The phase at maximum approach is essentially random because the angular frequency is very high. The elastic and gravitational energies therefore vary unpredictably, so the ratio cannot be pinned down to a single value. The author concludes that "estime" likely asks for an order-of-magnitude estimate, perhaps around 1 or 2. The author now doubts that the oscillatory phase argument is appropriate for this high school-level exam. They suspect there must be a simpler, more direct way to estimate the ratio. The author is now trying a different interpretation of the problem. They suspect the spring constant is so large that the particles barely move during approach, making the minimum distance 0.45 m. For part b), they are considering whether to estimate the ratio using the work done by the spring force, or if it should be approximately infinite. They plan to calculate the elastic energy directly. The author now suspects the 30° rotation changes the particles' heights because the system's orientation shifts, even though no kinetic energy is added. This would alter the gravitational potential energy variation in part b). The author now wonders whether the gravitational potential energy variation in part b) should be measured relative to the initial configuration, since the 30° rotation changes absolute heights even though positions remain fixed during the giro. The 30° rotation happens after the particles reach maximum approach, so the ratio in part b) is still evaluated at the original orientation. The elastic and gravitational energy variations remain small at that moment. The author now doubts whether the candidate should use x = F/k, since that requires assuming the net force is zero. With a very large spring constant, even a small displacement can produce a large acceleration if the forces are not perfectly balanced. They suspect IME problems may expect subtle approximations, and they are now considering what short answer format the exam likely requires. The author now considers that a very large spring constant forces the compression to be nearly zero, so the minimum distance remains 0.45 m. They also wonder if part b) should be estimated by taking the limit of the ratio as the compression approaches zero. The author now suspects the answer 29/18 assumes static equilibrium at maximum approach, with both particles instantaneously at rest. In a harmonic oscillator, however, maximum compression occurs when acceleration is maximal and nonzero. This quasi-static assumption is likely what the exam expects for high school students. The author now turns to part c). They suspect the spring constant is so large that the collision can be treated as elastic. They are trying to determine exactly what happens after the maximum approach condition, and how P2 moves from there to reach a vertex of the dodecagon. The author now interprets the constraint in part c) as P2 sliding along the segment BC without friction. They note that reaching vertex C is the endpoint of this path, and they are considering whether the velocity at that point should be calculated as √2. The author now suspects the answers are a) 0,45 m, b) a large ratio, and c) √2 m/s. They are trying to estimate that ratio by calculating the energy lost by P1 at maximum approach, or by using virtual displacement. The author now suspects the ratio is exactly 29/18, and that the word "aproximada" simply reflects the high spring constant approximation. They also wonder whether the question intends the absolute variation of the gravitational energy rather than its signed change. The author now suspects the squares in the figure mark right angles, and that side AB is not horizontal. They also notice the 30° rotation makes the top of the dodecagon flat, matching the configuration shown in Figure 3. The author now suspects the examiners deliberately chose the numbers so that the angle is exactly 30°, making sinθ = 1/2 and yielding the clean result r_min = 0,45 m. This explains why part a) is exactly 0,45 m. The same reasoning likely applies to part c), where the nice answer √2 m/s suggests the spring is removed at maximum approach, leaving both particles momentarily at rest before they fly apart. The author now doubts that P2 starts from rest when the spring is removed. In the limit of a very large spring constant, the spring acts as a rigid constraint, and removing it suddenly unbalances the net force on P2. Because the 30° rotation changes the component of gravity along the slope, the particle may already have a velocity before it leaves the vertex. The author now resolves the doubt about P2's initial velocity. When the spring is removed, no energy is released because there is no displacement. The particle therefore starts from rest. Any subsequent motion comes from the unbalanced forces of gravity and Coulomb as the particles move. The author confirms the velocity at vertex C is √2 m/s, directed along BC. The distance from the maximum approach point to C is exactly 0,15 m. The author verified the distances: P1 is 0,55 m from B and P2 is 0,15 m from C, so P2 reaches the vertex first. They now suspect part b) should yield a simple fraction, but ruled out a ratio near zero because the elastic energy remains non-zero. The author now suspects the ratio 29/18 is correct. They also tested a simpler approximation using only the spring force divided by twice the gravitational force, but that yields 10/9, which is too small. The author now suspects the ratio in part b) might be infinite. Because the spring constant is very large, the particles barely move during approach, so the variation of P2's gravitational potential energy is approximately zero. If the elastic energy variation is also non-zero, the ratio becomes extremely large. The author now suspects the ratio in part b) is simply 29/18. Because both energy variations are negligible, neither can be ignored, so the estimate must include both terms. The author now suspects the ratio in part b) is exactly 29/18, and that the word "aproximada" simply refers to the high spring constant approximation. They are also considering whether to take the absolute value of the gravitational energy variation, since the text does not specify a sign convention. The author now suspects the ratio in part b) should be negative. Because P2 moves downhill toward C, its gravitational potential energy decreases, so the variation is negative. They are re-reading the question to confirm whether the answer should be −29/18. The author now suspects the ratio in part b) is approximately −1. They reason that if both particles come to rest at maximum approach, energy conservation implies the elastic energy equals the negative of P2's gravitational potential variation, so U_el / ΔU_g_P2 ≈ −1. They are checking whether this makes sense. The author is now deriving the exact limit of the ratio in part b). They note that at maximum approach, energy conservation implies U_el = −ΔU_electric − ΔU_g1 − ΔU_g2. Because x is small and r stays near 0,45 m, the elastic energy must balance the small variations from the electric and gravitational terms. They are checking whether this yields exactly −1. The author now realizes that for both particles to momentarily come to rest at the same point, the configuration must be a true equilibrium. Otherwise the gradients would not vanish and the particles could not simultaneously have zero velocity. The author now suspects the equilibrium distance is r = √0,45 ≈ 0,67 m, not the 0,45 m found earlier. They realize that for a very large spring constant, P2 would remain essentially fixed while P1 oscillates between r = 1 m and r = 0 m. This means the particles cannot both momentarily come to rest at the same point during the dynamic approach, so the assumption of a true equilibrium at maximum approach is incorrect. The author now realizes that the anteparo can absorb momentum, so total momentum is not conserved. This means both particles could momentarily come to rest at the minimum distance if the configuration coincides with a potential maximum or turning point. The possibility of a true equilibrium at maximum approach is therefore not ruled out. The author now suspects the ratio in part b) is exactly −1. They reason that for a very large spring constant, the fast variable x is slaved to r, so the motion reduces to a slow degree of freedom with an effective potential. At the minimum distance, the total energy equals its initial value and the slow velocity vanishes; the elastic energy is then included in this effective potential. The author now derives the exact energy balance at the minimum distance. The sum of elastic and gravitational potential changes equals (1/2 F_e − 1/2 mg sinθ) x, where x is the spring compression. This confirms the ratio in part b) is exactly 29/18. The author now questions whether vibrational energy is actually excited. If the spring is slaved to its equilibrium compression, no oscillation energy is produced, so the ratio remains exactly 29/18. The author now suspects the initial spring compression is non-zero, which would excite vibrational energy. They estimate the oscillation amplitude is comparable to x_eq(0), and since the elastic energy varies with x_eq while the oscillatory energy stays constant, the total elastic energy at the final configuration becomes a specific sum of both terms. The author suspects the examiners expect a simpler fixed solution for part b). They checked the numbers and found the spring force is 29/10, not a clean fraction. The oscillation amplitude would therefore be messy. The numbers do not suggest the problem intends a vibrational energy calculation. The author now suspects the examiners expect the simple fixed-spring approximation for part b). They verified that 29/18 is a clean fraction, and that the ratio can be derived by assuming the spring compresses from its initial equilibrium position to the final one. The word "aproximada" likely refers to neglecting small position variations in other calculations. The author now suspects the ratio in part b) is exactly −29/18. They note that both the electric and gravitational forces are external to the spring, so the spring must balance their sum statically. They also question whether the word "aproximada" fits this exact value. The author now believes the intended answer is exactly 29/18. They reason that for a very large spring constant, the center-of-mass acceleration is negligible and the spring force balances the sum of the electric and gravitational forces. The word "aproximada" simply reflects this quasi-static approximation, not a sign error. The author now suspects the ratio in part b) is exactly −29/18. They derive this by writing the elastic energy as (1/2)(F_e + mg sinθ)·x and the gravitational variation as −mg sinθ·x, then taking their quotient. Because the electric force was evaluated at the distance from part a), this expression is exact. The author now doubts the ratio in part b) should be negative. They reason that "variação da energia potencial" is a signed quantity, but in typical physics problems this phrase usually refers to the absolute loss. They cite falling-body examples where the ratio is typically positive, so the answer may need to be 29/18 rather than −29/18. The author now doubts the ratio should be negative. They suspect the examiner expects a simple positive number, perhaps around 1 or 2, because the problem asks to estimate it. They also wonder if part b) does not rely on the static balance x = F/k. The author recalculates the net force on P2 as 11/9, confirming the spring cannot balance only the electric force. They now suspect the examiner expects a very small ratio, perhaps because the weight component is ignored. The author now suspects the intended model for part b) is a rigid, position-fixed spring, so both energies are negligible and the ratio should be estimated as an upper bound. They also notice the charge and mass values may be designed for a reduced-mass calculation in part c). Their initial velocity result for part c) seems too simple, so they are re-reading the exact wording of part c) to verify. The author now reads part c) as asking for the velocity of P2 when it reaches a vertex of the dodecagon after the spring is removed. They note that removing the spring eliminates its elastic energy without transferring it elsewhere, and they question whether gravity should be included in this calculation. The author now suspects the ratio in part b) compares the spring energy to the gravitational potential variation of P2 caused by the 30° rotation during maximum approach. The spring energy would remain unchanged, while the rotation alters the particle heights and produces a ΔEp that could be used as the denominator. The rotation axis is unspecified, so the height change cannot be calculated. This rules out the giro as the source of the gravitational variation in part b). The ratio must instead refer to the energy balance at the maximum approach condition itself. The author now suspects the intended solution for part a) treats P2 as fixed because the spring constant is very large. Energy conservation for P1 alone would then yield r = 0.45 m. The author now applies the work-energy theorem to P2. Because its velocity returns to zero, the net work done on it is zero. This yields the equation W_grav + W_spring + W_coulomb = 0, which relates the gravitational potential change, the spring energy change, and the electric work along the path. The author now derives the ratio in part b) by applying the work-energy theorem to P2. Because its velocity returns to zero, the net work done on it is zero. This yields U_el = −ΔU_g + W_c, so the ratio becomes −1 + W_c / ΔU_g. The electric work divided by the gravitational potential change simplifies to −F_avg / (mg sinθ), giving the final expression −1 − F_avg / (mg sinθ). The author notes the ratio still depends on which average force is used, and whether P2 ends at its equilibrium position. They confirm that for a quasi-static process with zero final velocity, the work-energy theorem remains valid. The author now performs the explicit integration for the electric work. Because the spring force is kx and the net force is zero, the electric force equals kx minus mg sinθ. Integrating this from zero to the final displacement x_f gives (1/2)kx_f² − mg sinθ·x_f. Substituting into the earlier equation U_el = −ΔU_g + W_c then yields U_el = (1/2)kx_f². The author now accepts the ratio as exactly −29/18. They reason that because the gravitational potential change is negative, the ratio itself is negative, and Brazilian exams may accept this signed value. The author now doubts the ratio is −1. They calculated that U_el + ΔU_g equals 11/9·x, while |ΔU_g| equals 2·x, so the two terms do not cancel to −1. They are puzzled about how −1 could arise. The author realizes the electric work on P2 cannot be ignored, because it is of the same order as the other terms. The ratio −29/18 is approximately −1.61, which does not match the expected simple estimate near −1. The author now suspects the examiner may expect the ratio to be approximated as −1 by neglecting the electric work, but rejects this because the electric force is roughly twice the gravitational component and cannot be ignored. They also consider a simpler approximation where the elastic energy is replaced by F·x, though they do not pursue it. The author now accepts −29/18 as the rigorous quasi-static ratio, and suspects the examiner expects the absolute value 29/18 because the problem asks to estimate it. They will check online for confirmation. The author now suspects part c) is more complex than a simple velocity at the dodecagon vertex. They wonder if the motion continues past that point, or if the velocity must be evaluated when P2 reaches the next vertex after the spring is removed. The problem restricts the motion to segment BC, so P2 stops at vertex C. Its velocity there is √2 m/s. The author now checks whether the distance from the starting point to that vertex is L0. The author now suspects the examiner may expect an extreme ratio, such as zero or infinity. They note that if P2 is treated as fixed, its gravitational potential variation and elastic energy would both vanish, making the ratio indeterminate. Because the problem asks to estimate, they also consider whether the ratio might be approximated as zero. The author re-reads part b) and wonders if the ratio should compare the spring energy to the gravitational potential variation of P2 during maximum approach. Because the spring constant is very large, the compression is small, so they consider using a series expansion. The examiner likely expects the ratio to be independent of k, so both energies must be written in terms of x and k and then eliminated. The author is now deciding which forces should appear in that expression. The observation about P2's position only concerns P1, so the weight and electric force must both be included in the spring's static balance. The ratio is confirmed as −29/18. The 30° rotation is designed to make the side horizontal for part c), eliminating gravity and allowing a clean repulsion calculation. The author now suspects part b) was similarly designed so that F_e / (mg sinθ) equals 20/9, yielding the nice ratio 29/18. The author now suspects the examiner may expect −10/9 instead of −29/18. If the student forgets to include the weight in the force balance for P2, the spring would only counter the electric force. That yields a ratio of −10/9, which is a simpler fraction. The author now doubts the quasi-static assumption. The maximum approach is the moment of minimum distance, not necessarily when P2's acceleration is zero or its displacement equals the equilibrium value. This suggests the spring force may not balance the weight and electric force at that instant. The minimum distance occurs at the turning point of the slow motion, when the fast oscillation is at an arbitrary phase and the spring is not at equilibrium. The ratio therefore cannot be exactly −29/18. Because the problem asks for an approximate estimate, the quasi-static value remains a reasonable approximation. The author now analyzes the initial motion. Because the spring is initially relaxed, P2 starts with a finite acceleration of 7.25 m/s². The oscillation amplitude shrinks as k increases, and its velocity scales as F/√(k·m), vanishing in the limit of infinite stiffness. Consequently, the oscillatory kinetic energy behaves like O(1/k), which may explain why the quasi-static approximation remains valid. The author briefly considers whether the examiner expects an approximate ratio near zero, reasoning that with a very stiff spring the compression is tiny and both energies vanish. They then realize this is incorrect because P2 is almost at rest, so its gravitational potential change is negligible, while the elastic energy remains finite. The author now suspects part b) is designed to show that the elastic energy is only about 1.6 times larger than the gravitational potential change, so it cannot be approximated as negligible. They also note that a larger angle θ would make the ratio much larger. The author now doubts the 30° rotation is the slope angle of BC. They re-examine the figure and suspect the square in Figura 2 might indicate a different angle, so they will re-read that part carefully. The 30° rotation is the slope angle of BC. The figure shows the side horizontal after that turn, so the original angle must have been 30°. The author now settles on the final answers: a) 0.45 m, b) −29/18, and c) √2 m/s. They briefly reconsidered whether part b) might expect an approximate zero, but concluded that both energies remain finite because P2 barely moves. They also note that the problem text does not explicitly ask for the absolute value of the ratio. The author re-reads the Portuguese wording and suspects the examiner may expect a positive ratio. Because the problem asks for the variation of P2's energy, they consider whether the gravitational potential change should be written as a positive amount, which would make the ratio approximately 29/18. The author briefly wonders if the spring could compress while P2 moves away from C, but the problem text places the anteparo at C and the relaxed spring at distance L0 from it, so P2 is inside the segment. The author now doubts that the spring energy must vanish in the infinite-stiffness limit. Because the total energy is finite, the compression can remain finite and the elastic energy may be a significant fraction of that total. This raises the question of whether the spring can store substantial energy even when k becomes very large. The author resolves the prior doubt. Because the electric force is finite and bounded, the spring cannot compress indefinitely even as k grows large. The compression scales as O(1/k), so the elastic energy vanishes in the infinite-stiffness limit. This confirms that the ratio of spring energy to gravitational potential change approaches zero. The author re-derives the equilibrium force balance for P2 along the incline. The spring compresses by x, so the elastic energy is (1/2)kx² and the gravitational potential change is −(mg sinθ)x. This confirms the ratio is −k/(mg sinθ). The author now doubts the examiner expects an exact fraction. Because the problem asks to estimate, they suspect a rough order-of-magnitude answer may be more appropriate. The examiner explicitly instructs the candidate to ignore P2's position variation for P1's motion calculation. This confirms that part a) is solved by treating P2 as fixed, while part b) separately estimates the ratio of spring energy to gravitational potential change. The author now suspects part b) might be designed so that P2's position variation is negligible, making the gravitational potential change zero and the ratio infinite. They also wonder if the maximum approach distance might not actually be 0.45 m. The author confirms that part a) is solved by treating P2 as fixed, yielding the minimum distance of 0.45 m. For part b), they note that k and x are not given, so only their ratio can be estimated; this is done by eliminating both variables from the energy expressions. The examiner expects this approximate ratio to be calculated using the force derived from the quasi-static approximation. The spring is fully removed, so its elastic energy vanishes and the particles start from rest. Because the ratio of elastic energy to gravitational potential change is finite, the error scales as O(1/k) and vanishes in the infinite-stiffness limit. This confirms the velocity reaches exactly √2 m/s. The ratio remains negative because the gravitational potential change is negative. The author now doubts the spring energy is uniquely defined. Because the problem mentions no damping or friction, the oscillation persists indefinitely and the elastic energy keeps varying with time. This makes the ratio of spring energy to gravitational potential change ambiguous. The examiner explicitly asks for an approximate estimate, so the answer should be the exact ratio −29/18. The oscillatory spring energy is irrelevant because the problem isolates a single instant of maximum approach. The rotation is instantaneous upon reaching the minimum distance, so the energy comparison in part b) applies to that exact moment before the turn. The gravitational potential change should therefore reflect only the pre-rotation height difference. The author is now drafting the solution in Portuguese. They begin with the geometry of the regular dodecagon and the 30° incline of BC, noting that Sin30° = 1/2. For part a), they treat P2 as fixed because k is very large, and set up the energy conservation equation for P1 using zero initial gravitational potential. The author derives the exact quadratic equation for part a) by equating initial and final energies. The resulting equation 20r² − 29r + 9 = 0 has discriminant 121, confirming the roots are rational. The author verifies the quadratic solution r = 0,45 m by substituting it back into the energy conservation equation. They then compute the electric force at that distance as 40/9 N and begin the quasi-static analysis for part b), noting that P2 is effectively at rest. After the 30° turn, the particles are at rest with a 0.45 m separation and P2 0.15 m from vertex C. The spring is removed, so the particles are now free to repel on the frictionless horizontal rail. Because both masses are equal and the repulsive forces are equal in magnitude, their accelerations are identical; starting from rest, their velocities remain equal in magnitude but opposite in direction. The author computes the final separation as 0,75 m and checks that the total kinetic energy is 0,8 J, confirming the velocity is exactly √2 m/s. They also decide to include the negative sign in part b), giving the ratio −29/18. The author suspects the examiner deliberately chose the numerical values so that the ratio in part b) simplifies to exactly 29/18. They note that the mass, charge, and spring constant were selected to yield a clean quadratic solution for part a), a perfect square velocity for part c), and a tidy ratio for part b). The author now doubts whether the examiner intended the acceleration to be zero. They note that omitting the gravitational term would yield a ratio of 10/9, which is equally clean, and question which value is more likely. They also seek a simpler energy argument that avoids assuming zero acceleration. The author calculates the oscillation amplitude as 2,9/k and the equilibrium position as roughly 6,444/k. Because the spring compresses by a positive amount at every instant, the gravitational potential change is always negative. This confirms the ratio in part b) remains negative even during oscillation. The author confirms the final answers: a) 0,45 m; b) −29/18; c) √2 m/s. They verify the energy conservation calculation for part c) and note that including gravity in the equilibrium estimate is appropriate. The author is reconsidering the sign of the ratio in part b). They note that if the gravitational potential change is calculated as negative, the ratio becomes negative. They also question whether P2 actually moves upward during compression, and re-read the figure to verify the direction of rotation. The author is now drafting the final solution with exact values and g = 10 m/s². They notice that the relaxed spring length L₀ = 0,15 m appears in part c). They also reconsider whether part b) might expect the ratio to be approximately zero, since P2's position variation could make the gravitational potential change negligible. However, they note that asking for an estimate of a ratio that would vanish would be inconsistent. The author is now questioning whether the original BC segment was actually inclined upward. If C were higher than B, compressing P2 toward it would raise its height and make the gravitational potential change positive. This would alter the force balance in part b) and could explain why the examiner chose the specific numerical values. The author rejects the idea that the original BC segment was inclined upward. If C were higher than B, P1 would not move toward P2 because the net initial force would point away from it. The author performs a numerical check to confirm the motion. With C lower than B, the net force on P1 points down the incline toward P2, and the spring force exactly balances gravity at r ≈ 0,67 m. Beyond that point the net force reverses, so P1 stops at the calculated 0,45 m. This validates the original downward incline and confirms the ratio in part b) is negative.

We need answer physics problem in Portuguese, likely from IME 2027. Need solve item a b c. Need parse geometry. We need understand setup. Dodecágono regular (12 sides). Figures: Fig1 regular dodecagon with horizontal bottom side? Top side? Points A, B, C? Actually Fig1: regular dodecagon, top side? It shows A B C on upper right? Let's inspect. Image: left figure 1: octagon? Wait text says dodecágono regular (12 sides), but drawing in Fig1 looks maybe 12-sided? I see polygon with sides, top segment A-B? Then B-C? It might be a regular dodecagon with top side? Let's parse: "Nas Figuras 1 e 2, a partícula P1 ... abandonada no vértice B de um dodecágono regular e pode deslizar sobre o segmento BC. No vértice C, encontra-se fixado um anteparo..." So B and C are adjacent vertices? Segment BC is a side of dodecagon? Dodecagon regular has 12 sides. Figure 1 shows A,B,C consecutive? The diagram: polygon has top side horizontal with A-B? Then B-C slope downward? Let's see. In Fig1, label A at top left? B at right of A? C further right on edge? Actually "A B" on top horizontal side, C at next vertex down. A and B maybe vertices? B and C adjacent? Let's zoom mentally: Fig 1 octagon-like polygon with top edge A-B, then B-C sloping down right, C next vertex. Yes. Fig2 enlarged segment? It shows A, B, P1 on top segment? Actually A at left, B at right of horizontal segment, then P1 somewhere on segment B? Then C on sloped segment? Hmm Fig2: line from A to B (horizontal), then from B to C downward sloping. P1 on horizontal segment? label P1 below P? P2? There is a particle P1 at vertex B? Actually text says P1 abandoned from rest at vertex B and can slide on segment BC. So P1 starts at B? Figure 2 shows P1 at B? It says "P1" near B? Then B-C segment contains P2? There is an anteparo at C with spring, P2 fixed at other end? Initially spring relaxed? "No vértice C, encontra-se fixado um anteparo que prende uma das extremidades de uma mola, inicialmente relaxada. Na outra extremidade da mola, foi fixada a partícula P2..." So spring along BC between C and P2, initially relaxed length L0=0.15m. P2 can slide on BC. P1 slides from B toward C/C? Wait P1 starts at B and can slide on segment BC. So P1 moves down BC toward P2? P2 initially near C? Because spring initially relaxed, distance from C to P2 = L0? But P2 fixed to spring end; spring at C fixed, length L0. So P2 initially at distance L0 from C along BC (toward B). P1 starts at B, slides down BC toward C under gravity and repulsion from P2? Both positive charges repel. P1 moves until maximum approximation with P2. Spring compresses? P2 may move? "constante elástica muito alta, despreze variação da posição de P2 para efeitos de cálculo no movimento de P1" -> stiffness very high, so spring doesn't compress much; P2 position approximately fixed? But P2 can slide? "mola muito alta, despreze variação da posição da P2 para efeitos de cálculo no movimento da P1". So during approach, P2 approximately stationary at initial position? Maybe P2 initially at distance L0 from C? Wait spring relaxed L0=0.15 m, and P2 fixed to spring other end; P2 can slide on BC. If spring is relaxed, distance between C and P2 = L0. Because no force along BC? Actually electrostatic repulsion pushes P2 toward C? But spring at C pulls? If relaxed, P2 at length L0 from C; but if repulsion from P1 pushes P2 toward C (left? Need orientation), spring may compress if stiffness finite. Since stiffness very high, variation negligible. So P2 position remains approximately at distance L0 from C along BC. But P1 moves. Maximum approximation means distance between P1 and P2 minimum. Since P2 approx fixed at distance L0 from C, distance P1-P2 = distance along BC from P1 to P2. If P1 starts at B, distance B to P2 initially = total side length? Wait d0 initial distance between particles = 1.0 m. That may be length between B and P2? Let's determine geometry. Dodecagon regular side length? Need perhaps from data d0 = distance initial between particles = 1.0 m. Since P1 at B, P2 fixed at distance L0 from C? Then initial distance between P1 and P2 along BC = side length s minus L0? If d0 = 1.0 m, then s = d0 + L0 = 1.15 m? But maybe d0 is initial distance between particles after release? Let's verify. Problem asks: a) na condição de aproximação máxima entre as partículas, determine a distância entre P1 e P2; b) ... estime razão aproximada entre energia potencial elástica da mola e variação da energia potencial gravitacional de P2; c) após a condição da Figura 2, determine a velocidade de P2 ao atingir um vértice do dodecágono. Need understand full process: Initially P1 at B, P2 at other end of relaxed spring near C? P1 slides along BC under gravity and electrostatic repulsion; spring maybe compressed by P2? Maximum approximation occurs; then system instantaneously rotated 30° anti-clockwise (around what? maybe around center?) without adding kinetic energy. Then spring disconnected and removed, particles interact freely. Figure 3 shows after rotation: A, B, P1, P2 on two adjacent sides? Let's parse geometry. Dodecagon regular: central angle 30°. Adjacent vertices separated by side length. If rotate by 30° counterclockwise around center? Since dodecagon has 30° rotational symmetry? Actually regular dodecagon has central angle 30°, so rotating by 30° maps vertices to next vertices. "todo o sistema será instantaneamente girado no sentido anti-horário, de 30°, sem adicionar energia cinética às partículas. A partir dessa nova configuração, a mola é desacoplada ... e da partícula P2, sendo que as duas partículas passam a interagir entre si livremente. Essa situação está representada na Figura 3." So after maximum approximation (P1 and P2 close), rotate entire system by 30° CCW, keeping relative positions unchanged. This rotation maybe changes which segment particles lie on? Figure 3 shows particles on two distinct adjacent sides? Let's interpret. Figure 2 (before rotation): Segment A-B horizontal? B at vertex, segment B-C is side sloping down. P1 at B (or on BC? Actually "P1 abandonada no vértice B ... pode deslizar sobre o segmento BC"). P1 moves along side BC. P2 fixed to spring end, initially at distance L0 from C along BC. So both on same side BC, with B at top, C at lower right. Maximum approximation occurs somewhere on BC; P1 closer to P2 (near C maybe). Then rotate system 30° CCW. Since dodecagon has symmetry, side BC maybe rotates to next side? But Figure 3 shows A-B-C? Let's identify coordinates. Dodecagon with a horizontal top side A-B? In Fig1: top side is A-B, then B-C is next side sloping down. The 12 sides around: after top, right side? Let's assign. A and B are vertices of top side (horizontal). C is vertex adjacent to B on right. So segment B-C is one side sloping downward to right. After rotating counterclockwise by 30° around center, the side B-C might rotate to a side in different orientation. Figure 3: It shows A at left? Labels: "B C" top? There are two little squares? Let's read: Figure 3 has "A" on left? Actually image: Fig3: left portion peak? It shows "B" at apex? Letters: "B" on top, "C" to right of "B"? Let's examine: In Fig3, the shape: a tent-like with a flat top? Actually labels: At top vertex? There is "B" above a small segment? Then "C" maybe at top right? Wait text in caption: Figure 3 likely shows after rotation, with particles P1 and P2 on a horizontal top side? Let's zoom: In Fig3, I see a polygon segment A-P1-P2-C? Let's parse labels: There is "B" at apex? Actually maybe labels: "A" left, "B" apex? Then "C" top right? Hmm The letters in image: In Fig3, at top there is "B" above? There is "C" maybe to right. Let's zoom: It shows top horizontal segment between two vertices? There is "B" at left of a horizontal side, "C" at right of horizontal side? Then below left, label A? Let's inspect: Fig3: top side has two small square particles? There is a label "B" at left vertex, "C" at right vertex? Actually under the top side? The letters: "B" at top left vertex, "C" at top right vertex. Between them, two particles P1 and P2 are on the horizontal top side. Left of B is a vertex A? It is labeled A on the left sloping edge? Wait: In Fig3, shape is like a wide roof: left side rising, top horizontal side from B to C, right side descending. A is lower left? Maybe labels: "A" near left vertex, "B" at top left vertex, "C" at top right vertex. P1 and P2 are on top segment B-C. The text: "após a condição da Figura 2, determine a velocidade de P2 ao atingir um vértice do dodecágono." So after rotation, particles interact freely; P2 may slide on dodecagon and reach a vertex. Need determine velocity of P2 when it reaches a vertex. Need understand geometry after rotation: The system rotated 30° CCW, so side BC (sloping down right) maybe becomes horizontal top side? Since rotating a sloped side by 30° CCW could align it horizontally? Let's check: A regular dodecagon has exterior angle 30° (central angle 30°). If side BC initially sloping downward at angle -30°? Let's calculate. Dodecagon regular: each exterior angle 30°. If top side AB horizontal, then as traverse clockwise or counterclockwise? Let's set coordinates. Regular dodecagon with one side horizontal at top: Since sum exterior angles 360°, each turn 30°. Starting at A=(?,?) and B=(?,?) top side horizontal. Next side BC turns exterior angle 30° downward/right, so its direction is -30° from horizontal? Actually interior angle 150°, exterior turn 30°. If going clockwise around polygon, at B you turn right 30°, so direction from B to C is 30° below horizontal (angle -30°). Thus BC slope length s, vertical drop s sin30 = s/2, horizontal run s cos30. So side BC is not horizontal; angle -30°. If rotate entire system counterclockwise by 30° (not rotate the whole polygon? "girado no sentido anti-horário, de 30°" around maybe central axis? likely around dodecagon center). Side BC direction becomes -30 + 30 = 0°, i.e., horizontal. Thus in Figure 3, the side that was BC becomes horizontal top side? But labels change? Figure 3 shows top side labeled? It might be the same physical side BC, rotated to horizontal, now containing P1 and P2 on horizontal top side. Yes! The label B and C are at ends of that same side after rotation? In Figure 3, top side is B-C (rotated side BC? But labels might be B left, C right). The side BC now horizontal (top side). Particles P1 and P2 on it. A is adjacent vertex to B (left of B) on sloping side. So Figure 3 shows after rotation, the original side BC is now horizontal at top, with B left, C right? Actually original B at left end (upper), C at right end (lower). After rotation CCW, original side BC becomes horizontal; B maybe left or right? Let's determine rotation by +30° (CCW) maps original side BC direction -30° to 0° (horizontal), but does it map B to left or right? Original BC vector from B to C is (cos(-30), sin(-30)) = (√3/2, -1/2). Rotate +30°: (cos0, sin0) = (1,0), so B to C points right. Thus B left, C right. P1 and P2 on this horizontal side, P1 closer to B maybe? In Fig3, labels B left, C right, P1 near left, P2 near right? Actually Fig3: "B" at left vertex, "C" at right vertex, P1 and P2 on top side. Yes. Now need solve physics. Let's define geometry needed. Dodecagon side length s =? Given initial distance between particles d0 = 1.0 m. In initial configuration (before rotation), P1 at vertex B, P2 at distance L0 = 0.15 m from C along side BC (spring relaxed anchored at C). So distance between P1 at B and P2 = s - L0? Wait if P2 is between B and C, distance from B to P2 = s - L0. But question says "distância inicial entre as partículas: d0 = 1,0 m" meaning initial distance between particles at start? Yes, P1 at B, P2 at spring end near C, so d0 = s - L0, where s is side length. Thus s = d0 + L0 = 1.0 + 0.15 = 1.15 m. But need check: d0 could be initial distance after release? They give "distância inicial entre as partículas: d0 = 1,0 m". At rest at vertex B? P1 abandoned from rest, P2 at one end of relaxed spring. The distance between them initially is side length minus L0 if P2 at distance L0 from C. So s=1.15m. Then dodecagon side length 1.15 m, central radius R? R = s / (2 sin(π/12)) = s / (2 sin15°). sin15 = (√6-√2)/4 ≈0.258819; 2 sin15 ≈0.517638; R ≈2.222 m. Maybe needed for height? Height of vertices relative to some baseline? For gravitational potential changes. P1 slides down from B to max approximation along BC. Need determine distance between P1 and P2 at maximum approximation. This is an equilibrium/minimum separation due to repulsion and gravity? P1 starts at B (higher gravitational potential) and slides down; positive charges repel. Spring with very high stiffness means P2 nearly fixed, but spring force may balance some? The spring is initially relaxed; if P1 repels P2, P2 pushed toward C compressing spring; since k very high, position of P2 remains near initial? But energy potential elastic stores due to infinitesimal displacement? Hmm "como constante elástica é muito alta, despreze a variação da posição de P2 para efeitos de cálculo no movimento de P1" means when computing P1 motion and final, take P2 as fixed at initial position. But item b asks ratio between elastic potential energy of spring and variation of gravitational potential energy of P2; need consider elastic energy due to P2 displacement? They ask estimate approximate ratio. Since k very high, elastic potential energy maybe large? Wait high k means small displacement but energy 1/2 k x^2 could be finite? Need compute from balance? Need find approximate ratio maybe using equilibrium? Let's explore. Part a: max approximation distance between P1 and P2. P1 moves along BC; forces: gravity component along incline, electrostatic repulsion from P2 (along side? Both on same side, so repulsion along side away from P2), maybe spring? P1 not attached to spring; spring only attached to P2 and C. But P2 may move due to spring/repulsion; however high k reduces movement. For P1's motion, P2 acts as fixed charge. P1 slides down incline due to gravity component m g sin θ (θ angle of BC relative horizontal = 30°? Since dodecagon exterior angle; yes side BC makes 30° with horizontal). Repulsion from P2 opposes motion? P1 starts at B, slides down toward C; repulsion from P2 (positive) points along BC from P2 to P1, i.e., up toward B? Actually both positive, P1 at B, P2 further down (closer to C). Coulomb force on P1 is repulsive, directed away from P2, i.e., up along BC (toward B), opposing downward gravity. As P1 approaches P2, repulsion increases, eventually balances gravity component; at that point P1 reaches closest distance (maybe equilibrium) then maybe stops? Since no friction, it would oscillate? But maximum approximation when velocity zero? Let's formulate energy. P1 moves from B (distance from P2 initial D = d0? Wait distance between particles along BC = s - L0 = 1.0 m) to distance x (final distance between P1 and P2). Let coordinate x = distance from B along BC? Or distance between particles r. Initially r0 = d0 = 1.0 m. P1 at B, P2 fixed at distance L0 from C, so distance P1-P2 along BC = side length s - L0 = 1.0 m. P1 slides downward by distance Δ = r0 - r (if final distance r). The vertical drop of P1: side BC angle below horizontal = 30°, so vertical drop = Δ sin30 = Δ/2. P1 loses gravitational potential m g Δ/2. Initial kinetic zero; final at maximum approximation maybe kinetic zero? If "aproximação máxima" could be turning point where velocity zero? Need consider work done by electrostatic repulsion: electric potential energy increases from initial to final. Spring? P1 not attached. Energy conservation for P1: ΔEp_grav + ΔEp_electric = 0 (if final kinetic zero). But maybe due to repulsion, P1 stops before equilibrium? Actually if released from rest, gravity accelerates downward; as it approaches P2, repulsion increases; velocity may still positive until net force reverses? The closest approach occurs when velocity zero, not necessarily force balance? It could be after passing equilibrium point? In a conservative 1D system with potential U(x) = gravity (linear decreasing with x? Actually going down lowers gravitational potential, so potential decreases; electric potential increases as r decreases, with 1/r). Potential total = -mg x sinθ + k_e q1 q2 / x (depending). It has a minimum; starting from x = r0 =1.0, if initial potential? The turning point at total energy equals initial: -mg Δ/2 + k q1 q2 (1/r - 1/r0) = 0. So r determined by balance of energy, not force. But maximum approximation maybe equilibrium? If released at B and allowed to slide, it might first accelerate, reach equilibrium (force balance) where potential minimum, then continue due to kinetic energy, reach closer turning point where kinetic energy zero, then reverse. "Aproximação máxima" likely the turning point, not equilibrium. Need decide. But "assim que as partículas alcançam a aproximação máxima entre elas, todo o sistema será instantaneamente girado" maybe no kinetic energy mentioned "sem adicionar energia cinética às partículas" — maybe at maximum approximation they are momentarily at rest? It says "sem adicionar energia cinética às partículas" after rotation; maybe because they have no kinetic energy at that instant? It says "sem adicionar energia cinética" could mean rotation doesn't add kinetic, but could have existing kinetic? Hmm "assim que as partículas alcançam a aproximação máxima entre elas, todo o sistema será instantaneamente girado ... sem adicionar energia cinética às partículas." If they are at maximum approximation, in a conservative oscillation between release and turning point, velocity zero at maximum approximation. So no kinetic energy. The phrase "sem adicionar energia cinética" likely means the rotation process doesn't change speeds (which are zero). So part a via energy conservation (including electrostatic potential) and maybe spring? Wait spring initially relaxed and compressed by P2 due to repulsion. Energy could be shared with spring; but in part a they may ignore spring for P1 motion? They only say "constante elástica muito alta, despreze a variação da posição da P2 para efeitos de cálculo no movimento da P1"; so P1 moves as if P2 fixed, but spring energy? Since P2 displacement negligible but force on P2 huge? P1's energy: electrostatic potential between P1 and P2 + gravitational potential of P1. Spring energy not part of P1? But if P2 fixed, no spring displacement? However P2 experiences force; spring compresses slightly, storing energy. For P1, force from P2 equals Coulomb; P2 position fixed, so potential energy of P1-P2 system includes Coulomb energy. Spring? Since spring is attached to P2 and C; if P2 is fixed approximately, spring energy maybe zero? But if P2 doesn't move, no spring compression; but P1-P2 repulsion would push P2 into spring; if k high, P2 moves infinitesimal but force is balanced by spring, so spring energy finite? Actually at static equilibrium, spring force equals electrostatic force; displacement x = F/k. If k→∞, x→0, spring energy 1/2 k x^2 = F^2/(2k)→0. So can ignore spring energy for P1. But item b asks estimate ratio between elastic potential energy and gravitational potential energy variation of P2; if we ignore displacement, both maybe zero? But they ask ratio, perhaps need consider P2's motion due to spring while P1 moves, with P2 position variation maybe small but not negligible compared to? Need understand. Maybe there is another process: P2 is initially fixed at spring; as P1 approaches, P2 is pushed toward C compressing spring. Since spring constant very high, but P2 may still move a tiny bit. The maximum approximation of P1 and P2 occurs when spring is maximally compressed? Actually if spring constant high, P2 position almost fixed near C; P1 reaches closest distance to P2. The elastic potential energy of spring equals work done by P1-P2 repulsion? Could be approximated by Coulomb force times? Wait P2's displacement Δx due to repulsion from P1. The spring force balances Coulomb at maximum? But if P2 moves, it also changes distance. Since high k, displacement small; ratio may be expressed using energy balance for P2: work done by electrostatic force on P2 equals elastic potential energy? At maximum approximation, P1's velocity zero; P2 also maybe momentarily at rest? "sem adicionar energia cinética" suggests both at rest? Need examine. Let's consider system of P1, P2, spring. External? No friction, vacuum; gravity acts on P1 (and P2 also has gravity? "massas das partículas P1 e P2: m = 0,4 kg" both have same mass. P2 also slides on BC with gravity and maybe normal; gravity acts on P2 too. Its gravitational potential variation could be significant because P2 moves down slope? Initially P2 at distance L0 from C; as P1 pushes P2 toward C, P2 moves upward along slope? Wait coordinate: B top left, C bottom right. P2 is between B and C near C. If P1 moves from B down toward C, it approaches P2; repulsion pushes P2 down toward C (away from P1), i.e., in direction of gravity (down slope). Actually P1 is above P2, so repulsive force on P2 points down along slope toward C (same as gravity component). So P2 would slide down toward C if not for spring? But spring anchored at C and relaxed, so spring can pull P2 back toward C? Wait spring at C attached to P2; if P2 moves toward C, spring compresses? Let's define orientation: segment B-C, B at higher end (left/up), C at lower end (right/down). Spring anchored at C extends left/up along BC to P2. P2 is at distance L0 from C, so spring length L0. If P1 above P2 repels P2, force on P2 is down the slope (away from P1 toward C), which would compress spring? Since C is the anchor; moving P2 toward C decreases distance between P2 and C, compressing spring (length < L0). Spring force on P2 would point away from C (up slope, toward B) to restore. Yes. So P2 tends to be pushed toward C; spring compresses. At maximum approximation, P1 and P2 maybe both have zero velocity? P2 might have moved closer to C; spring elastic energy. Also P2 gravitational potential decreases if it moves toward C (lower height). Item b: ratio between elastic potential energy of spring and variation of gravitational potential energy of P2. Since high k, displacement small; can estimate via work/energy balance for P2. But note P2 also experiences electrostatic force from P1; and possibly kinetic energy? At maximum approximation both may be instantaneously at rest? It says "sem adicionar energia cinética às partículas" after rotation, but before rotation at maximum approximation maybe no kinetic? Actually if P2 has moved and spring compressed, it might have kinetic energy? Hmm system conservative, starting from rest. Maximum approximation might be turning point where both relative motion stops? Need model. Given "constante elástica da mola: muito alta; despreze a variação da posição da partícula P2 para efeitos de cálculo no movimento da partícula P1" indicates for P1's motion, we can neglect P2 displacement, so treat P2 fixed. But for item b, need compute elastic energy using small displacement of P2 due to force from P1 (likely at maximum approximation). Maybe we can find from force balance? Actually if P2 is at rest at maximum approximation? Hmm. Let's read original problem maybe from IME? It might have a solution known? Let's derive carefully. Geometry: - Dodecagon regular side length s = d0 + L0 = 1.0 + 0.15 = 1.15 m. - Side BC inclined at 30° below horizontal; vertical drop per unit length = sin30 = 0.5. Part a: Maximum approximation distance between P1 and P2. Need use conservation of mechanical energy for P1 (with P2 fixed) and electric potential. Let final distance r between P1 and P2. Initial distance = d0 = 1.0 m. P1 drops vertical h = (d0 - r) * sin30 = (1.0 - r)/2. But note if P2 fixed at distance L0 from C, then as P1 moves from B to distance r from P2, distance along BC traveled = d0 - r (since initial P1-P2 distance = d0). Yes. Final P1 height lower by h. Electric potential energy between two point charges: U_e = k q1 q2 / r (with zero at infinity). Initial U_e0 = k q1 q2 / d0. Since both positive, U_e increases as r decreases. Energy conservation: mg h + (U_e_final - U_e_initial) = 0? Gravity does positive work, electric does negative work (repulsion), so ΔK = W_grav + W_electric. If final K=0: mg h + [k q1 q2 (1/r - 1/d0)] = 0? Wait work electric = -ΔU_e = -(U_f - U_i) = U_i - U_f = k q1 q2 (1/d0 - 1/r). This is negative. So ΔK = mg h + k q1 q2 (1/d0 - 1/r) = 0. Equivalently mg h = k q1 q2 (1/r - 1/d0). Yes. Compute: m=0.4 kg, g=10 -> mg = 4 N. h = (1.0 - r)/2 (m). Electric term: k q1 q2 = 9e9 * 20e-6 * 5e-6 = 9e9 * 100e-12 = 900? 9e9 * 1e-10 = 0.9? Let's compute: 20 μC = 20e-6 C; 5 μC = 5e-6 C; product = 100e-12 = 1e-10 C^2; times 9e9 = 9e9 * 1e-10 = 0.9 N·m^2? Actually 9e9 * 1e-10 = 9e-1 = 0.9 N·m^2? Units: k N m^2/C^2 * C^2 = N m^2; divided by r (m) gives N m = J. So constant A = k q1 q2 = 0.9 J·m? Wait 0.9 N m^2? Actually N m^2 / m = N m = J. So A = 0.9 J·m? Numerically 0.9. Then mg h = 4 * (1 - r)/2 = 2(1 - r) J. Electric: A(1/r - 1/1) = 0.9(1/r - 1). Set 2(1 - r) = 0.9(1/r - 1) = 0.9(1 - r)/r. If r ≠ 1, divide (1 - r): 2 = 0.9 / r => r = 0.9 / 2 = 0.45 m. Interesting. So final distance r = 0.45 m. This seems plausible. But wait if r < 1, yes. But is this too simple? Need consider initial height difference? P1 starts at B; P2 fixed at distance L0 from C; initial distance = d0 = 1.0. Fine. But side length? We didn't need side length except to know d0; okay. But maybe need include spring effect on P2? For part a "na condição de aproximação máxima entre as partículas, determine a distância entre P1 e P2". Since variable r solved as 0.45 m. However if P2 moves, distance different? But observation says neglect P2 variation. So yes. Part b: "na condição de aproximação máxima entre as partículas, estime a razão aproximada entre a energia potencial elástica da mola e a variação da energia potencial gravitacional de P2". Need compute ratio E_el / ΔE_g(P2). We need calculate P2's displacement at maximum approximation. Since P1 at distance r=0.45 from P2; P2 fixed? But spring compresses due to force? How to estimate? Maybe using energy balance for P2: At maximum approximation, P2 may not be at rest? But "sem adicionar energia cinética" later? Let's parse: "Considera que, assim que as partículas alcançam a aproximação máxima entre elas, todo o sistema será instantaneamente girado ... sem adicionar energia cinética às partículas. A partir dessa nova configuração, a mola é desacoplada do sistema e totalmente retirada do vértice C e da partícula P2, sendo que as duas partículas passam a interagir entre si livremente." It doesn't say P2 at rest before rotation; but "sem adicionar energia cinética" after rotation maybe means rotation doesn't add; but if P2 had kinetic energy, it would persist? However "sem adicionar energia cinética" not "sem que as partículas tenham energia cinética". But maybe at maximum approximation, both are instantaneously at rest? Need likely. For part b, estimate ratio. Since k very high, P2 displacement x is small; elastic energy = 1/2 k x^2; gravitational potential variation of P2 = m g Δh_P2 = m g x sin30 = m g x/2 if P2 moves up? Wait direction: P2 pushed toward C (down slope) by repulsion, so it loses gravitational potential (moves down), so ΔE_g_P2 = - m g x sin30 = - m g x/2? Variation could be negative. But they ask razão entre energia potencial elástica e variação da energia potencial gravitacional de P2; maybe ratio of magnitudes? They might want compare magnitudes. Since elastic energy positive, gravitational variation negative; "variação" might be absolute? Usually "variação" can be final - initial, negative; ratio would be negative if elastic positive? Maybe likely they ask "razão aproximada entre a energia potencial elástica da mola e a variação da energia potencial gravitacional de P2" and expect positive? Need infer. To estimate x, use force balance? At maximum approximation, if P2 is at rest? Hmm for P2, forces along BC: gravity component mg sin30 = 4*0.5=2 N down slope; electrostatic repulsion from P1; spring force. If P2 at rest at max? If maximum approximation occurs when P1 and P2 relative distance minimal; but P2 may have acceleration. Could use work-energy for P2: Initially at rest, P2 compressed by electrostatic force; if kinetic zero final, work from electrostatic and gravity equals elastic energy? Actually for P2, work done by electrostatic force from P1 (which varies with P1 position) plus work by gravity = change in kinetic + elastic potential. If at final (maximum) P2 maybe velocity zero? Then W_electric_on_P2 + W_grav_on_P2 = ΔE_el. But we don't know W_electric_on_P2 easily because force from P1 varies. However maybe approximate ratio using impulse? Alternatively, at maximum approximation, P1's electric force on P2 equals spring force plus gravity? At equilibrium? Wait if P2 is at turning point (max displacement), its velocity zero, but acceleration nonzero. Not force balance. But maybe "aproximação máxima" means relative velocity zero, not necessarily P2 velocity? Actually if P1 and P2 both on same line, relative distance r. Maximum approximation = minimum r. Then relative velocity dr/dt = 0. But individual velocities may not be zero? Initially both zero, so maybe both? Let's derive equations for two particles with gravity and spring. P1 free (no spring), P2 attached to spring. External gravity on both. At minimum r, relative velocity zero. It doesn't require each velocity zero. But because initial total momentum? Gravity external, so total momentum not conserved. Rotation later "sem adicionar energia cinética" maybe they have kinetic energy? Hmm. Let's set coordinates. Let s along BC downward from B to C. P1 coordinate x1 (distance from B), P2 coordinate x2 (distance from B). Initially x1=0 (at B), x2 = s - L0 = d0 = 1.0 m? Wait x2 = distance from B to P2 = s - L0 = d0 =1.0 m (since side length 1.15, L0=0.15). P1-P2 distance r = x2 - x1 (if P1 above P2; x1 < x2). Initially r0=1.0. Equations (using prime for time derivative): m x1'' = m g sinθ - F_e (if F_e repulsion pushes P1 up) = mg sinθ - F_e (since positive direction down; F_e on P1 points up, negative). Here θ=30°, g sinθ = 5 m/s²? Wait g=10, sin30=0.5 -> component along slope = 5 m/s². So mg sinθ = 2 N. Let's use accelerations: a1 = g sinθ - F_e/m = 5 - F_e/0.4? Actually mg sinθ = 4*0.5=2 N, /0.4=5 m/s². For P2: m x2'' = m g sinθ + F_e - F_spring (if positive down). Spring force on P2: if compressed by y = x2 - x2_initial? Need define x2_initial = d0 =1.0 m? But P2 initial at x2=1.0, spring length L0 = s - x2 = 0.15 if s=1.15. If P2 moves down (x2 increases), spring length decreases: L = s - x2. Compression = L0 - L = L0 - (s - x2) = x2 - (s - L0) = x2 - d0. So y = x2 - d0 (positive if P2 moves down). Spring force restoring on P2 points up (negative): -k y. Thus m x2'' = m g sinθ + F_e - k y. At maximum approximation, relative coordinate r = x2 - x1. r'' = x2'' - x1'' = (g sinθ + F_e/m - k y/m) - (g sinθ - F_e/m) = 2F_e/m - k y/m. At minimum r, r'=0 and r'' >=0. Not necessarily zero. But if k very high, y small. Also if P2 almost fixed, x2'' small? Since k large, x2'' = -k y/m + ... may be large? Actually if k high, y tiny; acceleration may not be small? For high k, natural frequency sqrt(k/m) large; displacement y can be small but acceleration = -ω² y *? Finite? Hmm. Energy method maybe. Total mechanical energy (including elastic and gravitational and electric): E = 1/2 m x1'^2 + 1/2 m x2'^2 + m g (-x1 sinθ) + m g (-x2 sinθ) + 1/2 k y^2 + U_e(r) = constant? Let's define potential: gravity potential = -mg x sinθ for both (since x down slope lowers height). Electric potential = k_e q1 q2 / r (zero at infinity). Spring = 1/2 k y^2. Initially all zero kinetic, spring y=0, electric U_e(r0), gravitational potential = -mg x2_initial sinθ (P1 at x1=0, P2 at x2=d0). Since arbitrary constant, we can set initial E = -mg d0 sinθ + U_e(r0). At maximum approximation, r'=0, but also y'? y = x2 - d0; x2' = y'; x1' = x2' - r' = x2'. So both velocities equal x2'? Since r' = x2' - x1' =0 => x1'=x2'=v. So they move together at same velocity. Kinetic energy = 1/2 m v^2 + 1/2 m v^2 = m v^2. v may not zero. So maximum approximation not necessarily zero kinetic energy. Oh! Important. Because both particles can move together; relative distance minimum when velocities equal, not necessarily zero. Since P2 attached to spring and can move, and P1 can move; at closest approach they have same velocity along slope. There may be kinetic energy. But problem says after maximum approximation "sem adicionar energia cinética" maybe they have some kinetic energy; rotation doesn't add. Then after spring removed, they interact freely and P2 eventually reaches vertex; need know initial kinetic energy. So we must solve dynamics including kinetic energy at closest approach. The observation "despreze a variação da posição da P2 para efeitos de cálculo no movimento da P1" maybe simplifies: P2 approximately fixed (x2 ≈ d0, v2 ≈ 0) for P1's motion, so at closest approach P1 has zero velocity? But if P2 fixed, then P1's velocity zero at turning point; but if P2 moves, P1 and P2 may have common velocity nonzero. Which is intended? Let's examine. The problem statement: "Considera que, assim que as partículas alcançam a aproximação máxima entre elas, todo o sistema será instantaneamente girado ... sem adicionar energia cinética às partículas. A partir dessa nova configuração, a mola é desacoplada ... e as duas partículas passam a interagir entre si livremente." This suggests at that moment they have kinetic energy; after spring removed, they interact freely, so their subsequent motion includes that kinetic energy. If at max approximation they were at rest, trivial? Part c asks determine velocity of P2 ao atingir um vértice. Need use energy conservation after spring removed, starting from configuration after rotation. We need know initial kinetic energy of particles at that moment. That kinetic energy likely comes from part a/b? Let's model. We need solve full problem likely including P2 motion due to spring? But observation says neglect P2 position variation for P1 motion; but for energy balance maybe include P2 kinetic? Hmm. Let's search memory? IME 2027? This might be a recent exam? I don't know. Need infer from numbers. Let's carefully parse the sequence: 1. P1 abandoned from rest at B; P2 initially at rest at distance L0 from C (spring relaxed). P1 slides on segment BC; P2 can also slide on BC but spring high stiffness. 2. "assim que as partículas alcançam a aproximação máxima entre elas" -> at that instant, rotate system 30° CCW without adding kinetic energy. "A partir dessa nova configuração, a mola é desacoplada... e da partícula P2, sendo que as duas partículas passam a interagir entre si livremente." So at the instant after rotation and removal of spring, both particles have velocities as they had at maximum approximation (maybe not zero). Then they move under mutual electrostatic repulsion and gravity on the dodecagon? Wait "interagir entre si livremente" means after spring removed, the only forces on particles are gravity and electrostatic? They slide on dodecagon? The text says "podem deslizar sobre o segmento BC" earlier; after rotation, they are on a different side? Figure 3 shows particles on a horizontal side; they may slide along it; at some point P2 reaches a vertex of dodecagon. Need determine velocity of P2 when it reaches a vertex. So indeed they have initial velocities. Need determine velocities at maximum approximation from dynamics. Maybe because P2's displacement negligible, but its velocity not negligible? If spring high stiffness, elastic potential energy might be negligible? But part b asks ratio elastic energy to gravitational variation of P2; if elastic energy negligible, ratio small; part c maybe P2 velocity due to electrostatic repulsion? Let's explore. Let's derive full two-particle system with high k. Since spring is very stiff, P2 undergoes small oscillations? Initially relaxed; when P1 approaches, repulsion pushes P2 toward C, compressing spring; P2 accelerates, then spring force decelerates; maximum compression occurs when P2 velocity zero? Wait maximum approximation between P1 and P2 might coincide with maximum compression? Let's analyze. Let x1, x2 along BC. Let x2 = d0 + y (y small). P1 starts at x1=0, x2=d0. Repulsion F_e = k_e q1 q2 / r^2 where r = x2 - x1. P1 moves down, x1 increases; P2 may move down slightly y>0; r = d0 + y - x1. Initially r=d0. For P1: x1'' = g sinθ - F_e/m = 5 - F_e/0.4? Actually F_e/m = k_e q1 q2/(m r^2) = 0.9/(0.4 r^2)? Wait k_e q1 q2 = 0.9 J m? Let's compute: F_e = k q1 q2 / r^2 = 0.9 / r^2 N? Since k q1 q2 = 0.9 N m^2; divide by r^2 (m^2) gives N. Yes F_e = 0.9 / r^2 N. Then acceleration = F_e/m = 0.9/(0.4 r^2)=2.25/r^2 m/s². Good. For P2: m x2'' = mg sinθ + F_e - k y. Since k high, y small. To leading order, if k→∞ with y finite? Let y = O(1/k) maybe. Then P2 acceleration? k y ≈ mg sinθ + F_e? Actually if P2 quasi-static, spring force balances external force: k y = mg sinθ + F_e. Then y = (mg sinθ + F_e)/k small. Then P2 acceleration negligible (since k y finite, m x2'' ≈ 0? Actually if x2'' = (mg sinθ + F_e - k y)/m, with k y approx external, then x2'' small). So P2 position nearly fixed, velocity small; but displacement y small. Good. So at closest approach, P2 velocity maybe small but not zero? Since y changes slowly, x2' = y' ~ O(1/k)? So P2 velocity negligible compared to P1? Then relative velocity r' ≈ -x1' (since x2' small) so maximum approximation when P1 velocity zero (turning point). Thus P1's kinetic zero; P2 kinetic also negligible. So initial kinetic energy after removal might be zero? But part c then P2 starts from rest on horizontal side? Wait if P1 velocity zero at turning point, but then gravity continues? At turning point, P1 would reverse; but before reversal, rotate/remove spring. After rotation, P1 and P2 have zero velocity; they are on horizontal top? Then only electrostatic repulsion pushes them apart, and gravity? On horizontal top, gravity component along side is zero? Actually after rotation, side is horizontal, so gravitational potential constant for motion along side? But they may slide along horizontal side and then down adjacent sides? "ao atingir um vértice do dodecágono" maybe P2 moves horizontally due to repulsion, then leaves? Hmm If initial velocities zero, then after spring removed, P1 and P2 at distance r=0.45; electrostatic repulsion will push them apart; P2 (or P1) may slide on horizontal side to a vertex, then down? The question asks determine velocity of P2 ao atingir um vértice do dodecágono. That suggests P2 starts from rest maybe, accelerated by electric repulsion, but also gravity? On horizontal side, no gravity component; but once past vertex, maybe P2 enters next side and gravity component? Need parse Figure 3: after rotation, side containing particles is top horizontal. P2 is near right end? It may move rightwards due to repulsion from P1 (since P1 left; repulsion pushes P2 away from P1, i.e., right toward C vertex) or maybe P1 left; P2 to right of P1? In Fig3, P1 near left, P2 near right. Repulsion pushes P1 left toward B and P2 right toward C. So P2 moves right along horizontal side BC toward C. "ao atingir um vértice do dodecágono" could be vertex C? It reaches C? But if it has initial kinetic maybe. Maybe after rotation, P2 is on a horizontal top side and will slide until vertex C; when reaching vertex, velocity asked. Since spring removed, electrostatic force only along horizontal; gravity no component on horizontal side; so energy conservation: electric potential energy initial (r=0.45) plus kinetic; final at vertex C? But as P2 moves right, its distance from P1 increases (P1 maybe also moves left), electric potential decreases, kinetic increases. Need compute. But we need know initial kinetic energy after rotation. Perhaps system was rotated, "sem adicionar energia cinética às partículas" but at maximum approximation, they might have kinetic energy due to motion before? We need calculate it. The earlier simplified energy without kinetic might be wrong for part c; must include dynamics. Let's solve more accurately. Let's read observation: "como a constante elástica da mola é muito alta, despreze a variação da posição da partícula P2 para efeitos de cálculo no movimento da partícula P1". This likely means treat P2 fixed for P1's motion, so ignore P2 displacement. But for part c, maybe need know velocity of P2? If P2 displacement neglected, its velocity also? They ask velocity of P2 after reaching vertex, so need its motion; but maybe use conservation of energy including electrostatic energy: P2 initially at rest after max approximation? Hmm. Maybe the phrase "sem adicionar energia cinética às partículas" at rotation: "assim que as partículas alcançam a aproximação máxima entre elas, todo o sistema será instantaneamente girado ... sem adicionar energia cinética às partículas." If at max approximation they have zero kinetic energy (because relative velocity zero and P2 fixed), then rotation doesn't add kinetic. After removal, they start from rest. That would be consistent. But why would P1 have zero velocity at max approximation if P2 fixed? P1 released from rest at B; goes down, repulsion increases; when electric force equals gravity component, it reaches equilibrium and would continue? Let's check if energy solution r=0.45 yields what velocity? If at r=0.45, force balance? Electric force at r=0.45: F_e=0.9/(0.45^2)=0.9/0.2025=4.444 N. Gravity component along slope = mg sin30 = 2 N. After equilibrium? Actually net force on P1 upward (F_e > gravity) at r=0.45, so it would be decelerating; velocity might still be zero? Wait if at equilibrium force zero at r? Solve F_e = mg sinθ ⇒ 0.9/r² = 2 ⇒ r² =0.45 ⇒ r≈0.6708 m. So equilibrium at r=0.6708, closer than 1.0 but farther than 0.45. At r=0.45, electric force > gravity, so net force up; P1 has passed equilibrium; velocity zero at turning point before? If released at r=1.0, it accelerates down until r=0.6708 (equilibrium), then kinetic max? Actually at equilibrium net force zero, velocity maximum; then net force up decelerates until r=0.45 where velocity zero, then reverse. So r=0.45 is turning point, yes. It has zero velocity at r=0.45. So energy conservation with kinetic zero at turning point holds. Good. Thus part a r=0.45 m. Part b: At max approximation, P2? Since P1 is at rest momentarily at r=0.45; P2 maybe also at rest? If P2 quasi-static, its velocity negligible; but maybe P2 also at max compression? Let's examine P2 displacement y at that instant. If P2 had been moving slightly, at r=0.45 turning point, P1's velocity zero, but P2 might have nonzero velocity? However if k very high, P2 oscillates? Let's derive quasi-static equilibrium for P2: It is pushed by F_e + mg sinθ; spring force k y balances; y = (F_e + mg sinθ)/k = (0.9/r² + 2)/k. At r=0.45, y = (4.444+2)/k = 6.444/k. As P1 moves, r decreases, F_e increases, y increases; P2 moves slowly toward C. Its velocity y' = d/dt[(0.9/r² +2)/k] = (-1.8 r'/r³)/k? Since r decreasing (r'<0), y' positive (P2 moving down) and small. At r=0.45, r'=0 (P1 turning point), so y'=0. So P2 also has zero velocity instantaneously. Great. So both at rest. Elastic energy = 1/2 k y² = 1/2 k [(F_e + mg sinθ)/k]² = (F_e + mg sinθ)^2/(2k). This tends to zero as k→∞, but ratio with gravitational variation? Gravitational variation of P2 = mg y sinθ = mg sinθ * y = mg sinθ * (F_e+mg sinθ)/k. Ratio = (E_el)/(ΔE_g_P2?) = [(F_e+mg sinθ)^2/(2k)] / [mg sinθ (F_e+mg sinθ)/k] = (F_e+mg sinθ)/(2 mg sinθ) = (F_e + 2)/(4)? Wait mg sinθ = 4*0.5=2 N. Actually F_e + mg sinθ = F_e + 2. Ratio = (F_e + 2)/(2*2?) Let's compute: denominator = mg sinθ * y = 2*y. Numerator = 1/2 k y² = 1/2 k y * y = 1/2 (F_ext) y, where F_ext = F_e+mg sinθ = k y. So ratio = [1/2 F_ext y] / [mg sinθ y] = F_ext / (2 mg sinθ) = (F_e + mg sinθ)/(2 mg sinθ). Since mg sinθ=2, ratio = (F_e +2)/4. At r=0.45, F_e=0.9/0.2025=4.444; ratio = (6.444)/4 = 1.611. So approx 1.6. But if they ask "estime" maybe use approximation: as k→∞, displacement y→0, both energies→0, ratio finite. So part b = 1.6 (or 8/5? compute exactly maybe). Let's verify exact expression maybe ratio = (k_e q1 q2 / r^2 + mg sinθ)/(2 mg sinθ). With values: A/r² = 0.9/0.45² = 0.9/0.2025 = 900/202.5? Simplify: 0.9/(0.45^2)=0.9/0.2025 = 4.444... = 40/9? 0.2025=81/400? 0.45=9/20; square=81/400; 0.9=9/10; divide = (9/10)/(81/400)= (9/10)*(400/81)=3600/810=40/9 ≈4.444. mg sinθ=2=18/9. Ratio = (40/9+2)/(4) = (40/9+18/9)/4 = (58/9)/4 = 58/36 = 29/18 ≈1.611. So part b ≈29/18. Maybe they ask "razão aproximada" maybe 1.6. Need decide format. We'll derive exact. Part c: After rotation, spring removed; particles interact freely on dodecagon. Need determine velocity of P2 ao atingir um vértice. Need model motion after rotation. At moment after rotation, particles are on a horizontal side (formerly BC) with P1 and P2 separated by r=0.45 m? Need confirm positions relative to vertices after rotation. Original side BC length s=1.15 m; initial P1 at B, P2 at distance L0=0.15 from C, so P1-P2 distance r0=1.0. At max approximation, r=0.45 m. Since P2 barely moved, P1 moved downward by distance d0 - r = 0.55 m from B? Wait initial P1 at B, P2 at x2=d0=1.0 from B? Actually P2 fixed at distance from C = L0=0.15, so from B to P2 = s-L0 =1.15-0.15=1.0. P2 at x2=1.0 from B; P1 at x1=0. At closest, distance r = x2 - x1 =0.45. If P2 fixed x2≈1.0, then x1=0.55 m from B. So P1 has moved 0.55 m down BC; P2 at 1.0 m from B, so P1 is above P2; distance 0.45. After rotation, side BC becomes horizontal with B at left? Let's map: P1 at x1=0.55 from B, P2 at x2=1.0 from B. Since side length 1.15, C at x=1.15 from B. Thus P2 is 0.15 m from C (consistent L0); P1 is 0.55 from B, 0.60 m from C. P1 left (closer to B), P2 right (closer to C). In Fig3, P1 left, P2 right, yes. Now after spring removed, both positive charges repel: P1 pushed left toward B, P2 right toward C. Both can slide on the horizontal side. Since horizontal, gravity has no component along motion (but normal? The dodecagon side is horizontal top; normal vertical supporting weight; no friction). After P2 reaches vertex C? But P2 is 0.15 m from C; P1 is 0.60 m from C. Repulsion pushes P2 toward C. It will reach C first? It might then enter adjacent side? The question "determine a velocidade de P2 ao atingir um vértice do dodecágono" — likely P2 reaches vertex C (or some vertex). It may leave the horizontal side at C and then? But as it reaches vertex, it may still be on side BC? Actually "ao atingir um vértice do dodecágono" = when reaching vertex C, at the instant it arrives at C, before possibly moving along next side. So energy conservation along horizontal side from initial separation r=0.45 to instant P2 at C. Need consider P1 also moves left toward B? Since repulsion pushes P1 left; it might reach B? But distance from P1 to C when P2 at C? Let's compute. Let coordinate along horizontal side after rotation: B at x=0 (left), C at x=s=1.15 (right). P1 at x1=0.55, P2 at x2=1.0 initially. Repulsion: P1 moves left (x1 decreases), P2 moves right (x2 increases). They both accelerate. At final when P2 reaches C, x2=s=1.15. What is x1 at that instant? Since no external horizontal force, electrostatic forces internal equal and opposite; so center of mass of the two-particle system (along horizontal line) remains at rest? Initially both at rest after rotation? Wait at maximum approximation they were at rest? Yes after rotation initial velocities zero. Then no external horizontal force (gravity vertical, normal; no friction), so total momentum along horizontal conserved (zero). Thus m x1' + m x2' =0 => x1 + x2 = constant = x1_0 + x2_0 = 0.55 + 1.0 = 1.55 m (from B). At final x2=1.15, so x1 = 1.55 - 1.15 = 0.40 m from B. Thus P1 also moved left by 0.15 m; P2 moved right by 0.15 m. Separation final r_f = x2 - x1 = 1.15 - 0.40 = 0.75 m. Initial r0=0.45 m. Electric potential energy decreases (charges separate), converted into kinetic energy of both particles. Since masses equal, kinetic energy split equally: each gets half of ΔU_e. P2 velocity magnitude = sqrt(2 * (ΔU_e/2)/m) = sqrt(ΔU_e / m)? Wait total kinetic = 2*(1/2 m v^2)=m v^2 (if both same speed magnitude; yes). So m v^2 = ΔU_e = k q1 q2 (1/r0 - 1/r_f) = A(1/0.45 - 1/0.75). Then v = sqrt(A/m * (1/0.45 - 1/0.75)). But need include gravity? On horizontal side no gravitational potential change. After P2 reaches C, maybe it drops down next side? But "ao atingir um vértice do dodecágono" means at instant at vertex, no further height change. So okay. Compute: A = 0.9 J·m? Actually k q1 q2 = 0.9 (units? Let's verify earlier: 9e9 * 20e-6 * 5e-6 = 9e9 * 100e-12 = 0.9 J*m? N m^2? Let's ensure: k q1 q2 / r has units J; numerator unit N m^2; denominator m. So constant A = 0.9 N m^2, not J m? But F_e = A/r² has units N; U_e = A/r has units N m = J. So A = 0.9 N m^2. For energy difference A(1/r0 - 1/rf) units: 0.9 N m² * (1/m) = 0.9 N m = J. Good. ΔU_e = 0.9 * (1/0.45 - 1/0.75) = 0.9 * (2.222... - 1.333...) = 0.9 * 0.888... = 0.8 J? Compute: 1/0.45 = 20/9? 0.45=9/20; inverse=20/9=2.222. 1/0.75 = 4/3=1.333. Difference = 20/9 - 4/3 = 20/9 -12/9 =8/9 ≈0.8889. Times 0.9 = 0.9*8/9 = 0.8 J. Yes ΔU_e = 0.8 J. Then m v² = 0.8 J, m=0.4 kg => v² = 0.8 / 0.4 = 2.0 (m²/s²). v = √2 ≈1.414 m/s. Is that the answer? Need consider if both move? Yes. But wait: Is P2 initially at rest after rotation? We assumed yes because at max approximation P1 velocity zero and P2 velocity zero due to quasi-static. If not, initial kinetic energy might add. Need maybe compute more accurately: The spring might have imparted kinetic energy to P2; but due to high k, velocity small? Let's estimate: y = (F_ext)/k, velocity y' = - (F_e')/(k?) Actually F_e = 0.9/r², F_e' = -2*0.9 r'/r³ = -0.8e? derivative wrt time. At turning point r'=0, so y'=0. So yes P2 velocity zero. But before turning, P2 had small velocity; at turning zero. So initial K=0. But maybe after rotation, "sem adicionar energia cinética" means the rotation itself doesn't add kinetic, but the particles may already have kinetic energy from motion before; at maximum approximation, maybe not zero? Let's revisit: The closest approach might not be a turning point if P2 moves? But with high k, it's quasi-static; P2 velocity small but P1 velocity zero at turning point; so total K ~0. Thus okay. Let's check part c if there is another interpretation: After spring removal, particles interact freely, but they are on the dodecagon; "ao atingir um vértice do dodecágono" could mean P2 moves down along the dodecagon under gravity plus electric? Actually after rotation, the side BC is horizontal top. P2 pushed toward C reaches vertex C. At that instant, maybe velocity? We computed. But maybe "um vértice do dodecágono" could refer to vertex B? P1 might reach B? But asks velocidade de P2, so P2 reaches C. Good. Need maybe include gravitational potential energy variation for P2 during motion after rotation? On horizontal top, no change. But after reaching a vertex, if continues? No, at vertex velocity. However, we must ensure initial separation after rotation is indeed r=0.45 and positions x. The rotation by 30° may change which side; but relative positions preserved. The question says "durante o giro, as posições relativas das partículas não se alteram." So separation along same physical side preserved. So r=0.45. The side after rotation is horizontal; particles remain on that side. Yes. Let's read item c wording: "após a condição da Figura 2, determine a velocidade de P2 ao atingir um vértice do dodecágono." "após a condição da Figura 2" maybe means after the situation of Figure 2 (max approximation and rotation?) Actually "após a condição da Figura 2" could mean after the instant described in Figure 2? Figure 2 is initial? Let's parse: "Na Figura 2 ... P1 abandonada ... Considera que, assim que ... girado ... A partir dessa nova configuração, a mola é desacoplada ... Essa situação está representada na Figura 3." Then item c: "após a condição da Figura 2, determine..." Hmm maybe they mean after the condition of Figure 2 (initial release? No, after the maximum approximation? Actually "condição da Figura 2" maybe the configuration at maximum approximation before rotation? But item c asks velocity after reaching vertex, likely after Figure 2 process? Let's parse: - Fig1: dodecagon with A,B,C. - Fig2: P1, P2 on BC with spring. (Initial? Actually maybe P1 abandoned at B, spring relaxed; P2 at C? But Fig2 shows P1 near B, P2 near C? It might represent initial condition? The text says "Nas Figuras 1 e 2, a partícula P1 ... é abandonada..." Not necessarily at max. - Fig3: after rotation and spring removed, new configuration. Then item a: "na condição de aproximação máxima ... determine distância" refers to before rotation? Yes. Item b: same condition. Item c: "após a condição da Figura 2" maybe "após a condição [descrita na Figura 2, i.e., após a aproximação máxima e o giro?]" It says "após a condição da Figura 2, determine a velocidade de P2 ao atingir um vértice do dodecágono." Could be after the situation of Figure 2 (maximum approximation) and subsequent rotation/removal? We think. Let's consider if part c maybe uses energy conservation including gravitational potential from initial configuration through entire process? But rotation before removal, then free motion; no additional energy. We used initial at max approximation (zero K) then electric potential energy to final at vertex. But what about gravitational potential change between initial release and max approximation? That energy was dissipated? No, stored as electric potential; at max approximation, system has gravitational potential lower, electric potential higher, kinetic zero. The total energy at max approximation equals initial energy? We can calculate total energy at max = initial. After rotation and spring removal, no energy added; so total energy after rotation = initial total energy (relative to some reference) including gravitational potential of both at final orientation? Rotation doesn't change gravitational potential? Since rotating around center? It might change heights of particles! Wait! The system is rotated as a whole in anti-clockwise by 30°; positions of particles rotate with dodecagon. This changes their heights relative to gravity! Important. The rotation of the entire dodecagon by 30° will change the gravitational potential energy of both particles because their vertical coordinates change. But text "sem adicionar energia cinética" not "sem realizar trabalho"; they rotate instantaneously maybe with some external torque, but no kinetic energy added? The gravitational potential energy after rotation might differ from before due to height changes, so total mechanical energy may change if external work done by rotation? "sem adicionar energia cinética às partículas" but the rotation could change potential energy and maybe work done by external agent? Hmm "instantaneamente girado ... sem adicionar energia cinética" means external work might be done to rotate, but no kinetic added; energy conservation not trivial? But maybe positions relative and heights? If rotate dodecagon about its center, the heights of particles change; gravitational potential energy changes. However "sem adicionar energia cinética" implies the external torque does work? Actually "sem adicionar energia cinética" means the rotation doesn't impart kinetic energy, but could change potential; then subsequent motion might include gravitational energy change from rotation? Need account. We need determine heights before and after rotation. Since dodecagon rotates 30° CCW; side BC originally sloping down at 30°; after rotation, side BC horizontal. The vertical position of the side changes? Let's calculate. Regular dodecagon: original top side AB horizontal. Side BC is one of the upper sides. Rotating the whole dodecagon by 30° CCW around its center maps vertex B? Let's assign coordinates to determine height of B and C before/after. Let dodecagon centered at origin. Circumradius R = s/(2 sin15°). For a regular dodecagon with top side horizontal, let's set vertices. There are 12 vertices. If top side AB is horizontal, its midpoint at top. The vertices can be parameterized by angle φ_k = 90° - k*30°? Let's derive. A regular dodecagon with a horizontal top side: The top side AB is horizontal; there are two vertices with same maximum y. The central angles from vertical axis to vertices? For a regular n-gon with one side horizontal, the side's midpoint is at top. The vertices adjacent to top side are at angles ±15° from vertical? Actually side length s subtends central angle 30°. The top side is perpendicular to radius to its midpoint; so its endpoints are at angles 15° from vertical? Let's set center O at (0,0). The top side midpoint is at (0,R). The top side endpoints A and B have y = R cos15°? Wait coordinates: vertex at angle θ from vertical (positive y axis upward): coordinates (R sinθ, R cosθ). The side between vertices at angles -15° and +15° from vertical? Let's check: If vertices have same y? For top side horizontal, both endpoints have same y = R cos15°, and their x coordinates ±R sin15°. Distance = 2R sin15° = s. So yes. Top side is from angle -15° to +15°? Let's set B maybe at angle +15°? A at -15°? Actually side AB top horizontal. B is right endpoint? C adjacent to B on right? The next vertex after B around CCW? If traverse from A (left) to B (right) top side, then next side B-C turns clockwise? Let's use standard: vertices in clockwise order? Let's set A at angle -15°, B at angle +15° from vertical (y-axis). Then vector from B to C: C is adjacent after B going around polygon clockwise? The central angle between vertices is 30°. If top side is between angles -15° and +15°; next vertex clockwise from B (decreasing angle? Actually around polygon from A(-15°) to B(+15°) via top side, then to C? Let's draw: top side horizontal; the right endpoint B is upper right. Going clockwise around polygon, next vertex C should be at even smaller angle? Wait if top side at top, right side slopes down. From B, the polygon turns exterior angle 30° downward; C has angle +? Let's parameter. Let's set center origin, top side endpoints have polar angles 90° ± 15°? If using standard angle from +x axis: top side is at 90°? The midpoint at 90° (0,R). Endpoints at 90°±15° = 75° and 105°. So B right endpoint at 75°, A left endpoint at 105°? But both y=R sin75°? sin75°=cos15° ~0.966. Yes. Next vertex after B clockwise: exterior angle 30° clockwise => polar angle decreases? From B=75°, next C =45°? Actually a regular dodecagon has vertices every 30°. Starting at 105° (A), then 75° (B), then 45° (C), 15°, -15°, -45°, etc. Thus B=75°, C=45°. Segment B-C: coordinates difference: B=(R cos75, R sin75), C=(R cos45, R sin45). Vector B->C = R(cos45-cos75, sin45-sin75). cos45=0.707, cos75=0.259 => dx=0.448R; sin45=0.707, sin75=0.966 => dy=-0.259R. Angle = atan2(-0.259,0.448) = -30°? Yes. Good. So side BC slopes down-right at -30°. Top side AB horizontal. Now rotate entire system 30° CCW about center? The vector B->C originally angle -30°; after +30° becomes 0° (horizontal). The side BC becomes horizontal at top? Its midpoint? B and C polar angles increase by 30°: B from 75° to 105°, C from 45° to 75°. So after rotation, B is at 105° (upper left), C at 75° (upper right)? Wait B originally right endpoint of top side; after rotation, B moves to angle 105° (left of top?), C to 75° (right). So the side BC becomes horizontal at top, with B at left (105°) and C at right (75°). That matches Figure 3: B left, C right. But note A originally at 105°, after rotation A moves to 135° (left upper?). So Figure 3: A at left? Let's see: A after rotation at 135° (upper left), B at 105° (top left), C at 75° (top right). The top horizontal side is B-C. Good. Now gravitational potential of particles: They are on side BC. Originally before rotation (initial configuration? The particles are on side BC at angles 75° to 45°? B at 75°, C at 45°). After rotation, side BC becomes at angles 105° to 75°, with B at 105°, C at 75°; so the entire side is higher? Let's compute vertical heights. Initial (Fig2) before rotation at maximum approximation: P1 and P2 on side BC; B at 75°, C at45°? Actually side BC from B (75°) to C (45°); P1 at some point between B and C. After rotation, side BC from B (105°) to C (75°). The y-coordinate of any point on the side is higher because central angles increased by 30° (closer to top? Actually 105° is lower than 90°? sin105=0.966; sin75=0.966; midpoint maybe? Wait C after rotation at75° has same y as B before rotation; B after at105° has same y as A before (top side). So the whole side shifted? The midpoint of side BC before: angle (75+45)/2=60°, y=R sin60=0.866R. After: (105+75)/2=90°, y=R sin90=R. So the side moved higher, increasing gravitational potential by R*? This rotation changes gravitational potential energy of both particles if external work done. But "sem adicionar energia cinética" maybe means the rotation is done by external forces that also change potential energy; but after rotation, the total energy relative to new configuration includes increased gravitational potential. However since they rotate the whole system instantaneously, the external work may be whatever; but we need energy conservation from after rotation. To compute final velocity, we need total mechanical energy after rotation: kinetic zero (if at rest) plus gravitational potential (new) plus electric potential (same separation). Then as particles slide on horizontal side, gravitational potential remains constant? On horizontal side, yes. At final when P2 reaches vertex C? But note after rotation, side BC is horizontal, not changing height. However after P2 reaches vertex C, if it continues to next side? The question says "ao atingir um vértice" at vertex, so still at same height as C? But the horizontal side's height is constant from B to C; so electric potential energy change only converts to kinetic. So gravity not involved after rotation on that side. But what about the initial kinetic energy? We assumed zero; but does gravitational potential increase due to rotation affect? It doesn't convert during sliding because horizontal. But wait, the initial total energy before rotation (at max approximation) had P1 and P2 at lower heights. After rotation, they are higher; total energy increased by external work. But item c asks velocity after rotation; we need use energy after rotation only. So we can ignore initial. We only need initial positions and energy after rotation: kinetic zero + electric potential at r=0.45 (same) + gravitational potential (constant during horizontal sliding). At final electric potential lower; gravitational same; kinetic equals decrease in electric potential. So our calculation for part c did not use gravitational potential change. Good. But perhaps the rotation by 30° is not around center but around a vertex? "todo o sistema será instantaneamente girado no sentido anti-horário, de 30°" Maybe could be around vertical axis through B? But "posições relativas não se alteram" means rigid rotation. The axis could affect external work and kinetic? But if no kinetic added, and potential changes, the velocity after would depend on gravitational potential after rotation; but since side horizontal, no further conversion; but if axis different, the height of side might be different. Need determine if rotation around which point? "todo o sistema" likely dodecagon and particles; rotating it around its center? Since dodecagon has symmetry; to align side BC horizontal, a 30° rotation about the center indeed maps BC to horizontal top. Let's verify: Center rotation by +30° (CCW) maps B at75° to105°, C at45° to75°, so side BC becomes top side between 105° and75°, yes horizontal at y=R sin75? Wait midpoint at90°, y=R. Top side AB before was y=R? Actually top side AB has endpoints at 105° and75°, midpoint 90° y=R. So after rotation, BC becomes exactly the original top side position. Thus the side BC moves to where AB was, at top. So gravitational potential of particles increases by difference between side heights: from y around R sin60 (mid) to R (top). But since after rotation side is top horizontal, yes. However, item c: "após a condição da Figura 2, determine a velocidade de P2 ao atingir um vértice do dodecágono." If after rotation, the side B-C is horizontal top; P2 at distance from B? It moves right along horizontal side; at vertex C, it is at the right endpoint of top side. At that instant, maybe its height is same as initial after rotation; no change. So no gravitational energy. Good. Let's double-check part a and b with possible inclusion of rotation's effect on part b? Part b asks "na condição de aproximação máxima entre as partículas" likely before rotation; no rotation. So okay. Now need compute exact final velocity with masses and charges. Let's derive using symbols. Let: - k_e = 9e9 - q1 = 20e-6 C, q2 = 5e-6 C - A = k_e q1 q2 = 0.9 N·m²? Wait units: 9e9 N m²/C² * (20e-6 C)(5e-6 C) = 9e9 * 1e-10 N m² = 0.9 N m². Yes. - m = 0.4 kg. Part a: distance r. Use energy from release at B to max approximation. Need include both particles? Since P2 fixed? Observation neglect P2 displacement for P1 motion, but P2 also has gravitational potential? P2 gravitational potential change due to its tiny displacement negligible; so energy: P1 initial gravitational potential relative to final: m g Δh = m g (d0 - r)/2. Electric potential energy change: A(1/r - 1/d0). Set equal: m g (d0 - r)/2 = A(1/r - 1/d0). Solve: m g (d0 - r)/2 = A (d0 - r)/(r d0). Cancel (d0 - r): m g /2 = A/(r d0) => r = 2 A / (m g d0?) Wait let's solve: m g /2 = A/(r d0) -> r d0 m g /2 = A -> r = 2 A / (m g d0) = 2*0.9/(4*1.0)=1.8/4=0.45 m. Yes. Good. Part b: Ratio elastic potential energy to variation of gravitational potential energy of P2. Need derive at max approximation using small displacement y of P2. At max approximation, P1's net force? Since P1 velocity zero, acceleration? At turning point, not necessarily equilibrium; P2's spring force? For P2, if its velocity zero and displacement y, the spring force might not equal external? But if k high, we can set quasi-static; at max approximation, P2 is at rest; acceleration? Since P2's velocity zero, but its acceleration maybe nonzero. However if k high, spring force nearly equals external force plus inertial term; let's derive more rigorously for ratio. For P2: m x2'' = m g sinθ + F_e(r) - k y. At max approximation, r'=0; but x2'' maybe not zero. If k large and y small, k y = m g sinθ + F_e - m x2''. If x2'' ~? P2 displacement y due to external load; but due to sudden change? The external load grows slowly as P1 approaches; P2 follows quasi-static equilibrium, so x2'' ≈ 0. At turning point, P1's velocity zero, so time derivative of external force zero (since F_e depends on r, r'=0), so x2'' =0 as well. Thus force balance holds: k y = m g sinθ + F_e. Then ratio as above. Potential issue: The elastic potential energy of spring initially relaxed, and P2 has moved down by y; gravitational potential variation of P2 = -m g y sinθ (negative). Ratio = (1/2 k y²) / [m g y sinθ] = k y / (2 m g sinθ). Since k y = m g sinθ + F_e, ratio = (m g sinθ + F_e)/(2 m g sinθ) = 1/2 + F_e/(2 m g sinθ). With numbers F_e=40/9 N, m g sinθ=2, ratio = 1/2 + (40/9)/(4) = 1/2 + 40/36 = 1/2 + 10/9 = (9+20)/18? Wait 1/2=9/18, 10/9=20/18, total 29/18. Yes. Approx 1.61. But if "variação da energia potencial gravitacional de P2" is considered positive magnitude, same. If negative, ratio negative; but ratio of energies likely positive. Part c: Need more carefully account for electrostatic potential energy change from initial r=0.45 to final when P2 reaches vertex. But after rotation, positions on side: Need determine final distance r_f when P2 at vertex C. We assumed center of mass fixed: x1+x2 = constant because no external horizontal force. But wait there is also normal force and gravity vertical; no horizontal external force; however electrostatic force on each is internal; so total momentum conserved. But are there other forces along side? The particles slide on dodecagon; no friction; normal perpendicular; so yes. Initial velocities zero after rotation? We believe yes. But if P2 had small kinetic? It's zero at max approximation due to high k. So total momentum zero initially. Thus x1+x2 constant. But need know initial coordinates relative to B,C after rotation, especially because rotation may not change relative positions: P1 at distance from B? In original side, P1 moved from B by Δ = r0 - r =1.0 -0.45 =0.55 m. So P1 is 0.55 m from B along side; P2 is 1.0 m from B (since initial P2 at 1.0 from B). After rotation, B left, C right, so yes. Thus initial x1=0.55, x2=1.0. At final P2 at vertex C = x=1.15 (since side length s=1.15). Then x1 = (0.55+1.0) -1.15 =0.40 m. So final separation r_f = x2 - x1 =1.15 -0.40 =0.75 m. Yes. But wait side length s = d0 + L0 =1.15. If P2 at vertex C, spring removed; but maybe P2 could reach a different vertex? Let's see direction: P2 moves right from x=1.0 to C=1.15 (0.15 m) and reaches vertex C. P1 moves left from 0.55 to 0.40; still on side BC (not at B=0); so P2 reaches vertex first. So yes. Now energy conservation after rotation: Initial electric potential energy U_i = A / r_i = 0.9 /0.45 =2 J? 0.9/0.45=2 J. Final U_f = A / r_f =0.9/0.75 =1.2 J. ΔU =0.8 J. Initial kinetic zero. So final kinetic of both particles =0.8 J. Since masses equal, momentum zero, velocities equal and opposite: let P2 velocity v to right, P1 velocity v to left (magnitudes same). Total K = 1/2 m v² +1/2 m v² = m v². Thus m v² =0.8 => v²=0.8/0.4=2.0 => v=√2 m/s ≈1.41 m/s. If vector? "velocidade de P2" maybe speed? Need indicate direction? Along side away from P1, horizontal right. However, let's not rush: What about gravitational potential of P2 after rotation? The side after rotation is horizontal at top; P2 motion along horizontal; gravity no work. But if P2 reaches vertex C, it may have fallen from original vertical position? Wait after rotation, vertex C is at top right; no change in height while moving along side. So okay. But maybe after rotation, the dodecagon itself is fixed? The particles slide on the segment after rotation? The text: "A partir dessa nova configuração, a mola é desacoplada... e da partícula P2, sendo que as duas partículas passam a interagir entre si livremente." It says free interaction, but they still are constrained to slide on the dodecagon? "podem deslizar sobre o segmento BC" earlier; after rotation, "situação representada na Figura 3" shows them on new side. They likely slide on dodecagon sides. "ao atingir um vértice" means reach end of side. Yes. Let's verify if P1 might also be affected by gravity after P2 reaches vertex? At final at vertex, P1 not at vertex yet; no. If P2 continues past vertex, maybe next side slope? But question asks at vertex; no. Now, potential complication: The two particles have different charges (q1=20 μC, q2=5 μC). Electric force magnitudes equal (Newton's third law) but accelerations different? They have same mass 0.4 kg, so accelerations equal magnitudes. So velocities equal magnitudes if start from rest and internal forces; yes. Let's consider if after rotation, the distance between them initially might not be r=0.45 due to P1's previous motion? Yes it is. But part a maybe should include gravitational potential energy of P2? The statement "despreze a variação da posição da P2 para efeitos de cálculo no movimento da P1" means P2's gravitational potential change not considered in P1's motion? But if P2 moves slightly, its gravitational potential change is negligible; okay. Let's maybe compute part a more generally: m g (d0 - r) sin30 = k_e q1 q2 (1/r - 1/d0) Actually earlier: mg h = A(1/r - 1/d0). Here h = (d0 - r) sin30 = (d0 - r)/2. So m g (d0 - r)/2 = A(1/r -1/d0) = A(d0 - r)/(r d0) => r = 2A/(m g d0) = 2*0.9/(4*1)=0.45. Good. Part b: Let's ensure F_e at r=0.45 is 0.9/r² = 0.9/0.2025 =4.444 N. mg sin30=2 N. Ratio = (F_e + mg sinθ)/(2 mg sinθ) = (4.444+2)/4=1.611. Could also derive from total force on P2 at max: spring force = electric + gravity component along slope. Since P2's velocity zero? Actually at max approximation, P1's velocity zero, but P2 may have velocity? We argued negligible. But if we use force balance at equilibrium? No. Let's test if "aproximação máxima" maybe occurs when P1 and P2 have same velocity, not P1 zero; but with high k, P2 small. The energy solution for part a ignored P2 kinetic; if we include P2, part a might differ? Let's see if need more rigorous two-body energy: At max approximation, total kinetic maybe not zero; then part a r would be different. But observation allows neglect P2 displacement for P1 calculation; since P2 fixed, P1's final velocity zero. So part a okay. Part c: If we use total energy conservation from initial release through final vertex? Let's see if maybe after rotation, the external work from rotation changes gravitational potential; but if we choose reference after rotation, kinetic zero at start. We computed using only electric potential. But how do we know initial kinetic zero after rotation? The problem says "sem adicionar energia cinética às partículas" not "as partículas estão em repouso". If at max approximation they had zero kinetic, okay. If not, rotation could add none but they'd still have previous kinetic. However due to high k, previous kinetic negligible; perhaps intended. Let's maybe derive part c using conservation of energy including total potential from before rotation? But rotation external work unknown; however "sem adicionar energia cinética" maybe means the total mechanical energy (including gravitational potential) is conserved during rotation? Wait "instantaneamente girado ... sem adicionar energia cinética" could be interpreted as the rotation does not change kinetic energy but may change potential, but if external work? Actually "sem adicionar energia cinética" not "sem realizar trabalho"; so energy conservation might not hold across rotation if potential changes. But we don't need. Let's look at possible official answer? I might search memory? IME 2027 Física? The problem might have answers: a) 0,45 m; b) 29/18; c) √2 m/s? Seems plausible. Let's consider if part c final velocity should be 2 m/s? Let's recalc with gravity potential after reaching vertex? Maybe P2 after rotation is on a horizontal top side; but if it reaches vertex C, maybe it then starts down next side and gravity accelerates; but "ao atingir um vértice" exactly at vertex, velocity already includes any previous drop? The side BC after rotation is not necessarily horizontal at top? Wait rotation by 30° CCW might make side B-C vertical? Let's verify orientation more carefully. Original side AB top horizontal; side BC direction -30° (down-right). Rotate whole system by 30° CCW about center. Direction of BC becomes 0° (right horizontal). So yes horizontal. But the side BC after rotation is located where AB was originally? Since B from 75° to105°, C from45° to75°; yes side B-C top horizontal between angles 105 and75. However note if the dodecagon rotates by 30° CCW, the top side AB moves to angles 135° and105° (left upper). So BC becomes the new top side. Thus both particles are on top horizontal side; okay. Now, if after spring removal, P2 is pushed toward vertex C (right). But note C after rotation is the right endpoint of the top side, at angle75°. If P2 reaches C, it would then enter the next side CD which slopes down-right at -30°? The dodecagon adjacent side after C (angle75°) clockwise? Next vertex D at45° (since C 75°, D45°); side CD direction? From C to D vector: angles decrease, direction maybe -60°? Let's see if P2 at C may continue down; but question asks at reaching vertex, no. Let's maybe compute if P2 reached a vertex B instead? Repulsion pushes P2 away from P1; P1 left of P2, so P2 pushed right toward C. So vertex C. Let's also compute using energy including P1's final position: If P1 also moves left, final separation r_f =? At final, x2=1.15; x1 = COM - x2 = (0.55+1.0) -1.15 =0.40. So yes r_f=0.75. If we instead assumed P2 reaches vertex C but P1 remains at same position? Then r_f =1.15 -0.55=0.60, ΔU =0.9(1/0.45 -1/0.60)=0.9*(2.222-1.667)=0.9*0.555=0.5 J; v = sqrt(0.5/0.4)=1.118 m/s. But P1 must move due to repulsion; so 0.75 and √2. Let's derive with momentum conservation: Initially both at rest, no external horizontal force, so COM stationary. Good. But wait P2 has a spring attached initially; after rotation/removal, spring removed; before removal during rotation, spring still attached and maybe transient? But after rotation, spring removed before interaction; so yes. However, the text: "A partir dessa nova configuração, a mola é desacoplada do sistema e totalmente retirada do vértice C e da partícula P2, sendo que as duas partículas passam a interagir entre si livremente." It might be that the spring is removed after rotation, but the particles are still on the side; no external force horizontal. Potential issue: The rotation of the entire system by 30° might orient the side BC horizontally, but the particles are on it. The dodecagon is a physical regular polygon; after rotation, the side formerly BC is now at top; but what about the rest of polygon? The particles may be constrained to slide on the entire dodecagon? They are on segment BC; after rotation, segment BC is still a side. So yes. Let's maybe include the gravitational potential energy change of P1 after rotation? On horizontal side, no. But if P2 moves right and P1 moves left, both remain on horizontal side; no height change. So part c = √2 m/s. Let's check item b: "estime a razão aproximada entre a energia potencial elástica da mola e a variação da energia potencial gravitacional de P2". If exact with quasi-static, ratio 29/18. Maybe they expect using "constante elástica muito alta" to approximate the spring force as constant? Let's derive in a way accessible: At maximum approximation, the force on P2 due to gravity + electric is F = mg sin30 + F_e. Since spring is very stiff, compression x ≈ F/k (desprezível). Elastic energy E_el = 1/2 k x² = F²/(2k). Variação da energia potencial gravitacional de P2 = m g Δh = mg sin30 * x = mg sin30 * F/k. Ratio = F/(2 mg sin30). Substitute F = mg sin30 + F_e = 2 + 0.9/r² at r=0.45: F_e = 0.9/(0.45²)=40/9 N; F = 2 + 40/9 =58/9 N; ratio = (58/9)/(4)=58/36=29/18≈1.61. Let's ensure r=0.45 in part a not 0.45 m? Yes. Let's consider if electric force at max between particles maybe should be calculated at final distance r, not equilibrium; yes. Part a maybe they expect using potential energy with charges: k q1 q2 = 9e9 *20e-6*5e-6 = 0.9; then equation as above. Good. Let's maybe provide units: r = 0,45 m; b) 29/18 ≈1,6; c) v = √2 m/s ≈1,4 m/s. But if part c final velocity of P2 after reaching vertex, maybe also include direction "afastando-se de P1" or "para a direita". We'll state. Need craft final answer in Portuguese with development item by item, as a candidate would. Should show formulas and calculations. Let's write solution steps. Detailed solution: 1. Geometry: - side length \(L = s\): initial separation \(d_0 = L - L_0\) ⇒ \(L = d_0 + L_0 = 1,15\) m. - side BC inclined 30° below horizontal; vertical component per unit length = sin30 = 1/2. - electric constant \(K = 9\cdot10^9 \cdot (20\cdot10^{-6})(5\cdot10^{-6}) = 0,9\ \mathrm{N\,m^2}\). a) Energy conservation for P1 from release to maximum approximation (P2 fixed due to high k). If final separation is \(x\): \[ mg\frac{d_0-x}{2}=Kq_1q_2\left(\frac1x-\frac1{d_0}\right) \] \[ 4\cdot\frac{1-x}{2}=0,9\left(\frac1x-1\right) \] Careful: mg = 0.4*10=4 N. Left: 4*(1-x)/2 =2(1-x). Right: 0.9(1/x -1)=0.9(1-x)/x. Divide: 2 =0.9/x ⇒ x=0.45 m. Good. So a) 0,45 m. b) At max approximation, let P2 compression be y (tiny). Spring force balances component along BC plus electric force: \(k y = mg\sin30 + \frac{Kq_1q_2}{x^2}\). Elastic energy \(E_{el}=\frac12 k y^2 = \frac{(mg\sin30 + F_e)^2}{2k}\). Gravitational variation of P2: \(\Delta E_{g,P2}=mg\,y\sin30 = mg\sin30 \frac{mg\sin30+F_e}{k}\) (negative, but magnitude). Ratio: \[ R=\frac{E_{el}}{|\Delta E_{g,P2}|}=\frac{mg\sin30+F_e}{2mg\sin30} =\frac{2+\frac{0,9}{0,45^2}}{2\cdot2} =\frac{2+\frac{40}{9}}{4}=\frac{58/9}{4}=\frac{29}{18}\approx1,61. \] Need maybe mention \(F_e=0,9/x^2=40/9\) N. c) After rotation, side BC is horizontal, spring removed. Initial state: P1 at distance from B \(x_{1i}=0,55\) m? Actually P1 has descended \(d_0-x=0,55\) m from B, so yes distance from B along side \(x_{1i}=0,55\) m. P2 at \(x_{2i}=1,0\) m from B (since initially at 1,0 m from B? Let's verify: initial distance between P1 and P2 d0=1.0; P1 at B, P2 at d0 from P1, so yes P2 at 1.0 from B; spring relaxed length 0.15 means P2 at 0.15 from C, so side length 1.15). - At instant P2 reaches vertex C, \(x_{2f}=L=1,15\) m. - Since no external horizontal force and both started at rest, center of mass fixed: \(x_{1f}=x_{1i}+x_{2i}-x_{2f}=0,55+1,0-1,15=0,40\) m. Thus final separation \(r_f=x_{2f}-x_{1f}=1,15-0,40=0,75\) m. - Initial separation \(r_i=0,45\) m. Electric potential energy decrease: \[ \Delta U_e = Kq_1q_2\left(\frac1{r_i}-\frac1{r_f}\right) =0,9\left(\frac1{0,45}-\frac1{0,75}\right) =0,9\left(\frac{20}{9}-\frac43\right) =0,9\cdot\frac89=0,8\ \mathrm{J}. \] - This is total kinetic energy of both particles. Since masses equal, \(v_1=v_2=v\) (magnitudes) and total kinetic \(=\frac12mv^2+\frac12mv^2=mv^2\). So \[ mv^2=0,8 \Rightarrow 0,4v^2=0,8 \Rightarrow v^2=2 \Rightarrow v=\sqrt2\ \mathrm{m/s}\approx1,41\ m/s. \] The velocity of P2 is directed away from P1, i.e., toward C (right) at the instant it reaches the vertex. Need maybe mention "ao atingir um vértice" = C. Let's ensure part c formula for x_{2f} if P2 reaches vertex C = 1,15 m; but the side length 1.15; yes. P2 initially at x=1.0, so moves 0.15 m; P1 moves 0.15 m left; COM preserved. Potential nuance: P2 initially after rotation at rest? We assumed. If not, initial kinetic might be nonzero. But we can state "desprezando a pequena velocidade residual de P2, pois a mola é muito rígida, as partículas partem do repouso nessa nova etapa." Actually at maximum approximation, P2 also has zero velocity because P1's velocity zero and spring quasi-static. We'll mention. Let's maybe include "sem adicionar energia cinética" means initial kinetic after rotation zero; "despreze a variação da posição da P2" supports. Let's consider if part b ratio "variação da energia potencial gravitacional de P2" maybe positive variation = m g Δh = mg y sin30. If P2 moves down slope, its height decreases, so variation = -? But "variação" could be final - initial = -mg y sin30. Ratio would negative. But "razão aproximada entre ... e a variação" usually they might ask magnitude; we'll provide positive and mention em módulo. In final answer maybe "b) 29/18 (em módulo)". Good. Let's double-check part b using exact values: F_e = K q1q2 / x² = 0.9 /0.45² = 0.9 /0.2025 =4.4444 N = 40/9 N. F_total = 2 + 40/9 = (18+40)/9=58/9 N. Ratio = F_total / (2*mg sin30) = (58/9)/(4)=58/36=29/18=1.611. Yes. Potential alternative: If elastic energy vs variation gravitational of P2 maybe they don't use force balance but energy balance at maximum approximation: The work done on P2 by electric + gravity equals elastic energy? Work = integral F_net_on_P2 dx. If quasi-static, work done on spring equals? Actually if P2 quasi-static, net force ~0, so work done by external (electric+gravity) equals spring energy? Let's verify: For spring, infinitesimal work by external force = F_ext dx = k x dx; integrate = 1/2 k x². So same. Ratio = average? At final x = F_ext_final/k; energy =1/2 F_final x. Gravitational variation = mg sinθ x; ratio = F_final/(2 mg sinθ). Same. Now, final answer in Portuguese. Should I show "Resposta final:" line exactly. Let's craft final response: ``` a) ... b) ... c) ... Resposta final: a) 0,45 m; b) 29/18 ≈ 1,61; c) √2 m/s (≈1,41 m/s), afastando-se de P1. ``` Need maybe include the direction of P2 velocity? "ao atingir um vértice" maybe magnitude only; but can say "para a direita (afastando-se de P1)". Let's ensure units: b ratio dimensionless. Let's also maybe mention "dodecágono" not "octógono". Good. Let's write final in a clear step-by-step. Potential issue: In part a, I used electric potential energy constant \(Kq_1q_2=0.9\). But if \(K=9e9\), \(q1=20e-6\), \(q2=5e-6\): product q1 q2 = 100e-12 =1e-10; times 9e9 =0.9. Yes. Let's include "adotando o nível de referência no vértice B"? Not necessary. Let's maybe include a note about the vertical drop sign: - P1 desce \(d_0-r\) ao longo de BC, então perda de energia potencial gravitacional = \(mg(d_0-r)\sin30\). - Ganho de energia potencial elétrica = \(Kq_1q_2(1/r -1/d_0)\). - Igualam porque \(v=0\) na aproximação máxima. Part c: "A energia potencial elétrica inicial é \(Kq_1q_2/r_i\); a final é \(Kq_1q_2/r_f\). A diferença converte-se em energia cinética." Good. Let's maybe be mindful: After rotation, side BC is horizontal; but "ao atingir um vértice do dodecágono" P2 may reach C after moving 0.15 m while P1 moves left. We computed x1f using COM. Need justify: "Como não há força externa horizontal e as partículas partem do repouso, o centro de massa permanece parado." Good. Let's now produce final answer.

**Analyzing Physics Problem** I am currently focusing on understanding the initial conditions and setup of the physics problem. I've begun reading the provided text and am noting that a positively charged particle, $P_1$, is released from rest at vertex B of a dodecagon, as depicted in Figures 1 and 2. **Processing Interactions** I've moved on to analyzing the interactions between the charged particles and the spring. I'm noting that particle $P_1$ is released from rest and that the spring connects a screen at C to particle $P_2$. The system then rotates, and the spring is removed, leaving only electrostatic interactions. **Calculating Particle Distances** I am now processing the quantitative details provided. I see that $q_1 = +20\ \mu\text{C}$ and $q_2 = +5\ \mu\text{C}$, with a very high spring constant and $K_e = 9 \cdot 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$. I'm focusing on part (a), determining the maximum approach distance between $P_1$ and $P_2$. **Examining Dodecagon Geometry** I'm now scrutinizing the geometric properties of the regular dodecagon. My current focus is on determining the orientation and angles of its sides, especially how the horizontal top edge, AB, relates to the overall structure given the internal angles and exterior angles of 150° and 30° respectively. **Rotating Dodecagon Geometry** I'm now processing the effect of the counterclockwise $30^\circ$ rotation. What was initially a $30^\circ$ downward inclination for BC is now horizontal. This significantly alters the positions and relative angles of the subsequent edges and vertices in Figure 3. **Revising Dodecagon Orientation** I'm re-evaluating the dodecagon's orientation after the rotation. It appears my previous interpretation of AB's position was incorrect; the clockwise sequence starting A, B, C and the subsequent rotation clarify AB tilts downwards left. **Confirming Geometric Alignment** My latest thoughts confirm the dodecagon's geometry aligns perfectly with Figure 3 after the rotation. The vector BA now correctly points down and left from B. This means I can proceed to analyze the physics problem's initial conditions. **Analyzing Spring Constraint** I am now processing the significant implication of the infinitely high spring constant. This constraint implies particle $P_2$ remains stationary at its initial position while $P_1$ moves towards it. Therefore, $P_2$ is fixed at a distance $d_0 = 1.0\text{ m}$ from $P_1$'s starting point. **Calculating Initial Forces** I am now calculating the initial forces acting on $P_1$. I see gravity pulls it down the BC incline with $2\text{ N}$, while electrostatic repulsion from $P_2$ acts upwards along BC. I am confirming the absence of friction as stated. **Determining Closest Approach** I'm now focused on the definition of "closest approach." My analysis indicates this occurs when the speed of particle $P_1$ momentarily becomes zero, not necessarily when acceleration is zero. This is a crucial point for applying energy conservation. **Applying Energy Conservation** I am now applying the principle of conservation of mechanical energy. I've set up the equation by equating initial kinetic and potential energies (gravitational and electrostatic) to the final energies. The critical part is defining the gravitational potential energy change based on the distance moved along the inclined plane. **Solving Quadratic Equation** I have successfully derived a quadratic equation, $20 d^2 - 29 d + 9 = 0$, to determine the closest approach distance. My calculations for the discriminant ($\Delta = 121$) indicate two real roots, which I am now proceeding to solve. **Finding Closest Approach** I have successfully solved the quadratic equation for the closest approach distance, yielding two roots. One root corresponds to the initial separation, while the other, 0.45 m, represents the minimum distance. I've also confirmed that the particle's kinetic energy remains positive throughout this motion. **Analyzing Spring Attachment** I am currently analyzing the attachment of the spring. It's connected to vertex C and particle $P_2$. The prompt mentions the spring is initially relaxed, and I need to understand $P_2$'s behavior and potential gravitational potential energy changes. **Considering Spring Effects** I am now re-evaluating the "infinitely high spring constant" statement and its implication that $P_2$ is essentially stationary relative to $P_1$'s motion. This clarifies that $P_2$'s position is fixed for the purpose of $P_1$'s calculation, simplifying the energy conservation analysis. **Analyzing Spring Forces** I am currently grappling with the implications of the "infinitely high" spring constant on particle $P_2$'s behavior and the spring's energy. I need to determine if $P_2$ is effectively stationary or in quasi-static equilibrium, and how that affects the elastic and gravitational potential energies. **Calculating Energy Ratio** I am now focusing on part (b) and the ratio of elastic potential energy to the change in gravitational potential energy for $P_2$. The cancellation of $\Delta x$ is a significant realization, indicating the ratio is independent of the spring constant, which is key to solving this without its explicit value. **Analyzing Spring Equilibrium** I am re-evaluating whether particle $P_2$ is in static equilibrium. I'm considering if the force from $P_1$ combined with gravity and the spring force means $P_2$ is stationary. The calculation of electrostatic force at the closest approach distance is now a focus. **Revising Spring Dynamics** I'm now re-evaluating the behavior of particle $P_2$ given the "infinitely high" spring constant. My current thought is that $P_2$ experiences forces from $P_1$, gravity, and the spring, but due to the extreme stiffness, it likely remains in a quasi-static equilibrium, meaning its displacement is minimal. **Estimating Energy Ratio** I am now considering the phrase "estimate the approximate ratio" and its implications for how to approach the problem. This suggests a quasi-static approximation where particle $P_2$ is in approximate equilibrium. My current focus is on calculating the elastic potential energy and the change in gravitational potential energy for $P_2$ under this assumption, comparing $E_{el}$ with $\Delta E_{grav}$. **Calculating Energy Ratio** I am now focused on part (b), the ratio of elastic potential energy to the change in gravitational potential energy for $P_2$. I've realized that $\Delta x$ cancels out. My current calculation yields a ratio of approximately 1.61. I'm double-checking if this calculation only considers $P_2$'s potential energy change. **Estimating Energy Ratio** I am now refining my estimation of the energy ratio. My previous calculation of the elastic potential energy and the change in gravitational potential energy for $P_2$ under a quasi-static approximation is being re-checked. I am verifying the electrostatic force at the closest approach distance of 0.45 m to ensure accuracy. **Revising Energy Ratio Interpretation** I am re-evaluating part (b), specifically the energy ratio. I'm considering if the sign of the gravitational potential energy change for $P_2$ should be factored into the ratio, as it's a decrease. I am also questioning if $P_2$'s potential motion from a zero displacement needs further analysis rather than assuming a constant force. **Verifying Adiabatic Approximation** I'm confirming the quasi-static approximation's validity. Given the vast spring constant and approaching time significantly longer than oscillation periods, the system should indeed remain in near-equilibrium. This means the spring force will continuously adjust to balance electrostatic and gravitational forces on $P_2$. **Calculating Energy Ratio** I have confirmed the energy ratio calculation for part (b). My current thoughts indicate the ratio of elastic potential energy to the change in gravitational potential energy for $P_2$ is indeed $\frac{29}{18}$, which simplifies to approximately 1.61. This result aligns perfectly with the expected elegant solution. **Revising Energy Ratio Calculation** I'm re-examining the energy ratio calculation for $P_2$. My understanding is that $P_2$ is in quasi-static equilibrium, meaning its kinetic energy is negligible. I'm now focusing on whether the gravitational potential energy change for $P_2$ should be positive or negative in the ratio. **Confirming Initial Distances** I'm now confirming the initial configuration and distances. My thoughts indicate that particle $P_2$ remains at its relaxed spring position, 0.15 m from vertex C. This means the initial separation between $P_1$ and $P_2$ is the sum of the initial distance and the relaxed spring length, setting a crucial parameter for further analysis. **Confirming Final Positions** My current thoughts confirm the consistency of all calculated positions and distances after the closest approach. The final configuration, with $P_1$ at 0.55 m from B, $P_2$ at 0.45 m from $P_1$, and $P_2$ at 0.15 m from C, perfectly sums to the total length $L$, indicating a fully resolved state. **Analyzing Gravitational Force** I'm re-examining the impact of the dodecagon segment BC being horizontal. My recent thoughts reveal that gravity, acting vertically downwards, is now perpendicular to the segment. This means there's no gravitational component acting *along* BC, simplifying the force analysis to primarily electrostatic interactions. **Analyzing System Momentum** I am now considering the momentum of the $P_1$ and $P_2$ system. Since both particles are on a horizontal segment and only internal electrostatic forces act between them, the center of mass of the system must remain stationary. This implies that any movement of $P_1$ is directly related to $P_2$'s movement, but in opposite directions. **Conserving System Momentum** I've realized that due to equal masses and internal electrostatic forces, the linear momentum of $P_1$ and $P_2$ along the horizontal segment BC is conserved. This implies their velocities and displacements are equal in magnitude, a brilliant simplification I'm now exploring fully. **Conserving System Momentum** My latest thoughts confirm that with equal masses and only internal electrostatic forces acting, the total momentum of $P_1$ and $P_2$ along the horizontal segment BC is conserved. This means their velocities and displacements are equal in magnitude and opposite in direction. **Confirming System Dynamics** I am now confirming the system's behavior based on equal masses and internal electrostatic forces. This implies that the total linear momentum along the horizontal segment BC is conserved, leading to equal magnitudes of velocities and displacements, but in opposite directions. **Confirming Energy Conservation** I am now meticulously re-verifying every calculation step. The conservation of energy principle has yielded an exact kinetic energy of 0.8 J for the system and a velocity of $\sqrt{2}$ m/s for $P_2$. The elegance of these results is striking, and I am performing a deep check of all geometric and kinematic details. **Analyzing Initial Conditions** I am re-examining the initial conditions. My thoughts now clarify that BC becomes horizontal after rotation, and gravity's component along it is therefore zero. This is a crucial simplification for subsequent force calculations. **Confirming Energy Equation** My latest calculations confirm the derivation of the quadratic equation $2d^2 - 2.9d + 0.9 = 0$ from energy conservation. The roots, 1.0 m and 0.45 m, accurately represent the initial and closest approach distances for $P_1$. **Estimating Energy Ratio** I'm now focusing on part (b) and the energy ratio. My current thoughts suggest that $P_2$ remains in quasi-static equilibrium due to the infinitely high spring constant. I'm calculating the elastic potential energy and the change in gravitational potential energy for $P_2$ under this assumption, and I've determined the ratio is approximately 1.61. **Calculating Energy Ratio** My current focus is on part (b), the ratio of elastic potential energy to the change in gravitational potential energy for $P_2$. I've confirmed that $\Delta x$ cancels out and the ratio is indeed $\frac{29}{18}$, approximately 1.61, when considering magnitudes. I'm now exploring the sign convention for the gravitational potential energy change. **Confirming Gravitational Potential Energy** I am confirming my interpretation of the gravitational potential energy variation for $P_2$. The change from the initial relaxed state to maximum approach, $\Delta U_{g2}$, is correctly calculated as $-m g \Delta x \sin(30^\circ)$, which aligns with the value I used. **Confirming Center of Mass** My current thoughts focus on the center of mass of the $P_1$ and $P_2$ system. Given their equal masses and the absence of external horizontal forces, the center of mass will remain stationary. This implies that their velocities and displacements along BC will be equal in magnitude and opposite in direction. **Analyzing Particle Displacements** I'm now focusing on the displacement of particles $P_1$ and $P_2$ from their positions at maximum approach. Due to equal masses and conservation of momentum, their displacements along BC will be equal in magnitude and opposite in direction, a crucial insight for the next step. **Calculating Final Velocities** My current thoughts confirm that the total linear momentum of $P_1$ and $P_2$ along BC is conserved due to equal masses and internal forces. This implies their final velocities are equal in magnitude and opposite in direction, a key outcome for system dynamics. **Confirming Final Displacements** I'm now verifying the consistency of the final configuration. The equal magnitudes of displacements from the closest approach point along BC, with opposite directions, are confirmed. This ensures the conservation of momentum for the system. **Verifying Final Energies** My latest calculations confirm the total energy of the system at the point when $P_2$ reaches vertex C. I've equated initial energy to final energy and determined the resulting kinetic energy term for $P_2$ is 0.8 J, leading to a velocity of $\sqrt{2}$ m/s. **Calculating Final Velocities** I am now confirming the total momentum of the system. Given equal masses and only internal forces, the center of mass is stationary. This means $P_1$ and $P_2$ have equal and opposite velocities and displacements from the closest approach point, ensuring system momentum conservation. **Confirming Calculation Details** My latest thoughts confirm the successful calculation of the closest approach distance as 0.45 m using the derived quadratic equation. I am now focusing on part (b), calculating the ratio of elastic potential energy to the change in gravitational potential energy for particle $P_2$, yielding approximately 1.61 in magnitude.

**Analyzing Physics Problem Details** I'm currently dissecting the physics problem from the IME entrance exam. My immediate focus is on thoroughly understanding the provided figures and the initial conditions of the positively charged particle $P_1$. **Exploring Spring Behavior** My thoughts are now zeroing in on the spring's behavior. Given its "very high" spring constant, I'm contemplating how this effectively means its compression will be negligible, allowing us to treat it as if it starts at its relaxed length. **Considering Spring Stiffness** My focus shifts to the spring's properties. Its extremely high spring constant implies negligible compression, allowing me to assume it begins at its relaxed length, simplifying calculations for $P_1$'s motion. **Confirming Geometric Orientations** I've confirmed that side AB is indeed horizontal. This implies side BC makes a $30^\circ$ angle with the horizontal in Figures 1 and 2, aligning with the provided description of the system's rotation. **Analyzing Energy Conservation** I'm now focusing on applying energy conservation to particle $P_1$ as it slides down incline BC. I'll neglect $P_2$'s movement and consider the loss in $P_1$'s gravitational potential energy to be converted into electrostatic potential energy between the particles. My current calculations involve $m g \sin 30^\circ$ and the electrostatic constant term. **Calculating Maximum Approach** I've confirmed particle $P_1$ accelerates downwards, as the gravitational component exceeds electrostatic repulsion at $d_0=1$ m. This allows me to divide by $(1-d)$, yielding a clean result for maximum approach distance $d=0.45$ m. **Calculating Spring Interaction** My current focus is on the spring's elastic potential energy and $P_2$'s gravitational potential energy change. The spring, initially relaxed, compresses as $P_1$ approaches, requiring careful calculation of these energy components to determine the required ratio. **Investigating Spring Equilibrium** I'm realizing that the product $kx$, representing the elastic force, is what matters, not $k$ or $x$ individually. This suggests $P_2$ is in quasi-static equilibrium, simplifying the calculation of forces without needing the exact spring constant or compression. **Revisiting Spring Dynamics** I'm re-evaluating the spring's behavior, specifically how its initial relaxed state interacts with the very high spring constant. My current thought is that $P_2$ remains in quasi-static equilibrium, with the spring force always balancing the total forces acting on it, simplifying the energy calculations by allowing $x$ and $x_0$ to cancel. **Analyzing Forces on P2** I'm now scrutinizing the forces acting on $P_2$. Specifically, I'm confirming whether it's in true equilibrium or experiencing negligible acceleration at the point of maximum approach between $P_1$ and $P_2$. This distinction is crucial for the next steps. **Confirming Spring Force Calculation** I've confirmed the calculation for the spring force, finding it to be approximately $58/9$ N. This value arises from balancing gravity and electrostatic forces. Now, I am considering alternative interpretations of $P_2$'s motion and its equilibrium state. **Revising System Energy Balance** I'm reassessing the energy conservation for the entire system at maximum approach. Considering the coupled motion, the gravitational potential energy changes for both particles and the electrostatic potential energy must be accounted for, especially when accounting for $P_2$'s displacement and the spring's influence. **Revising Spring Force Calculation** I'm realizing the product $kx$ is key, simplifying $P_2$'s equilibrium analysis. The problem's phrasing "estime a razão aproximada" suggests the quasi-static equilibrium assumption, where spring force balances electrostatic and gravitational forces on $P_2$, is the intended approximation for this step. **Calculating Spring to Gravity Ratio** I'm now focused on the ratio of spring's elastic potential energy to the absolute value of $P_2$'s gravitational potential energy change. The phrasing implies using the magnitude for the ratio, though I will also note the negative sign for clarity on the actual change. **Confirming Algebraic Solution** I've discovered a remarkably clean algebraic simplification $\frac{1}{2}(1 + \frac{d_0}{d})$. This expression, derived from previous steps, strongly suggests it's the intended solution, confirming my approach to calculating the spring-to-gravity ratio. **Analyzing Initial Conditions** I am carefully examining the transition to Figure 3. My current focus is determining the exact positions and velocities of both $P_1$ and $P_2$ at the start of this phase, recognizing that the BC segment becomes horizontal and initial kinetic energy is zero. **Analyzing Initial Conditions** I am carefully examining the transition to Figure 3. My current focus is determining the exact positions and velocities of both $P_1$ and $P_2$ at the start of this phase, recognizing that the BC segment becomes horizontal and initial kinetic energy is zero. **Analyzing Particle Motion** My focus is now on the horizontal segment BC in Figure 3. Given equal masses and no external horizontal forces, the system's center of mass remains at rest. This means $P_1$ and $P_2$ will move with equal speeds, covering equal distances. I've confirmed $P_2$ reaches vertex C first. **Calculating Particle Velocities** I'm now calculating the final velocities of $P_1$ and $P_2$ as $P_2$ reaches vertex C. Using conservation of momentum and energy on the horizontal segment, I've determined that their speeds are equal, and I'm plugging these values into the derived energy equation to find $v_2$. **Confirming Velocity Calculations** I've successfully substituted the numerical values into the derived equations. The resulting calculations for $v_2$ yielded an exact $\sqrt{2}$ m/s, which, along with the clean fractional arithmetic, strongly validates the entire approach taken so far.

**Analyzing max condition correction** I'm looking into the max condition correction time, which is O(1/√k). If I opt for a sufficiently high k, I can choose the ω phase at the event effectively. It's intriguing how the initial ωT phase and dense initial phase variations could lead to a mapping with multiple stationary maxima. I wonder how these mathematical elements interplay and what implications they might have in practical scenarios. There’s a lot to unpack here!**Calculating energy dynamics** I'm trying to calculate the distance and energy involved in this scenario. I find that the distance d comes out to 0.75 and the energy change is 0.8 J. I'm concerned that there's no dissipation at P2 before equilibrium is reached. Since the figure states P1 is released but doesn't explicitly mention P2, I assume it must be released at the same time. Then, in part b, I will analyze free oscillations assuming P2 starts at rest. Overall, it looks good!