GnitGnit
ContestsProblemsBlogSubmit a ProblemMy PageSign In

© 2026 Gnit. All rights reserved.

Terms of ServicePrivacy PolicyThird-Party SoftwareContactOfficial X
Contests/Gnit Sunday Open 002 (GSO002)/Problem 16 Center-of-Mass Frame 2D Harmonic Oscillation: Trajectory of a Puck Under Special Attractive Force
Problem 16

Center-of-Mass Frame 2D Harmonic Oscillation: Trajectory of a Puck Under Special Attractive Force

Finished
500 ptsLv.8 IntermediateMechanics
2026/03/22 19:30〜2026/03/22 21:00
Author: admin

Problem Statement

On an infinitely wide smooth horizontal surface (xyxyxy-plane), there are two pucks A and B treated as point masses.

Between the two pucks, a special attraction mechanism of negligible mass is attached. When the distance between the pucks is rrr, this mechanism exerts an attractive force of magnitude KrKrKr (KKK is a positive constant) on each puck along the line connecting them. This is mechanically equivalent to being connected by a special spring with natural length 000.

At time t=0t = 0t=0, puck A is at the origin (0,0)(0, 0)(0,0) and puck B is at point (0,d)(0, d)(0,d) on the yyy-axis. Simultaneously, puck A is given initial velocity v⃗A(0)=(vAx,vAy)\vec{v}_A(0) = (v_{Ax}, v_{Ay})vA​(0)=(vAx​,vAy​) and puck B is given initial velocity v⃗B(0)=(vBx,vBy)\vec{v}_B(0) = (v_{Bx}, v_{By})vB​(0)=(vBx​,vBy​).

The two pucks then move under mutual attraction, tracing complex trajectories combining the translational motion of the center of mass and relative motion about the center of mass.

At time t1t_1t1​ (t>0t > 0t>0) when the distance between pucks A and B is first maximum, find the kinetic energy EKE_KEK​ of puck B in the stationary coordinate system fixed to the horizontal surface.

Constraints

  • Mass of puck A: mA=12 kgm_A = 12 \text{ kg}mA​=12 kg
  • Mass of puck B: mB=4 kgm_B = 4 \text{ kg}mB​=4 kg
  • Proportionality constant of attraction: K=75 N/mK = 75 \text{ N/m}K=75 N/m
  • yyy-coordinate of puck B at t=0t = 0t=0: d=6 md = 6 \text{ m}d=6 m
  • Velocity of puck A at t=0t = 0t=0: v⃗A(0)=(5,0) m/s\vec{v}_A(0) = (5, 0) \text{ m/s}vA​(0)=(5,0) m/s
  • Velocity of puck B at t=0t = 0t=0: v⃗B(0)=(−20,10) m/s\vec{v}_B(0) = (-20, 10) \text{ m/s}vB​(0)=(−20,10) m/s

Input Format

The kinetic energy EKE_KEK​ [J\text{J}J] is expressed as an irreducible fraction AB\frac{A}{B}BA​ with A,BA, BA,B coprime positive integers. Give the value of A+BA + BA+B as a positive integer.

Submit Answer

Please sign in to submit an answer

Sign In

Calculator

0
View Scoreboard
1234567891011121314151617181920
Read Solution Blog