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JEE Advanced Physics Center of Mass Momentum and Collision 2024 JEE Advanced 2024 (Paper 2)

JEE Advanced Physics Question (2024) — Solution

Question

Paragraph II: Two particles, 1 and 2 , each of mass m, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x_0, are oscillating with amplitude a and angular frequency . Thus, their positions at time t are given by x_1(t)= (x_0+d )+a t and x_2(t)= (x_0-d )-a t, respectively, where d>2 a. Particle 3 of mass m moves towards this system with speed u_0=a / 2, and undergoes instantaneous elastic collision with particle 2 , at time t_0. Finally, particles 1 and 2 acquire a center of mass speed v_ cm and oscillate with amplitude b and the same angular frequency . If the collision occurs at time t_0=0, the value of v_ cm /(a ) will be ________ .

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

At t _0=0 Before collision After collision aligned v _ CM & = m a 2 + m a m + m \\ v _ CM & = 3 a 4 \\ V _ CM a & = 3 4 \\ V _ CM a & =0.75 aligned

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