JEE Advanced 2024 · Paper 2 · Physics Q16

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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 $\omega$ . Thus, their positions at time $t$ are given by $x_1(t) = (x_0 + d) + a\sin\omega t$ and $x_2(t) = (x_0 - d) - a\sin\omega t$ , respectively, where $d > 2a$ . Particle 3 of mass $m$ moves towards this system with speed $u_0 = a\omega/2$ , and undergoes instantaneous elastic collision with particle 2, at time $t_0$ . Finally, particles 1 and 2 acquire a center of mass speed $v_{\text{cm}}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$ .

On a horizontal frictionless surface: particle 3 on the left moving right with speed u₀ toward particle 2; particles 2 (left) and 1 (right) are joined by a spring, and the dashed mark x₀ above the spring shows their centre of mass.

If the collision occurs at time $t_0 = 0$ , the value of $v_{\text{cm}}/(a\omega)$ will be _____.

Your answer

Answer and worked solution

Official answer: 0.74

Velocities before the collision. $v_1=a\omega\cos\omega t$ and $v_2=-a\omega\cos\omega t$ .

At $t_0=0$ , particle 2 moves at $-a\omega$ . Particle 3 comes from the left at $+\frac{a\omega}{2}$ .

Collision. An elastic collision between equal masses exchanges their velocities, so particle 2 leaves at $+\frac{a\omega}2$ . Particle 1 is unaffected and moves at $a\omega$ .
$v_{cm}=\frac{a\omega+\frac{a\omega}{2}}{2}=\frac34a\omega\implies\frac{v_{cm}}{a\omega}=0.75.$

Topics:Elastic and inelastic collisions Centre of mass Spring-mass systems

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