JEE Advanced 2024 Paper 2 · Q16 · Work-Energy-Power

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 $\omega$. 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 $\omega$.

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

SolveKar AI · Accepted Answer

Velocities just before collision at $t_0 = 0$:
- $\dot x_1(0) = a\omega\cos 0 = a\omega$ (particle 1 moving in $+x$).
- $\dot x_2(0) = -a\omega\cos 0 = -a\omega$ (particle 2 moving in $-x$).
- Particle 3 incoming with speed $u_0 = a\omega/2$ towards particle 2 (i.e., velocity $+a\omega/2$ since 3 is to the left of 2).

Elastic collision between particle 3 (mass $m$, $v_3 = +a\omega/2$) and particle 2 (mass $m$, $v_2 = -a\omega$). For equal masses elastic collision, velocities exchange:
After: $v_3' = -a\omega$ (particle 3 takes 2's old velocity), $v_2' = a\omega/2$ (particle 2 takes 3's old velocity).

After collision, particles 1 and 2 are the bound oscillating system. Their centre-of-mass velocity:
$v_{	ext{cm}} = \dfrac{m\dot x_1(0^+) + m v_2'}{2m} = \dfrac{a\omega + a\omega/2}{2} = \dfrac{3a\omega/2}{2} = \dfrac{3a\omega}{4}$.

$\dfrac{v_{	ext{cm}}}{a\omega} = \dfrac{3}{4} = 0.75$.

Units: $a$ in m, $\omega$ in rad/s ⇒ $a\omega$ in m/s; ratio dimensionless. Answer: $0.75$.

JEE Advanced 2024 Paper 2 · Q16 · Work-Energy-Power

Solution