JEE Advanced 2024 — Paper 2 — PHY-IV solution

JEE Advanced2024Paper 2PHY-IV

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 $ω$ . Thus, their positions at time $t$ are given by $x_{1} (t) = (x_{0} + d) + a sin ω t$ and $x_{2} (t) = (x_{0} - d) - a sin ω t$ , respectively, where $d > 2 a$ . Particle $3$ of mass $m$ moves towards this system with speed $u_{0} = aω /2$ , and undergoes instantaneous elastic collision with particle $2$ , at time $t_{0}$ . Finally, particles $1$ and $2$ acquire a center of mass speed $v_{c m}$ and oscillate with amplitude $b$ and the same angular frequency $ω$ .

Q1. If the collision occurs at time $t_{0} = 0$ , the value of $v_{c m} / (aω)$ will be _________.

Your answer

Solution

At $t_{0} = 0$ , particle $2$ is at its mean position ( $x_{0} - d$ ) moving with velocity $\frac{d x _{2}}{d t}_{0} = - aω$ (i.e. of magnitude $aω$ directed leftward).

Particle $3$ arrives from the left with speed $u_{0} = aω /2$ moving rightward.

In an elastic collision between equal masses, velocities are exchanged. After collision:

$v_{2} = + aω /2$ (now moves rightward at $aω /2$ ), and particle $3$ takes the original velocity of particle $2$ .

Particle $1$ is not touched, so its velocity at $t = 0$ remains $+ aω$ (rightward).

Centre-of-mass velocity of particles $1$ and $2$ after collision:

$v_{c m} = \frac{v _{1} + v _{2}}{2} = \frac{aω + aω /2}{2} = \frac{3 aω}{4} .$

$∴ \frac{v _{c m}}{aω} = \frac{3}{4} = 0.75.$

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