A stationary particle breaks into two parts of masses m A and m B which move with velocities v A and v B respectively. The ratio of their kinetic energies ( K B : K A ) is :

Physics · System Of Particles

JEE Main 2024 — 8 April, Shift 1 — Question 46

A stationary particle breaks into two parts of masses $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$ which move with velocities $\mathrm{v}_{\mathrm{A}}$ and $\mathrm{v}_{B}$ respectively. The ratio of their kinetic energies $\left(\mathrm{K}_{\mathrm{B}}: \mathrm{K}_{\mathrm{A}}\right)$ is :

Option A:
$v_{B}: v_{A}$
Correct
Option B:
$m_{B}: m_{A}$
Option C:
$m_{B} v_{B}: m_{A} v_{A}$
Option D:
$1: 1$

Answer: A

Step-by-step solution

Initial momentum is zero.

Hence $\left|P_{A}\right|=\left|P_{B}\right|$

$\Rightarrow \mathrm{m}_{\mathrm{A}} \mathrm{V}_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}$

$\frac{(\mathrm{KE})_{\mathrm{A}}}{(\mathrm{KE})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}}{\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^{2}}=\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}$

$\frac{(\mathrm{KE})_{\mathrm{B}}}{(\mathrm{KE})_{\mathrm{A}}}=\frac{\mathrm{v}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{A}}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: System Of Particles
Topic: Conservation of Linear Momentum

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Step-by-step solution

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