Physics · System Of Particles

JEE Main 2024 — 4 April, Shift 2 — Question 53

In a system two particles of masses $\mathrm{m}_{1}=3 \mathrm{~kg}$ and $\mathrm{m}_{2}=2 \mathrm{~kg}$ are placed at certain distance from each other. The particle of mass $m_{1}$ is moved towards the center of mass of the system through a distance 2 cm . In order to keep the center of mass of the system at the original position, the particle of mass $\mathrm{m}_{2}$ should move towards the center of mass by the distance $\qquad$ cm .

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{m}_{1}=3 \mathrm{~kg} \quad \mathrm{~m}_{2}=2 \mathrm{~kg}$

$\Delta \mathrm{X}_{\text {C.O.M. }}=\frac{\mathrm{m}_{1} \Delta \mathrm{x}_{1}+\mathrm{m}_{2} \Delta \mathrm{x}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}$

$\Rightarrow 0=\frac{3 \times 2+2(-x)}{3+2}$

$\Rightarrow \mathrm{x}=3 \mathrm{~cm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Physics
Chapter: System Of Particles
Topic: Motion of Center of Mass & Frame of CM