Physics · System Of Particles

JEE Main 2026 — 2 April, Evening Shift — Question 3

Two blocks of masses $2\mathrm{kg}$ and $1\mathrm{kg}$ respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in $2\mathrm{s}$ is $\mathrm{m}$ . (Take $\mathrm{g} = 10\mathrm{m / s}^2$ )

Option A:
0
Option B:
3.12
Option C:
2.22
Correct
Option D:
1.42

Answer: C

Step-by-step solution

$\mathrm{a} = \frac{(2-1)}{3}\times 10 = \frac{10}{3}\mathrm{m/s}^2$ , $\mathrm{a}_{\mathrm{cm}} = \frac{1\cdot10/3 - 2\cdot10/3}{3} = -\frac{10}{9}\mathrm{m/s}^2$ , $\mathrm{S}_{\mathrm{cm}} = \frac{1}{2}\cdot\frac{10}{9}\cdot 4 = \frac{20}{9}=2.22\mathrm{m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: System Of Particles
Topic: Conservation of Linear Momentum