Consider two blocks A and B of masses m 1 =10 kg and m 2 =5 kg that are placed on a frictionless table. The block A moves with a constant speed v=3 m / s towards the block B kept at rest. A spring with spring constant k=3000 N / m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B , along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)

Physics · System Of Particles

JEE Main 2025 — 3 April, Evening Shift — Question 47

Consider two blocks $A$ and $B$ of masses $m_{1}=10 \mathrm{~kg}$ and $m_{2}=5 \mathrm{~kg}$ that are placed on a frictionless table. The block $A$ moves with a constant speed $v=3 \mathrm{~m} / \mathrm{s}$ towards the block $B$ kept at rest. A spring with spring constant $k=3000 \mathrm{~N} / \mathrm{m}$ is attached with the block $B$ as shown in the figure. After the collision, suppose that the blocks $A$ and $B$ , along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)

Option A:
0.2 m
Option B:
0.3 m
Option C:
0.4 m
Option D:
0.1 m
Correct

Answer: D

Step-by-step solution

COLM $10 \times 3+50 \times 0=15 V_{C}$

$v_{C}=2 \mathrm{~m} / \mathrm{s}$

COME $\frac{1}{2} k x^{2}=\left|\frac{1}{2}(m+m) v_{C}^{2}-\frac{1}{2} m v_{0}^{2}\right|$

$1500 x^{2}=\left|\frac{1}{2} \times 15 \times 2^{2}-\frac{1}{2} \times 10 \times 3^{2}\right|$

$1500 x^{2}=|30-45|$

$x^{2}=\frac{15}{1500}$

$x=\frac{1}{10}=0.1 \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: System Of Particles
Topic: Collisions in One Dimension