JEE Main 2025 Apr 3 Shift 2, Physics Q10: Collision

JEE Main2025Apr 3, Shift 2Physics

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)

A
$0.2$ m
B
$0.4$ m
C
$0.1$ m
D
$0.3$ m

Solution

When both blocks move together with the spring in steady compression, they share a common velocity equal to the centre-of-mass velocity (momentum conservation):

$v_{c m} = \frac{m _{1} v _{1}}{m _{1} + m _{2}} = \frac{10 \cdot 3}{15} = 2$ m/s.

The kinetic energy lost from the original motion is stored as elastic PE in the spring:

$\frac{1}{2} k x^{2} = \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}$

$= \frac{1}{2} (10) (9) - \frac{1}{2} (15) (4) = 45 - 30 = 15$ J.

$x^{2} = \frac{2 \cdot 15}{3000} = \frac{1}{100} \Rightarrow x = 0.1$ m.

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JEE Main 2025 Apr 3 Shift 2, Physics Q10: Collision

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