Consider two blocks A and B of masses \(\mathrm{m}_{1}=10 \mathrm{~kg}\) and \(\mathrm{m}_{2}=5 \mathrm{~kg}\) 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 \(\mathrm{k}=3000 \mathrm{~N} / \mathrm{m}\) 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)

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Conservation of Momentum

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✓ Correct
(D)0.3 m

Answer: (C)

Solution

While the spring holds a fixed compression the two blocks share one common velocity, found from momentum conservation as

v = \frac{m _{1} v _{0}}{m _{1} + m _{2}} = \frac{10 \times 3}{15} = 2 m/s

. The kinetic energy lost in reaching this common motion is stored in the spring:

\frac{1}{2} (10) (3)^{2} - \frac{1}{2} (15) (2)^{2} = 45 - 30 = 15 J

. Equating this to

\frac{1}{2} k x^{2}

with

k = 3000 N/m

gives

x^{2} = 0.01

, so the compression in the spring is

0.1 m

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Conservation of Momentum

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