Consider two blocks A and B of masses m₁=10 kg and m₂=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)

Solution

Physics sir · 3 steps

Momentum gives the common speed, and the kinetic energy that goes missing is stored in the spring.

1. Common velocity. No external horizontal force acts, so momentum is conserved. With both blocks moving together, 10 × 3 = (10 + 5) V, so V = 2 m s⁻¹.
2. Kinetic energy lost. Before: 1/2 × 10 × 3² = 45 J. After: 1/2 × 15 × 2² = 30 J. The table is frictionless, so the missing 15 J is held in the compressed spring.

   1/2kx² = 15 ⇒ x² = (2 × 15)/3000 = 0.01

3. So. x = √0.01 = 0.1 m, which is option (b).

The idea to keep. When two bodies end up moving together through a spring, use momentum for the common speed. The kinetic energy lost in that perfectly inelastic step is exactly the spring energy 1/2kx².

Answer from the NTA final answer key. Solution by PrepWiser.