Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is

Solution

Physics sir · 4 steps

C hands all its speed to A; maximum squeeze comes when A and B move together at the same speed.

1. The collision. C and A have equal masses and collide elastically, so they swap velocities. C stops, and A starts moving with speed v while B is still at rest.
2. Maximum compression. The spring keeps shortening as long as A moves faster than B. It is shortest when both move with the same velocity. Momentum is conserved, so mv = 2m v_c, which gives v_c = v/2.

   1/2mv² = 1/2(2m)(v/2)² + 1/2Kx²

3. Energy balance. The right side has kinetic energy 1/4mv², so the spring stores 1/2mv² - 1/4mv² = 1/4mv². Setting 1/2Kx² = 1/4mv² gives x² = mv²/2K.
4. So. The maximum compression is x = v√(m/2K), which is option (d).

The idea to keep. In a two-block spring system, maximum compression or extension happens at the moment of common velocity. Use momentum for that velocity, then energy for the spring.

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