As per the given figure, two blocks each of mass 250 g are connected to a spring of spring constant 2 Nm⁻¹. If both are given velocity v in opposite directions, then maximum elongation of the spring is :

Solution

Physics sir · 4 steps

The centre of mass stays at rest, so both blocks stop together and all their KE goes into the spring.

1. Centre of mass. The blocks have equal mass and equal speeds in opposite directions, so the total momentum is zero. The centre of mass stays at rest the whole time.
2. At maximum stretch. At maximum elongation the blocks stop moving relative to each other. With the centre of mass at rest, that means both blocks are momentarily at rest. All the kinetic energy is now stored in the spring.

   2 × 1/2 m v² = 1/2 k x²

3. Put in numbers. With m = 0.25 kg and k = 2 N m⁻¹, the left side is 0.25 v² and the right side is 1/2 × 2 × x² = x². So x² = 0.25 v².
4. So. x = v/2, which is option (b).

The idea to keep. In a two-block spring problem, maximum stretch comes when the blocks share a common velocity. If the total momentum is zero, that common velocity is zero, and all the KE becomes spring energy.

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