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ExamsJEE MainPhysics

A block A of mass 5 kg rests on a smooth horizontal surface, attached to a spring (k = 200 N/m) that is compressed by 10 cm and held by a string. A block B of mass 5 kg moving at 2 m/s collides with A; during the collision the string breaks and afterward the spring returns to its natural length, releasing its stored energy into the blocks. Taking rightward as positive and treating the collision as elastic-type with the spring energy added, with final velocities v1 (of A) and v2 (of B), which statement is correct?

  1. v1 + v2 > 2
  2. Initial kinetic energy of system = final kinetic energy of system
  3. v1² + v2² = 4.4 (m/s)²
  4. v1 - v2 = 2

Correct answer: v1² + v2² = 4.4 (m/s)²

Solution

Momentum gives v1 + v2 = 2 m/s; energy conservation including the 1 J released by the spring gives (1/2)(5)(v1²+v2²) = (1/2)(5)(2²)/... leading to v1² + v2² = 4.4 (m/s)².

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