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NEETPhysicsCenter of Mass, Momentum and Collision

Block A of mass 2 kg moving at 10 m s ⁻¹ on a smooth horizontal surface hits Block B of mass 2 kg initially at rest. The collision is perfectly elastic. Immediately after the collision, Block B enters a rough horizontal surface with a coefficient of kinetic friction 0.5 . Taking g = 10 m s ⁻² , the distance Block B slides before coming to rest is:

Options

  1. A2.5 m
  2. B20 m
  3. C10 m
  4. D5 m

Correct answer

C. 10 m

Step-by-step solution

Since the collision is perfectly elastic and the masses of Block A and Block B are equal, they exchange their velocities. Velocity of Block B just after the collision, v = 10 m s ⁻¹ . As Block B slides on the rough surface, it experiences a retardation a = g . a = 0.5 10 = 5 m s ⁻² Using the equation of motion v_f^2 = v^2 - 2as for the stopping distance s (where final velocity v_f = 0 ): 0 = (10)^2 - 2 5 s 100 = 10 s s = 10 m. Answer: 10 m

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