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JEE Main201912 Jan 2019Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual

An alpha- particle of mass m suffers 1 - dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64 % of its initial kinetic energy. The mass of the nucleus is

Options

  1. A1.5 m
  2. B4 m
  3. C3.5 m
  4. D5 m

Correct answer

B. 4 m

Step-by-step solution

From momentum conservation, m v + 0 = - m v 1 + M v 2       . . . ( 1 ) Using kinetic energy conservation, 1 2 m v 2 = 1 2 m v 1 2 + 1 2 M v 2 2 ⇒ 1 2 m v 2 = 0 . 36 × 1 2 m v 2 + 1 2 M v 2 2 ⇒ 0 . 64 m v 2 = M v 2 2 ⇒ v 2 = 0 . 8 v m M As it retain 36 % , ⇒ 0 . 36 × 1 2 m v 2 = 1 2 m v 1 2   ⇒ 0 . 36 v 2 = v 1 2 Hence, v 1 = 0.6 v From ( 1 ) , m v = - m 0 . 6 v + M 0 . 8 v m M ⇒ 1 = - 0 . 6 + 0 . 8 M m On solving, we get M = 4 m

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