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
- A1.5 m
- B4 m
- C3.5 m
- 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