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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

A bullet of mass m strikes an obstruction and deviates off at 60 ° to this original direction. If its speed also changed from u to v , find the magnitude of the impulse acting on the bullet.

Options

  1. Am u 2 - uv + v 2
  2. Bm u 2 + uv - v 2
  3. Cm u 4 - uv + v 4
  4. Dm u 4 + uv - v 4

Correct answer

A. m u 2 - uv + v 2

Step-by-step solution

impulse = Δ P x i ^ + Δ P y j ^ Δ P x = mv cos 60 o - mu ⇒ Δ P X = ( mv 2 - u ) Δ P y = mv sin 60 o - 0 ⇒ Δ P y = mv 3 2 Δ P → = P x 2 + P y 2 ⇒ Δ P → = m ( v 2 - u ) 2 + ( 3 v 2 ) 2 Δ P → = m v 2 4 + u 2 - uv + 3 v 2 4 impulse = m v 2 + u 2 - uv

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