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JEE Main20206 Sep 2020Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual

Particle A of mass m 1 moving with velocity 3 i ^ + j ^ ms − 1 collides with another particle B of mass m 2 which is at rest initially. Let v → 1 and v → 2 be the velocities of particles A and B after collision respectively. If m 1 = 2 m 2 and after collision v → 1 − i ^ + 3 j ^ ms − 1 , the angle between v → 1 and v → 2 is :

Options

  1. A15 °
  2. B60 °
  3. C− 45 °
  4. D105 °

Correct answer

D. 105 °

Step-by-step solution

m 1 u → 1 + m 2 u → 2 = m 1 v → 1 + m 2 v → 2 2 m 2 ( 3 i ^ + j ^ ) + m 2 × 0 = 2 m 2 ( i ^ + 3 j ^ ) + m 2 × v → 2 2 3 i ^ + 2 j ^ = 2 i ^ + 2 3 j ^ + v → 2 v → 2 = ( 3 - 1 ) i ^ - ( 3 - 1 ) j ^ v → 1 = i ^ + 3 j ^ Angle between v → 1 and v → 2 is 105 °

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