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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

A particle of mass m is projected with a velocity v 0 at an angle α with the horizontal. The coefficient of restitution for any of its impact with the smooth ground is e . The time after which the particle will stop bouncing off the ground is

Options

  1. AT = 2 v 0 s i n α ( 1 + e ) g ( 1 - e )
  2. BT = 2 v 0 s i n α ( 1 - e ) g ( 1 + e )
  3. CT = 2 v 0 s i n α g ( 1 + e )
  4. DT = 2 v 0 s i n α g ( 1 - e )

Correct answer

D. T = 2 v 0 s i n α g ( 1 - e )

Step-by-step solution

After each collision, the particle loses some part of its vertical velocity while its horizontal velocity remains unchanged. T = 2 v 0 s i n α g + 2 e v 0 s i n α g + 2 e 2 v 0 sin α g + . . . . . . . . . . ⇒ T = 2 v 0 s i n α g 1 + e + e 2 + . . . . . . . . . T = 2 v 0 s i n α g ( 1 - e )

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