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