AP EAMCET201824 Apr 2018Morning ShiftPhysicsCenter of Mass, Momentum and CollisionActual
A ball falls freely from a height of 180 ~m on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is 0.5 , the average speed and average velocity of the ball before it ceases to rebound are respectively (acceleration due to gravity =10 ~ms ⁻² )
Options
- A10 ~ms ⁻¹, 10 ~ms ⁻¹
- B50 ~ms ⁻¹, 50 3 ~ms ⁻¹
- C50 3 ~ms ⁻¹, 10 ~ms ⁻¹
- D20 3 ~ms ⁻¹, 50 3 ~ms ⁻¹
Correct answer
C. 50 3 ~ms ⁻¹, 10 ~ms ⁻¹
Step-by-step solution
When ball dropped from height h , then time taken to reach the ground t₀= 2 h g and speed, v₀= 2 g h After first collision, its speed will become v₁=e v₀=e 2 g h where, e= coefficient of restitution. Now, the ball will go up and will take time t₁ , when it stops aligned & v=u+a t & 0=v₁-g t₁ & t₁= v₁ g aligned It will come down and take same time t₁ before second collision. So, time taken between first and second collision is 2 t₁ . Similarly, time taken between second and third collision will be 2 t₃= 2 v₂ g Total