A smooth ball of mass `1 kg` is projected with velocity `7 m//s` horizontal from a tower of height `3.5 m`. It collides elastically with a wedge of mass 3 kg and inclination of `45^(@)` kept on ground. The ball collides with the wedge at a height of `1 m` above the ground. Find the velocity of the wedge and the ball after collision. (Neglect friction at any contact.)

Answer

`v_(1)=4m//s,v=5sqrt(2)m//s`

Explanation

`v_(x)=7m//s` velocity along `y`-axis of the ball just before collision.
`v_(y)=sqrt(2xx9.8x2.5)=7m//s`
As `v_(x)=v_(y)` so it strikes the plane of incline perpendicularly. Let the ball rebound with velocity `V` and `v_(1)` be the velocity of the wedge

Applying the principle conservation of momentum in horizonal direction, we get
`1xx7=1xxv/(sqrt(2))+3v_(1)`
`7sqrt(2)=3sqrt(2)v_(1)-v`...........i
Applying the equation for coefficient of restitution, we get `e=1=(v_(1)//sqrt(2)+v)/(7sqrt(2)),7sqrt(2)=v_(1)/sqrt(2)+v`.........ii
Solving eqn i and ii `v_(1)=4m//s` and `v=5sqrt(2)m//s`