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.)

A smooth ball of mass 1 kg is projected with velocity 7 m//s horizonta

1.	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» SOLUTION :`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`