A smooth ball is released from rest from a height h as shown in figure. It slides down the first inclined plane and collides with the second inclined plane.
a. If `e = 0`, find the speed of the ball just after leaving the inclined plane 1.
b. If the particle mioves horizontally just after the collision find `e`.

Explanation

The ball collides with second inclined plane.
Since `e=0`, along the velocity of the ball is zero.
Conserving linear momentum along the tantent to the 2nd inclined plane, we have

`v=v_(0)cos30^(@)=(sqrt(3))/2v_(0)`
`=(sqrt(3))/2 sqrt(2gh)=sqrt((3gh)/2)`
b. The components of velocity are
`v_(x)=v_(0)cos30^(@)=(sqrt(3))/2v_(0)`

`v_(y)=ev_(0)sin30^(@)=(ev_(0))/2`
since the ball moves horizontally
`(ev_(0))/2cos30^(@)=(sqrt(3))/2v_(0)sin30^(@)`
`e=1`