A particle of mass `m` strikes a smooth floor with speed `u` at angle of incidence `theta` with the normal. The coefficient of resultant is `e`. Find the magnitude and direction of velocity with which the particle rebounds. Also, find the impulse and loss in `K.E.`

Explanation

Before the collision:
After the collision:
Since there is no force along the surface , hence the velocity component remains same.
`v sin alpha = u sin theta` (i)
Due to the normal force between the particle and the floor , the velocity component becomes `e times` in opposite direction
`v cos alpha = e u cos theta` (ii)
`v = sqrt((v sin alpha)^(2) + (v cos alpha)^(2)) = u sqrt(sin^(2) theta + e^(2) cos^(2) theta)`
`tan alpha = ( v sin alpha)/(v cos alpha) = (1)/(e) tan theta`
If the collision is elastic , `e = 1 , v = u , alpha = theta`
`J_(x) = 0`
`J_(y) = m(v cos alpha + u cos theta)`
`= m ( e u cos theta + u cos theta) = mu cos theta ( 1 + e)`
`J = mu cos theta ( 1 + e)`
`Delta K = (1)/(2) mu^(2) - (1)/(2) m u^(2) = (1)/(2) mu^(2) cos^(2) theta ( 1 - e^(2))`
`Delta` K: loss in `K.E.`