A ball strikes a smooth horizontal floor obliquely and rebounds inelastically.

Options

A. The kinetic energy of the ball just after hitting the floor is equal to the potential energy of the ball at its maximum height alter rebound.
B. Total energy of the ball is not conserved.
C. The angle of rebound with the vertical is greater than the angle of incidence.
D. None of the above.

Answer

B, C

Explanation

Let the ball stirke the floor at angle alpha with the vertical and velocity at the instant of striking the floor be `u` and let the ball bound with velocilty `v` at angle `beta` with the vertical as shown in figure

Since the floor is smooth, therefore horizontal component of velocity remains unchange. Hence
`vsinbeta=usinalpha`...............i
Since, the floor is inelastic, therfore normal component of velocity just after the collision is less than that just before the collision.hence.
`vcosbetaltcosalpha`........ii
Diving eqn i by eqn ii `tanbetagttanalpha` or `betagtalpha`
hence option c correct.
Since the ball has a horizontal component of velocity, therefore the at highes point it kinetic energy is not equal to zero.
it means, at highest point, potential energy of the ball is less than the kinetic energy just after reboud. Hence option a is incorrect.
Since the floor is inelastic, therefore there is some loss of energy during the collision. hence, the total energy of the ball does not remain conserved. So, option b is also correct.