A particle of mass `m` strikes a horizontal smooth floor with velocity `u` making an angle `theta` with the floor and rebound with velocity `v` making an angle `theta` with the floor. The coefficient of restitution between the particle and the floor is `e`. Then

Options

A. the impulse delivered by the floor to the body is `mu(1+e)sintheta`
B. `tanphi=e"tan"theta`
C. `v=usqrt(1-(1-e)^2sin^(2)theta)`
D. the ratio of final kinetic energy to the initial kinetic energy is `(cos^(2)theta+e^(2)sin^(2)theta)`

Answer

A, B, D

Explanation

`vsinphi=eusintheta, v cos phi=ucostheta`

`v=usqrt(cos^(2)theta+e^(2)sin^(2)theta)=usqrt(1-sin^(2)theta+e^(2)sin^(2)theta)`
`=usqrt(1-(1-e^(2))sin2theta)`
`I=m(vsinphi+usintheta)="mu"sintheta(1+e)`
Ratio of `KE` `=(1/2mv^(2))/(1/2"mu"^(2))=cos^(2)theta+e^(2)sin^(2)theta`