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A particle of mass m strikes a horizonta...

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

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`u cos theta=v cos phi` …(i)
`vsinphi=eusintheta`
or `eusintheta=vsinphi` …(ii)

From Eqs. (i) and (ii), we can see that,
`tan phi=e tan theta`
Momentum or velocity changes only in vertical direction.
`:. |Impu lse|=|Deltap|`
`=m(usintheta+eusintheta)`
`=m(1+e)usintheta`
`v=sqrt((vcosphi)^2+(vsinphi)^2)`
`=sqrt((ucostheta)^2+(eusintheta)^2)`
`=sqrt(u^2(cos^2theta+e^2sin^2theta))`
`=usqrt(1-(1-e^2)sin^2theta)`
`K_f/K_i=(1/2mv^2)/(1/2m u^2)=v^2/u^2`
`=cos^2theta+e^2sin^2theta`
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