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Q.

A particle strikes a horizontal smooth floor with a velocity ‘u’ making an angle θ with the floor and rebounds with velocity ‘v’ making an angle ϕ with the floor. If the coefficient of restitution between the particle and the floor is ‘e’, then

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a

the impulse delivered by the floor to the body is mu (1 + e) sinθ

b

tanϕ =e tanθ

c

v=u1−(1−e2)sin2θ

d

The ratio of the final kinetic energy to the initial kinetic energy is (cos2θ+e2sin2θ)

answer is Բ.

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Detailed Solution

vcosϕ=ucosθ ……. (1)vsinϕ=eusinθ …….. (2)∴(1)2+(2)2→v=ucos2θ+e2sin2θv=ucos2θ+sin2θ−sin2θ+e2sin2θ=u1−(1−e2)sin2θ(2)(1)⇒tanϕ=etanθImpulse delivered by the floor to body = mvsinϕ+musinθ=meusinθ+musinθ=mu(1+e)sinθKEFKEI=12mv212mu2=cos2�+e2sin2�

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