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Question

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

A
the impulse delivered by the floor to the body is mu(1+e)sinθ.
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B
tanϕ=etanθ
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C
v=u1(1e2)sin2θ
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D
The ratio of the final kinetic energy to the initial kinetic energy is (cos2θ+e2sinθ)
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Solution

The correct options are
A the impulse delivered by the floor to the body is mu(1+e)sinθ.
B tanϕ=etanθ
C v=u1(1e2)sin2θ
D The ratio of the final kinetic energy to the initial kinetic energy is (cos2θ+e2sinθ)

Impulse (J)=ΔP = mvsinϕm(usinθ)
=m(vsinϕ+usinθ)=m(Vsep+Vapp)=m(eVapp+Vapp)
J=m(usinθ)(1+e)
In horizontal direction, momentum is conserved:
ucosθ=vcosϕ v=ucosθcosϕ
e=VsepVapp=vsinϕusinθ=tanϕtanθ
tanϕ=etanθ
In vertical direction ,
e=vsinϕusinθ
vsinϕ=eusinθ
v=(eusinθ)2+(ucosθ)2=ue2sin2θ+cos2θ
v=u1(1e2)sin2θ
Final kinetic energy = 12mv2
Initial kinetic energy = 12mu2
Ratio =v2u2=e2sin2θ+cos2θ

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