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=u√1−(1−e2)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. Btanϕ=etanθ Cv=u√1−(1−e2)sin2θ D The ratio of the final kinetic energy to the initial kinetic energy is (cos2θ+e2sinθ)
Impulse (J)=ΔP = mvsinϕ−m(−usinθ) =m(vsinϕ+usinθ)=m(Vsep+Vapp)=m(eVapp+Vapp) J=m(usinθ)(1+e) In horizontal direction, momentum is conserved: ucosθ=vcosϕv=ucosθcosϕ e=VsepVapp=vsinϕusinθ=tanϕtanθ tanϕ=etanθ In vertical direction , e=vsinϕusinθ vsinϕ=eusinθ v=√(eusinθ)2+(ucosθ)2=u√e2sin2θ+cos2θ v=u√1−(1−e2)sin2θ Final kinetic energy = 12mv2 Initial kinetic energy = 12mu2 Ratio =v2u2=e2sin2θ+cos2θ