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Question

A ball of mass 'm' moving horizontally which velocity 'u' hits a wedge of mass 'M'. The wedge is situated on a smooth horizontal source. If after striking with wedge the ball starts moving in vertical direction and the wedge starts moving in horizontal plane. Calculateda) The velocity of wedge V.b) The velocity (v) at which the ball moves in vertical direction. c) The impulse imparted by the ball on the wedge.d) The coefficient of restitution e=?

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Solution

a) As no external force is acting on the system in horizontal direction the linear momentum should be constant and conserved in horizontal direction.Let velocity of wedge after collision be V.Then mu=MV⟹V=muM (i)b) As there is no impulse on the ball in the direction parallel to slopping side hence the velocity of ball along the slope (tangent direction) should remain unchanged.⟹ucosθ=νsinθwhich gives ν=ucotθ.c) We can find the values ν and V by impulse approach, when ball hits the wedge the impulse is generated between ball an wedge in the direction perpendicular to sloping surface. (normal direction)For ball : mu−Jsinθ=0Jsinθ=muJcosθ=mνFrom Eqs. (i) and (ii), ν=ucotθFrom Eq. (iii), J=mucosecθJsinθ=MV⟹V=JsinθMwhich gives V=muMd) the coefficient of restitution is given as(ν2−ν1)n=e(u1−u2)n[Vsinθ−(−νcosθ)]=e[usinθ−0]e=Vsinθ+νcosθusinθe=Vu+νucotθ=mM+(cotθ)cotθe=mM+cot2θ

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