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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 surface. If after striking wedge the ball starts moving in vertical direction and the wedge starts moving in horizontal plane. Calculate the coefficient of restitution e.

A

Mm+tan2θ

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B

mM+cot2θ

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C

mM+tan2θ

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D

Mm+cot2θ

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Solution

## The correct option is C mM+tan2θ As no external force are acting on the system in horizontal direction the linear momentum should be constant conserved in horizontal direction. Let velocity of wedge after collision be V. Then mu = MV ⇒ V = muM. . . . . . (i) As there is no impulse on the ball in the direction parallel to sloping side hence the velocity of ball along the slope (tangent direction) should remain unchanged. ⇒ucosθ = v sin θ which gives v = u cot θ . . . . . . (ii) The coefficient of restitution is given as (v2−v1)normal = e(u1−u2)normal [V sin θ−(−v cosθ)] = e[u sin θ−0] e = v sinθ+vcosθusinθ e = Vu+vucot θ = mM+(cot θ)cot θ (from(i)&(ii)) e = mM+cot2θ

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