A block of mass m is released from rest at the top of a smooth wedge of mass M and inclination θ . The wedge is placed on a smooth horizontal surface. What is the velocity of the wedge when the block

Question

A block of mass m is released from rest at the top of a smooth wedge of mass M and inclination θ . The wedge is placed on a smooth horizontal surface. What is the velocity of the wedge when the block slides down a vertical height h ?

TestHub · Accepted Answer

Let V be the speed of the wedge and v the speed of the block relative to the wedge along the incline. The system starts at rest and no external horizontal force acts, so horizontal momentum is conserved:

$M V = m (v cos θ - V) \Rightarrow v cos θ = \frac{( M + m ) V}{m}$

The ground-frame velocity components of the block are vx = MV/m and vy = (M+m)V tanθ/m. By conservation of mechanical energy:

$m g h = \frac{1}{2} M V^{2} + \frac{1}{2} m [(\frac{M V}{m})^{2} + (\frac{( M + m ) V tan θ}{m})^{2}]$

Solving: 2m²gh = V²(M+m)[M + (M+m) tan²θ] = V²(M+m)(M + m sin²θ)/cos²θ, which gives:

$V = \frac{2 m ^{2} g h cos ^{2} θ}{( M + m ) ( M + m sin ^{2} θ )}$

Hence option (B) is correct.

Physics - Center of Mass Question with Solution | TestHub

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