A block of mass m is released from rest from the top of a movable wedge of mass M and height h. The displacement of centre of mass of the wedge when the block reaches the bottom of wedge is (Ignore any friction)

Engineering

Physics

Constrained Relation

Center of Mass

Relative Motion

Question

A block of mass m is released from rest from the top of a movable wedge of mass M and height h. The displacement of centre of mass of the wedge when the block reaches the bottom of wedge is (Ignore any friction)

$\frac{mhcotθ}{M + m}$

$\frac{mh}{M + m}$

zero

$\frac{Mh}{M + m}$

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Since there is no external force horizontally, the center of mass (COM) of the system (wedge + block) remains fixed horizontally. The wedge moves left as the block slides right. The horizontal displacement of the block relative to the wedge is h cotθ. Let the wedge's displacement be X. The block's horizontal displacement relative to ground is (h cotθ - X). For the COM to remain stationary: M*X + m*(h cotθ - X) = 0. Solving for X:

$X = \frac{m h \cot θ}{M + m}$

Final Answer: $\frac{m h \cot θ}{M + m}$