A ball of mass m is allowed to roll down the wedge of mass M = 2m as shown in the figure. What is the displacement of wedge (in m) when the ball reaches from A to B? Take $θ = 45^{0}, h = 1 m, d = 4 m$

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answer is 2.

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Detailed Solution

Displacement of Wedge is ${ΔX}_{wedge}$ towards left.

${MΔX}_{wedge} = m (2 hcotθ + d - x)$
$\begin{array}{l} {ΔX}_{wedge} = \frac{m (2 hcot θ + d)}{(M + m)} \\ = (\frac{m}{2 m + m}) (4 + 2 \times 1 \times 1) \\ = \frac{1}{3} \times 6 = 2 m \end{array}$

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