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A block of mass ‘m’ is moving towards a movable wedge of mass $M = km$ and height ‘h’ with velocity ‘u’ (All surfaces are smooth). If the block just reaches the top of the wedge, the value of ‘u’ needed is $\sqrt{\frac{ngh (1 + k)}{k}}$ where ‘ $n$ ’ is

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Detailed Solution

When the particle just reaches the top of the wedge mu = (m + km)V

Apply conservation of energy $\frac{1}{2} {mu}^{2} = \frac{1}{2} (M + m) V^{2} + mgh$

Solve we get $u = \sqrt{2 gh (\frac{1 + k}{k})}$

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