A block of mass m is moving with a velocity u | vedclass

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(A) Let the mass of the block be $m$ and the mass of the wedge be $m$. Initially, the block has velocity $u$ and the wedge is at rest. Since the surfa

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$u$

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(A) Let the mass of the block be $m$ and the mass of the wedge be $m$. Initially, the block has velocity $u$ and the wedge is at rest. Since the surface is smooth and there are no external horizontal forces, the linear momentum of the system is conserved in the horizontal direction. When the block reaches the maximum height $h$ relative to the wedge, both the block and the wedge move with the same horizontal velocity $v$. By conservation of momentum: $mu = (m + m)v$, which gives $v = u/2$. However, the question asks for the final velocity of the wedge after the block has moved back down to the horizontal plane. Since the collision is elastic (smooth surfaces, no energy loss), the system behaves like a one-dimensional elastic collision between two equal masses. In an elastic collision between two bodies of equal mass where one is initially at rest, they exchange their velocities. Therefore, the block comes to rest and the wedge moves with the initial velocity $u$ of the block.

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