The inclined surfaces of two movable wedges o | vedclass

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(C) Since there is no friction, external forces do not act on the system in the horizontal direction. Let $v$ be the velocity of the left wedge and $u

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$h{\left( {\frac{M}{{M + m}}} \right)^2}$

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(C) Since there is no friction, external forces do not act on the system in the horizontal direction. Let $v$ be the velocity of the left wedge and $u$ be the velocity of the washer immediately after the descent. Using the conservation of energy and horizontal momentum:$\frac{1}{2}M v^2 + \frac{1}{2}m u^2 = mgh$$Mv = mu \implies v = \frac{m}{M}u$Substituting $v$ into the energy equation:$\frac{1}{2}M(\frac{m}{M}u)^2 + \frac{1}{2}mu^2 = mgh \implies \frac{1}{2}u^2(\frac{m^2}{M} + m) = mgh \implies u^2 = \frac{2ghM}{M+m}$At the moment of maximum ascent $h_{max}$ on the right wedge, the washer and the right wedge move with a common horizontal velocity $V$. By conservation of horizontal momentum:$mu = (M+m)V \implies V = \frac{mu}{M+m}$Using conservation of energy for the second stage:$\frac{1}{2}mu^2 = \frac{1}{2}(M+m)V^2 + mgh_{max}$$mgh_{max} = \frac{1}{2}mu^2 - \frac{1}{2}(M+m)(\frac{mu}{M+m})^2 = \frac{1}{2}mu^2(1 - \frac{m}{M+m}) = \frac{1}{2}mu^2(\frac{M}{M+m})$Substituting $u^2 = \frac{2ghM}{M+m}$:$mgh_{max} = \frac{1}{2}m(\frac{2ghM}{M+m})(\frac{M}{M+m}) = \frac{mghM^2}{(M+m)^2}$$h_{max} = h(\frac{M}{M+m})^2$

The inclined surfaces of two movable wedges of same mass $M$ are smoothly conjugated with the horizontal plane as shown in the figure. $A$ washer of mass $m$ slides down the left wedge from a height $h$. To what maximum height will the washer rise along the right wedge? Neglect friction.

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