The inclined surfaces of two movable wedges o | vedclass

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Question

since there is no friction, external forces do not act on the system under consideration in the

horizontal direction $(Fig.\, 160 ).$ In order to d

Vedclass · Accepted Answer

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

Vedclass · Answer

since there is no friction, external forces do not act on the system under consideration in the

horizontal direction $(Fig.\, 160 ).$ In order to determine the velocity $v$ of the left wedge and the velocity $u$ of the washer immediately after the descent, we can use the energy and momentum conservation laws$:$

$\frac{M v^{2}}{2}+\frac{m u^{2}}{2}=m g h$

$M v=m u$

since at the moment of maximum ascent $h_{\max }$ of the washer along the right wedge, the velocities of the washer and the wedge will be equal, the momentum conservation law can be written in the form

$\mathrm{mu}=(\mathrm{M}+\mathrm{m}) \mathrm{V}$

where $V$ is the total velocity of the washer and the right wedge. Let us also use the energy conservation law:

$\frac{m u^{2}}{2}=\frac{M+m}{2} V^{2}+m g h_{\mathrm{max}}$

The joint solution of the last two equations leads to the expression for the maximum height $\mathrm{h}_{\mathrm{max}}$ of the ascent of the washer along the right wedge$:$

$h_{\max }=h \frac{M^{2}}{(M+m)^{2}}$

The inclined surfaces of two movable wedges of same mass $M$ are smoothly conjugated with the horizontal plane as shown in 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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