A smooth right-angled wedge of mass M is kept | vedclass

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(C) Let $V_w$ be the velocity of the wedge of mass $M$ when the block of mass $m$ reaches the ground. Since there is no external horizontal force on t

Vedclass · Accepted Answer

$\frac{-1}{2} \frac{m^2v^2}{M}$

Vedclass · Answer

(C) Let $V_w$ be the velocity of the wedge of mass $M$ when the block of mass $m$ reaches the ground. Since there is no external horizontal force on the system, the horizontal component of the momentum of the system is conserved.Initially, the system is at rest, so the total initial horizontal momentum is $0$.Thus, $m v_x + M V_w = 0$, where $v_x$ is the horizontal component of the velocity of mass $m$. This implies $v_x = -\frac{M V_w}{m}$.By the work-energy theorem for the wedge, the work done by the normal force $N$ on the wedge equals the change in its kinetic energy:$W_{N'} = \Delta K_w = \frac{1}{2} M V_w^2$.Since the normal force on the block is equal and opposite to the normal force on the wedge, the work done by the normal force on the block is $W_N = -W_{N'} = -\frac{1}{2} M V_w^2$.From the conservation of horizontal momentum, $V_w = \frac{m |v_x|}{M}$. However, we need the answer in terms of $V$. The kinetic energy of the system is $K = \frac{1}{2} M V_w^2 + \frac{1}{2} m V^2$. Since $mgh = K$, and $W_N = \Delta K_{block} - W_g = (\frac{1}{2} m V^2) - mgh = -\frac{1}{2} M V_w^2$.Using $V_w = \frac{m v_x}{M}$, the work done is $W_N = -\frac{1}{2} M (\frac{m v_x}{M})^2 = -\frac{1}{2} \frac{m^2 v_x^2}{M}$. Given the options, the correct expression is $C$.

$A$ smooth right-angled wedge of mass $M$ is kept on a smooth horizontal surface as shown. $A$ mass $m$ is released from the top of the wedge. When $m$ reaches the ground, its speed is $V$. Find the work done by the normal contact force on $m$ while it comes down to the ground.

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