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(A) Let the velocity of the wedge be $v$ and the velocity of the ball relative to the wedge be $v_r$.$1$. Conservation of linear momentum in the horiz

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$(\frac{gR}{3\sqrt{2}})^{1/2}$

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(A) Let the velocity of the wedge be $v$ and the velocity of the ball relative to the wedge be $v_r$.$1$. Conservation of linear momentum in the horizontal direction:Since there are no external horizontal forces,the net horizontal momentum is conserved. Initially,the system is at rest,so the final horizontal momentum must be zero.$m v_{ball,x} + m v = 0$$v_{ball,x} = -v$The velocity of the ball relative to the ground is $\vec{v}_b = \vec{v}_r + \vec{v}_w$. At point $B$,the ball moves at an angle of $45^{\circ}$ to the horizontal relative to the wedge. $v_{ball,x} = v_r \cos 45^{\circ} - v = -v \implies v_r \cos 45^{\circ} = 0$ (This implies the horizontal component of the ball's velocity relative to the wedge must cancel the wedge's velocity).Actually,using the constraint: $v_{ball,x} = v_r \cos 45^{\circ} - v = -v \implies v_r = 0$ is incorrect. Let's use energy.$2$. Conservation of Mechanical Energy:The loss in potential energy of the ball equals the gain in kinetic energy of the ball and the wedge.$m g R \sin 45^{\circ} = \frac{1}{2} m v^2 + \frac{1}{2} m v_b^2$Using momentum conservation $m v_r \cos 45^{\circ} - m v = 0 \implies v_r \cos 45^{\circ} = v$.Solving the system leads to $v = (\frac{gR}{3\sqrt{2}})^{1/2}$.

$A$ ball of mass $m$ is released from inside a smooth wedge of mass $m$ as shown in the figure. What is the speed of the wedge when the ball reaches point $B$?

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