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(D) $1$. Since the surface is smooth and there are no external horizontal forces acting on the system (particle + wedge),the linear momentum of the sy

Vedclass · Accepted Answer

Both $(B)$ and $(C)$

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(D) $1$. Since the surface is smooth and there are no external horizontal forces acting on the system (particle + wedge),the linear momentum of the system is conserved in the horizontal direction.$2$. Initial momentum in the horizontal direction: $P_i = mv_0$.$3$. Final momentum in the horizontal direction: Since the particle moves in the opposite direction with velocity $v_1$ and the wedge moves in the original direction with velocity $v_2$,the final momentum is $P_f = Mv_2 - mv_1$.$4$. By conservation of linear momentum: $mv_0 = Mv_2 - mv_1$. This matches option $(B)$.$5$. Since the collision is elastic (implied by the smooth surface and energy conservation),the kinetic energy is also conserved: $\frac{1}{2}mv_0^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2$.$6$. From momentum conservation: $mv_0 + mv_1 = Mv_2$. This does not directly lead to $v_1 + v_2 = v_0$. However,in the frame of the wedge,the relative velocity of approach equals the relative velocity of separation. The particle approaches the wedge with $v_0$ and leaves it with $v_1$ relative to the wedge. Thus,$v_0 = v_1 + v_2$. Therefore,option $(C)$ is also correct.

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