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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 $(C)$.

Vedclass · Answer

(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.$2$. At the maximum height $h$ relative to the wedge,the particle and the wedge move with the same horizontal velocity $V$. By conservation of momentum: $mv_0 = (m + M)V$,so $V = \frac{mv_0}{m+M}$.$3$. In the centre of mass $(COM)$ frame,the total momentum is zero. At the maximum height $h$,the relative velocity between the particle and the wedge is zero (they move together horizontally). Since the total momentum in the $COM$ frame is zero,both must be stationary with respect to the $COM$ frame.$4$. In the $COM$ frame,the total energy is the sum of the kinetic energy of the system relative to the $COM$ and the potential energy. At the maximum height $h$,the relative kinetic energy is zero,meaning all the initial kinetic energy relative to the $COM$ has been converted into potential energy $mgh_{eff}$.$5$. Thus,both statements $(B)$ and $(C)$ are correct.

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