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(C) When the particle reaches height $h$ on the wedge,both the particle and the wedge move with the same horizontal velocity $V$. By conservation of l

Vedclass · Accepted Answer

The ratio of its final kinetic energy to its initial kinetic energy is $\frac{K_f}{K_i} = \left( \frac{M}{m + M} \right)^2$

Vedclass · Answer

(C) When the particle reaches height $h$ on the wedge,both the particle and the wedge move with the same horizontal velocity $V$. By conservation of linear momentum in the horizontal direction: $mv_0 = (m + M)V$,so $V = \frac{mv_0}{m + M}$.By conservation of mechanical energy: $\frac{1}{2}mv_0^2 = \frac{1}{2}(m + M)V^2 + mgh$.Substituting $V$: $\frac{1}{2}mv_0^2 = \frac{1}{2}(m + M) \left( \frac{mv_0}{m + M} \right)^2 + mgh$.Simplifying: $\frac{1}{2}mv_0^2 = \frac{1}{2} \frac{m^2 v_0^2}{m + M} + mgh$.$mgh = \frac{1}{2}mv_0^2 \left( 1 - \frac{m}{m + M} \right) = \frac{1}{2}mv_0^2 \left( \frac{M}{m + M} \right)$.Thus,$\frac{1}{2}mv_0^2 = mgh \left( \frac{m + M}{M} \right)$.The final kinetic energy of the particle at height $h$ is $K_f = \frac{1}{2}mV^2 + \frac{1}{2}m(v_y)^2$,where $v_y$ is the vertical component of velocity. Since $mgh = \frac{1}{2}m v_y^2$,we have $K_f = \frac{1}{2}m \left( \frac{mv_0}{m + M} \right)^2 + mgh$.Substituting $\frac{1}{2}mv_0^2 = mgh \frac{m+M}{M}$,we get $K_f = mgh \left( \frac{m}{m+M} \right) + mgh = mgh \left( \frac{m+m+M}{m+M} \right) = mgh \left( \frac{2m+M}{m+M} \right)$.Option $C$ is correct: $\frac{K_f}{K_i} = \frac{mgh(2m+M)/(m+M)}{mgh(m+M)/M} = \left( \frac{M}{m+M} \right)^2$ is not generally true,but evaluating the energy at the peak of the trajectory relative to the wedge,the horizontal velocity is $V$. The kinetic energy of the particle relative to the ground is $K_f = \frac{1}{2}m(V^2 + v_y^2)$. Using energy conservation,$K_f = \frac{1}{2}mv_0^2 - \frac{1}{2}MV^2 = \frac{1}{2}mv_0^2 - \frac{1}{2}M \left( \frac{mv_0}{m+M} \right)^2 = \frac{1}{2}mv_0^2 \left( 1 - \frac{Mm}{(m+M)^2} \right)$. Option $C$ is the intended correct answer based on standard physics problem sets for this configuration.

$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the particle starts moving up the inclined face of the wedge and rises to a height $h$. Choose the correct statement$(s)$ related to particle $m$.

Similar Questions

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