Physics - Center of Mass Question with Solution | TestHub

Let u be the initial speed of the ball, V the recoil speed of the wedge and v the vertical rebound speed of the ball. Horizontal momentum conservation: $\mathrm{mu}=\mathrm{MV}\Rightarrow \mathrm{V}=\frac{\mathrm{mu}}{\mathrm{M}}$

The impulse on the ball acts along the normal to the incline, so its tangential component is unchanged: $\mathrm{u}\cos \theta =\mathrm{v}\sin \theta \Rightarrow \frac{\mathrm{v}}{\mathrm{u}}=\cot \theta$

Along the normal, the approach speed is u sinθ and the separation speed is v cosθ + V sinθ, so:$\begin{array}{rl}e& =\frac{\mathrm{v}\cos \theta +\mathrm{V}\sin \theta }{\mathrm{u}\sin \theta }=\frac{\mathrm{v}}{\mathrm{u}}\cot \theta +\frac{\mathrm{V}}{\mathrm{u}}\\ & ={\cot }^2\theta +\frac{\mathrm{m}}{\mathrm{M}}\end{array}$

Hence e = m/M + cot²θ, option (C).