A wedge having a vertical slot in it is placed on smooth horizontal surface. Two blocks are arranged as shown in Figure. The system is released from rest. The speed of the wedge when block 1 comes down a distance ‘h is $\sqrt{\frac{2\mathrm{gh}}{\alpha }}$ . Calculate $\alpha$

Assuming wedge, block 1 and block 2 as a system. Since no external force acts on the system in horizontal direction, so both the linear momentum and mechanical energy will be conserved for the system.

Final step : Let velocity of block 1 in vertical direction be v. Hence velocity of block 2 w.r.t wedge will be v towards left. Let velocity of wedge towards right be V.

Hence velocity of block 2 towards left (w.r.t ground) will be (v – V)

Applying conservation of linear momentum for the system in horizontal direction, we get

$(\mathrm{M}+\mathrm{m})\mathrm{V}-\mathrm{m}(\mathrm{v}-\mathrm{V})=0$

$\text{ Since }M=3\mathrm{m}\Rightarrow \mathrm{v}=5\mathrm{V}$

Applying conservation of mechanical energy for the system, we get

$\mathrm{\Delta K}+\mathrm{\Delta U}=0$

$\Rightarrow \frac{(\mathrm{M}+\mathrm{m})}{2}{\mathrm{V}}^2+\frac{\mathrm{m}}{2}(\mathrm{v}-\mathrm{V}{)}^2+\frac{\mathrm{m}}{2}{\mathrm{v}}^2-0+(-\mathrm{mgh})=0$

$\Rightarrow \frac{1}{2}(3\mathrm{m}+\mathrm{m}){\mathrm{V}}^2+\frac{1}{2}\mathrm{m}(5\mathrm{V}-\mathrm{V}{)}^2+\frac{1}{2}\mathrm{m}(5\mathrm{V}{)}^2=\mathrm{mgh}$

$\Rightarrow \mathrm{V}=\sqrt{\frac{2\mathrm{gh}}{45}}\Rightarrow \alpha =45$