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Centre of Gravity

A particle of...

Question

A particle of mass $m_{1}$ is projected to the right with a speed $v_{1}$ onto a smooth wedge of mass $m_{2}$ which is simultaneously projected due to the left with a speed $v_{2}$ . Highest point on the wedge attained by the particle is $\frac{m_{2} (v_{1} + v_{2})^{2}}{x g (m_{1} + m_{2})}$ . Find $x$ .

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Solution

Applying work - energy theorem
$W_{t o t a l} = Δ K_{b l o c k s, c m} + Δ K_{c m}$
As no external forces acting, the velocity of centre of mass will be constant. Hence $Δ K_{c m} = 0$ .
Hence, $W_{t o t a l} = Δ K_{b l o c k s, c m}$
$W_{g r a v i t y} = {[\frac{1}{2} μ v_{r e l}^{2}]}_{f i n a l} - [\frac{1}{2} μ v_{r e l}^{2}]$
$- m_{1} g h = 0 - \frac{1}{2} (\frac{m_{1} m_{2}}{m_{1} + m_{2}}) (v_{1} + v_{2})^{2}$
$- m_{1} g h = - \frac{1}{2} (\frac{m_{1} m_{2}}{m_{1} + m_{2}}) (v_{1} + v_{2})^{2}$
$h = \frac{1}{2} \frac{m_{2} (v_{1} + v_{2})^{2}}{g (m_{1} + m_{2})}$
Thus, $x = 2$

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