Question 112 of 183

A wedge of mass $M=4m$ lies on a frictionless plane. A particle of mass $m$ approaches the wedge with speed $v.$ There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by:

A$\dfrac{{v}^{2}}{g}$B$\dfrac{{v}^{2}}{2g}$C$\dfrac{2{v}^{2}}{5g}$D$\dfrac{2{v}^{2}}{7g}$

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Answer: (C) $\dfrac{2{v}^{2}}{5g}$

From conservation of linear momentum,

$mv=\left(4m+m\right){v}_{f}$
${v}_{f}=\dfrac{v}{5}$
From conservation of mechanical energy,
$\dfrac{1}{2}m{v}^{2}=\dfrac{1}{2}\left(4m+m\right){v}_{f}^{2}+mgh$
${v}^{2}=5{v}_{f}^{2}+2gh$
${v}^{2}=5\times {\left(\dfrac{v}{5}\right)}^{2}+2gh$
$h=\dfrac{2{v}^{2}}{5g}$