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A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed $ν$ . 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 :

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$\frac{2 ν^{2}}{5 g}$

$\frac{ν^{2}}{2 g}$

$\frac{2 ν^{2}}{7 g}$

$\frac{ν^{2}}{g}$

answer is A.

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Detailed Solution

By applying linear momentum conservation,

$mv = (m + M) v_{c} \Rightarrow v_{c} = \frac{v}{5}$

Applying the work energy theorem,

$- mgh = \frac{1}{2} (m + M) {v_{c}}^{2} - \frac{1}{2} {mv}^{2} \Rightarrow h = \frac{2 v^{2}}{5 g}$

Hence the correct answer is $\frac{2 v^{2}}{5 g} .$

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