A wedge of mass M = 4 m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v…

1. $\frac{v^2}{g}$
2. $\frac{v^2}{2g}$
3. $\frac{2v^2}{5g}$
4. $\frac{2v^2}{7g}$

Solution

From conservation of linear momentum,

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

Asked in: JEE Main 2019 (09 Apr Shift 2)

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