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

  $\frac{v^2}{g}$
• B

  $\frac{v^2}{2g}$
• Correct answer
  C

  $\frac{2v^2}{5g}$
• D

  $\frac{2v^2}{7g}$

Solution

From conservation of linear momentum,

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

Hence, the correct answer is option: 'C'.