Q. 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 :
Solution:
Applying Linear momentum conservation
$mv = (m + M) v_c$
$v_c = \frac{v}{5}$
applying work energy theorem
$ - mgh = \frac{1}{2} (m + M) v_c^2 - \frac{1}{2} mv^2$
solve , $h = \frac{2v^2}{5g}$
