# A wedge of mass M=4 m lies on a frictionless plane.

Question:

A wedge of mass $M=4 \mathrm{~m}$ 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:

1. (1) $\frac{v^{2}}{g}$

2. (2) $\frac{2 v^{2}}{7 g}$

3. (3) $\frac{2 v^{2}}{5 g}$

4. (4) $\frac{v^{2}}{2 g}$

Correct Option: , 3

Solution:

(3) $m v=(m+M) V^{\prime}$

or $v=\frac{m v}{m+M}=\frac{m v}{m+4 m}=\frac{v}{5}$

Using conservation of ME, we have

$\frac{1}{2} m v^{2}=\frac{1}{2}(m+4 m)\left(\frac{v}{5}\right)^{2}+m g h$

or $h=\frac{2}{5} \frac{v^{2}}{g}$