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:
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 $h=\frac{2}{5} \frac{v^{2}}{g}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.