Q 11-05-188JEE MainJEE Main 2019 (9 Apr, Shift 2)Medium
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
Answer: (C) $\dfrac{2v^2}{5g}$
At the maximum height the particle moves with the wedge. Horizontal momentum is conserved:
$$mv = (m + 4m)V \Rightarrow V = \frac v5$$
Energy conservation (no friction):
$$\frac12mv^2 = \frac12(5m)\frac{v^2}{25} + mgh \Rightarrow gh = \frac{v^2}{2}\left(1 - \frac15\right) = \frac{2v^2}{5}$$
$$h = \frac{2v^2}{5g}$$
Solution by Sreeraj P, M.Sc Physics