Question

A particle of mass $m=0.1kg$ is released from rest from a point $A$ of a wedge of mass $M=2.4kg$ free to slide on a frictionless horizontal plane. The particle slides down the smooth face $AB$ of the wedge. When the velocity of the wedge is $0.2m/s$ what is the velocity of the particle in $m/s$ relative to the wedge.

Answer 1

By momentum conservation,

$Mu=m(v\cos 60°-u)2.4\times 0.2=0.1(\frac{v}{2}-0.2)0.48=\frac{0.1}{2}v-0.020.5=\frac{0.1}{2}v$

$v=10$m/s velocity of particle w.r.t wedge.