  Question

# A man of mass 60 kg is standing on a platform of mass 140 kg kept on a smooth horizontal surface. The man starts moving on the platform with a velocity 2 m/s relative to the platform. Find the recoil velocity of the platform. 1 m/s0.5 m/s0.6 m/s2 m/s

Solution

## The correct option is C 0.6 m/sGiven, mass of man, (m1)=60 kg mass of platform, (m2)=140 kg Let the man move at a speed u towards right and the platform recoil at a speed v towards left, relative to the smooth surface.  Thus, speed of the man relative to platform is  u+v=vr ⇒u=vr−v   ...(1) Here, man and platform form a system. There is no external horizontal force on the system. So, linear momentum of the system remains constant in horizontal direction.  Initially, both man and platform were at rest. So, applying conservation of linear momentum (taking rightward +ve) 0=−m2v+m1u ⇒m2v=m1u ⇒m2v=m1(vr−v) (from (i)) ⇒v=m1vrm1+m2 ⇒v=60×260+140=0.6 m/s So, recoil velocity of the platform is 0.6 m/s.  Suggest corrections   