Question

A man (mass =50 kg) and his son (mass=20
kg) are standing on a frictionless surface
Touching each other. The man pushes his son,
so that he starts moving at a speed of 0.70
m/s with respect to the man. The speed of
the man with respect to the surface is

Answer 1

The speed of the man with respect to the surface is equal to 0.2 m/s

• taking the man and his son to be a system, there is no external force acting on this system in the horizontal direction, since the floor is friction less and the push force is an internal force, hence we can apply the conservation of momentum on the system in horizontal direction.
• Let v1 and v2 be the velocity of man and son with respect to surface respectively.
• By conservation of momentum, 50.v1+20.v2=0  ⇒ v1= -0.4(v2)
• Velocity of son with respect to man = v2-v1 = v2-(-0.4v2) = 1.4v2
• Given, 1.4 v2 = 0.7 ⇒v2 = 0.5 m/s
• Hence, v1=0.4.v2 ⇒ v1 = 0.2 m/s