Question

A man weighs 60 kg and runs along the rails with a velocity of 18 m/s. He jumps into a train bogey of mass 1 quintal standing on the rails. Calculate the velocity with which the car will start moving along the rails.

Answer 1

Initial momentum of man=m*v=60*18=1080 kg m/s
Initial momentum of train bogey=m*v=500*0=0 kg m/s      [1 quintal=100 kg]
total initial momentum of man and train bogey=(1080+0)kg m/s =1080 kg m/s 
According to conservation of momentum
Total initial momentum=Total final momentum
1080=(M+m)v
1080=(60+100)v
1080=160v
6.75m/s=v

Answer 2

Answer: 6.75m/s=v

Explanation:

Initial momentum of man=m*v=60*18=1080 kg m/s

Initial momentum of train bogey=m*v=500*0=0 kg m/s      [1 quintal=100 kg]

total initial momentum of man and train bogey=(1080+0)kg m/s =1080 kg m/s 

According to conservation of momentum

Total initial momentum=Total final momentum

1080=(M+m)v

1080=(60+100)v

1080=160v

6.75m/s=v