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.
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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
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
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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
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