A man wearing a bulletproof vest stands still on roller skates. the total mass is 80 kg. a bullet of mass 20g is fired at 400 m/s. it is stopped by the vest and falls to the ground. what is then the velocity of the man ?
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Mass of the bullet= 20g=0.02kg (m1)
Velocity of the bullet before it hits the man= 400m/s (v1)
Momentum=m1v1= 400*0.02=8kgm/s
Since the bullet comes to rest after hitting the man, the entire momentum of the bullet is transferred to the man.( according to the law of conservation of linear momentum)
Mass of the man= 80kg (m2)
Velocity of the man after being hit= v2
Momentum of the man after being hit= 8 kgm/s (p)
Therefore, v2=p/m2= 8/80=1/10= 0.1 m/s
Since the person is on roller skates, the friction between him and ground is negligible.
ANSWER= Velocity of the man will be 0.1m/s.
Velocity of the bullet before it hits the man= 400m/s (v1)
Momentum=m1v1= 400*0.02=8kgm/s
Since the bullet comes to rest after hitting the man, the entire momentum of the bullet is transferred to the man.( according to the law of conservation of linear momentum)
Mass of the man= 80kg (m2)
Velocity of the man after being hit= v2
Momentum of the man after being hit= 8 kgm/s (p)
Therefore, v2=p/m2= 8/80=1/10= 0.1 m/s
Since the person is on roller skates, the friction between him and ground is negligible.
ANSWER= Velocity of the man will be 0.1m/s.
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