A man wearing a bullet-proof vest stands on roller skates. The total mass is 80 kg. A bullet of mass 20 g 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?
Answers
Answer:
Given:
Mass of the man,m
1
=80kg
Speed of man after collision, v
1
Mass of bullet m
2
=20g=0.02kg
Speed of bullet v
2
=400m/s
Acc to law of conservation of momentum
m
1
v
1
=m
2
v
2
80×v
1
=0.02×400
v
1
=
80
400×0.02
=0.1m/s
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Given : A man wearing a bulletproof vest stands on roller skates.
The total mass is 80 kg.
A bullet of mass 20 g is fired at 400 m/s.
It is stopped by the vest and falls to the ground.
To Find: the velocity of the man
Solution:
mass of man = m₁ = 80 kg
Initial velocity of man = u₁ = 0 m/s
Final velocity of man = v₁ = 0 m/s
mass of bullet = m₂= 20 g = 20/1000 kg = 0.02 kg
Initial velocity of bullet = u₂ = 400 m/s
Final velocity of bullet = v₂ = 0 m/s
Law of conservation of momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
=> 80(0) + (0.02)400 = 80(v₁ ) + 0.02(0)
=> 0 + 8 = 80(v₁ ) + 0
=> 8 = 80(v₁ )
=> v₁ = 8/80
=> v₁ = 0.1 m/s
Velocity of man is 0.1 m/s
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