A man fires a bullet of mass 200 g at a speed of 5m/s. the gun is of one kg mass, by how much velocity the gun rebound backwards.
Answers
m1 - mass of bullet = 200 g = 0.2 kg
m2 - mass of gun = 1 kg
V1 - velocity of bullet = 5 m/s
V2 - velocity of gun = ??
According to law of conservation of linear momentum,
Linear momentum=Linear momentum
before firing after firing
0 = m1 V1 + m2 V2
V2 = - m1 V1/m2
= - 0.2×5×1
V2 = - 1 m/s
Velocity of the gun = -1 m/s.
Answer:
m1 - mass of bullet = 200 g = 0.2 kg
m2 - mass of gun = 1 kg
V1 - velocity of bullet = 5 m/s
V2 - velocity of gun = ??
According to law of conservation of linear momentum,
Linear momentum=Linear momentum
before firing after firing
0 = m1 V1 + m2 V2
V2 = - m1 V1/m2
= - 0.2×5×1
V2 = - 1 m/s
Velocity of the gun = -1 m/s.
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