A bullet of mass 20 gram is horizontally fired with a velocity 150 metre per second from a pistol of 2 kilogram what is the recoil velocity of the stock

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Solution

mass1=0.02kg
u1=0m/s
v1=150m/s
mass2=2kg
u2=0m/s
v2=v.
∴by the law of conservation of momentum,
m1u1+m2u2=m1v1+m2v2
0.020+20=0.02150+2v
⇒0+0=3+2v
⇒3=2v
⇒3/2=v
⇒1.5=v
∴recoil velocity = 1.5 m/s

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A bullet of mass 20 gram is horizontally fired with a velocity 150 metre per second from a pistol of 2 kilogram what is the recoil velocity of the stock

mass1=0.02kg u1=0m/s v1=150m/s mass2=2kg u2=0m/s v2=v. ∴by the law of conservation of momentum, m1u1+m2u2=m1v1+m2v2 0.02*0+2*0=0.02*150+2*v ⇒0+0=3+2v ⇒3=2v ⇒3/2=v ⇒1.5=v ∴recoil velocity = 1.5 m/s

mass1=0.02kg
u1=0m/s
v1=150m/s
mass2=2kg
u2=0m/s
v2=v.
∴by the law of conservation of momentum,
m1u1+m2u2=m1v1+m2v2
0.020+20=0.02150+2v
⇒0+0=3+2v
⇒3=2v
⇒3/2=v
⇒1.5=v
∴recoil velocity = 1.5 m/s