a bullet has a muzzle velocity of 300m/s. The gun of mass 3kg has a recoil velocity of 'v'. Calculate 'v'. Mass of bullet is 30 grams
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The momentum of the bullet:
p₁ = m₁*V₁ = 0,030*300 = 9 kg*m/s
The momentum of the gun (recoil):p = m*V = 3*V
According to the law of conservation of momentum:
p = p₁3*V = 9
V = 9/3 = 3 m/s
(The impact is directed against the speed of a bullet !)
p₁ = m₁*V₁ = 0,030*300 = 9 kg*m/s
The momentum of the gun (recoil):p = m*V = 3*V
According to the law of conservation of momentum:
p = p₁3*V = 9
V = 9/3 = 3 m/s
(The impact is directed against the speed of a bullet !)
Anonymous:
it should be negative , right ?
Yes!
so V = -3 m/s
(The impact is directed against the speed of a bullet !) :)))
Answered by
18
Hi !
mass of gun = m₁ = 3 kg
mass of bullet = m₂ = 30 g = 0.03 kg
Initial velocity of gun = u₁ = 0
Initial velocity of bullet = u₂ = 0
(NOTE : initial velocities in such qns are zero , as they were at rest initially )
Final velocity of gun/recoil velocity = v₁ = ?
Final velocity of bullet = v₂ = 300 m/s
Law of conservation of momentum :-
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0 + 0 = 3v₁ + 0.03*300
0 = 3 v₁ + 9
3v₁ = -9
v₁ = -3
recoil velocity = v₁ = v = -3 m/s
mass of gun = m₁ = 3 kg
mass of bullet = m₂ = 30 g = 0.03 kg
Initial velocity of gun = u₁ = 0
Initial velocity of bullet = u₂ = 0
(NOTE : initial velocities in such qns are zero , as they were at rest initially )
Final velocity of gun/recoil velocity = v₁ = ?
Final velocity of bullet = v₂ = 300 m/s
Law of conservation of momentum :-
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0 + 0 = 3v₁ + 0.03*300
0 = 3 v₁ + 9
3v₁ = -9
v₁ = -3
recoil velocity = v₁ = v = -3 m/s
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