Question

a bullet of mass 60g is fired from a gun of mass 3kg. with a velocity of 600m/s.find the recoil velocity of the gun.
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Answer 1

Answer:

-12 m/s

Explanation:

Given:

Mass of the bullet = 60 grams = 0.06kg (m₁)

Mass of the gun = 3kg (m₂)

velcoity of bullet after firing = 600 m/s (v₁)

Recoil velocity of gun = (v₂)

The velocity of bullet before firing = 0 m/s (u₁)

The velocity of gun before firing = 0 m/s (u₂)

According to the law of conservation of momentum:

m₁u₁+m₂u₂=m₁v₁+m₂v₂

Substituting the above values, we get:

0.06×0+3×0=0.06×600+3×v₂

0+0=36+3v₂

0=36+3v₂

-36=3v₂

v₂ = -12 m/s

The recoil velocity of the gun is equal to -12m/s

Answer 2

Answer:

Explanation:

Given :-

Mass of the bullet, m₁ = 60 grams = 0.06 kg

Mass of the gun, m₂ = 3 kg

Velcoity of bullet after firing, v₁ = 600 m/s

Velocity of bullet before firing, u₁ = 0 m/s

Velocity of gun before firing, u₂ = 0 m/s

To Find :-

Recoil velocity of the gun, v₂ = ??

Law to be used :-

law of conservation of momentum,

i.e, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Solution :-

Putting all the values, we get

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

⇒ (0.06 × 0) + (3 × 0) = (0.06 × 600) + (3 × v₂)

⇒  0 + 0 = 36 + 3v₂

⇒  0 = 36 + 3v₂

⇒  - 36 = 3v₂

⇒ - 36/3 = v₂

⇒ v₂ = - 12 m/s

Hence, the recoil velocity of the gun is - 12 m/s.