Question

A bullet of mass 30g is
Fired
from on of gun 9 kg
iF the
velocity of bullet
is 400m/s then the recoil velocity of the gun will be​

Answer 1

Answer :-

Recoil velocity of the gun is 1.33 m/s .

Explanation :-

We have :-

→ Mass of the bullet (m) = 30 g = 0.03 kg

→ Mass of the gun (M) = 9 kg

→ Velocity of the bullet (m) = 400 m/s

To find :-

→ Recoil velocity of the gun (V) .

________________________________

After the bullet is fired, the gun will recoil in the opposite direction (backwards) . So the value of recoil velocity will be -V m/s .

According to "Law of Conservation of Momentum", total momentum of the system will remain conserved. So, by using this concept, we have :-

⇒ M(-V) + mv = 0

⇒ -MV = -mv

⇒ V = -mv/(-M)

⇒ V = mv/M

Putting values in the obtained formula, we get :-

V = mv/M

⇒ V = (0.03 × 400)/9

⇒ V = 12/9

⇒ V = 1.33 m/s

Answer 2

REQUIRED EXPLANATION:-

Given :-

• Mass of the bullet is 30g
• Mass of the gun is 9kg
• Velocity of the bullet is 400 m/s

To find :-

• Recoil of the velocity of gun

Solution:-

We can find the velocity of the gun by the law of conservation of momentum that is :-

m₁ v₁ = m₂ v₂

• m₁ = 30g = 3/100 kg
• m₂ = 9kg
• v1 = 400 m/s
• v₂ = ?

Converting 30g - kg :-

1kg = 1000 g

1g = 1/1000 kg

30g = 30/1000 kg

30g = 3/100 kg

3/100 × 400 = 9 × v₂

3 × 4 = 9 × v ₂

12 = 9v₂

v₂ = 12/9

v₂ = 4/3

v₂ = 1.33 m/s

So, the Recoil of velocity of the gun is 1.33 m/s