Question

A bullet of mass 40gm is fired from a gun of mass 8kg with a velocity of 800 m/s, calculate the recoil velocity of gun

Answer 1

Recoil velocity of gun is -4 m/s .

Explanation

Given that mass of bullet (m') = 40 gm = (40/1000) = 0.04 kg. Mass of gun (m") = 8 kg. Velocity of bullet after firing (v') = 800 m/s. As bullet and gun are at rest before firing, so velocity of bullet and gun before firing = 0 m/s.

We have to find recoil velocity of gun(v").

Solution

Momentum of system before firing = Momentum of bullet before firing + Momentum of gun before firing.

→ Momentum of system before firing(p') = (0.04 × 0) + (8 × 0)

→ Momentum of system before firing(p') = 0 + 0

→ Momentum of system before firing(p') = 0 m/s

____________________

Momentum of system after firing = Momentum of bullet after firing + Momentum of gun after firing.

→ Momentum of system after firing(p'') = (0.04 × 800) + (8 × v")

→ Momentum of system after firing(p'') = 32 + 8v"

→ Momentum of system after firing(p'') = (32 + 8v") m/s

____________________

Now according to law of conservation of momentum :

Momentum of system before firing(p') = Momentum of system after firing(p'')

→ 0 = 32 + 8v"

→ -32 = 8v"

→ v" = -32/8

→ v" = -4 m/s

Therefore, recoil velocity of gun = -4 m/s

Answer 2

Answer :-

4 m/s  

 

Explanation :-

Given :-

m1 = 40 g = 0.04 kg  

m2 = 8 kg  

v1 = 800 m/s  

To find :-

v2 = ?  

Fomula : -

m1v1 = m2v2  

Solution :-

According to law of conservation of momentum  

m1v1 = m2v2  

= 0.04 × 800 = 8 × v2  

= v2 = 0.04 × 800/8  

v2 = 0.04 × 100  

v2 = 4  

v2 = 4 m/s  

Therefore, the recoil velocity of the gun is 4 m/s.

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