Question

a bullet of mass 40 grams is fired from a gun of mass 8 kg with a velocity of 800 metre per second calculate the recoil velocity of the gun​

Answer 1

Momentum of system before firing:

Momentum of bullet:

We know,

$\qquad \quad \; \boxed{p = mv}$

Given:

• mass of bullet, m = 40 g or 0.04 Kg
• velocity of bullet, u = 0 m/s ( Because it is at rest )

So,

$\qquad \quad P = mv \\ \qquad \quad \Rightarrow 0.04 \times 0 \\ \qquad \quad \Rightarrow 0 \: Kg.m/s$

Momentum of Gun:

Given:

• mass of gun, M = 8 Kg
• velocity of gun, u = 0 m/s ( Because it is at rest )

So,

$\qquad \quad P = Mv \\ \qquad \quad \Rightarrow 8 \times 0 \\ \qquad \quad \Rightarrow 0\: Kg.m/s$

Total Momentum of System before firing = Momentum of Bullet before firing + Momentum of Gun before firing

=> 0 + 0

Momentum of System before firing = 0 Kg.m/s

Momentum of System after firing:

Momentum of Bullet:

Given:

• mass of bullet, m = 40 g or 0.04 Kg
• velocity of bullet, v = 800 m/s

So,

$\qquad \quad P = mv \\ \qquad \quad \Rightarrow 0.04 \times 800 \\ \qquad \quad \Rightarrow 32 \: Kg.m/s$

Momentum of Gun:

Given:

• mass of Gun, M = 8 Kg
• velocity of Gun,v = v m/s

So,

$\qquad \quad P = Mv \\ \qquad \quad \Rightarrow 8 \times v \\ \qquad \quad \Rightarrow 8v \: Kg.m/s$

Total Momentum of system after firing = Momentum of Gun after firing + Momentum of bullet after firing

=> 32 + 8v

Momentum of System after firing = 32 + 8v Kg.m/s

Momentum of System before firing = Momentum of System after firing ( By Law of Conservation of Momentum )

$\Rightarrow \quad 0 = 32 + 8v \\ \Rightarrow \quad 8v = -32 \\ \Rightarrow \quad v = -4 m/s$

Therefore, Recoil Velocity of Gun is $\bold{-4 \: m/s}$

Answer 2

Answer:

-4

Explanation: