Question

A bullet of mass 40gm is fired from a gun of mass 2 kg with a velocity of 300m/s.Calculate the recoil velocity of gun.




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Answer 1

Given :

Bullet mass,m =40g = 0.04 kg

Velocity of bullet, u = 300 m/s

Mass of Gun ,M = 2kg

Theory :

Conservation of linear momentum :

If no external force acts on the system of constant mass the total momentum of the system remains constant .

Solution :

Let recoil velocity be v

Intital both bullet and gun are at rest

Thus , Intital momentum before firing = 0

As no external force acts on the system

So According to the law of conservation of Momentum .

Total momentum before firing = Total momentum after the firing

$\sf\implies\:0=mu+Mv$

$\sf\implies\:Mv=-mu$

Now , Put the given values

$\sf\implies\:2\times\:v=-0.04\times300$

$\sf\implies\:v=\dfrac{-300\times0.04}{2}$

$\sf\implies\:v=-6ms^{-1}$

Here , The negative sign shows that v and u are in opposite directions i.e ,the gun gives a kick in the backward direction or gun recoils with velocity v.

Answer 2

Answer:

v = -6m/s

Explanation:

Let velocity of the gun is v

There is no external force acting, so linear momentum will conserve

Before the firring, linear momentum = 0

after the firring, linear momentum = 0.04 * 300 +2v

0.04*300+2v=0

12+2v=0

2v=-12

v=-6m/s

Hence 6m/s opposite to direction of the bullet