Question

A bullet of mass 45 gm is fired from a gun of mass 9 kg with a velocity of 600 m/s, Then the recoil velocity of the gun is​

Answer 1

Answer:

Let velocity of gun is v.

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

Before the firing, Linear Momentum =0

After the firing, Linear momentum =0.04×400+10×v

⇒0.04×400+10×v=0 ⇒v=−1.6m/s

Hence, 1.6m/s opposite to the direction of the bullet.

Answer 2

Solution:-

Mass of bullet ,

$\sf = 45gm = \dfrac{45}{1000} kg = \dfrac{9}{200} kg$

Velocity of bullets , v = 600 m/s

Mass of gun = 9kg

Let , Recoil of gun be x

We know initial momentum of system = 0

We can write,

Final momentum of the system = momentum of bullet + momentum of gun

$\to \: \sf \: mv + MV = 0$

Because initial momentum = 0

$\to \sf \: \dfrac{9}{200} \times 600 + 9 \times v = 0$

$\to \rm \: \sf9 \times 3 + 9v = 0$

$\sf \to \: 27 = - 9v$

$\sf \: \to \: v = - 3ms {}^{ - 1}$

=> negative value show that direction of gun Recoil is opposite to the velocity of bullet