Question

A bullet of mass 0.16 kg fired from a gun with velocity 80 m/s. Mass of gun is 16 kg. Find
the recoil velocity of gun.​

Answer 1

Answer:

M = mass of gun =16 kg U = initial velocity of gun = 0 m/s

m = mass of bullet = 0.16 kg u = initial velocity of bullet = 0 m/s

V = final velocity of gun (recoil velocity)=x

v = final velocity of bullet = 80 m/s

By using the conservation of momentum

=> MU + mu = MV +mv

=> (16 × 0 ) + (0.16 × 0) = (16 × x) + (0.16 × 80)

=> 0 + 0 = 16x + 12.8

=> 16x = - (12.8)

=> x = -( 0.8 )

So, the recoil velocity of gun is 0.8 m/s (- sign shows the opposite direction of the velocity of bullet)

Answer 2

Answer:

A bullet of mass 0.16 kg was fired from a gun with a velocity of 80 m/s. Mass of the gun is 16 kg.  The recoil velocity of the gun is 0.8 m/s.​

Explanation:

Given,

Mass of the bullet  ( $m_{1}$ )  =  0.16 kg

Mass of the gun     ( $m_{2}$ )  =  16 kg

velocity of bullet     ( $v_{1}$ )   =  80 m/s

Let recoil velocity of gun be $v_{2}$ ,

We know,

Momentum (p) = mv

Then,

The momentum of bullet  =  $m_{1} v_{1}$

The momentum of gun     =  $m_{2}v_{2}$

According to the law of conservation of momentum,

The momentum of bullet   =  momentum of the gun

That is,

$m_{1} v_{1} = m_{2}v_{2}$

$v_{2} = \frac{m_{1}\ v_{1}}{m_{2}}$

$v_{2} = \frac{0.16\times80}{16}$

     = 0.8 m/s

Ans :

Recoil velocity of the gun = 0.8 m/s