Question

A Gun of mass 8 kg fires a bullet of mass 100 gram with velocity of 200 m/s. What is the recoil velocity of the gun

Answer 1

Answer:Mass of bullet, m=100g=  

1000

100

​  

kg=  

10

1

​  

kg

Velocity of bullet, v=100ms  

−1

 

Mass of gun, M=20kg

Let recoil velocity of gun =V

Before firing, the system (gun+bullet) is at rest, therefore, initial momentum of the system =0

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

=mv+MV=  

10

1

​  

×100+20V=10+20V

Applying law of conservation of momentum, Final momentum = Initial momentum

i.e., 10+20V=0

20V=−10

or V=−  

20

10

​  

=−0.5ms  

−1

.

Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet.

Answer 2

Answer:

A Gun of mass 8 kg fires a bullet of mass 100 gm with a velocity of 200 m/s. What is the recoil velocity of the gun is 2.5 m/s.

Explanation:

We can express momentum (P) as,

P = mass (m) $\times$ velocity(v)   ...(1)

Given,

Mass of the bullet  ( $m_{1}$ )  =  100g = 0.1 kg

Mass of the gun     ( $m_{2}$ )  =  8 kg

velocity of bullet     ( $v_{1}$ )   =  200 m/s

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

From equation (1),

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

Then,

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

Rearranging the above equation to find out $v_{2}$

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

$v_{2} = \frac{0.1\times200}{8}$

     = 2.5 m/s

Ans :

Recoil velocity of the gun = 2.5 m/s