Question

A bullet of mass 0.01 kg is horizontally fired with a velocity 132 m/s
from a pistol of mass 3 kg. What is the recoil velocity of the pistol?

Answer 1

Given, the mass of the bullet$(m_{1}) = 0.01kg$

The mass of the pistol $(m_{2}) = 3kg$

The initial velocity of the bullet $(u_{1} )=0m/s$

The initial velocity of the pistol $(u_{2} )=0m/s$

The final velocity of the bullet $v_{1} =132m/s$

Let, the recoil velocity of the pistol be $v_{2}$.

From, the conservation of momentum=

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

The initial momentum of the system is zero, as both the bullet and the pistol are at rest.

$0+0=(0.01\times 132)+(3\times v_{2})\\ -1.32=3\times v_{2} \\v_{2}=-\frac{1.32}{3} =-0.44m/s$

Hence, the recoil velocity of the pistol is 0.44m/s.

Answer 2

Recoil velocity of the pistol is 0.44 m/s

Given:

• Bullet mass 0.01 kg
• Horizontally fired with velocity 132 m/s
• Pistol mass  3 kg

To Find:

• Recoil Velocity of the pistol

Solution:

• Law of conservation of linear momentum
• Initial Momentum = Final Momentum
• Momentum = Mass x Velocity

Step 1:

Find Initial Momentum ( Velocity of both bullet and pistol are zero as at rest)

0.01 x 0  + 3 x 0   =  0

Step 2:

Find Final Momentum

( Velocity of  bullet = 132 m/s and  velocity of pistol = v)

0.01 x 132  + 3 x v  

= 1.32 + 3v

Step 3:

Equate both Momentum and solve for v

1.32 + 3v = 0

3v = -1.32

v = - 0.44

-ve sign indicates that velocity is in opposite direction hence Recoil Velocity

So Recoil velocity of the pistol is 0.44 m/s