Question

A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 seconds to move through the barrel of the gun, and acquires a velocity of 100 m/s. Calculate:a) the velocity at which the gun recoilsb) the force exerted on the gunman due to recoil of the gun.I want the solution as well as the answer.

Answer 1

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Answer 2

The velocity of the gun is 1 m/s.

The Force exerted on the gunman while the recoiling of gun is 1000N

Given:

Mass of gun =3kg= $m_{1}$

Mass of bullet=30g=0.03kg= $m_{2}$

Velocity of bullet=100m/s =$v_{2}$

Time taken by the bullet=0.003s

To find:

The velocity of the gun and the force of the gun man:

Solution:

According to the law of conservation of momentum we have

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

Substitute the values of m_{1}, m_{2}, v_{2} in the above expression

$3 \times v_{1}=0.03 \times 100$

$3 v_{1}=3$

$v_{1}=\frac{3}{3}=1$

Hence, $v_{1} =1m/s$

Let us find the

Acceleration,$a=v-\frac{u}{t}$

$=100-\frac{0}{0.003}$

=33333.33m/s

The velocity of the given gun is 33333.33 m/s.

To find the force we have  

$F=m \times a$

Where, m=0.03

a=33333.33

By substituting the values

$=(0.03) \times(33333.33)$

=1000N.