Question

A machine gun of mass 10 kg fires 20 g bullets with a speed of 500 ms^-1 at the rate of 10 bullets per second. To hold the gun steady in its position, how much force is necessary.

• A. 200 N
• B. 500 N
• C. 100 N
• D. 250 N

Answer 1

The correct option is C

100 N

Step 1: Given data

Mass of bullet $m=20g=\frac{20}{1000}kg=\frac{1}{50}kg$ [Since, $1g=\frac{1}{1000}kg$]

Velocity of bullet $v=500m{s}^{-1}$

The total bullet fired in 1 second$=10$

Step 2: Finding change in momentum for 1 bullet

The momentum of a body is defined as the product of its mass and velocity.

Initially, the bullet is at rest.

So, the initial momentum of the bullet$=0kgm{s}^{-1}$

The momentum of the bullet after getting fired$=mv$

$=\frac{1}{50}kg\times 500m{s}^{-1}$

$=10kgm{s}^{-1}$

Change in momentum for 1 bullet$=Finalmomentum-Initialmomentum=10-0=10kgm{s}^{-1}$

Step 3: Finding the force to hold the gun steady in its position

Change in momentum for 10 bullet$=10\times 10=100kgm{s}^{-1}$

10 bullets are fired in 1 second. So, $t=1$.

Rate of change of momentum$=\frac{changeinmomentum}{t}=\frac{100}{1}=100kgm{s}^{-2}$

We know that rate of change of momentum is equal to force.

So, total force applied to bullets$=100kgm{s}^{-2}=100N$

So, the force applied by the gun on the bullets is $100N$

So, by newton's third law of motion, if then force applied by the bullets on the gun will also be equal to $100N$ in backward direction.

So, to hold the gun steady in its position, we need to apply 100 N of force.

Thus, option (c) is correct.