Question

A machine gun fires a bullet of mass 40g with a velocity$1200m{s^{ - 1}}$. The man holding it can exert a maximum force of 144N on the gun. How many bullets can he fire per second at the most?
A) 2
B) 4
C) 1
D) 3

Answer 1

Hint:Imagine that you are firing a bullet and you can exert a maximum force of 144N. When you are firing a bullet there many variables that come into play, such as the time taken to fire a bullet, the mass of the bullet, the force exerted, the velocity at which the bullet is going and the number of bullets you are firing, to find out how many bullets you fired at a certain time period, we need to all the information of the above variables and also we have to establish a relation between all of them. We have to apply the concept of impulse according to it, impulse is the force acted in a short time and it is also equal to the change in momentum. Equate force into time to the change in momentum and solve.

Formula used:The formula used is:
$I = F.\Delta t = {\text{ change in momentum = }}Mv - 0$……(initial momentum of the bullet is zero)
I = Impulse;
F = Force;
$\Delta t$= Time (1s);
M = Mass ($m \times n$);
v = Velocity;

Complete step by step answer:
 Apply the concept of impulse and solve:
A man fired bullet n times in t seconds of mass M: $m \times n$
Now we have been given a force of 144 N;
Apply the concept of Impulse,
$I = F.\Delta t = {\text{ change in momentum = }}Mv$
Here, $M = m \times n$
$F.\Delta t = mnv$;
Put value in the above equation:
$144 \times 1 = 0.04 \times n \times 1200$
Solve for n, where n = number of bullets fired in one second.
$\dfrac{{144}}{{1200 \times 0.04}} = n$
The number of bullets fired per second is
$n = 3$

The number of bullets that he can fire per second at the most is 3.

Note:Here, we have to find a relation between force, mass, velocity and time. To make the relation between all the variables we use the concept of impulse which forces into time and which is also equal to the change in momentum. The time $\Delta t$ asked here is 1 second.