Question

A machine gun fires a bullet of mass 65g with a velocity of 1300m/s. The man holding it can exert a maximum force of 169N on the gun. The number of bullets he can fire per second will be
(A) 1
(B) 2
(C) 3
(D) 4

Answer 1

In this problem initially there are two bodies and when bullets are fired as per Newton’s third law there must be two opposite and equal forces. Moreover, no external force acts on the system so the law of conservation of momentum will be applicable. Alternatively, we can also use impulse to solve this problem.

Complete step by step answer:
Mass of bullet,
${{m}_{1}}$=65g
=$\dfrac{65}{1000}kg$
=0.065kg
Initial velocity of bullet, ${{u}_{1}}=0m/s$
Final velocity of bullet, ${{v}_{1}}=1300m/s$
Force being exerted, F=169N
Impulse= change in momentum
$j=m{{v}_{1}}-m{{u}_{1}}$
$j=m({{v}_{1}}-{{u}_{1}})$

& j=0.065(1300-0) \\\ & j=84.5kgm/s \\\ \end{aligned}$$ Also, impulse is given by impulse= force$$\times $$time $$j=F\times t$$ Putting the value of j and F we get, $$84.5=169\times t$$ This gives us t=2 Thus, the number of bullets he can fire per second will be 2. Hence, the correct option is (B) Additional Information: The law of conservation of momentum states that the momentum of an isolated system remains constant. It means the total momentum before collision must be equal to total momentum after the collision. **Note:** We have solved this problem using the Impulse momentum theorem that the change in momentum of an object equals the impulse applied to it. While solving such problems we have to keep in mind that all the quantities are taken in standard units that are SI else our answer will come wrong.