Question

A machine gun fires n bullets per second and the mass of each bullet is m. If v is the speed of each bullet, then the force exerted on the machine gun is:
A.nmV
B.\[m{V^2}g\]
C.\[m{V^2}n\]
D.mgV

Answer 1

Hint: Force exerted is equal to change in momentum divided by the change in time.

Complete step by step solution:
Assuming,
Initial velocity of the bullet, u=0
Final velocity of the bullet is v
Therefore,
Change in momentum of one bullet is: \[\Delta {P_1}\]

So, change in momentum is equal to: \[\Delta {P_1} = mv - 0 = mv\]
In one second, the number of bullets fired is n.
So change in momentum of the bullets in one second is equal to: \[\Delta P = n\Delta {P_1} = mnv\]
Therefore,
According to the definition of speed.
Force exerted on the machine gun is equal to; \[F = \dfrac{{\Delta P}}{{\Delta t}}\]
On putting in the values we get.
\[F = \dfrac{{mnv}}{1} = mnv\]

Therefore, the correct answer is option (a)

Additional information: Before solving the question, we need to know about a few concepts. Firstly, Momentum is a measurement of mass in motion: that is how much mass in how much motion. It is usually denoted by the letter p. Where m is the mass and v are the velocity of the object. The standard unit is kgm/s and momentum is always a vector quantity.
There are two kinds of momentum, linear and angular. For example, a rotating object has an angular momentum and an object travelling in rectilinear motion with a velocity has linear momentum.

Note: Students often do not know the force for force exerted on an object with respect to momentum and time. Students also need to keep in mind the meaning of recoil of a gun and they need to understand it to be able to solve the sum.