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Question: A machine gun fires \(50\,g\) bullets at a speed of \(1000\,m{s^{ - 1}}\). The gunner holding the gu...

A machine gun fires 50g50\,g bullets at a speed of 1000ms11000\,m{s^{ - 1}}. The gunner holding the gun in his hands can exert an average force of 180N180\,N against the gun. Determine the maximum number of bullets that can be fired per minute.
(A) 6060
(B) 120120
(C) 216216
(D) 300300

Explanation

Solution

In this problem, two formulas are used to determine the solution. First the momentum of the one bullet is found by using the given data. And take the number of bullets as nn and by using Newton's second law of motion, then the number of bullets nn can be determined.

Formula Used:
Momentum,
P=m×vP = m \times v
Where,
PP is the momentum of the bullet, mm is the mass of the bullet and vv is the velocity of the bullet.

Complete step by step answer:
Given that,
The mass of the bullet, m=50gm = 50\,g
The velocity of the bullet, v=1000ms1v = 1000\,m{s^{ - 1}}
By taking momentum,
P=m×v...................(1)P = m \times v\,...................\left( 1 \right)
Substitute the mass and velocity value in the equation (1),
P=50×103×1000P = 50 \times {10^{ - 3}} \times 1000
In the above equation, the term 103{10^{ - 3}} which is used for unit conversion from grams to kilograms, the mass value is given in terms of grams but here it is substituted in kilograms only, so the unit conversion takes place.
On multiplying the above equation, then
P=50kgms1P = 50\,kgm{s^{ - 1}}
The above equation shows the momentum of one bullet.
We have to find the number of bullets nn, so the momentum of the nn number of bullets is 50nkgms150n\,kgm{s^{ - 1}}
According to Newton’s second law of motion,
F=Pt.................(2)F = \dfrac{P}{t}\,.................\left( 2 \right)
Here, we are going to determine the number of bullets that can be fired in one minute, so take Δt=60s\Delta t = 60\,s.
Now substitute the momentum and time in equation (2), then the above equation is written as,
F=50×n60F = \dfrac{{50 \times n}}{{60}}
The force value is given as 180N180\,N and substitute this value in the above equation, then,
180=50×n60180 = \dfrac{{50 \times n}}{{60}}
On further calculation, the above equation is written as,
180×60=50×n180 \times 60 = 50 \times n
On multiplying the LHS, then
10800=50×n10800 = 50 \times n
Then keep nn in one side and other terms in other side, then the above equation is written as,
n=1080050n = \dfrac{{10800}}{{50}}
On dividing the above equation, the number of bullets is,
n=216n = 216
Thus, the 216216 can be fired in one minute.
Hence, the option (C) is the correct answer.

Note: The momentum value we determined is only for one bullet, because the mass value is given for one bullet. And then the number of bullets is assumed as nn and the momentum is multiplied by the number of bullets nn. And then by using Newton’s second law of motion, the number of bullets can be determined.