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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 beA. 1B. 2C. 3D. 4

Last updated date: 20th Sep 2024
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Hint: As we have to find out the number of bullets he can fire per second. So we can use here the formula of impulse which is equal to the product of change of momentum. On substituting the values we can find easily the value of the number of bullets he can fire per second.

It is given that mass of bullet is $M = 65kg = 0.065g$
Initially the bullet was it at rest so its initial velocity is $U = 0$
Machine fires the bullet with velocity of 1300m/s, therefore
Final velocity of bullet is $V = 1300m/s$
Force exerted on gun $F = 169N$
Now, as we know that impulse is equal to the change of momentum, therefore we will find change in momentum first. As change is momentum is the difference of final momentum and the initial momentum, therefore
Change in momentum $P = MV - MU = M\left( {V - U} \right)$
For N bullets the change in momentum is $\Delta P = NM\left( {V - U} \right)$
Now Impulse= change in momentum that is $I = \Delta P$…………………………(1)
And we also know that impulse is the product of force and time. So $I = F \times T$
Put this value in equation (1)
⇒ $F \times T = NM\left( {V - U} \right)$
Now substituting the values we get,
⇒$168 \times 1 = N \times 0.065\left( {1300 - 0} \right)$
⇒ $N = \dfrac{{169}}{{0.065 \times 1300}} = 2$
⇒ Number of bullets fire per second = 2

Hence, B option is correct.

Note: Impulse is the integral of the force over the time interval for which it acts. This is also a vector quantity like force. Impulse applied to any object produces an equivalent vector change in its linear momentum. Here, we need to find the number of bullets fired per second so when we write impulse as the product of force and time so we must take time equal to one second.