An average force of 125 N is applied on a machine gun firing bullets each of mass 10 g at the speed of 250 m/s to keep i

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An average force of 125 N is applied on a machine gun firing bullets each of mass 10 g at the speed of 250 m/s to keep it in position. The number of bullets fired per second by the machine gun is:

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Given: Average force on the machine gun is $125 \, \text{N}$ . Mass of each bullet is $10 \, \text{g} = 0.01 \, \text{kg}$ and speed of each bullet is $250 \, \text{m/s}$ .

Find: The number of bullets fired per second.

The force required to keep the machine gun in position is equal to the rate of change of momentum of the bullets.

F = \frac{d(mv)}{dt}

Momentum of each bullet is

mv = 0.01 \times 250 = 2.5 \, \text{kg m/s}

If $n$ bullets are fired per second, then

F = n \times 2.5

Using $F = 125 \, \text{N}$ ,

125 = 2.5n

n = \frac{125}{2.5} = 50

Therefore, the number of bullets fired per second is $50$ . The correct option is D.

Given: A machine gun fires bullets of mass $10 \, \text{g}$ with speed $250 \, \text{m/s}$ and an average opposing force of $125 \, \text{N}$ keeps it in position.

Find: Bullets fired per second.

For one bullet, convert mass into SI unit:

10 \, \text{g} = \frac{10}{1000} = 0.01 \, \text{kg}

Then momentum carried by one bullet is

p = mv

p = 0.01 \times 250 = 2.5 \, \text{kg m/s}

If $n$ bullets are fired each second, total momentum given to bullets per second is

2.5n

This equals the force on the gun in magnitude:

F = 2.5n

125 = 2.5n

n = 50

Thus, the machine gun fires $50$ bullets per second.

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