JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 12)

An average force of $$125 \mathrm{~N}$$ is applied on a machine gun firing bullets each of mass $$10 \mathrm{~g}$$ at the speed of $$250 \mathrm{~m} / \mathrm{s}$$ to keep it in position. The number of bullets fired per second by the machine gun is :
25
50
5
100

Explanation

To find the number of bullets fired per second, we can use the concept of momentum. When the machine gun fires bullets, it experiences a backward force due to the conservation of momentum. The force applied on the machine gun is used to balance this backward force.

First, let's find the momentum of each bullet:

Momentum = Mass × Velocity

Bullet mass = $$10 \mathrm{~g} = 0.01 \mathrm{~kg}$$ Bullet velocity = $$250 \mathrm{~m/s}$$

Momentum per bullet = $$0.01 \mathrm{~kg} \cdot 250 \mathrm{~m/s} = 2.5 \mathrm{~kg \cdot m/s}$$

Now let's find the momentum per second that needs to be balanced by the applied force:

Force = $$125 \mathrm{~N}$$

Momentum per second = Force × Time

Since we are considering a time interval of 1 second, the momentum per second is equal to the applied force:

Momentum per second = $$125 \mathrm{~N}$$

Now, let's find the number of bullets fired per second:

Number of bullets = Momentum per second / Momentum per bullet

Number of bullets = $$\frac{125 \mathrm{~N}}{2.5 \mathrm{~kg \cdot m/s}}$$

Number of bullets = 50

Therefore, the machine gun fires 50 bullets per second.

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