An average force of 125 N is applied on a machine gun firing bullets each of mass 10 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 :

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Conservation of Momentum

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 :

(A)5
(B)50✓ Correct
(C)100
(D)25

Answer: (B)

Solution

The force needed to hold the gun equals the rate at which momentum is carried off by the bullets,

F = n m v

, with

n

the number fired each second. Substituting

F = 125

m = 0.01

kg and

v = 250

m/s gives

125 = n (0.01) (250) = 2.5 n

. Hence

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

bullets per second.

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option

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