First slide

Law of conservation of momentum and it's applications

Question

A machine gun fires a bullet of mass 40 g with a velocity 1200 ms^-1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per
second at the most?

Moderate

A

One
B

Four
C

Two
D

Three

Solution

u = Velocity of bullet
$\frac{d m}{d t} =$ Mass fired per second by the gun
$Mass of bullet (m_{B}) \times Bullets fired per \sec (N)$
Maximum force that man can exert, $F = u (\frac{d m}{d t})$
$\begin{array}{l} ∴ F = u \times m_{B} \times N \\ \Rightarrow N = \frac{F}{m_{B} \times u} = \frac{144}{40 \times 10^{- 3} \times 1200} = 3 \end{array}$

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