A machine gun fires bullets of mass $30 \mathrm{~g}$ with velocity of $1000 \mathrm{~m} / \mathrm{s}$. The…
A machine gun fires bullets of mass $30 \mathrm{~g}$ with velocity of $1000 \mathrm{~m} / \mathrm{s}$. The man holding the gun can exert a maximum force of $300 \mathrm{~N}$ on it. How many bullets can he fire per second at most?
3
6
10
9
Solution
Momentum of bullets per second:
$\mathrm{P}=\mathrm{nmv}$
According to Newton's second law of motion,
$\begin{aligned}
\mathrm{F} & =\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{nmv})}{\mathrm{dt}} \\
\therefore \quad \frac{\mathrm{dn}}{\mathrm{dt}} & =\frac{\mathrm{F}}{\mathrm{mv}} \\
& =\frac{300}{0.03 \times 1000} \\
& =10
\end{aligned}$