MHT CET · Physics · Center of Mass Momentum and Collision
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?
- A 3
- B 6
- C 10
- D 9
Answer & Solution
Correct Answer
(C) 10
Step-by-step Solution
Detailed explanation
Momentum of bullets per second:
\(\mathrm{P}=\mathrm{nmv}\)
According to Newton's second law of motion,
\(\begin{aligned}
\mathrm{F} & =\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{nmv})}{\mathrm{dt}} \\
\therefore \quad \frac{\mathrm{dn}}{\mathrm{dt}} & =\frac{\mathrm{F}}{\mathrm{mv}} \\
& =\frac{300}{0.03 \times 1000} \\
& =10
\end{aligned}\)
\(\mathrm{P}=\mathrm{nmv}\)
According to Newton's second law of motion,
\(\begin{aligned}
\mathrm{F} & =\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{nmv})}{\mathrm{dt}} \\
\therefore \quad \frac{\mathrm{dn}}{\mathrm{dt}} & =\frac{\mathrm{F}}{\mathrm{mv}} \\
& =\frac{300}{0.03 \times 1000} \\
& =10
\end{aligned}\)
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