Question

A robber fires $100$ bullets of mass $100\text{gm}$ each from a machine gun, each at a speed of $500\text{m/s}$. If the first $40$ bullets are fired in $10\text{s}$, what is the average force exerted on the machine gun by the robber during this period?

• A. $250\text{N}$
• B. $50\text{N}$
• C. $2000\text{N}$
• D. $200\text{N}$

Answer 1

The correct option is D $200\text{N}$

Mass of each bullet is $m=100\text{gm}=0.1\text{kg}$
Considering bullets were initially at rest inside machine gun,
Change in momentum per bullet $\Delta P=(m{v}_f-m{v}_i)=m{v}_f-0$
$\therefore \Delta P=0.1\times 500=50\text{N-s}$

Total number of bullets fired in $t=10\text{s}$ is $n=40$
Hence average force imparted by the machine gun to the bullets is:
$F_{\text{avg}}=n\times \frac{\Delta P}{\Delta t}$
$\Rightarrow F_{\text{avg}}=40\times \frac{50}{10}=200\text{N}$

By Newton's $3^{\text{rd}}$ law, the gun will also experience the same amount of force in a direction opposite to the firing direction.
i.e the robber will have to apply a force $200\text{N}$ in order to hold the machine gun.