Question

The hero of a stunt film fires $50\text{gm}$ bullets from a machine gun, each at a speed of $1.0\text{km/s}$. If he fires $20$ bullets in $4\text{s}$, what average force does he exert against the machine gun during this period?

• A. $125\text{N}$
• B. $200\text{N}$
• C. $250\text{N}$
• D. $400\text{N}$

Answer 1

The correct option is C $250\text{N}$

From Newtons's $2^{nd}$ law,
$F=\frac{\Delta p}{\Delta t}=\frac{m\Delta v}{\Delta t}...\left(i\right)$
$F=$ total average force on body
Number of bullets fired, $n=20$
Change in momentum of each bullet,
$m\Delta v=(50\times {10}^{-3}\text{kg})\times (1\times {10}^3\text{m/s})=50\text{N.s}$
$\Delta t=4\text{s}$
From Eq $\left(i\right)$,
$F=n\times \frac{m\Delta v}{\Delta t}$
$\Rightarrow F=20\times \frac{50}{4}$
$\therefore F=250\text{N}$