Question

A bullet is fired from a gun. The force on the bullet is given by $F=600-(2\times {10}^5)t$,where $F$ is in $N$ and $t$ in $sec$. The force on bullet becomes zero as soon as it leaves barrel. What is average impulse imparted to bullet at $t=3\times {10}^{-3}$?

• A. $9Ns$
• B. $zero$
• C. $0.9Ns$
• D. $1.8Ns$

Answer 1

The correct option is C $0.9Ns$

The force is given $F=600-(2\times {10}^5)t$

For zero force, $0=600-(2\times {10}^5)t\Rightarrow t=3\times {10}^{-3}s$

Thus, average impulse, $I={\int }_0^{t=3\times {10}^{-3}}Fdt$

or $I={\int }_0^{3\times {10}^{-3}}[600-(2\times {10}^5\left)t\right]dt=600\times 3\times {10}^{-3}-{10}^5\times (3\times {10}^{-3}{)}^2=1.8-0.9=0.9Ns$