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 newton and $t$ in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

• A. $9N-s$
• B. $0$
• C. $0.9N-s$
• D. $1.8N-s$

Answer 1

Answer: (C)

$0.9N-s$

Force on the bullet: $F=600-2\times {10}^{5}t$

Let the time at which $F=0$ be $t^{\prime }$.

$\therefore t^{\prime }=\frac{600}{2\times {10}^5}=3\times {10}^{-3}s$

The impulse imparted to bullet

$I={\int }_0^{t^{\prime }}Fdt={\int }_0^{t^{\prime }}(600-2\times {10}^{5}t)dt=[600t-{10}^5{t}^2{]}_0^{t^{\prime }}$

$\Rightarrow I=600\times t^{\prime }-{10}^5{t}^{\prime 2}=600\times 3\times {10}^{-3}-{10}^5(3\times {10}^{-3}{)}^2=1.8-0.9=0.9Ns$