Question

A bullet is fired from a gun. The force on the bullet is given by $F=600-\left(2\times 10^5\right)t$ , where $F$ is in newton and $t$ is in seconds. If the force on the bullet becomes zero as soon as it leaves the barrel, then the average impulse imparted to the bullet is

• A. $1.8Ns$
• B. $18Ns$
• C. $9Ns$
• D. $0.9Ns$

Answer 1

Answer: (D)

Given $F=600-\left(2\times 10^{5}t\right)$
The force is zero at time, $t$, given by
$\begin{array}{l}0=600-2\times 10^{5}t\\ \Rightarrow t=\frac{600}{2\times 10^5}=3\times 10^{-3}\text{ seconds }\\ \therefore \text{ Impulse }={\int }_0^{t}Fdt={\int }_0^{3\times 10^{-3}}\left(600-2\times 10^{5}t\right)dt\\ ={\left[600t-\frac{2\times 10^5{t}^2}{2}\right]}_0^{3\times 10^{-3}}\\ =600\times 3\times 10^{-3}-10^5{\left(3\times 10^{-3}\right)}^2\\ =1.8-0.9=0.9Ns\end{array}$