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} F d t=\int_{0}^{3 \times 10^{-3}}\left(600-2 \times 10^{5} t\right) d t \\ =\left[600 t-\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=0.9 N s \end{array} $