Question

A bullet is fired from a gun. The force on the bullet is given by $F=600-(2\times10^5)t$ where, F is in newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

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

Answer 1

Answer: (D)

0.9 Ns

Answer 2

When $F= 0, 600 - 2\times10^5t = 0$
$\therefore t=\frac{600}{2\times10^5}=3\times10^{-3}s.$
Now, impulse, $I=\int\limits^{t}_{0} Fdt=\int\limits^{t}_{0} (600-2\times10^5t)dt$
$600t-2\times10^5\frac{t^2}{2}=600\times3\times10^{-3}-10^5\times(3\times10^{-3})^2$
or, $I= 1.8-0.9 = 0.9\, Ns$.