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 newtons 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. $9\text{Ns}$
• B. Zero
• C. $0.9\text{Ns}$
• D. $1.8\text{Ns}$

Answer 1

The correct option is C $0.9\text{Ns}$

Given
$F=600-2\times {10}^{5}t$
${}^{\prime }{F}^{\prime }$ is zero
$\Rightarrow 0=600-2\times {10}^{5}t$
$\frac{\text{/}{6}^3}{\text{/}{2000}_{1000}}=t$
$t=\frac{3}{1000}$
Now impulse
$=\underset{0}{\overset{t}{\int }}Fdt$
$\Rightarrow \underset{0}{\overset{t}{\int }}(600-2\times {10}^{5}t)dt$
$=600t-{10}^5{t}^2$
$=600\times \frac{3}{1000}-{10}^5\times \frac{9}{{10}^6}$
$=\frac{18}{10}-\frac{9}{10}$
$=\frac{9}{10}=0.9\text{Ns}$