Given  $F=600-(2\times {10}^{5}t)$

The force is zero at time, t, given by

                                      $0=600-2\times {10}^{5}t$

 $\Rightarrow t=\frac{600}{2\times {10}^5}=3\times {10}^{-3}\mathrm{seconds}$

 $\therefore \mathrm{Impulse}={\displaystyle \underset{0}{\overset{t}{\int }}Fdt}={\displaystyle \underset{0}{\overset{3\times {10}^{-3}}{\int }}(600-2\times {10}^{5}t)dt}$

 $\begin{array}{l}={\left[600t-\frac{2\times {10}^5{t}^2}{2}\right]}_0^{3\times {10}^{-3}}\\ =600\times 3\times {10}^{-3}-{10}^5{(3\times {10}^{-3})}^2\end{array}$

  $=1.8 – 0.9 = 0.9 \mathrm{N} \mathrm{s}$