91Ó°ÊÓ

We need to find the impulse on the particle that experiences th force.

A force for a short time when a object recives is called impulse. it is the area under the curve of the force-versus-time graph and it is the same as the change in momentum. The impulse is the quantity $j_x$and it is given by equation in the form

$$\begin{gathered} impulse=J_x={∫}_{t_i}^{t_f}{F}_x\left(t\right)dt \\ =areaunderthe{F}_x\left(t\right)curvebetween{t}_{i}and{t}_f\left(1\right) \end{gathered}$$

The graph we have is force versus time in figure, so the force is between the range of time from $t_i=2msto{t}_f=8ms$. the impulse take the shape of triangle and the height is $h=2000N$and the base is $(b=8ms-2ms=6ms)$. the area of triangle is half product of the height and the base.

$A=\frac{1}{2}\left(height\right)\left(base\right)$

To get impulse we can use equation $\left(1\right)$

localid="1649418166522" $$\begin{gathered} J_x=\frac{1}{2}\left(height\right)\left(base\right) \\ =\frac{1}{2}\left(2000N\right)(6×{10}^{-3}s) \\ =6N×s \end{gathered}$$