91Ó°ÊÓ

We need to find that the object’s speed and direction after the force ends.

For a short time a object receives a force is known as impulse. it is the area under the curve of the force-versus-time graph and it is the same as momentum. The impulse is the quantity and it is given by equation in the form

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

In a graph of force versus time the force is in the range of time $â–³t=0.50s$. the impulse takes the rectangle shape, where length is $l=0.50s$and width is $w=+8N$. The area of the rectangle is the product of the width and the length

$A=\left(width\right)\left(lenght\right)$

Using equation $\left(1\right)$to get impulse

$$\begin{gathered} J_x=\left(width\right)\left(length\right) \\ =(8N)(0.50s) \\ =(4kg)×(m/s) \end{gathered}$$

The object has mass $m$and moves with speed $v$that has momentum , Vector is a product of object's mass and its velocity.The momentum is given by equation in the form

$p=mv\left(2\right)$

Initial velocity $v_i=1.0m/s$To get initial momentum $p_{ix}$of the object using equation $\left(2\right)$and putting values for $2kgand1.0m/s$.

$$\begin{gathered} p_{iv}=m{v}_i \\ =(2kg)(1.0m/s) \\ =(2kg)×(m/s) \end{gathered}$$

The impulse changes its momentum to and it is given by equation in the form

$p_{fx}=p_{ix}+jx\left(3\right)$

Putting values for role="math" localid="1649762340199" $p_{ix}and{J}_x$in equation $\left(3\right)$to get $p_{fx}$

$$\begin{gathered} p_{fx}=p_{ix}+ j_x \\ =(2kg)×(m/s)+(4kg)×(m/s) \\ =(6kg)×(m/s) \end{gathered}$$

To get $v_f$of the object using equation $\left(2\right)$

$v_f=\frac{p_{fx}}{m}\left(4\right)$

The values for $m=2kgand{p}_{fx}$putting into equation$\left(4\right)$to get $v_f$

$$\begin{gathered} v_f=\frac{p_{fx}}{m} \\ =\frac{\left(6kg\right)\left(m/s\right)}{2kg} \\ =3m/s \end{gathered}$$

The direction of the object is rightbecause the velocity is positive.