A $2.0kg$object is moving to the right with a speed of $1.0m/s$when it experiences the force shown in FIGURE EX11.8. What are the object’s speed and direction after the force ends?

We need to find 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 $J_x$and it is given by equation in the form

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

In a graph of force versus time the force is in the range of time $△t=1.0s$. the impulse takes the rectangle shape, where length is $l=1.0s$and width is $w=-2N$. 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

$$\begin{gathered} J_x=\left(width\right)\left(length\right) \\ =(-2N)\left(1s\right) \\ =(-2kg)\times (m/s) \end{gathered}$$

The object has mass $m$and moves with speed $v$that has momentum $p$, 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 $2kgand1.0m/s$.

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

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 forv$p_{ix}and{j}_x$in equation $\left(3\right)$to get $p_{fx}$

$$\begin{gathered} p_{fx}=p_{ix}+\hspace{0.17em}{j}_x \\ =(2kg)\times (m/s)+(-2kg)\times (m/s) \\ =(0kg)\times (m/s) \end{gathered}$$

Zero is the final momentum means objective has been stopped, and final velocity is zero

$v_f=0m/s$.