A $2kg$object is moving to the right with a speed of $1m/s$when it experiences an impulse of $-4Ns$. What are the object’s speed and direction after the impulse?

We have given that a$2kg$object is moving to the right with a speed of $1m/s$when it experiences an impulse of $-4N/s$.

We need to find the object's speed and direction after the impulse.

The force with a short time when an object receives a force is called the impulse, so 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 is given as

impulse$=J_x={\int }_{t_i}^{t_f}{F}_x\left(t\right)dt$

=area under the $F_x\left(t\right)$curve between $t_i$and $t_f$ $\left(1\right)$

When an object that has mass $m$and moves with speed $v$has momentum $p$, this momentum is a vector and it is the product of the object's mass and its velocity. The momentum is given by equation $(11.3)$in the form

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

The initial velocity is $v_i=1.0m/s$. We can use equation $\left(2\right)$and plug the values for $2kg$and $1.0m/s$to get the initial momentum $p_{ix}$of the object by

$$\begin{gathered} p_{ix}=m{v}_i \\ =\left(2kg\right)(1.0m/s) \\ =2kg(m/s) \end{gathered}$$

The impulse is the quantity $J_x$and we a force deliver an impulse to an object with an initial momentum$p_{ix}$it changes momentum to $p_{ix}$and is given as

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

Now, putting the values for $p_{ix}$and $J_x$into equation $\left(3\right)$to get $p_{fx}$

$$\begin{gathered} p_{fx}=p_{ix}+J_x \\ =2kg(m/s)+(-4kg(m/s\left)\right) \\ =-2kg(m/s) \end{gathered}$$

Now, using equation $\left(2\right)$to get $u_f$of the object by

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

Putting the values for $m=2kg$and $p_{fx}$into equation $\left(4\right)$to get $u_f$

$$\begin{gathered} u_f=\frac{p_{fx}}{m} \\ =\frac{-2kg(m/s)}{2kg} \\ =-1m/s \end{gathered}$$

The velocity is negative, so the direction of the object is to the left.