91Ó°ÊÓ

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

We need to find that 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 it is given by equation $(11.6)$in the form

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

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

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 $v_i=1.0m/s$. Using equation $\left(2\right)$and putting the values for $2kg$and $10m/s$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 force a deliver an impulse to an object with an initial momentum $p_{ix}$it changes its momentum to $p_{ix}$and is given by equation as

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

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

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

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

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

Now, putting the values $m=2kg$and $p_{fx}$in equation $\left(4\right)$to get $u_f$

$u_f=\frac{p_{fx}}{m}$

$$\begin{gathered} =\frac{6kg(m/s)}{2kg} \\ =3m/s \end{gathered}$$

The velocity is positive, so the direction of the object is to the right.