A $600g$air-track glider collides with a spring at one end of the track. FIGURE EX11.12 shows the glider’s velocity and the force exerted on the glider by the spring. How long is the glider in contact with the spring?

We need to find that How long is the glider in contact with the spring.

We have to find $△t$because of the glider is in contact with spring at the time of impulse localid="1649920534377" $△t$.

For a short time a object receives a force is known as impulse. and under the area 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 $△t$. the impulse takes the rectangle shape, where height is $h=36N$and and the base is $b=△t$. The area of the rectangle is half the product of the heightand the base

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

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

$$\begin{gathered} J_x=\frac{1}{2}\left(height\right)\left(base\right) \\ =\frac{1}{2}(36N)△t \\ =18△t \end{gathered}$$

After finding impulse $J_x=18△t$.An impulse is the quantity $J_x$that is delivered by a force to an object with an initial momentum $p_{ix}$that changes its momentum $p_{fx}$, and it is given by equation in the form of a force.

$p_{fx}=p_{ix}+J_x\left(2\right)$

THe momentum is given by equation in the form

$p=mv$

By using the momentum $p$and impulse $J_x$expressions in equation (2), we can find an expression for the time $△t$as follows

$$\begin{gathered} p_{fx}=p_{ix}+J_x \\ m{v}_f=m{v}_i+18△t \\ △t=\frac{m(v_f-v_i)}{18}\left(3\right) \end{gathered}$$

In a graph of velocity versus time the initial speed is $v_i=-3m/s$and final velocity $v_f=3m/s$.

So puttin values into equation $\left(3\right)$for the mass $m,v_{i}and{v}_f$ to get $△t$

$$\begin{gathered} △t=\frac{m(v_f-v_i)}{18} \\ =\frac{\left(0.60kg\right)\left(3.0m/s-\left(-3.0m/s\right)\right)}{18} \\ =0.2s \end{gathered}$$