A ball rebounding between two walls located between at $x=0$ and $x=2 \,cm ;$ after every

$2$ $s$, the ball receives an impulse of magnitude $0.08 \times 10^{-2} \,kg\, m / s$ from the walls

The given graph shows that a body changes its direction of motion after every $2$ $s$. Physically, this situation can be visualized as a ball rebounding to and fro between twe stationary walls situated between positions $x=0$ and $x=2 \,cm .$ since the slope of the $x-$ graph reverses after every $2$ $s$, the ball collides with a wall after every $2$ $s$. Therefore, bal receives an impulse after every $2$ $s$. Mass of the ball, $m=0.04 \,kg$

The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity $(u)$ as:

$u=\frac{(2-0) \times 10^{-2}}{(2-0)}=10^{-2}\, m / s$

Velocity of the ball before collision, $u=10^{-2} \,m / s$ Velocity of the ball after collision, $v=-10^{-2} \,m / s$

(Here, the negative sign arises as the ball reverses its direction of motion.) Magnitude of impulse $=$ Change in momentum $=|m v-m u|$

$=|0.04(v-u)|$

$=\left|0.04\left(-10^{-2}-10^{-2}\right)\right|$

$=0.08 \times 10^{-2} \,kg\, m / s$