(N/A) Physical context: $A$ ball rebounding between two walls located at $x = 0$ and $x = 2 \; cm$.
$1$. Time between two consecutive impulses: The graph shows that the body changes its direction of motion after every $2 \; s$. Since the slope of the $x-t$ graph reverses after every $2 \; s$, the ball collides with a wall after every $2 \; s$. Therefore, the time between two consecutive impulses is $2 \; s$.
$2$. Magnitude of each impulse:
Mass of the ball, $m = 0.04 \; kg$.
The slope of the graph gives the velocity of the ball.
Initial velocity $(u) = \frac{\Delta x}{\Delta t} = \frac{(2 - 0) \times 10^{-2} \; m}{(2 - 0) \; s} = 10^{-2} \; m/s$.
Velocity before collision, $u = 10^{-2} \; m/s$.
Velocity after collision, $v = -10^{-2} \; m/s$ (negative sign indicates reversal of direction).
Magnitude of impulse = Change in momentum = $|mv - mu| = |m(v - u)|$.
Magnitude of impulse = $|0.04 \times (-10^{-2} - 10^{-2})| = |0.04 \times (-2 \times 10^{-2})| = 0.08 \times 10^{-2} \; kg \cdot m/s$.