1. Analyzing the Graph Structure

The graph shows Kinetic Energy ($K$) versus Height ($h$). We can identify two distinct phases of motion:

• Region $h = 2.0 \to 1.5$ m (Straight Line): The ball is in free fall. $K$ increases linearly as height decreases. The potential energy is converting to kinetic energy.
• Point $h = 1.5$ m: The graph changes slope . This indicates the ball hits the spring.
  $\therefore$ Relaxed Length of Spring = 1.5 m.
• Region $h = 1.5 \to 1.0$ m (Curve): The spring is compressing. The ball slows down until it stops momentarily at $h = 1.0$ m.

2. Determining Mass of the Ball (m)

Consider the free fall phase ($h = 2.0 \to 1.5$).
Change in height $\Delta h = 2.0 – 1.5 = 0.5$ m.
Loss in Potential Energy = Gain in Kinetic Energy. $$ m g \Delta h = K_{impact} $$ From the graph, at $h = 1.5$, the Kinetic Energy corresponds to $2.5$ J (based on grid extrapolation between 2.0 and 3.0). $$ m (10) (0.5) = 2.5 $$ $$ 5m = 2.5 $$ $$ m = 0.5 \text{ kg} $$

3. Determining Stiffness of the Spring (k)

Consider the entire motion from drop ($h=2.0$) to max compression ($h=1.0$).
At both points, Kinetic Energy is zero ($K=0$).
Work done by Gravity + Work done by Spring = 0.
Gravity: Moves down by $2.0 – 1.0 = 1.0$ m. $$ W_g = m g (1.0) = 0.5 \times 10 \times 1 = 5 \text{ J} $$ Spring: Compresses by $x = 1.5 – 1.0 = 0.5$ m. $$ W_s = -\frac{1}{2} k x^2 = -0.5 k (0.5)^2 = -0.125 k $$ Energy Balance: $$ W_g + W_s = 0 $$ $$ 5 – 0.125 k = 0 $$ $$ k = \frac{5}{0.125} = 40 \text{ N/m} $$

Final Results