1. Analyzing Vertical Motion
The vertical motion of the ball is independent of the car’s horizontal motion. The ball behaves as a projectile under gravity ($g = 10 \text{ m/s}^2$).
We are given the time interval between the first and second bounce as $T = 2.0 \text{ s}$. For a projectile returning to the same level, the time of flight is: $$ T = \frac{2u_y}{g} $$ Substituting the values: $$ 2.0 = \frac{2u_y}{10} \implies u_y = 10 \text{ m/s} $$
Calculating Drop Time:
Since the collision is perfectly elastic, the ball rebounds with the same vertical speed it struck the ground with ($10 \text{ m/s}$). Since the ball was dropped from rest ($u=0$), the time taken to fall to the ground for the first bounce ($t_1$) is:
$$ v = u + gt_1 \implies 10 = 0 + 10(t_1) \implies t_1 = 1.0 \text{ s} $$
Thus, the first bounce occurs at $t = 1.0 \text{ s}$, and the second bounce occurs at $t = 3.0 \text{ s}$.
2. Analyzing Relative Horizontal Motion from Graph
The car accelerates forward (positive direction), while the ball has zero horizontal acceleration relative to the ground. Therefore, relative to the car, the ball accelerates backward. $$ \vec{a}_{rel} = \vec{a}_{ball} – \vec{a}_{car} = 0 – a = -a $$ This negative relative acceleration means the ball appears to move backward (to the left) relative to the car.
Interpreting the Tangent:
The graph shows the “First Bounce” at the right end ($x=8$) and the “Second Bounce” at the left end ($x=0$), consistent with backward relative motion.
- Vertical Scale: Max height $H = u_y^2 / 2g = 5 \text{ m}$. The graph peak is 5 grid units high, so 1 vertical unit = 1 m.
- Tangent Slope: The dashed tangent line is drawn at the start of the bounce (Right side). It starts at $(8, 0)$ and passes through the intersection of grid lines at $(6, 5)$.
The slope $m$ of the tangent in the Displacement-Time (or $y-x$) graph represents the ratio of initial vertical velocity to initial relative horizontal velocity: $$ \text{Slope } m = \frac{\Delta y}{\Delta x} = \frac{5 – 0}{8 – 6} = \frac{5}{2} = 2.5 $$ Also, physically: $$ m = \left| \frac{u_y}{u_{x,rel}} \right| $$ $$ 2.5 = \frac{10}{u_{x,rel}} \implies u_{x,rel} = 4 \text{ m/s} $$
So, at the instant of the first bounce ($t=1.0 \text{ s}$), the relative horizontal velocity of the ball is 4 m/s.
3. Calculating Car Displacement
The relative horizontal velocity at the first bounce is given by: $$ u_{x,rel} = u_{ball,x} – u_{car,x} $$ $$ u_{x,rel} = 0 – v_{car}(t_1) $$ Substituting magnitude $|u_{x,rel}| = 4 \text{ m/s}$ and $t_1 = 1 \text{ s}$: $$ 4 = a(t_1) = a(1) \implies a = 4 \text{ m/s}^2 $$
We need the displacement of the car between the first bounce ($t=1 \text{ s}$) and the second bounce ($t=3 \text{ s}$). $$ \Delta x_{car} = x_{car}(3) – x_{car}(1) $$ Using the equation $x = \frac{1}{2}at^2$: $$ \Delta x_{car} = \frac{1}{2}(4)(3)^2 – \frac{1}{2}(4)(1)^2 $$ $$ \Delta x_{car} = 2(9) – 2(1) $$ $$ \Delta x_{car} = 18 – 2 = 16 \text{ m} $$