Phase Space Diagram for Ball Thrown Vertically Upward

To determine the correct phase space diagram for a ball thrown vertically upward from the ground, we analyze the relationship between its position $x(t)$ and momentum $p(t)$ over time.

1. Governing Equations of Motion

Let:

• Ground level be $x = 0$, with the upward direction taken as positive ($x > 0$).
• $m$ be the mass of the ball.
• $v_0 > 0$ be the initial upward velocity, giving an initial momentum $p_0 = m v_0 > 0$.

By conservation of total mechanical energy $E$:
$E = \text{Kinetic Energy} + \text{Potential Energy} = \frac{p^2}{2m} + mgx$

At $t = 0$, the ball is at $x = 0$ with initial momentum $p_0$, so the total energy is:
$E = \frac{p_0^2}{2m}$

Equating the two expressions for total energy:
$\frac{p^2}{2m} + mgx = \frac{p_0^2}{2m}$

Solving for position $x$ as a function of momentum $p$:
$x = \frac{p_0^2 - p^2}{2m^2 g}$

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2. Analysis of the Curve in Phase Space ($x$ vs $p$)

• Shape of the Curve:
  The relationship $x(p) = \frac{p_0^2}{2m^2 g} - \frac{p^2}{2m^2 g}$ represents a parabola opening to the left (towards negative $x$) in the phase space plane, with its vertex located at:
  $(x, p) = \left( \frac{p_0^2}{2m^2 g}, 0 \right) = (h_{\text{max}}, 0)$

• Domain of Position:
  Since the ball is thrown from the ground and remains above ground throughout its flight, $x \ge 0$ at all times.

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3. Time Evolution (Direction of Arrows)

1. Ascent Phase ($0 \le t \le t_{\text{max}}$):

   • Initial State ($t = 0$): The ball starts at the ground ($x = 0$) with positive momentum ($p = p_0 > 0$).
   • Motion: As the ball moves upward, its position $x$ increases from $0$ to $h_{\text{max}}$, while its velocity and momentum $p$ decrease from $p_0$ to $0$.
   • Trajectory in Phase Space: The state moves through the first quadrant ($x > 0, p > 0$) from $(0, p_0)$ to the vertex $(h_{\text{max}}, 0)$. The arrow points downwards and to the right.

2. Maximum Height ($t = t_{\text{max}}$):

   • The ball reaches its peak height $x = h_{\text{max}}$ where $p = 0$.

3. Descent Phase ($t_{\text{max}} \le t \le T$):

   • Motion: As the ball falls back down, its position $x$ decreases from $h_{\text{max}}$ back to $0$. The momentum $p$ becomes negative, decreasing from $0$ to $-p_0$.
   • Trajectory in Phase Space: The state moves through the fourth quadrant ($x > 0, p < 0$) from $(h_{\text{max}}, 0)$ to $(0, -p_0)$. The arrow points downwards and to the left.

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Conclusion

The phase space diagram is a parabolic arc located entirely in the region $x \ge 0$, starting on the positive vertical axis at $(0, p_0)$, looping through positive $x$ to reach the horizontal axis at $x = h_{\text{max}}$, and returning to the negative vertical axis at $(0, -p_0)$ in a clockwise manner.

This description uniquely corresponds to Option D.