Physics 9702/43 — May/June 2016

Background Concept

Gravitational potential $\phi$ is potential energy per unit mass:

$$\phi = \frac{U}{m}$$

So a change in potential corresponds to a change in gravitational potential energy:

$$\Delta U = m\Delta \phi.$$

If only gravitational forces do work (no engines, no friction), mechanical energy is conserved:

$$\Delta K + \Delta U = 0 \quad \Rightarrow \quad \Delta K = -\Delta U.$$

Understanding the Question

A $180\ \text{kg}$ rock moves from $x=0.1\times 10^{12}\ \text{m}$ to $x=1.2\times 10^{12}\ \text{m}$. The graph gives $\phi$ at those points (in units of $10^8\ \text{J kg}^{-1}$). You must find how much the rock’s kinetic energy changes and state whether it increases or decreases.

Approach

1. Read $\phi$ at the two $x$ values.
2. Calculate $\Delta \phi = \phi_2 - \phi_1$.
3. Convert to $\Delta U = m\Delta \phi$.
4. Use $\Delta K = -\Delta U$.

Step-by-Step Reasoning

1. From the graph, at $x=0.1\times 10^{12}\ \text{m}$, $\phi_1 \approx -10\times 10^8\ \text{J kg}^{-1}$.
2. At $x=1.2\times 10^{12}\ \text{m}$, $\phi_2 \approx -14\times 10^8\ \text{J kg}^{-1}$.
3. The potential becomes more negative, so

$$\Delta \phi = -14\times 10^8 - (-10\times 10^8) = -4\times 10^8\ \text{J kg}^{-1}.$$

1. Multiply by mass to get the change in potential energy:

$$\Delta U = 180\times (-4\times 10^8) = -7.2\times 10^{10}\ \text{J}.$$

The negative sign means gravitational potential energy decreases (the rock has moved to a “lower” potential).

1. With conservation of mechanical energy (gravity only), the kinetic energy must increase by the same amount:

$$\Delta K = -\Delta U = +7.2\times 10^{10}\ \text{J}.$$

Key Takeaways

• $\phi$ is energy per unit mass, so multiply by $m$ to get $U$.
• A decrease in $U$ corresponds to an increase in $K$ when only conservative forces act.
• Always track signs carefully with negative potentials.

Common Mistakes

• Using $\Delta K = m\Delta \phi$ instead of $\Delta K = -m\Delta \phi$.
• Forgetting the graph’s scale factor ($10^8\ \text{J kg}^{-1}$).
• Stating “decrease” because $\phi$ decreases (more negative), without converting properly to energy changes.

Things to Be Careful About

• The graph values are approximate; sensible rounding (2–3 s.f.) is expected.
• Keep units consistent: $\phi$ in $\text{J kg}^{-1}$ leads to $U$ in $\text{J}$ after multiplying by kg.

Background Concept

In a conservative field (like gravity), total mechanical energy per unit mass is conserved:

$$\frac{K}{m} + \phi = \text{constant}, \quad \text{where } \frac{K}{m} = \frac{1}{2}v^2.$$

So between two points 1 and 2,

$$\frac{1}{2}v_1^2 + \phi_1 = \frac{1}{2}v_2^2 + \phi_2.$$

A key idea here is a “potential barrier”: if the potential increases to a maximum somewhere along the path, the object must have enough initial kinetic energy to reach that maximum.

Understanding the Question

The rock starts at $x=0.1\times 10^{12}\ \text{m}$ with speed $v$. It must travel toward star B and reach $x=1.2\times 10^{12}\ \text{m}$. The graph shows that $\phi$ first rises (becomes less negative) to a maximum around $x\approx 0.6\times 10^{12}\ \text{m}$, then falls.

The minimum starting speed is the one that just allows the rock to get over the highest point of $\phi(x)$; once it has passed that point, it will then move to lower potential and speed up again.

Approach

1. Read $\phi_1$ at the start.
2. Find the maximum value $\phi_{\max}$ on the graph between start and destination.
3. Set the kinetic energy at the top of the barrier to zero (minimum condition).
4. Use conservation of energy per unit mass to solve for $v$.

Step-by-Step Reasoning

1. Starting potential (from graph):

$$\phi_1 = -10\times 10^8\ \text{J kg}^{-1}.$$

1. The curve reaches a maximum (least negative) value of about

$$\phi_{\max} \approx -4.4\times 10^8\ \text{J kg}^{-1}.$$

1. To just reach this point, the rock arrives with $v=0$ there. Apply energy per unit mass conservation between start and the maximum point:

$$\frac{1}{2}v^2 + \phi_1 = 0 + \phi_{\max}.$$

Rearrange:

$$\frac{1}{2}v^2 = \phi_{\max} - \phi_1.$$

1. Substitute:

$$\phi_{\max} - \phi_1 = (-4.4 - (-10))\times 10^8 = 5.6\times 10^8\ \text{J kg}^{-1}.$$

1. Solve for $v$:

$$v = \sqrt{2(5.6\times 10^8)} = \sqrt{1.12\times 10^9} \approx 3.3\times 10^4\ \text{m s}^{-1}.$$

(The mass cancels automatically because we used energy per unit mass.)

Key Takeaways

• Minimum launch speed is determined by the highest potential along the route, not the potential at the final point.
• Use energy per unit mass: $\frac{1}{2}v^2$ pairs naturally with $\phi$.
• Reading the correct feature (the maximum) from the graph is the main physics skill here.

Common Mistakes

• Using $\phi$ at $x=1.2\times 10^{12}\ \text{m}$ instead of $\phi_{\max}$.
• Forgetting that $\phi$ values are negative and mishandling $\phi_{\max} - \phi_1$.
• Not including the factor of 2 when solving from $\frac{1}{2}v^2$.

Things to Be Careful About

• Ensure you pick the maximum of the curve (least negative value), not the minimum.
• Keep the $10^8$ scale factor from the axis.
• Quote $v$ to a sensible number of significant figures given the graph reading (typically 2 s.f. or 3 s.f.).