Physics 9702/21 — May/June 2022

Background Concept

For a velocity–time graph, the displacement over a time interval is the area under the graph:

$$\Delta s = \int v\,\mathrm{d}t.$$

Key sign idea:

• If $v$ is positive, the area contributes a positive displacement (upwards here).
• If $v$ is negative, the area contributes a negative displacement (downwards here).

Understanding the Question

The rock is thrown upwards with initial velocity $4.0\ \text{m s}^{-1}$. From the graph, $v$ decreases linearly, crosses $v=0$ at $t=2.5\ \text{s}$ (top of the motion), and becomes negative by $t=4.0\ \text{s}$ (rock is moving downward). You must find its height above the surface at $t=4.0\ \text{s}$, i.e. the net vertical displacement from $t=0$ to $t=4.0\ \text{s}$.

Approach

Find the area under the $v$–$t$ graph from $0$ to $4.0\ \text{s}$. Because the graph crosses the axis, split it into:

1. a positive triangle from $0$ to $2.5\ \text{s}$,
2. a negative triangle/trapezium from $2.5$ to $4.0\ \text{s}$,
   then add them (including the negative sign).

Step-by-Step Reasoning

1. From $t=0$ to $2.5\ \text{s}$, velocity goes from $4.0$ to $0$, making a triangle.

$$A_1 = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 2.5 \times 4.0 = 5.0\ \text{m}.$$

This is the upward displacement to the highest point.

1. From $t=2.5$ to $4.0\ \text{s}$, velocity goes from $0$ to $-2.4\ \text{m s}^{-1}$. The area is below the axis, so it is negative. Treat it as a trapezium (or a triangle here):

Time interval:

$$\Delta t = 4.0 - 2.5 = 1.5\ \text{s}.$$

Average velocity over that interval (since the graph is a straight line):

$$\bar v = \frac{0 + (-2.4)}{2} = -1.2\ \text{m s}^{-1}.$$

So displacement:

$$A_2 = \bar v\,\Delta t = (-1.2)\times 1.5 = -1.8\ \text{m}.$$

1. Net displacement (height above the surface at $t=4.0\ \text{s}$):

$$\Delta s = A_1 + A_2 = 5.0 + (-1.8) = 3.2\ \text{m}.$$

So at $4.0\ \text{s}$ the rock is still above the launch point, but is on its way back down.

Key Takeaways

• Area under a $v$–$t$ graph gives displacement.
• Areas below the time axis count as negative displacement.
• When the graph is linear, using average velocity over an interval is quick and accurate.

Common Mistakes

• Adding the magnitudes of areas and forgetting the negative sign after $v$ becomes negative.
• Using the gradient instead of the area (gradient gives acceleration, not displacement).
• Using $t=4.0\ \text{s}$ as the base for the first triangle (should be $2.5\ \text{s}$ where $v=0$).

Things to Be Careful About

• Read the correct intercepts from the graph: $v=0$ at $t=2.5\ \text{s}$ and $v=-2.4\ \text{m s}^{-1}$ at $t=4.0\ \text{s}$.
• Keep consistent units: area units are $(\text{m s}^{-1})\times \text{s} = \text{m}$.

Background Concept

Linear momentum $p$ is

$$p = mv$$

and is a vector (it has direction because $v$ has direction). The change in momentum is

$$\Delta p = p_\text{final} - p_\text{initial} = m(v_\text{final} - v_\text{initial}).$$

The unit $\text{kg m s}^{-1}$ is equivalent to $\text{N s}$.

Understanding the Question

You are asked for the change in momentum from $t=0$ to $t=4.0\ \text{s}$.

• Mass $m = 7.5\ \text{kg}$.
• From the graph: $v_0 = +4.0\ \text{m s}^{-1}$ and $v_4 = -2.4\ \text{m s}^{-1}$.
  The negative final velocity indicates the rock is moving downward at $t=4.0\ \text{s}$.

Approach

Use

$$\Delta p = m(v_4 - v_0)$$

and keep the sign of velocities to show direction.

Step-by-Step Reasoning

1. Find momentum at $t=0$:

$$p_0 = mv_0 = 7.5 \times 4.0 = 30\ \text{kg m s}^{-1}.$$

1. Find momentum at $t=4.0\ \text{s}$:

$$p_4 = mv_4 = 7.5 \times (-2.4) = -18\ \text{kg m s}^{-1}.$$

1. Subtract to get change in momentum:

$$\Delta p = p_4 - p_0 = -18 - 30 = -48\ \text{kg m s}^{-1} = -48\ \text{N s}.$$

The negative sign means the momentum has changed in the negative (downward) direction.

Key Takeaways

• Use $\Delta p = m\Delta v$ with correct signs.
• Negative velocity gives negative momentum (direction matters).

Common Mistakes

• Using $p_0 - p_4$ instead of $p_4 - p_0$.
• Ignoring the negative sign on $v_4$.
• Writing the unit as $\text{N}$ (momentum change is $\text{N s}$, not $\text{N}$).

Things to Be Careful About

• State or imply direction through the sign; if you choose to give a magnitude instead, you should state “downwards”.
• Do not round intermediate momenta too aggressively; here values are simple so it is fine.

Background Concept

On a planet with no air resistance (as stated), the only significant force on the rock in flight is its weight $W = mg$ acting downward.

The rock’s acceleration is then constant and equal to the gravitational field strength (taking upward as positive):

$$a = -g.$$

From a velocity–time graph:

• the gradient is acceleration:

$$a = \frac{\Delta v}{\Delta t}.$$

Understanding the Question

You are given a straight-line $v$-–$t$ graph, so acceleration is constant. You must determine the weight $W$ of the $7.5\ \text{kg}$ rock on this planet.

Because there is no atmosphere in this part, the only force is weight, so the acceleration found from the graph is the gravitational acceleration $g$ (in magnitude).

Approach

1. Calculate the gradient of the $v$-–$t$ graph to find $a$.
2. Use $a = -g$ (upwards taken as positive) to get $g$.
3. Calculate weight using $W = mg$.

Step-by-Step Reasoning

1. Choose two clear points on the straight line. From the graph:

• at $t=0$, $v=4.0\ \text{m s}^{-1}$,
• at $t=2.5\ \text{s}$, $v=0$.

Gradient:

$$a = \frac{0 - 4.0}{2.5 - 0} = \frac{-4.0}{2.5} = -1.6\ \text{m s}^{-2}.$$

So the magnitude of gravitational acceleration is

$$g = 1.6\ \text{m s}^{-2}.$$

1. Weight:

$$W = mg = 7.5 \times 1.6 = 12\ \text{N}.$$

(Weight acts downward.)

Key Takeaways

• Gradient of a $v$-–$t$ graph gives acceleration.
• With no air resistance, acceleration is constant and equals $-g$.
• Weight is $W=mg$.

Common Mistakes

• Using area under the graph (displacement) instead of gradient (acceleration).
• Forgetting that the gradient is negative; $g$ is the magnitude of acceleration due to gravity.
• Writing $W = ma$ with $a$ negative and giving a negative weight (weight is a force magnitude; direction is downward).

Things to Be Careful About

• Use two well-separated points for the gradient to reduce reading error.
• Quote weight in newtons, $\text{N}$, and use consistent significant figures (here 2 s.f. is appropriate).