2018 A/L Physics Past Paper · Question 16 (English Medium)

Answer: (2) 0, 1.5, 0

What the question asks

We want: the resultant force $F$ in each of the three time parts.

First read the acceleration from the shape of the graph.

Then use $F = ma$.

Reading the options

Each option gives three forces, one for each time part (in N).

Read the graph first. Then compare.

Relevant theory

gradient = velocity of the body

The rate of change of moments of an object is proportional to the unbalanced (resultant) force which acts on the object.

Required equations

$v = \frac{\Delta x}{\Delta t}$

$a = \frac{v_2 - v_1}{\Delta t}$

$F = ma$

Reasoning and calculation

The flat axis is time $t$ (s).

The up axis is the displacement $x$ (m).

One small square is 0.2 s across and 0.2 m up.

Key points: (0, 0), (2, 1), (4, 3), (5, 5).

From 0 to 2 s the line is straight.

From 2 to 4 s it curves.

From 4 to 5 s it is straight again.

$0.5$

$v_1 = \dfrac{1 - 0}{2 - 0}$

$2$

$v_2 = \dfrac{5 - 3}{5 - 4}$

The slope of the graph is the velocity.

In 0–2 s and 4–5 s the slope does not change. So the velocity does not change.

No change in velocity means no acceleration. So $F = 0$ there.

In 2–4 s the velocity grows from $0.5$ to $2\ \text{m s}^{-1}$. Only here is there a force.

$a = \dfrac{v_2 - v_1}{\Delta t} = \dfrac{2 - 0.5}{2}$

$0.75$

$a = \dfrac{1.5}{2}$

$1.5$

$F = 2 \times 0.75$

So the three forces are 0, 1.5 N and 0.

That is the row of option (2).

Checks

$m * (v2 - v1) / dt$

Check the unit.

$$\frac{\text{kg} \times \text{m s}^{-1}}{\text{s}} = \text{N}$$

The force comes out in newtons, like the table. Good.

A second way: use the distance.

If the acceleration is steady in 2–4 s, then $s = ut + \tfrac12 at^2$ must hold.

The graph says the block moves $s = 3 - 1 = 2$ m in that time.

$3 - 1 = 0.5 \times 2 + \tfrac{1}{2} \times a \times 2^2$

$2 = 1 + 2a$

$0.5$

$a = \dfrac{2 - 1}{2}$

$1$

$F = 2 \times 0.5$

The official answer sheet says: All. Every option was accepted.

Why? The two ways do not agree.

The slope way gives $a = 0.75\ \text{m s}^{-2}$, so $F = 1.5$ N.

The distance way gives $a = 0.5\ \text{m s}^{-2}$, so $F = 1$ N.

A steady force must give the same answer both ways. So the graph is drawn wrong, and every option got the mark.

What to remember: on the straight parts the force is zero. Only on the curved part is $F = ma$. In the exam, (2) from the slopes is the best answer.

Official ruling — all options accepted

Conclusion

From the slopes: $F = 0$ in 0–2 s, $1.5$ N in 2–4 s, $0$ in 4–5 s.

That is option (2).

The graph does not fit itself, so officially all options were accepted.

Explore

For the first $2$ s the block moves at a steady speed; no resultant force pushes it.

From $2$ s to $4$ s a push speeds it up, from $0.5$ to $2\ \text{m s}^{-1}$.

After $4$ s the push is gone and it moves at a steady speed again.

So there is a force only in the middle part, where the speed changes: $0$, $1.5$ N, $0$.