2023 A/L Physics Past Paper · Question 04 (English Medium)

Answer: (3) $4\ \text{m}$.

What the question asks

The two displacements are $9\ \text{m}$ and $6\ \text{m}$. Find the value that the resultant displacement can have.

What each letter means:

$P$ ($\text{m}$): the first displacement.

$Q$ ($\text{m}$): the second displacement.

$R$ ($\text{m}$): the resultant displacement of the two.

$\theta$ ($^\circ$): the angle between the two displacements.

Relevant theory

eg. Displacement, acceleration, impulse, moment, momentum, magnetic flux density, velocity, force, electric field intensity, gravitational field intensity, weight.

They have both magnitude and direction and obey the law of vector addition.

If two forces are represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant will be represented given by the corresponding diagonal in magnitude and direction.

$$R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$$

Reading the options

The first quote has displacement in its list of vectors. The second quote says that vectors obey the law of vector addition.

The last two quotes give that law, written for two forces. The same law adds two displacements.

Each option is one value of $R$.

The answer is the value that lies inside the range of $R$.

So first find the range.

Reasoning and calculation

For example, think of walking $9\ \text{m}$ east and then $6\ \text{m}$ more.

If you go on the same way, you end $9 + 6 = 15\ \text{m}$ from where you started.

If you turn round and walk back, you end $9 - 6 = 3\ \text{m}$ from where you started.

If you turn at some angle, you end somewhere in between.

$$|P - Q| \le R \le P + Q$$

In words: the resultant is not smaller than the difference of the two displacements, and not bigger than their sum.

So $R$ can be any value from $3\ \text{m}$ to $15\ \text{m}$.

The parallelogram law says the diagonal is the resultant.

Squash the parallelogram flat along one line.

Then the diagonal becomes the sum of the two displacements, or their difference.

Those are the two ends of the range.

Put the two displacements into the formula.

$$R^2 = P^2 + Q^2 + 2PQ\cos\theta$$

$$R^2 = 9^2 + 6^2 + 2 \times 9 \times 6\cos\theta$$

$$R^2 = 117 + 108\cos\theta$$

The cosine is never below $-1$ and never above $+1$.

Same direction, $\theta = 0^\circ$:

$$R^2 = 117 + 108 = 225 \quad\Rightarrow\quad R = 15\ \text{m}$$

Opposite directions, $\theta = 180^\circ$:

$$R^2 = 117 - 108 = 9 \quad\Rightarrow\quad R = 3\ \text{m}$$

So the resultant lies from $3\ \text{m}$ to $15\ \text{m}$.

$(16 - 117) / 108$

Why the other options fail

$1\ \text{m}$ is below $3\ \text{m}$, the difference of the two displacements. No angle gives it.

$2\ \text{m}$ is below $3\ \text{m}$, the difference of the two displacements. No angle gives it.

$16\ \text{m}$ is above $15\ \text{m}$, the sum of the two displacements. No angle gives it.

$20\ \text{m}$ is above $15\ \text{m}$, the sum of the two displacements. No angle gives it.

Conclusion

$R_{\min} = 9 - 6 = 3\ \text{m}, \quad R_{\max} = 9 + 6 = 15\ \text{m}$

The resultant can be from $3\ \text{m}$ to $15\ \text{m}$.

$4\ \text{m}$ is inside this range, so it can be the resultant.

The answer is (3).

For two displacements, the resultant is between their difference and their sum.

Find those two ends first. Then an option inside them can happen, and an option outside them cannot.

Exam tip