2022 A/L Physics Past Paper · Question 02 (English Medium)

Answer: (5) $5\ \text{N}$

What the question asks

The two forces are $11\ \text{N}$ and $5\ \text{N}$. Find the value that the resultant can never have.

What each letter means:

$P$ ($\text{N}$): the first force.

$Q$ ($\text{N}$): the second force.

$R$ ($\text{N}$): the resultant of the two forces.

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

Relevant theory

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

Each option is one possible value of $R$.

The answer is the value that falls outside the range of $R$.

So first find the range.

Reasoning and calculation

For example, think of two children pulling a cart with two ropes.

If both pull the same way, their pulls add up. The cart feels the biggest pull, $P + Q$.

If they pull in opposite ways, one pull works against the other. The cart feels only what is left over, $P - Q$.

If the ropes make some angle, the cart feels a pull in between.

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

In words: the resultant is never smaller than the difference of the two forces, and never bigger than their sum.

So $R$ can take any value between those two ends, and no value outside them.

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 forces, or their difference.

Those are the two ends of the range.

Put the two forces into the formula.

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

$$R^2 = 11^2 + 5^2 + 2 \times 11 \times 5\cos\theta$$

$$R^2 = 146 + 110\cos\theta$$

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

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

$$R^2 = 146 + 110 = 256 \quad\Rightarrow\quad R = 16\ \text{N}$$

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

$$R^2 = 146 - 110 = 36 \quad\Rightarrow\quad R = 6\ \text{N}$$

So the resultant lies from $6\ \text{N}$ to $16\ \text{N}$.

$(25 - 146) / 110$

Why the other options fail

$16\ \text{N}$ is the sum of the two forces. It can happen when both pull the same way.

$9\ \text{N}$ lies between $6\ \text{N}$ and $16\ \text{N}$. Some angle gives it.

$7\ \text{N}$ lies between $6\ \text{N}$ and $16\ \text{N}$. Some angle gives it.

$6\ \text{N}$ is the difference of the two forces. It can happen when they pull in opposite ways.

Conclusion

$R_{\min} = 11 - 5 = 6\ \text{N}$

The resultant can only be from $6\ \text{N}$ to $16\ \text{N}$.

$5\ \text{N}$ is below $6\ \text{N}$, so it cannot be the resultant.

The answer is (5).

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

Find those two ends first. Then the options outside them are the ones that cannot happen.

Exam tip