2020 A/L Physics Past Paper · Question 28 (English Medium)

Answer: (3) $c$

What the question asks

We must find: the path the $+Q$ sphere at A can follow on the smooth table, starting from rest.

Relevant theory

Like charges (+, + and -, -) repel each other while unlike charges (+, -) attract each other.

The mutual force between two point electric charges is directly proportional to the product of the two charges and inversely proportionatel to the square of the distance between them

The direction of the unbalanced (resultant) force is the same as the direction of the momentum change.

Applying it to this question

Direction convention: in the figure $BC$ is horizontal.

The direction from $C$ towards $B$ is "left", the direction from $BC$ towards $A$ is "up".

A is the top vertex of the triangle; $AB = AC = a$.

Vectors: two forces act on A.

The attraction $\vec F_B$ from the $-Q$ at B (from A towards B).

The repulsion $\vec F_C$ from the $+Q$ at C (from C towards A, continuing outward).

Since the table is smooth and horizontal, these alone determine the motion.

Reasoning and calculation

Direction rule: both forces are from charges of equal magnitude $Q$ at the same distance $a$.

So $F_B = F_C = \dfrac{kQ^2}{a^2} = F$.

$\vec F_B$ is $60^\circ$ below horizontal, towards the left.

$\vec F_C$ is $60^\circ$ above horizontal, towards the left.

The vertical components $F\sin 60^\circ$ cancel each other.

$F_x = F\cos 60^\circ + F\cos 60^\circ$

$F_x = 2F \times \frac{1}{2} = F$

$F_y = F\sin 60^\circ - F\sin 60^\circ = 0$

$F_{net} = \frac{kQ^2}{a^2}\ \text{(left, horizontal)}$

Eventual direction: starting from rest, the first velocity is in the direction of the resultant — horizontally left.

As A moves left, $AB$ shrinks and $AC$ grows.

$F_B \propto \dfrac{1}{AB^2}$ grows, $F_C$ shrinks, so a downward component develops.

Since the velocity turns towards the force, the path starts horizontal and gradually curves downward.

Reading the options

Comparing with the diagram's paths: only path $c$ starts horizontal and curves downward — option (3).

Why the other options fail

Curves upward.

Assumes the force direction stays fixed.

Considers only B's attraction.

Moves downward right from the start.

Checks

$0.5 * QB - 0.5 * QC$

Second way (a limiting case): if B also had $+Q$, the horizontal components of both repulsions would cancel, and A would be pushed straight upward.

Flipping one sign turns the resultant $90^\circ$ to point left — matching our result.

Also, paths that move away from A's starting direction ($d$, $e$) are impossible for an object starting from rest.

Conclusion

At the start the resultant $\dfrac{kQ^2}{a^2}$ points horizontally left; as A moves, B's attraction grows and the force turns downward: path $c$ — the answer is (3).

Assuming an object moves straight along a force line, or in the direction of the force forever.

A force changes velocity; the velocity turns slowly towards the force. Only when starting from rest is the initial direction the direction of the force.