How to Explain Algebraic Relationships in HSC Physics

A trolley is pulled with twice as much force. Should its acceleration double?

You might use Newton’s second law, \(F_{\mathrm{net}}=ma\), and say yes. But imagine increasing the pulling force by adding weights to a mass hanging from a pulley. Those weights also increase the mass being accelerated. The acceleration might not double after all.

The equation hasn’t failed. We’ve overlooked something that must stay constant.

To explain an algebraic relationship without substituting numbers, rearrange the equation to make the quantity you’re investigating the subject. Identify which other quantities stay constant, then describe how changing one variable affects the other. If several variables change, compare their ratios rather than assuming a simple proportional relationship.

For example, from \(a=F_{\mathrm{net}}/m\), acceleration is directly proportional to net force only when mass is constant. That’s a complete physical relationship, not just a description of the symbols.

01The trolley experiment: why the first prediction can fail

Suppose you’re investigating how the force pulling a trolley affects its acceleration.

You want to answer a specific question: if the pulling force increases, does the acceleration increase in the same proportion?

Setting up the apparatus

A trolley of mass \(M\) sits on a horizontal track. A light string connects the trolley to a mass hanger at the right-hand end of the track.

The string passes over a pulley and hangs vertically. Slotted masses placed on the hanger provide a downward gravitational force.

A motion sensor positioned to the left of the trolley measures its position as it moves towards the pulley. The position measurements can be used to determine its acceleration.

The important quantities are:

• \(M\): mass of the trolley and any masses carried on it, in kilograms (kg).
• \(m\): mass of the hanger and its suspended masses, in kilograms (kg).
• \(g\): gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\) near Earth’s surface.
• \(T\): tension in the string, in newtons (N).
• \(a\): acceleration of the trolley and hanging mass, in metres per second squared (\(\mathrm{m\,s^{-2}}\)).

We’ll take motion towards the pulley as positive for the trolley, and downward as positive for the hanging mass.

For the initial model, assume the track is level and frictionless, the string is taut and has negligible mass, and the pulley rotates without friction or significant rotational inertia.

These are modelling assumptions, not claims about real laboratory equipment.

First attempt: add more mass to the hanger

Imagine keeping the trolley mass \(M\) unchanged and doubling the hanging mass from \(m\) to \(2m\).

The gravitational force on the hanger doubles.

It seems reasonable to predict that the acceleration also doubles. After all, Newton’s second law connects force and acceleration.

But the hanging mass is not merely pulling the trolley. It must accelerate too.

We need to treat the trolley and hanger as one connected system.

For the trolley, the horizontal force is the string tension:

\[
T=Ma
\]

For the hanging mass, its weight acts downwards and tension acts upwards:

\[
mg-T=ma
\]

The string is taut and inextensible, so the trolley and hanging mass have equal acceleration magnitudes.

Adding the two force equations eliminates the internal tension:

\[
\begin{aligned}
T+mg-T&=Ma+ma\\
mg&=(M+m)a
\end{aligned}
\]

Therefore,

\[
a=\frac{mg}{M+m}
\]

The numerator \(mg\) is the external force driving the ideal system. The denominator \(M+m\) is the total mass that must accelerate.

Now the problem is visible.

When we increase \(m\), both the numerator and denominator increase.

Acceleration rises, but it is not directly proportional to the hanging mass while \(M\) stays fixed.

The revised experiment: transfer masses instead

How could you increase the driving force without increasing the total mass being accelerated?

Place several slotted masses on the trolley at the beginning.

Then transfer one slotted mass at a time from the trolley to the hanger.

Each transfer increases the hanging mass but decreases the mass carried by the trolley by the same amount.

The total moving mass remains constant.

Call this total mass \(S\):

\[
S=M+m
\]

Here, \(S\) is constant across the trials, even though \(M\) and \(m\) individually change.

Newton’s second law now gives:

\[
a=\frac{mg}{S}
\]

Because \(g\) and \(S\) are constant,

\[
a\propto m
\]

The symbol \(\propto\) means “is proportional to”.

In the ideal revised experiment, doubling the hanging mass doubles the acceleration. Tripling the hanging mass triples the acceleration.

More precisely, acceleration is directly proportional to the driving force \(mg\), which is also directly proportional to \(m\).

This is the relationship the original experiment was trying to isolate.

What would the measurements tell us?

The ideal model predicts a straight-line graph of acceleration against hanging mass, passing through the origin.

Real equipment is less obliging.

Wheel resistance, pulley friction, and other effects reduce the measured acceleration. Transferring masses can also change the trolley’s normal force and therefore its friction.

Consider the following entirely illustrative data. These are invented values for examining the method, not measurements from a real experiment.

Assume \(S=0.80\ \mathrm{kg}\) throughout, with \(g=9.8\ \mathrm{m\,s^{-2}}\).

Notice two things.

First, each transfer maintains the same total mass of \(0.80\ \mathrm{kg}\).

Second, the illustrative accelerations are lower than the ideal predictions, but they still increase approximately linearly with hanging mass.

A roughly straight graph alone does not establish that friction is negligible.

We can use one row for a small force estimate.

For the middle trial, the downward gravitational force is:

\[
F_{\mathrm{drive}}=mg
\]

Substituting the illustrative values:

\[
\begin{aligned}
F_{\mathrm{drive}}
&=(0.10)(9.8)\\
&=0.980\ \mathrm{N}
\end{aligned}
\]

The measured acceleration in this illustrative trial is \(0.88\ \mathrm{m\,s^{-2}}\). Therefore the system’s net driving force along its motion is:

\[
\begin{aligned}
F_{\mathrm{net}}&=Sa\\
&=(0.80)(0.88)\\
&=0.704\ \mathrm{N}
\end{aligned}
\]

The difference is approximately:

\[
\begin{aligned}
F_{\mathrm{resist}}
&=0.980-0.704\\
&=0.276\ \mathrm{N}
\end{aligned}
\]

So the illustrative result is consistent with approximately \(0.28\ \mathrm{N}\) of opposing force in that trial, under the assumed model.

The calculation doesn’t prove which part of the apparatus produced that resistance. Further tests would be needed to separate wheel friction, pulley effects, and other losses.

That’s an important distinction: an experiment can support an algebraic model without proving every assumption behind it.

02How to explain a relationship using symbols instead of numbers

The most useful technique is to compare two states of the same system.

Suppose a system initially experiences a net force \(F_1\) and has mass \(m_1\). Its acceleration is:

\[
a_1=\frac{F_1}{m_1}
\]

After something changes:

\[
a_2=\frac{F_2}{m_2}
\]

Dividing the second equation by the first gives:

\[
\frac{a_2}{a_1}
=
\frac{F_2}{F_1}
\frac{m_1}{m_2}
\]

Here, \(F_1\) and \(F_2\) are signed net forces in newtons, and \(m_1\) and \(m_2\) are the corresponding masses in kilograms.

We’ve assumed the first acceleration and net force are non-zero so that these ratios are defined.

This equation tells us exactly how acceleration changes without needing numerical values.

For example:

The final row is useful because it exposes a common mistake.

You cannot say acceleration triples just because the net force triples. You must also account for the mass.

Where this explanation would fail

Return to the original design, where we increased the hanging mass without removing anything from the trolley.

The total mass no longer remains constant.

If the hanger changes from \(m\) to \(2m\), with trolley mass \(M\) fixed:

\[
a_1=\frac{mg}{M+m}
\]

and

\[
a_2=\frac{2mg}{M+2m}
\]

Their ratio is:

\[
\frac{a_2}{a_1}
=
\frac{2(M+m)}{M+2m}
\]

For positive \(M\) and \(m\), this ratio is greater than one but less than two.

The acceleration increases, but by less than a factor of two.

When \(M\) is much larger than \(m\), the change approaches a doubling because the added mass makes little difference to the total. That is an approximation, not an exact direct proportion.

When \(m\) becomes much larger than \(M\), the acceleration approaches \(g\). Adding more hanging mass eventually has very little relative effect.

Both limits agree with the physical model.

03Choosing the right equation means checking what stays fixed

An equation can be mathematically valid and still support the wrong explanation if you assume that a variable is constant when it isn’t.

This happens in mechanics, electricity, waves, and gravitational motion.

A useful decision process is:

1. Identify the dependent variable. What are you trying to explain: acceleration, speed, force, wavelength, or power?
2. Identify the changing quantities. Read the physical setup, not just the list of symbols.
3. Choose an equation whose model applies. Check its assumptions, including whether acceleration is constant, forces are net forces, or a wave speed can be treated as fixed.
4. Rearrange and compare. Put the dependent variable on one side, then form a ratio between initial and final states if more than one quantity changes.
5. Translate the mathematics. Say what increases or decreases, by what factor, and under which conditions.

Some common relationships illustrate the final step.

In the wave relationship, \(v\) is wave speed in \(\mathrm{m\,s^{-1}}\), \(f\) is frequency in hertz (Hz), and \(\lambda\) is wavelength in metres (m).

In the electrical relationship, \(P\) is power in watts (W), \(V\) is potential difference in volts (V), and \(R\) is resistance in ohms (\(\Omega\)). The expression applies to the power dissipated by a resistor when its resistance and potential difference have the stated values.

In the gravitational relationship, \(F_g\) is gravitational force in newtons, \(G\) is the universal gravitational constant, \(M\) and \(m\) are the interacting masses in kilograms, and \(r\) is their centre-to-centre separation in metres. The inverse-square form applies to point masses and, externally, to spherically symmetric masses.

These conditions are part of the physics.

04Equation selection: known variables, assumptions, and scope

Sometimes two equations contain the same variable but answer different physical questions.

Consider these examples:

For the constant-acceleration relationship, \(u\) is initial velocity, \(v\) is final velocity, \(a\) is signed acceleration, and \(s\) is signed displacement. Velocities are measured in \(\mathrm{m\,s^{-1}}\), and displacement in metres.

If acceleration varies significantly throughout the motion, you cannot use that equation unchanged.

If the question asks about speed rather than velocity, remember that speed is a magnitude. A negative velocity indicates direction under the chosen sign convention, not a negative speed.

The physical situation determines which equation is useful. Having the right letters is not enough.

For help locating relationships in the examination reference material, see How to Use the HSC Physics Data and Formulae Sheets.

Which NSW Physics syllabus does this apply to?

As at November 2026, the Physics Stage 6 Syllabus (2017) remains the syllabus for the 2026 HSC. Its Year 11 Module 2, Dynamics, provides the mechanics foundation used in the trolley experiment.

The Physics 11-12 Syllabus (2025) begins implementation for Year 11 in 2027, with the first HSC examination under that syllabus in 2028. Its Year 11 Fundamentals of mechanics focus area provides the corresponding context for force and motion.

The symbolic comparison method is useful across both courses. However, the exact examination resources and content requirements should be checked against the syllabus for your cohort rather than assuming the 2017 and 2025 editions are identical. The transition is confirmed in NESA’s implementation timetable.

05Questions and solutions

Question 1

A trolley of constant mass \(M\) moves along a horizontal track.

In the first trial, a constant net horizontal force \(F\) produces acceleration \(a\). In the second trial, the net force becomes \(3F\).

Explain how the acceleration changes without substituting numerical values.

Solution 1

The acceleration triples because net force increases by a factor of three while mass remains constant.

Newton’s second law gives:

\[
a=\frac{F_{\mathrm{net}}}{M}
\]

Initially:

\[
a_1=\frac{F}{M}
\]

In the second trial:

\[
a_2=\frac{3F}{M}
\]

Therefore:

\[
\frac{a_2}{a_1}=3
\]

The result is \(a_2=3a_1\), in the same direction as the net force.

The key condition is constant mass. If the force increase had been achieved by adding mass to the moving system, a direct threefold change could not be assumed.

Question 2

A trolley on a horizontal track is connected to a hanging mass by a string passing over a pulley.

Masses are transferred from the trolley to the hanger so that the total moving mass \(S\) stays constant.

The system experiences a constant opposing resistive force \(F_r\), measured in newtons. Assume this resisting force has the same magnitude during both trials, and the system is already moving towards the pulley.

The hanging mass increases from \(m\) to \(2m\), and \(mg>F_r>0\).

(a) Derive expressions for the accelerations before and after the transfer.

(b) Determine whether the second acceleration is less than, equal to, or greater than twice the first acceleration.

(c) Predict the acceleration-versus-hanging-mass graph for this model and explain why extrapolating it to very small hanging masses requires care.

Solution 2

The second acceleration is greater than twice the first. The driving force doubles, but the constant opposing force does not.

Part (a): Deriving the accelerations

Taking motion towards the pulley as positive, the net driving force is:

\[
F_{\mathrm{net}}=mg-F_r
\]

Newton’s second law gives:

\[
Sa=mg-F_r
\]

Thus:

\[
a_1=\frac{mg-F_r}{S}
\]

After the hanging mass doubles:

\[
a_2=\frac{2mg-F_r}{S}
\]

Both expressions have acceleration units of \(\mathrm{m\,s^{-2}}\), because the numerators are forces in newtons and \(S\) is in kilograms.

Part (b): Comparing the accelerations

Twice the initial acceleration is:

\[
2a_1=\frac{2mg-2F_r}{S}
\]

Compare this with:

\[
a_2=\frac{2mg-F_r}{S}
\]

Subtracting:

\[
a_2-2a_1=\frac{F_r}{S}
\]

Because \(F_r>0\) and \(S>0\),

\[
a_2>2a_1
\]

The initial driving force must first overcome the opposing resistance. When the hanging mass doubles, the additional gravitational force contributes to acceleration without requiring an equivalent increase in the assumed constant resisting force.

That produces a greater-than-twofold change in acceleration.

Part (c): Predicting the graph

Rearranging gives:

\[
a=\frac{g}{S}m-\frac{F_r}{S}
\]

This has the form of a straight line, with:

• Gradient \(g/S\), measured in \(\mathrm{m\,s^{-2}\,kg^{-1}}\).
• Vertical intercept \(-F_r/S\), measured in \(\mathrm{m\,s^{-2}}\).
• A zero-acceleration point at \(m=F_r/g\) when extrapolated within the same force model.

However, the equation assumes a moving system experiencing the stated resistive force.

At rest, static friction may behave differently. It can prevent motion until the driving force exceeds a threshold.

Therefore, extrapolating the sliding-motion equation to hanging masses too small to initiate motion may give an invalid prediction.

The tempting mistake is to treat the full gravitational force \(mg\) as the net force. The missing constraint is the resisting force.

Question 3

Two resistors of equal resistance \(R\) are connected in series to an ideal battery with constant potential difference \(V\).

One resistor is replaced with a resistor of resistance \(2R\). The other resistor remains unchanged.

Assume the resistances have the stated values throughout, connecting wires have negligible resistance, and the battery has negligible internal resistance.

Without assigning numerical values to \(V\) or \(R\):

(a) Compare the total current before and after the replacement.

(b) Compare the potential difference across the replaced resistor before and after the change.

(c) Determine whether the power dissipated in that resistor increases or decreases, and find the exact ratio.

A student argues that doubling resistance must double power because \(P=I^2R\). Evaluate that reasoning.

Solution 3

The total current decreases to \(\frac23\) of its original value. The potential difference across the replaced resistor increases to \(\frac43\) of its original value, but the power dissipated in it decreases to \(\frac89\) of its original value.

This occurs because changing one resistor changes the current and voltage distribution throughout the series circuit.

Part (a): Compare the currents

Initially, the total resistance is:

\[
R_{\mathrm{total},1}=2R
\]

From Ohm’s law:

\[
I_1=\frac{V}{2R}
\]

After replacement:

\[
R_{\mathrm{total},2}=3R
\]

Therefore:

\[
I_2=\frac{V}{3R}
\]

The ratio is:

\[
\frac{I_2}{I_1}
=
\frac{V/(3R)}{V/(2R)}
=\frac23
\]

The total current falls to two-thirds of its original value.

Part (b): Compare the voltage across the replaced resistor

Initially, the resistor has resistance \(R\):

\[
V_1=I_1R=\frac V2
\]

After replacement, its resistance is \(2R\):

\[
V_2=I_2(2R)=\frac{2V}{3}
\]

Consequently:

\[
\frac{V_2}{V_1}
=
\frac{2V/3}{V/2}
=\frac43
\]

The potential difference across the replaced resistor increases by one-third.

Part (c): Compare the power

The electrical power dissipated by a resistor is:

\[
P=I^2R
\]

Initially:

\[
\begin{aligned}
P_1&=\left(\frac{V}{2R}\right)^2R\\
&=\frac{V^2}{4R}
\end{aligned}
\]

After replacement:

\[
\begin{aligned}
P_2&=\left(\frac{V}{3R}\right)^2(2R)\\
&=\frac{2V^2}{9R}
\end{aligned}
\]

Thus:

\[
\frac{P_2}{P_1}
=
\frac{2V^2/(9R)}{V^2/(4R)}
=\frac89
\]

The power decreases by one-ninth of its original value.

The student’s argument fails because \(P\propto R\) using \(P=I^2R\) requires constant current.

The current is not constant in this circuit. Doubling one resistor changes the total resistance and therefore changes the current through both resistors.

Likewise, using \(P=V^2/R\) and assuming the power must halve would be incorrect because the potential difference across the resistor also changes.

The result illustrates why identifying the fixed quantity is more important than memorising a proportional rule.

Question 4

Two satellites orbit the same planet in stable circular orbits.

Satellite A has mass \(m\) and orbital radius \(r\), measured from the planet’s centre.

Satellite B has mass \(4m\) and orbital radius \(2r\).

Assume the planet has mass \(M\), the satellites do not significantly affect each other, and gravity provides the required centripetal force.

(a) Compare the magnitudes of the gravitational forces on the satellites.

(b) Compare their centripetal accelerations.

(c) Compare their orbital speeds and orbital periods.

(d) Explain why equal gravitational forces do not imply equal accelerations or orbital periods.

Solution 4

The gravitational forces are equal, but satellite B has one-quarter of satellite A’s centripetal acceleration, \(1/\sqrt2\) of its orbital speed, and \(2\sqrt2\) times its orbital period.

Part (a): Gravitational force

Newton’s law of universal gravitation gives:

\[
F_g=\frac{GMm}{r^2}
\]

For satellite A:

\[
F_A=\frac{GMm}{r^2}
\]

For satellite B:

\[
\begin{aligned}
F_B&=\frac{GM(4m)}{(2r)^2}\\
&=\frac{4GMm}{4r^2}\\
&=F_A
\end{aligned}
\]

The greater satellite mass increases the force by a factor of four, while doubling the orbital radius reduces the force by a factor of four.

These effects cancel.

Part (b): Centripetal acceleration

Newton’s second law gives \(a=F/m\).

For satellite A:

\[
a_A=\frac{F_A}{m}
\]

For satellite B:

\[
a_B=\frac{F_B}{4m}
\]

Since the gravitational forces are equal:

\[
\frac{a_B}{a_A}=\frac14
\]

Satellite B accelerates towards the planet’s centre at one-quarter the rate of satellite A.

Part (c): Orbital speed and period

The centripetal acceleration for circular motion is:

\[
a=\frac{v^2}{r}
\]

Therefore:

\[
\frac{a_B}{a_A}
=
\frac{v_B^2/(2r)}{v_A^2/r}
\]

Using \(a_B/a_A=1/4\):

\[
\frac14
=
\frac{v_B^2}{2v_A^2}
\]

Hence:

\[
\frac{v_B}{v_A}=\frac1{\sqrt2}
\]

The orbital period \(T\), measured in seconds, is the circumference divided by orbital speed:

\[
T=\frac{2\pi r}{v}
\]

Therefore:

\[
\frac{T_B}{T_A}
=
\frac{2r}{r}\frac{v_A}{v_B}
\]

So:

\[
\frac{T_B}{T_A}=2\sqrt2
\]

Satellite B takes \(2\sqrt2\) times as long to complete one orbit.

Part (d): Explain the apparent contradiction

Equal force does not imply equal acceleration when the masses differ.

The gravitational force on B equals the force on A, but B has four times as much inertia. Its acceleration must therefore be four times smaller.

The orbital period also depends on radius and orbital speed. A larger circular orbit is not travelled at the same speed around a planet of fixed mass.

The tempting mistake is to use one equation while silently holding a variable constant that the question explicitly changes.

Question 5

A laboratory investigates a trolley-and-hanger system. Slotted masses are transferred between the trolley and hanger, keeping their total moving mass \(S\) constant.

In every trial the trolley is moving towards the pulley, and the recorded acceleration follows:

\[
a=\frac{mg}{S}
\]

where \(m\) is the hanging mass.

A student concludes that the trolley must be travelling on a horizontal, frictionless track because the acceleration is directly proportional to the hanging mass.

Another student proposes a different apparatus.

The trolley travels on a track inclined downwards towards a pulley at its lower end. The hanging mass descends as the trolley approaches the pulley.

The track makes an angle \(\theta\) with the horizontal, where \(0<\theta<20^\circ\). The trolley experiences kinetic friction with coefficient \(\mu_k>0\). The string remains taut and parallel to the inclined track.

The pulley and string are otherwise ideal.

Let the mass of the trolley, including any masses carried on it, be \(M=S-m\).

(a) Derive the acceleration for the inclined apparatus while the trolley moves towards the pulley.

(b) Determine whether there is a condition under which the inclined apparatus produces exactly the same acceleration relationship as the horizontal frictionless apparatus for every hanging mass in the tested range.

(c) Can the forward-motion acceleration data alone distinguish the two models?

(d) Suppose the trolley is given an initial push away from the pulley so that it moves up the incline, pulling the hanger upwards. Derive its acceleration while it is moving away, taking acceleration towards the pulley as positive.

Explain how this additional observation could distinguish the models.

Assume kinetic friction maintains its stated magnitude whenever the trolley slides, and that the string stays taut during the reverse-motion test. Do not analyse the instant when the trolley momentarily stops.

Solution 5

The student’s conclusion is not justified. An inclined track with non-zero kinetic friction can produce exactly the same forward-motion acceleration relationship as a horizontal frictionless track.

The two models become distinguishable when the direction of the trolley’s motion is reversed.

The key is that gravity along the incline and kinetic friction oppose one another while the trolley moves downhill, but act in the same direction while it moves uphill.

Part (a): Acceleration towards the pulley

Choose the positive direction towards the pulley for the trolley, and downward for the hanger.

On the inclined track, the trolley experiences:

• String tension \(T\), directed towards the pulley.
• A component of its weight \(Mg\sin\theta\), directed down the incline towards the pulley.
• Kinetic friction \(F_k=\mu_kMg\cos\theta\), directed up the incline, opposite its motion.

The normal reaction has magnitude \(Mg\cos\theta\), because the string acts parallel to the track and there is no acceleration perpendicular to it.

Newton’s second law for the trolley is:

\[
Ma=T+Mg\sin\theta-\mu_kMg\cos\theta
\]

For the hanging mass:

\[
ma=mg-T
\]

Adding the equations eliminates the tension:

\[
Sa=mg+Mg\sin\theta-\mu_kMg\cos\theta
\]

Therefore:

\[
a=
\frac{mg+Mg(\sin\theta-\mu_k\cos\theta)}{S}
\]

Part (b): Find the condition for identical acceleration relationships

For the inclined apparatus to match the ideal horizontal result, the extra contribution from the trolley’s motion along the slope must vanish.

This requires:

\[
Mg\sin\theta-\mu_kMg\cos\theta=0
\]

For positive trolley mass:

\[
\sin\theta=\mu_k\cos\theta
\]

Hence:

\[
\mu_k=\tan\theta
\]

Under this condition:

\[
a=\frac{mg}{S}
\]

This is exactly the same relationship as the horizontal frictionless model.

Notice why the cancellation persists when masses are transferred.

Both the downhill gravitational component and the kinetic friction force are proportional to the trolley mass \(M\). Changing \(M\) changes both contributions in the same proportion, so their cancellation remains exact.

The inclined trolley still experiences kinetic friction and dissipates mechanical energy through friction. It simply gains an equal amount of driving force from its component of gravity along the incline.

Part (c): Can the acceleration data identify the correct model?

No. The relationship \(a=mg/S\) during forward motion cannot, by itself, distinguish these two particular models.

Both predict exactly the same acceleration for every hanging mass while the stated conditions hold.

A straight graph through the origin is therefore insufficient evidence that the apparatus is frictionless.

This is more subtle than saying that experimental errors might hide friction. Even perfectly accurate forward-acceleration measurements would not separate the models under the cancellation condition.

Part (d): Reverse the trolley’s motion

Now give the trolley an initial velocity away from the pulley.

While the trolley moves uphill, the hanging mass rises. The acceleration is still taken as positive towards the pulley.

The component of gravity along the incline still points downhill, towards the pulley.

But kinetic friction reverses direction because it opposes the trolley’s velocity.

Kinetic friction now also points towards the pulley.

The trolley’s force equation becomes:

\[
Ma=T+Mg\sin\theta+\mu_kMg\cos\theta
\]

The hanger equation remains:

\[
ma=mg-T
\]

Adding them:

\[
Sa=mg+Mg\sin\theta+\mu_kMg\cos\theta
\]

Thus:

\[
a=
\frac{mg+Mg(\sin\theta+\mu_k\cos\theta)}{S}
\]

Using the cancellation condition from Part (b):

\[
\sin\theta=\mu_k\cos\theta
\]

we obtain:

\[
a=
\frac{mg+2\mu_kMg\cos\theta}{S}
\]

Because \(\mu_k>0\), \(M>0\), and \(\cos\theta>0\),

\[
a>\frac{mg}{S}
\]

The inclined apparatus now has a greater acceleration towards the pulley than the horizontal frictionless apparatus.

While the trolley is moving away from the pulley, that positive acceleration means it slows down.

For the frictionless horizontal apparatus, the acceleration remains:

\[
a=\frac{mg}{S}
\]

regardless of whether the trolley is moving towards or away from the pulley.

The reversal test separates the models because the direction of kinetic friction depends on the direction of motion, whereas the gravitational component along the incline does not.

The small-angle assumption and taut-string condition ensure the simplified connected-system model remains applicable during the moving portion of the reverse trial.

The deeper lesson is that even a perfectly matching algebraic relationship does not necessarily establish a unique physical explanation.

Two different mechanisms can produce the same measured pattern.

To distinguish them, you need an experiment that changes a condition under which their predictions diverge.

06The next step: test the physical model before trusting the algebra

Explaining a relationship without numerical substitution is not just an algebra exercise.

You’re identifying which physical quantities can vary independently, which quantities are constrained, and what the governing equation predicts when those conditions change.

A strong explanation therefore does three things: it states the proportional relationship, identifies what must remain constant, and explains why the physical system obeys that relationship.

Once you’ve done that, dimensional analysis provides a separate check. It can help reject an equation whose units are inconsistent, although it cannot establish that the equation’s physical assumptions are correct.

See How to Check a Physics Equation Using Dimensional Analysis for that next step.

In an unfamiliar HSC question, start by deciding which variables are actually free to change. Only then decide what the algebra allows you to conclude.