When Can You Cancel Mass in HSC Physics Equations?

In 1971, an astronaut dropped a hammer and a feather on the Moon. They hit the ground together. Yet on Earth, a hammer falls much faster than a feather.

The hammer has more mass, so gravity pulls on it with a greater force. Shouldn’t that make it accelerate faster?

Not necessarily. The hammer also has more inertia, meaning it takes more force to produce the same acceleration. In free fall without air resistance, these two effects balance exactly.

You can cancel mass from a physics equation when the same non-zero mass is a common factor of every term, and the physical model justifies treating those masses as the same quantity. You cannot remove mass simply because the symbol \(m\) appears on both sides.

The important distinction is between dividing an equation by mass and making mass disappear from the final relationship. The first is often possible. The second depends on the forces involved.

01The experiment: why did the hammer and feather fall together?

What was the experiment trying to show?

For centuries, the motion of falling objects raised a difficult question.

A heavy stone experiences a greater gravitational force than a light stone. Does that mean it must fall faster?

Galileo argued that, when resistance from the surrounding air is negligible, objects fall with the same acceleration regardless of their mass.

The problem was testing that prediction fairly.

On Earth, dropping a hammer and a feather in ordinary air gives a misleading result. The feather falls slowly because air resistance is large compared with its weight.

To investigate the effect of mass, we need to separate gravitational acceleration from the effects of air.

The Apollo 15 demonstration

During the Apollo 15 mission in 1971, astronaut David Scott performed a demonstration on the Moon.

He held a geological hammer and a feather at approximately the same height, around 1.6 metres above the lunar surface.

He released both objects at approximately the same time.

A television camera recorded their descent towards the ground.

Before revealing the result, consider two predictions:

• If greater gravitational force always produced greater acceleration, the hammer would arrive first.
• If gravitational force and inertia increased in the same proportion, both objects would accelerate equally.

The hammer and feather struck the lunar surface together, within the precision of the release and observation.

The Moon has essentially no atmosphere capable of producing significant aerodynamic drag. The demonstration therefore removed the main complication that makes the feather fall slowly on Earth.

NASA’s account of the Apollo 15 demonstration describes the result as consistent with the established prediction of equal gravitational acceleration.

This was a demonstration of an existing physical principle, not the discovery of that principle in 1971. It also wasn’t a precision measurement capable of proving that the accelerations were mathematically identical.

What would a controlled Earth experiment need to change?

Suppose we wanted to investigate the same relationship in a laboratory.

We could use a tall, transparent chamber containing two objects of different masses.

At the top, a release mechanism would hold both objects at the same height. A camera positioned beside the chamber would record their positions as they fell past a measured scale.

The release must not give either object an initial push.

First, we could release the objects with air inside the chamber.

Then we could remove most of the air using a vacuum pump and repeat the release.

The vacuum pump changes the air pressure. It doesn’t meaningfully change the objects’ masses or the gravitational field.

In the air-filled chamber, we predict that the feather will be delayed significantly by air resistance.

In the evacuated chamber, we predict that both objects will follow nearly the same vertical motion.

This comparison reveals why the original Earth demonstration is incomplete: changing the object’s mass may also change the importance of air resistance.

For an even more controlled comparison of mass alone, we could use two spheres with the same diameter and surface finish but different masses. They would experience approximately the same drag at the same speed, while having different weights.

The experiment tests a physical prediction. The equations explain why that prediction makes sense.

02Why mass cancels in ideal free fall

Imagine dropping two metal balls from the same height in an evacuated chamber.

One has a mass of 1.0 kg. The other has a mass of 4.0 kg.

Which ball experiences the greater gravitational force?

The 4.0 kg ball does. Near Earth’s surface, its weight is four times as large.

But a larger force doesn’t automatically mean a larger acceleration.

Newton’s second law says:

\[
F_{\text{net}}=ma
\]

Here:

• \(F_{\text{net}}\) is the net force in newtons (N).
• \(m\) is the object’s mass in kilograms (kg).
• \(a\) is its acceleration in metres per second squared (\(\mathrm{m\,s^{-2}}\)).

Near Earth’s surface, the gravitational force is:

\[
F_g=mg
\]

Here, \(g\) is the gravitational field strength, approximately \(9.8\ \mathrm{N\,kg^{-1}}\), numerically equivalent to \(9.8\ \mathrm{m\,s^{-2}}\) for gravitational acceleration.

Assume the object is in free fall, air resistance is negligible, and gravity is the only significant force.

Taking downward as positive:

\[
ma=mg
\]

Because \(m\) is non-zero, divide both sides by \(m\):

\[
\begin{aligned}
\frac{ma}{m}&=\frac{mg}{m}\\
a&=g
\end{aligned}
\]

Mass disappears completely.

The 4.0 kg ball experiences four times the gravitational force, but it also has four times the inertia.

The resulting acceleration is identical.

This gives an important rule:

A larger force does not necessarily produce a larger acceleration if the mass is also larger.

A small but important physical assumption

There’s a deeper detail hidden in the familiar expression \(mg=ma\).

Mass plays two roles:

• Gravitational mass determines how strongly an object responds to a gravitational field.
• Inertial mass determines how strongly an object resists acceleration.

If we represent these separately as \(m_g\) and \(m_i\), the equation becomes:

\[
m_i a=m_g g
\]

Therefore:

\[
a=\frac{m_g}{m_i}g
\]

The equal acceleration of freely falling objects is connected to the experimentally supported equivalence of gravitational and inertial mass.

In ordinary HSC Newtonian mechanics, we treat these masses as equal and use the same symbol \(m\).

The cancellation is therefore justified by a physical principle, not merely because two letters happen to look alike.

03Where mass cancels – and where it doesn’t

Free fall is a particularly clean example because the only force is proportional to mass.

Other situations require more care.

Consider a student pushing an empty shopping trolley and then pushing the same trolley loaded with textbooks.

If the student applies the same horizontal force in both cases, the loaded trolley accelerates less.

The applied force has stayed constant, while the mass has increased.

For the simplified model with no resistive forces:

\[
F_0=ma
\]

Dividing by mass gives:

\[
a=\frac{F_0}{m}
\]

Here, \(F_0\) is the fixed applied force.

We can divide both sides by mass, but mass does not disappear from the result.

Doubling \(m\) while keeping \(F_0\) constant halves the acceleration.

A comparison of common HSC equations

The symbols introduced here are:

• \(\theta\): the incline angle, measured from the horizontal.
• \(\mu_k\): coefficient of kinetic friction, which has no units.
• \(D\): magnitude of air resistance, measured in newtons.
• \(k\): spring constant, measured in \(\mathrm{N\,m^{-1}}\).
• \(x\): spring compression, measured in metres.
• \(v\): speed, measured in \(\mathrm{m\,s^{-1}}\).

The inclined-plane results assume a constant gravitational field, a stationary plane, and no additional forces affecting the normal reaction. The friction result also assumes sliding and a mass-independent coefficient of kinetic friction.

The spring relationship assumes no energy losses, an initially stationary object, and a spring ending at its natural length.

The point isn’t to memorise which rows say yes.

It’s to recognise which forces or energies change in proportion to mass.

Why friction doesn’t always prevent cancellation

A common prediction is that friction automatically stops mass from cancelling.

That sounds reasonable because friction is an additional force.

But ordinary sliding friction on a stationary incline is proportional to the normal reaction, which is itself proportional to mass.

For a block sliding down a slope:

\[
F_f=\mu_k N
\]

Here, \(F_f\) is kinetic friction and \(N\) is the normal reaction force, both measured in newtons.

With no additional perpendicular forces:

\[
N=mg\cos\theta
\]

Therefore:

\[
F_f=\mu_k mg\cos\theta
\]

Taking downhill as positive, Newton’s second law becomes:

\[
ma=mg\sin\theta-\mu_k mg\cos\theta
\]

Every term has a common factor of \(m\).

Dividing gives:

\[
a=g(\sin\theta-\mu_k\cos\theta)
\]

Mass disappears.

The heavier block has more friction, but it also has proportionally more gravitational force pulling it down the slope.

This cancellation only applies while the assumed sliding-friction model is valid. Static friction behaves differently: it adjusts as needed up to a maximum value.

04How to decide whether mass can cancel

When an HSC question contains several forces, use this process before doing any algebra.

1. Identify the object or system

Are you analysing one falling object, two connected blocks, a satellite, or a collision between different objects?

The masses must be defined clearly.

For example, \(m_1\) and \(m_2\) are not automatically interchangeable just because both are masses.

2. Write the physical equation before cancelling anything

For a falling object with upward air resistance \(D\), taking downward as positive:

\[
mg-D=ma
\]

Dividing by mass gives:

\[
a=g-\frac{D}{m}
\]

You cannot remove \(m\) from the drag term unless your force model establishes that \(D\) is also proportional to \(m\).

Air resistance generally depends on shape, size, speed, and the surrounding fluid. It doesn’t automatically increase in proportion to an object’s mass.

3. Look for one common non-zero factor

For example:

\[
ma=mg\sin\theta-\mu_k mg\cos\theta
\]

All terms contain the same \(m\).

Mass can disappear completely.

But:

\[
ma=mg-F_0
\]

contains a term without \(m\).

Dividing by mass is still valid:

\[
a=g-\frac{F_0}{m}
\]

It simply doesn’t remove mass from the prediction.

4. Check whether the model remains physically possible

Cancelling mass doesn’t validate the assumptions used to write the equation.

A rope can pull but cannot push. A surface can provide a normal reaction but cannot pull an unattached object towards itself. Kinetic friction only applies while surfaces are sliding.

If a calculation requires a negative rope tension or a negative normal reaction under the assumed directions, that may indicate that the contact or rope constraint has failed.

In that case, the equation needs to change.

What about two different masses?

Sometimes the whole equation contains a common mass scale, even though different objects have different masses.

Suppose two connected objects have masses \(m\) and \(3m\).

If the net force on their combined system is \(2mg\), then:

\[
4ma=2mg
\]

Mass cancels:

\[
a=\frac{g}{2}
\]

The acceleration is independent of the absolute value of \(m\), provided the mass ratio remains 1:3 and the force model remains valid.

But it is not independent of the mass ratio.

Changing only one object’s mass changes the ratio, so the cancellation no longer predicts the same acceleration.

Does mass cancel for orbiting satellites?

Yes, under the usual idealised model of a satellite in a circular orbit with gravity as its only significant force.

Newton’s gravitational force supplies the required centripetal force:

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

Here:

• \(G\) is the gravitational constant in \(\mathrm{N\,m^2\,kg^{-2}}\).
• \(M\) is the central body’s mass in kilograms.
• \(m\) is the satellite’s mass in kilograms.
• \(r\) is the orbital radius, measured from the central body’s centre in metres.
• \(v\) is orbital speed in \(\mathrm{m\,s^{-1}}\).

Cancelling the satellite’s mass and rearranging gives:

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

The satellite’s mass disappears, but the central body’s mass does not.

Two ideal satellites at the same orbital radius have the same required circular-orbit speed, regardless of their individual masses.

That doesn’t mean the satellites have the same momentum, kinetic energy, or gravitational force.

05NSW HSC Physics syllabus note

For students completing the HSC in 2026 or 2027, the relevant course is the Physics Stage 6 Syllabus (2017).

The Physics 11-12 Syllabus (2025) begins implementation with Year 11 in 2027, with the first HSC examination under that syllabus in 2028, according to the NSW Department of Education’s implementation information.

The basic relationships used here – Newton’s second law, gravitational force, and motion – are foundational mechanics. This article isn’t claiming that every historical demonstration or advanced variation is a separately prescribed syllabus requirement.

The hardest practice question below extends familiar Newtonian mechanics to a demanding constraint problem. It is intended as enrichment, not as a prediction of standard HSC examination difficulty.

06Questions and solutions

Use \(g=9.8\ \mathrm{m\,s^{-2}}\) unless another value is given. Take all numerical data as illustrative unless explicitly identified as historical.

Question 1

Two balls are released simultaneously from rest, 5.0 m above the floor of an evacuated chamber.

Ball A has a mass of 0.50 kg. Ball B has a mass of 2.0 kg.

Assume negligible air resistance and constant gravitational acceleration.

Calculate:

(a) The gravitational force on each ball.

(b) The acceleration of each ball.

(c) The time taken for each ball to reach the floor.

Explain how the forces can be different when the accelerations are the same.

Solution 1

Both balls accelerate downwards at \(9.8\ \mathrm{m\,s^{-2}}\) and reach the floor after approximately 1.0 s. Ball B experiences four times the gravitational force, but its mass is also four times as large.

(a) Gravitational forces

Using \(F_g=mg\):

For Ball A:

\[
\begin{aligned}
F_A&=0.50(9.8)\\
&=4.9\ \mathrm N
\end{aligned}
\]

For Ball B:

\[
\begin{aligned}
F_B&=2.0(9.8)\\
&=19.6\ \mathrm N
\end{aligned}
\]

Both forces act downwards.

(b) Accelerations

Because gravity is the only significant force:

\[
ma=mg
\]

For either non-zero mass:

\[
a=g=9.8\ \mathrm{m\,s^{-2}}
\]

The greater gravitational force on Ball B is accompanied by proportionally greater inertia.

(c) Fall times

Take downwards as positive.

With initial velocity zero:

\[
s=\frac12at^2
\]

Therefore:

\[
\begin{aligned}
t^2&=\frac{2(5.0)}{9.8}\\
&=1.02\ \mathrm{s^2}\\
t&\approx1.0\ \mathrm s
\end{aligned}
\]

Both balls take approximately 1.0 s to fall.

The tempting mistake is assuming that greater gravitational force means greater acceleration. Newton’s second law relates acceleration to force divided by mass, not force alone.

Question 2

Two blocks are connected by a light, inextensible rope passing over a frictionless pulley at the top of two identical 30° inclined planes.

The left block has a mass of 2.0 kg. The right block has a mass of 6.0 kg.

Both surfaces have a coefficient of kinetic friction of 0.20.

The blocks are already moving, with the 6.0 kg block sliding down its slope and the 2.0 kg block moving up its slope.

Assume the rope remains taut.

(a) Calculate the acceleration of the system, including its direction.

(b) Calculate the tension in the rope.

(c) Both masses are now doubled to 4.0 kg and 12.0 kg. The slope angles and coefficients of friction remain unchanged. Determine what happens to the acceleration and the tension.

Explain precisely which mass factor can be cancelled and why.

Solution 2

The 6.0 kg block accelerates downhill at approximately \(0.753\ \mathrm{m\,s^{-2}}\), and the rope tension is 14.7 N. Doubling both masses leaves the acceleration unchanged but doubles the tension to 29.4 N.

The key is that the mass ratio remains fixed at 1:3.

(a) Acceleration

Let the smaller mass be \(m=2.0\ \mathrm{kg}\).

The larger mass is \(3m=6.0\ \mathrm{kg}\).

Treat both blocks as one system. The internal rope tensions cancel when the force equations are added.

Take the direction of the 6.0 kg block’s motion as positive.

The downhill gravitational driving force from the larger block is \(3mg\sin30^\circ\).

The smaller block’s gravitational component and both friction forces oppose the motion.

Therefore:

\[
4ma=2mg\sin30^\circ-4\mu_k mg\cos30^\circ
\]

Every term has a common factor of \(m\).

Dividing by \(4m\):

\[
a=\frac12g\sin30^\circ-\mu_k g\cos30^\circ
\]

Using \(\sin30^\circ=0.500\) and \(\cos30^\circ\approx0.866\):

\[
\begin{aligned}
a&=9.8(0.250-0.1732)\\
&\approx0.753\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The positive result confirms that the acceleration is in the assumed direction: the 6.0 kg block accelerates downhill.

(b) Rope tension

Consider the 2.0 kg block separately.

It accelerates uphill.

Tension acts uphill, while gravity’s component along the slope and kinetic friction act downhill.

\[
T-mg\sin30^\circ-\mu_k mg\cos30^\circ=ma
\]

Rearranging:

\[
T=m(a+g\sin30^\circ+\mu_k g\cos30^\circ)
\]

The three acceleration terms are approximately \(0.753\), \(4.90\), and \(1.70\ \mathrm{m\,s^{-2}}\).

\[
\begin{aligned}
T&=2.0(0.753+4.90+1.70)\\
&\approx14.7\ \mathrm N
\end{aligned}
\]

(c) Doubling both masses

The original acceleration equation depends on the mass ratio and the coefficient of friction, not the overall mass scale.

If both masses are doubled, the driving gravitational force, friction forces, and total inertia all double.

The acceleration remains:

\[
a=0.753\ \mathrm{m\,s^{-2}}
\]

However, rope tension is proportional to the common mass scale:

\[
T_{\text{new}}=2(14.7)=29.4\ \mathrm N
\]

The mistake would be concluding that all physical quantities become independent of mass once the mass factor cancels from the acceleration equation.

The acceleration is unchanged, but the force required to maintain the connected motion is greater.

Question 3

Two spheres have the same diameter and surface finish but different masses.

Sphere A has a mass of 0.20 kg. Sphere B has a mass of 0.80 kg.

Both are falling vertically through air.

At one instant, both spheres have a downward speed of \(6.0\ \mathrm{m\,s^{-1}}\). At that speed, the upward air resistance on each sphere is 1.2 N.

Assume the air resistance on either sphere follows the same model:

\[
D=cv^2
\]

Here, \(c\) is a positive constant with units \(\mathrm{N\,s^2\,m^{-2}}\), and \(v\) is the magnitude of velocity.

(a) Calculate the instantaneous acceleration of each sphere.

(b) Determine the terminal speed of each sphere.

(c) Explain why the spheres would accelerate equally in a vacuum but have different terminal speeds in air.

(d) A student argues: “The heavier sphere must have four times the terminal speed because it has four times the mass.” Assess this claim.

Solution 3

At \(6.0\ \mathrm{m\,s^{-1}}\), Sphere A accelerates downwards at \(3.8\ \mathrm{m\,s^{-2}}\), while Sphere B accelerates downwards at \(8.3\ \mathrm{m\,s^{-2}}\). Their predicted terminal speeds are approximately 7.7 m/s and 15.3 m/s respectively.

Mass does not cancel completely because drag is not proportional to mass in the stated model.

(a) Instantaneous accelerations

Take downwards as positive.

For either sphere:

\[
ma=mg-D
\]

Therefore:

\[
a=g-\frac{D}{m}
\]

For Sphere A:

\[
\begin{aligned}
a_A&=9.8-\frac{1.2\ \mathrm N}{0.20\ \mathrm{kg}}\\
&=9.8-6.0\\
&=3.8\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

For Sphere B:

\[
\begin{aligned}
a_B&=9.8-\frac{1.2\ \mathrm N}{0.80\ \mathrm{kg}}\\
&=9.8-1.5\\
&=8.3\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Both spheres accelerate downwards, but not equally.

(b) Terminal speeds

At terminal speed, the net force is zero.

Therefore:

\[
mg=cv_t^2
\]

Here, \(v_t\) represents terminal speed.

Rearranging:

\[
v_t=\sqrt{\frac{mg}{c}}
\]

First calculate \(c\) using the given drag force and speed.

\[
\begin{aligned}
c&=\frac{D}{v^2}\\
&=\frac{1.2}{6.0^2}\\
&=0.0333\ \mathrm{N\,s^2\,m^{-2}}
\end{aligned}
\]

For Sphere A:

\[
\begin{aligned}
v_{t,A}&=\sqrt{\frac{0.20(9.8)}{0.0333}}\\
&\approx7.7\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For Sphere B:

\[
\begin{aligned}
v_{t,B}&=\sqrt{\frac{0.80(9.8)}{0.0333}}\\
&\approx15.3\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

(c) Why the vacuum and air results differ

In a vacuum, the equation is:

\[
ma=mg
\]

Mass cancels, giving \(a=g\).

In air, the equation is:

\[
ma=mg-cv^2
\]

After dividing by mass:

\[
a=g-\frac{cv^2}{m}
\]

Mass remains.

At the same speed, the spheres experience equal drag forces under the stated model, but different gravitational forces.

The heavier sphere must reach a greater speed before its drag grows enough to balance its weight.

(d) Assessing the student’s claim

The student’s claim is incorrect.

The terminal-speed equation is:

\[
v_t=\sqrt{\frac{mg}{c}}
\]

For identical \(c\) and \(g\):

\[
v_t\propto\sqrt{m}
\]

Sphere B has four times Sphere A’s mass, so its terminal speed is:

\[
\sqrt{4}=2
\]

times as large, not four times as large.

The subtle mistake is assuming that a fourfold change in weight requires a fourfold change in speed.

Because drag depends on the square of speed, doubling speed produces four times the drag.

Question 4

Two powered trolleys travel separately up the same rough incline, which makes an angle of 12° with the horizontal.

One trolley has a mass of 2.0 kg. The other has a mass of 4.0 kg.

Each trolley has an identical motor supplying the same unknown constant uphill force \(F\).

The coefficient of kinetic friction \(\mu_k\) is also unknown but is the same for both trolleys.

At an instant when both are moving uphill, their accelerations are measured as:

(a) Determine the motor force \(F\).

(b) Calculate the coefficient of kinetic friction.

(c) The 4.0 kg trolley eventually comes to rest. Its motor continues to supply the same force. The coefficient of static friction is 0.25.

Will the trolley remain stationary, start moving uphill, or begin sliding downhill? Justify your answer using forces.

Assume the motors maintain their stated forces and no other significant resistive forces act.

Solution 4

Each motor supplies approximately 12 N uphill, and the coefficient of kinetic friction is 0.15. When the 4.0 kg trolley stops, it can remain stationary because static friction is sufficient to balance the remaining uphill force.

The essential distinction is that negative acceleration while moving uphill means slowing down, not necessarily travelling downhill.

(a) Determine the motor force

Take uphill as positive.

For a trolley of mass \(m\):

\[
ma=F-mg\sin12^\circ-\mu_k mg\cos12^\circ
\]

Divide by \(m\):

\[
a=\frac{F}{m}-g\sin12^\circ-\mu_k g\cos12^\circ
\]

For the two trolleys, the gravitational and frictional acceleration terms are identical.

Subtract the acceleration equation for the 4.0 kg trolley from the equation for the 2.0 kg trolley.

Those common terms disappear:

\[
2.53-(-0.47)=\frac{F}{2.0}-\frac{F}{4.0}
\]

The masses in the denominators are measured in kilograms, and the accelerations are in \(\mathrm{m\,s^{-2}}\).

The left side is \(3.00\ \mathrm{m\,s^{-2}}\).

Therefore:

\[
3.00=\frac{F}{4.0\ \mathrm{kg}}
\]

Giving:

\[
F=12.0\ \mathrm N
\]

(b) Determine kinetic friction

Use the 4.0 kg trolley.

Its acceleration equation is:

\[
a=\frac{F}{m}-g\sin12^\circ-\mu_k g\cos12^\circ
\]

Rearrange:

\[
\mu_k=\frac{F/m-g\sin12^\circ-a}{g\cos12^\circ}
\]

Using:

\[
\begin{aligned}
F/m&=3.00\ \mathrm{m\,s^{-2}}\\
g\sin12^\circ&\approx2.04\ \mathrm{m\,s^{-2}}\\
g\cos12^\circ&\approx9.59\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

We obtain:

\[
\begin{aligned}
\mu_k&\approx\frac{3.00-2.04+0.47}{9.59}\\
&\approx0.15
\end{aligned}
\]

The coefficient of kinetic friction is approximately 0.15.

(c) What happens when the heavier trolley stops?

The trolley was moving uphill but accelerating downhill. Therefore, it was slowing down.

When it reaches zero velocity, we must replace the kinetic-friction model with the static-friction condition.

First, calculate the gravitational component downhill:

\[
\begin{aligned}
F_{g,\parallel}&=mg\sin12^\circ\\
&=4.0(9.8)\sin12^\circ\\
&\approx8.15\ \mathrm N
\end{aligned}
\]

The motor pulls uphill with 12.0 N.

Without friction, the trolley would tend to move uphill because:

\[
12.0-8.15=3.85\ \mathrm N
\]

Static friction must therefore act downhill with a magnitude of 3.85 N to hold the trolley stationary.

The maximum available static friction is:

\[
F_{s,\max}=\mu_s N
\]

Here, \(\mu_s=0.25\) and \(N=mg\cos12^\circ\).

\[
\begin{aligned}
F_{s,\max}&=0.25(4.0)(9.8)\cos12^\circ\\
&\approx9.59\ \mathrm N
\end{aligned}
\]

The required static friction is only 3.85 N, less than the available maximum of 9.59 N.

The trolley can therefore remain stationary.

The tempting approach is to keep using kinetic friction after the trolley stops or to assume that negative acceleration guarantees a reversal of motion.

Neither is valid.

The force law changes when sliding stops, and static friction adjusts to prevent motion as long as its maximum is not exceeded.

Question 5

A lift contains a light, frictionless pulley fixed to its ceiling.

Two masses, 1.0 kg and 3.0 kg, hang from opposite ends of a light, inextensible rope passing over the pulley.

Initially, both masses are held at rest relative to the lift by a catch, with the rope taut.

Three separate trials are performed.

In each trial, the catch is released at the same instant the lift begins accelerating vertically downwards at a specified constant acceleration \(A\).

The lift is powered, so it can accelerate downwards faster than gravitational free fall.

Assume the masses do not contact the floor, ceiling, or each other. Ignore air resistance. Use \(g=9.8\ \mathrm{m\,s^{-2}}\).

(a) For Trials I and II, determine the tension in the rope and the acceleration of each mass relative to the lift.

(b) In Trial II, the 1.0 kg mass accelerates upwards relative to the lift. Does it also accelerate upwards relative to the ground? Demonstrate your conclusion mathematically.

(c) In Trial III, determine whether the rope remains taut. Find the actual acceleration of each mass relative to the ground and relative to the lift.

(d) Explain why cancelling mass from the original equations does not guarantee that those equations remain valid in all three trials.

Solution 5

In Trial I, the masses accelerate relative to the lift at \(4.90\ \mathrm{m\,s^{-2}}\), with tension 14.7 N. In Trial II, their relative accelerations decrease to \(1.96\ \mathrm{m\,s^{-2}}\), and the tension becomes 5.88 N.

In Trial III, the rope goes slack. Both masses then accelerate downwards at \(g\) relative to the ground.

The crucial constraint is that a rope cannot exert negative tension.

Step 1: Write Newton’s second law in the ground frame

Let:

• \(m=1.0\ \mathrm{kg}\), the smaller mass.
• \(3m=3.0\ \mathrm{kg}\), the larger mass.
• \(a_1\), the smaller mass’s acceleration relative to the ground.
• \(a_3\), the larger mass’s acceleration relative to the ground.
• \(T\), the rope tension in newtons.
• \(A\), the lift’s downward acceleration.

Take downward as positive in the ground frame.

For the 1.0 kg mass:

\[
mg-T=ma_1
\]

For the 3.0 kg mass:

\[
3mg-T=3ma_3
\]

These equations describe both weights and the upward rope tensions.

Step 2: Use the rope-length constraint

If the rope remains taut, the two hanging sections must have a constant total length.

The distance from the pulley to the first mass plus the distance from the pulley to the second mass is constant.

Because the pulley accelerates downwards at \(A\), the accelerations satisfy:

\[
a_1+a_3=2A
\]

This follows from the geometry of the inextensible rope.

The equation is valid only while the rope remains taut.

Step 3: Derive an expression for tension

From Newton’s second law:

\[
\begin{aligned}
a_1&=g-\frac{T}{m}\\
a_3&=g-\frac{T}{3m}
\end{aligned}
\]

Add the equations:

\[
a_1+a_3=2g-\frac{4T}{3m}
\]

Apply the rope constraint:

\[
2A=2g-\frac{4T}{3m}
\]

Rearranging:

\[
T=\frac{3m}{2}(g-A)
\]

This is the tension required to maintain a taut rope under the assumed motion.

Notice that the common mass factor can be removed when calculating acceleration, but tension itself still depends on mass.

Step 4: Calculate the acceleration relative to the lift

Define the accelerations relative to the lift as:

\[
a_{1,\mathrm{rel}}=a_1-A
\]

and:

\[
a_{3,\mathrm{rel}}=a_3-A
\]

Substituting the tension expression gives:

\[
a_{1,\mathrm{rel}}=-\frac12(g-A)
\]

and:

\[
a_{3,\mathrm{rel}}=\frac12(g-A)
\]

These results are valid while the rope is taut.

The negative sign means acceleration upwards relative to the lift, because downwards was chosen as positive.

Step 5: Trial I – stationary lift

Here, \(A=0\).

The tension is:

\[
\begin{aligned}
T&=\frac{3(1.0)}2(9.8)\\
&=14.7\ \mathrm N
\end{aligned}
\]

The relative acceleration magnitude is:

\[
\frac12(9.8)=4.90\ \mathrm{m\,s^{-2}}
\]

Therefore:

• The 1.0 kg mass accelerates upwards at \(4.90\ \mathrm{m\,s^{-2}}\).
• The 3.0 kg mass accelerates downwards at \(4.90\ \mathrm{m\,s^{-2}}\).

This is the ordinary Atwood-machine result.

Step 6: Trial II – lift accelerating downwards at \(0.60g\)

The lift acceleration is:

\[
A=0.60(9.8)=5.88\ \mathrm{m\,s^{-2}}
\]

The difference \(g-A\) is:

\[
9.8-5.88=3.92\ \mathrm{m\,s^{-2}}
\]

Therefore:

\[
\begin{aligned}
T&=\frac{3(1.0)}2(3.92)\\
&=5.88\ \mathrm N
\end{aligned}
\]

The relative acceleration magnitude is:

\[
\frac12(3.92)=1.96\ \mathrm{m\,s^{-2}}
\]

Therefore, relative to the lift:

• The 1.0 kg mass accelerates upwards at \(1.96\ \mathrm{m\,s^{-2}}\).
• The 3.0 kg mass accelerates downwards at \(1.96\ \mathrm{m\,s^{-2}}\).

But this does not tell us their acceleration directions relative to the ground.

Step 7: Resolve the apparent contradiction in part (b)

The 1.0 kg mass accelerates upwards relative to the lift.

However, the lift itself accelerates downwards.

For the lighter mass:

\[
a_1=A+a_{1,\mathrm{rel}}
\]

Taking downward as positive:

\[
\begin{aligned}
a_1&=5.88-1.96\\
&=3.92\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The result is positive.

The lighter mass actually accelerates downwards relative to the ground.

For the heavier mass:

\[
\begin{aligned}
a_3&=5.88+1.96\\
&=7.84\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Both masses accelerate downwards in the ground frame.

Yet, because the heavier mass accelerates downwards faster than the lift while the lighter one accelerates downwards more slowly, they move in opposite directions relative to the lift.

The tempting mistake is treating an upward acceleration relative to the lift as an upward acceleration in every reference frame.

Acceleration directions depend on the chosen frame.

Step 8: Trial III – lift accelerating downwards at \(1.20g\)

Now:

\[
A=1.20(9.8)=11.76\ \mathrm{m\,s^{-2}}
\]

If we mechanically substitute into the taut-rope tension equation:

\[
\begin{aligned}
T&=\frac{3(1.0)}2(9.8-11.76)\\
&=-2.94\ \mathrm N
\end{aligned}
\]

This is the decisive result.

A negative tension would mean that the rope must push on the masses rather than pull them.

An ordinary flexible rope cannot do that.

Therefore, the taut-rope assumption is invalid in Trial III.

The rope goes slack.

Step 9: Find the actual accelerations after the rope becomes slack

With no tension and no air resistance, gravity is the only force on each mass.

Therefore:

\[
ma=mg
\]

The non-zero mass cancels, giving:

\[
a=g=9.8\ \mathrm{m\,s^{-2}}
\]

Both masses accelerate downwards at \(9.8\ \mathrm{m\,s^{-2}}\) relative to the ground.

The lift accelerates downwards faster, at \(11.76\ \mathrm{m\,s^{-2}}\).

Therefore, relative to the lift, each mass has acceleration:

\[
\begin{aligned}
a_{\mathrm{rel}}&=9.8-11.76\\
&=-1.96\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Both masses accelerate upwards relative to the lift at \(1.96\ \mathrm{m\,s^{-2}}\).

The ceiling and pulley move downwards faster than the freely falling masses, allowing the rope to become slack.

Step 10: Explain the deeper issue

The taut-rope derivation was algebraically valid under its assumptions.

But its assumptions stopped being physically possible when the lift’s downward acceleration exceeded \(g\).

Mass cancellation did not cause the mistake.

The mistake was continuing to apply a constraint that required a rope to exert an impossible force.

This reveals a broader principle:

An equation can be dimensionally correct, algebraically rearranged correctly, and still give an unphysical answer because its assumptions no longer describe the actual system.

The correct sequence is to establish the physical constraints, write the valid equations, and only then cancel common factors.

07The next useful check: does the answer still describe the physics?

Cancelling mass is a useful shortcut, but it should come near the end of your reasoning, not the beginning.

Before removing \(m\), establish whether the same mass factor appears throughout the equation, whether all relevant forces have been included, and whether the assumed physical conditions remain valid.

After cancelling, ask what the result actually means. An acceleration independent of mass doesn’t mean the forces, energies, or momenta are also independent of mass.

Two useful follow-up skills are checking that an equation has the correct physical units and deciding which equation describes the interaction you are modelling.

For the first, see How to Check a Physics Equation Using Dimensional Analysis.

For connected objects, collisions, and changing forces, see How to Choose Between Force and Momentum Equations in HSC Physics.

The habit worth developing is simple: decide what the forces are doing before deciding what the algebra allows you to remove.