How to Rearrange Physics Formulas When the Unknown Is Squared

A trolley has a mass of \(0.50\text{ kg}\) and a kinetic energy of \(16\text{ J}\). What is its speed?

You know the kinetic energy equation:

\[
E_k = \frac{1}{2}mv^2
\]

You can substitute the mass and kinetic energy, but the unknown, \(v\), is squared. How do you get it on its own?

The key is to **isolate the squared variable first, then take the square root of both sides**. You must also check whether the question asks for speed or velocity, because taking a square root can produce two mathematical answers even when only one is physically appropriate.

This method appears in several HSC Physics calculations. The important skill is recognising when a squared unknown can be isolated directly, when a quadratic equation needs a different approach, and which mathematical solution makes physical sense.

## 1. The basic method: isolate the square, then take the square root

Start with:

\[
E_k = \frac{1}{2}mv^2
\]

Here:

- \(E_k\) is kinetic energy, measured in joules (J).
- \(m\) is mass, measured in kilograms (kg).
- \(v\) is velocity, measured in metres per second (\(\text{m s}^{-1}\)).

Suppose you know the kinetic energy and mass, and need to calculate the speed.

The unknown appears as \(v^2\). Your first job is not to take the square root. It is to get \(v^2\) by itself.

**Step 1: Remove the fraction.**

Multiply both sides by 2:

\[
2E_k = mv^2
\]

**Step 2: Remove the mass from the right-hand side.**

Divide both sides by \(m\):

\[
\frac{2E_k}{m}=v^2
\]

**Step 3: Take the square root of both sides.**

\[
v = \sqrt{\frac{2E_k}{m}}
\]

You have now rearranged the formula to make the unknown the subject.

Notice the order. You divide by the mass before taking the square root. Taking the square root too early makes the algebra unnecessarily complicated.

[IMAGE: An annotated kinetic energy equation showing E_k = 1/2 mv^2 transforming step by step into v = square root of 2E_k divided by m, with a small moving trolley illustrating mass and speed as the physical quantities.]

### Worked example: Finding the speed from kinetic energy

A trolley has a mass of \(0.50\text{ kg}\) and a kinetic energy of \(16\text{ J}\). Calculate its speed.

**Step 1**: Write down the known quantities and the unknown.

\[
E_k = 16\text{ J}, \qquad m = 0.50\text{ kg}
\]

We need to find \(v\).

**Step 2**: Select the equation and rearrange it.

\[
E_k = \frac{1}{2}mv^2
\]

Rearranging:

\[
v = \sqrt{\frac{2E_k}{m}}
\]

**Step 3**: Substitute the values, including their units.

\[
\begin{aligned}
v &= \sqrt{\frac{2(16)}{0.50}}\\
&= \sqrt{64}\\
&= 8.0\text{ m s}^{-1}
\end{aligned}
\]

**Step 4**: State what the result means.

The trolley's speed is:

\[
\boxed{v = 8.0\text{ m s}^{-1}}
\]

The square root gives the magnitude of the velocity. Kinetic energy tells you how much energy the trolley has because of its motion, but it does not tell you which direction the trolley is travelling.

## 2. Why you cannot always just take the square root

There is an important catch.

Consider these two equations:

\[
v^2 = u^2 + 2as
\]

and

\[
s = ut + \frac{1}{2}at^2
\]

Both contain a squared quantity. However, they require different rearranging methods if you want to solve for the unknown.

In the first equation, \(v^2\) appears on its own. You can isolate it and take the square root.

In the second equation, \(t\) appears twice: once as \(t\), and again as \(t^2\). You cannot isolate \(t^2\) by simply moving the other terms to the opposite side. You must treat the expression as a quadratic equation.

This distinction is particularly important in HSC Physics. Seeing a square does not automatically mean you can use the square-root method.

[IMAGE: Side-by-side comparison of two physics equations. On the left, v squared is isolated in v^2 = u^2 + 2as. On the right, t appears in both the linear and squared terms of s = ut + 1/2 at^2. Annotate why only the left equation permits direct isolation followed by a square root.]

### Case A: The unknown appears only as a squared term

Consider:

\[
v^2 = u^2 + 2as
\]

This is one of the constant-acceleration equations.

If you know the initial velocity \(u\), acceleration \(a\), and displacement \(s\), you can calculate the final velocity \(v\).

Rearranging is straightforward because \(v^2\) is already isolated:

\[
v = \pm\sqrt{u^2+2as}
\]

The plus-or-minus sign deserves attention. Mathematically, both roots may be possible. Physically, you must use the chosen positive direction and the motion described in the question to determine the correct velocity.

If the question asks for speed, use the non-negative magnitude. If it asks for velocity, the direction must also be specified.

### Case B: The unknown appears as both a linear and squared term

Now consider:

\[
s = ut + \frac{1}{2}at^2
\]

Suppose you need to find the time \(t\).

You cannot simply move \(ut\) to the left, take a square root, and finish. The \(ut\) term also contains the unknown.

Instead, rearrange into quadratic form:

\[
\frac{1}{2}at^2+ut-s=0
\]

This has the structure:

\[
At^2+Bt+C=0
\]

where \(A\), \(B\), and \(C\) are coefficients.

You can then use the quadratic formula:

\[
t=\frac{-B\pm\sqrt{B^2-4AC}}{2A}
\]

This is a different method, even though the original physics equation contains a squared unknown.

For a detailed explanation of this particular motion equation, see [Using \(s=ut+\frac{1}{2}at^2\) in HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-the-equation-s-ut-1-2-at-squared/).

## 3. Choosing the right physics equation before rearranging

Before doing any algebra, make sure you have selected an equation that actually describes the situation.

A formula can be rearranged perfectly and still produce a meaningless answer if its assumptions are not satisfied.

For example, the constant-acceleration equation \(v^2=u^2+2as\) assumes constant acceleration over the displacement being considered. It is not automatically suitable for an object whose acceleration changes throughout the motion.

Use the known quantities, the unknown, and the conditions in the question to guide your choice.

| Equation | Useful when you know | Common squared unknown |
|---|---|---|
| \(E_k=\frac{1}{2}mv^2\) | Mass and kinetic energy | Speed |
| \(v^2=u^2+2as\) | Initial velocity, acceleration, and displacement | Final velocity |
| \(a_c=\frac{v^2}{r}\) | Radius and centripetal acceleration | Speed |
| \(s=ut+\frac{1}{2}at^2\) | Initial velocity, acceleration, and time | Time, but requires quadratic methods |

Here, \(a_c\) is centripetal acceleration, \(r\) is the radius of circular motion, and the other symbols have their usual constant-acceleration meanings.

[IMAGE: A compact annotated decision diagram showing three equation-selection routes: kinetic energy when mass and energy are known, constant-acceleration motion when u, a and s are known, and circular motion when centripetal acceleration and radius are known. Show the squared unknown in each equation and distinguish isolated squares from quadratic expressions.]

A useful way to organise your thinking is:

1. Identify what the question asks you to calculate.
2. List the quantities supplied in the question.
3. Choose an equation containing those quantities and the unknown.
4. Check the assumptions behind the equation.
5. Rearrange symbolically before substituting numbers.
6. Check the mathematical sign and physical meaning of your answer.

Do not choose an equation simply because it contains a squared variable. Choose it because it models the physical situation.

## 4. When should you use the positive or negative square root?

The square-root step has a detail that is easy to overlook.

If:

\[
x^2=25
\]

then:

\[
x=+5 \quad \text{or} \quad x=-5
\]

Both answers are mathematically correct because:

\[
(+5)^2=25
\]

and

\[
(-5)^2=25.
\]

In general, if \(x^2=A\), then:

\[
x=\pm\sqrt{A}
\]

provided \(A\) is non-negative.

But physics determines which answers are physically possible.

### Speed and velocity are not interchangeable here

Consider the kinetic energy equation:

\[
K=\frac{1}{2}mv^2
\]

In this equation, \(v\) represents speed, which is the magnitude of velocity. Speed cannot be negative, so you take the positive root:

\[
v=\sqrt{\frac{2K}{m}}
\]

The situation is different for the constant-acceleration equation:

\[
v^2=u^2+2as
\]

Here, \(v\) can represent signed velocity along a chosen axis.

If east is positive, a velocity of \(+5\text{ m s}^{-1}\) means travelling east, while \(-5\text{ m s}^{-1}\) means travelling west.

You may need either root, or both.

[IMAGE: A straight horizontal track with a trolley moving east towards a marked position, then reversing direction under a westward acceleration. Show east as the positive direction, with separate labelled velocity arrows illustrating eastward and westward motion at the same position.]

### A useful way to choose the root

| Physical quantity | Which root? | Reason |
|---|---|---|
| Speed | Positive | Speed is a magnitude |
| Radius | Positive | A physical radius cannot be negative |
| Mass | Positive | Physical mass is non-negative |
| Signed velocity | Positive or negative | Direction determines the sign |
| Displacement component | Positive or negative | Depends on the chosen axis |

There are two other cases worth recognising.

**If the expression under the square root is zero**, the unknown is zero. For example, a ball at its highest point in vertical motion has zero vertical velocity.

**If the expression under the square root is negative**, there is no real solution. In a physical calculation, check your substitution, signs, units, and assumptions. You may have selected an inappropriate equation or described an impossible situation.

One final algebra warning: square roots do not distribute across addition.

For example:

\[
\sqrt{9+16}\neq\sqrt{9}+\sqrt{16}
\]

The left-hand side is \(5\), while the right-hand side is \(7\).

Keep the entire expression inside the square root until you have evaluated it.

### Worked example: A trolley passes the same position twice

A trolley starts at position \(x=0\), travelling east at \(6.0\text{ m s}^{-1}\). It experiences a constant acceleration of \(2.0\text{ m s}^{-2}\) westwards.

Calculate every possible velocity of the trolley when its displacement is \(8.0\text{ m}\) east of its starting position.

**Step 1**: Choose a positive direction and identify the known quantities.

Take east as positive.

\[
u=+6.0\text{ m s}^{-1}
\]

\[
a=-2.0\text{ m s}^{-2}
\]

\[
s=+8.0\text{ m}
\]

We need to calculate the final velocity \(v\).

**Step 2**: Select an equation that does not require time.

We know initial velocity, acceleration, and displacement.

Use:

\[
v^2=u^2+2as
\]

**Step 3**: Substitute the values.

\[
\begin{aligned}
v^2&=(6.0)^2+2(-2.0)(8.0)\\
&=36-32\\
&=4\text{ m}^2\text{ s}^{-2}
\end{aligned}
\]

**Step 4**: Take the square root.

\[
v=\pm\sqrt{4}
\]

Therefore:

\[
v=+2.0\text{ m s}^{-1}
\]

or

\[
v=-2.0\text{ m s}^{-1}
\]

**Step 5**: Decide whether both answers are physically possible.

Both answers are possible.

The trolley initially travels east but accelerates west. It slows down, stops momentarily, and then starts travelling west.

We can calculate where it stops by setting \(v=0\):

\[
\begin{aligned}
0&=u^2+2as\\
0&=36-4s\\
s&=9.0\text{ m}
\end{aligned}
\]

The trolley turns around at \(9.0\text{ m}\) east of its starting position.

It therefore passes the \(8.0\text{ m}\) position twice:

- First, travelling east at \(+2.0\text{ m s}^{-1}\).
- Later, travelling west at \(-2.0\text{ m s}^{-1}\).

This is why taking only the positive root would lose a physically valid answer.

The equation determines the possible velocity magnitudes. The motion determines which direction applies at each instant.

## 5. Check the units before trusting your answer

A square-root rearrangement should also pass a dimensional check.

Take the kinetic energy equation:

\[
v=\sqrt{\frac{2K}{m}}
\]

Kinetic energy is measured in joules:

\[
1\text{ J}=1\text{ kg m}^2\text{ s}^{-2}
\]

Therefore:

\[
\begin{aligned}
[v]&=\sqrt{\frac{\text{J}}{\text{kg}}}\\
&=\sqrt{\frac{\text{kg m}^2\text{ s}^{-2}}{\text{kg}}}\\
&=\sqrt{\text{m}^2\text{ s}^{-2}}\\
&=\text{m s}^{-1}
\end{aligned}
\]

The result has the correct units for speed.

You can perform the same check on:

\[
v=\sqrt{a_cr}
\]

Centripetal acceleration \(a_c\) has units of \(\text{m s}^{-2}\), while radius \(r\) has units of metres.

Their product has units of:

\[
(\text{m s}^{-2})(\text{m})
=\text{m}^2\text{ s}^{-2}
\]

Taking the square root gives \(\text{m s}^{-1}\), again the correct units for speed.

Dimensional analysis is especially useful when you have rearranged a complicated equation and want to catch an algebra error before moving on. See [How to Check a Physics Equation Using Dimensional Analysis](https://tutorgum.com/learn/resources/hsc-physics-how-to-check-a-derived-physics-equation-using-dimensions/).

## 6. Which squared-variable equations should you recognise?

The following relationships are particularly useful when working through HSC Physics calculations.

| Original equation | Rearranged for speed | What to watch |
|---|---|---|
| \(K=\frac{1}{2}mv^2\) | \(v=\sqrt{\frac{2K}{m}}\) | \(v\) is speed |
| \(v^2=u^2+2as\) | \(v=\pm\sqrt{u^2+2as}\) | Check direction and constant acceleration |
| \(a_c=\frac{v^2}{r}\) | \(v=\sqrt{a_cr}\) | \(a_c\) is inward acceleration |
| \(F_c=\frac{mv^2}{r}\) | \(v=\sqrt{\frac{F_cr}{m}}\) | \(F_c\) is the net inward force |

A few important distinctions:

**Kinetic energy**

The equation \(K=\frac{1}{2}mv^2\) applies to classical, non-relativistic motion. It uses speed, not a signed velocity component.

**Constant-acceleration motion**

The equation \(v^2=u^2+2as\) assumes constant acceleration along the direction being analysed. The quantities \(u\), \(v\), \(a\), and \(s\) must use a consistent sign convention.

**Circular motion**

The equation \(a_c=v^2/r\) describes the inward, centripetal component of acceleration for circular motion. If an object is also speeding up or slowing down around the circle, it has a tangential acceleration as well.

Similarly, \(F_c=mv^2/r\) describes the resultant inward force. It is not an additional force that you should automatically add to a force diagram.

[IMAGE: A cyclist travelling around a circular bend, viewed from above. Show the circular path, centre of curvature, radius r, tangent velocity v, and inward centripetal acceleration a_c. A second small inset shows the same bend with a larger speed and a visibly greater inward acceleration, without implying a change in radius.]

### A quick note about the NSW Physics syllabus

The examples in this resource use familiar relationships from the Physics Stage 6 Syllabus (2017) formulae sheet, including kinetic energy, constant-acceleration motion, and centripetal acceleration.

As at November 2026, the 2017 syllabus remains relevant to the current Year 12 HSC cohort. The Physics 11-12 Syllabus (2025) begins implementation for Year 11 in Term 1, 2027, and for Year 12 in Term 4, 2027. Its first HSC examination is scheduled for 2028.

The algebraic rearranging method remains useful across both editions. However, students should check the equations and examinable content against the syllabus and formulae sheet applicable to their cohort.

You can find the [official NESA Physics syllabus and examination resources](https://www.nsw.gov.au/education-and-training/nesa/curriculum/science/physics-stage-6-2017).

## Questions and solutions

### Question 1

A \(0.75\text{ kg}\) ball has a kinetic energy of \(27\text{ J}\).

Calculate its speed.

### Solution 1

The ball's speed is \(\boxed{8.5\text{ m s}^{-1}}\).

Start with the kinetic energy equation:

\[
K=\frac{1}{2}mv^2
\]

Rearrange to make speed the subject:

\[
v=\sqrt{\frac{2K}{m}}
\]

Substitute the known values:

\[
\begin{aligned}
v&=\sqrt{\frac{2(27)}{0.75}}\\
&=\sqrt{72}\\
&=8.49\text{ m s}^{-1}
\end{aligned}
\]

To two significant figures:

\[
\boxed{v=8.5\text{ m s}^{-1}}
\]

Only the positive root is appropriate because the question asks for speed. The kinetic energy equation cannot tell you whether the ball is moving north, south, or in another direction.

### Question 2

A cyclist travels around a circular bend with a radius of \(24\text{ m}\). The cyclist experiences a centripetal acceleration of \(6.0\text{ m s}^{-2}\).

Calculate the cyclist's speed.

The cyclist then doubles their speed while following the same circular path. Determine the new centripetal acceleration and explain why it changes by this amount.

[IMAGE: A cyclist following a circular road bend viewed from above, with the centre of the bend marked, radius labelled 24 m, and inward centripetal acceleration labelled 6.0 m s^-2. Show the tangent direction of travel without supplying the unknown speed.]

### Solution 2

The cyclist's original speed is \(\boxed{12\text{ m s}^{-1}}\). After doubling their speed, their centripetal acceleration becomes \(\boxed{24\text{ m s}^{-2}}\).

**Part A: Calculate the original speed.**

Use the centripetal acceleration equation:

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

Rearrange:

\[
v=\sqrt{a_cr}
\]

Substitute:

\[
\begin{aligned}
v&=\sqrt{(6.0)(24)}\\
&=\sqrt{144}\\
&=12\text{ m s}^{-1}
\end{aligned}
\]

The cyclist's original speed is \(12\text{ m s}^{-1}\).

**Part B: Calculate the new centripetal acceleration.**

The new speed is:

\[
v_{\text{new}}=2(12)=24\text{ m s}^{-1}
\]

The radius remains \(24\text{ m}\).

Using:

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

gives:

\[
\begin{aligned}
a_{c,\text{new}}&=\frac{24^2}{24}\\
&=24\text{ m s}^{-2}
\end{aligned}
\]

The acceleration has increased from \(6.0\) to \(24\text{ m s}^{-2}\), which is four times its original value.

This happens because centripetal acceleration is proportional to the square of speed:

\[
a_c\propto v^2
\]

Doubling the speed multiplies its square by four.

A common mistake is to assume that doubling speed doubles centripetal acceleration. That would only be true if the relationship were directly proportional to \(v\), rather than \(v^2\).

### Question 3

A ball is projected vertically upwards from ground level at \(18.0\text{ m s}^{-1}\). Ignore air resistance and take gravitational acceleration as \(9.8\text{ m s}^{-2}\) downwards.

Take upwards as positive.

The ball passes a height of \(13.0\text{ m}\) above its launch point.

Calculate every possible vertical velocity the ball can have at this height. Explain why more than one answer is physically possible.

[IMAGE: A ball travelling vertically upwards from ground level, following a curved-free vertical trajectory to a highest point and then descending. Mark the 13.0 m height as a horizontal reference level crossed on both the upward and downward journeys. Show upwards as the positive direction, without labelling the unknown velocities.]

### Solution 3

The ball can have either \(\boxed{+8.3\text{ m s}^{-1}}\) or \(\boxed{-8.3\text{ m s}^{-1}}\) at the specified height.

Take upwards as positive.

The known quantities are:

\[
u=+18.0\text{ m s}^{-1}
\]

\[
a=-9.8\text{ m s}^{-2}
\]

\[
s=+13.0\text{ m}
\]

Use the equation:

\[
v^2=u^2+2as
\]

Substitute:

\[
\begin{aligned}
v^2&=(18.0)^2+2(-9.8)(13.0)\\
&=324-254.8\\
&=69.2\text{ m}^2\text{ s}^{-2}
\end{aligned}
\]

Take the square root:

\[
\begin{aligned}
v&=\pm\sqrt{69.2}\\
&=\pm8.32\text{ m s}^{-1}
\end{aligned}
\]

Therefore:

\[
\boxed{v=+8.3\text{ m s}^{-1}}
\]

or:

\[
\boxed{v=-8.3\text{ m s}^{-1}}
\]

Both answers are physically possible.

The ball first passes the \(13.0\text{ m}\) height while travelling upwards. It later passes the same height while travelling downwards.

You can check that the ball actually reaches this height by calculating its maximum displacement:

\[
\begin{aligned}
0&=u^2+2as_{\text{max}}\\
s_{\text{max}}&=\frac{18.0^2}{2(9.8)}\\
&=16.5\text{ m}
\end{aligned}
\]

Since \(13.0\text{ m}\) is below the maximum height of \(16.5\text{ m}\), the ball crosses that height twice.

The tempting mistake is to assume the ball's initial upward direction determines the sign of its velocity throughout the motion. Gravity eventually reverses the ball's direction, so both roots are needed.

### Question 4

A trolley starts at position \(x=0\), travelling east at \(8.0\text{ m s}^{-1}\). It experiences a constant acceleration of \(2.0\text{ m s}^{-2}\) westwards.

Take east as positive.

Determine:

(a) The times at which the trolley is \(12.0\text{ m}\) east of its starting position.

(b) The velocity of the trolley at each of these times.

(c) Why solving for time requires a different algebraic method from simply taking the square root of an isolated squared variable.

[IMAGE: A trolley on a straight horizontal track, starting at x = 0 and moving east, with a target position marked at x = +12.0 m. Show the initial velocity arrow pointing east and the constant acceleration arrow pointing west. Do not mark the turning point or the times at which the trolley passes the target.]

### Solution 4

The trolley reaches the \(12.0\text{ m}\) position at \(\boxed{2.0\text{ s}}\) and \(\boxed{6.0\text{ s}}\). Its corresponding velocities are \(+4.0\text{ m s}^{-1}\) and \(-4.0\text{ m s}^{-1}\), respectively.

**Part A: Find the two times.**

The known quantities are:

\[
u=+8.0\text{ m s}^{-1}
\]

\[
a=-2.0\text{ m s}^{-2}
\]

\[
s=+12.0\text{ m}
\]

Use the displacement equation:

\[
s=ut+\frac{1}{2}at^2
\]

Substitute:

\[
12=8t+\frac{1}{2}(-2)t^2
\]

Simplify:

\[
12=8t-t^2
\]

Rearrange into quadratic form:

\[
t^2-8t+12=0
\]

Factorise:

\[
(t-2)(t-6)=0
\]

Therefore:

\[
\boxed{t=2.0\text{ s or }6.0\text{ s}}
\]

Both times are positive and physically possible.

**Part B: Calculate the velocity at each time.**

Use:

\[
v=u+at
\]

At \(t=2.0\text{ s}\):

\[
\begin{aligned}
v&=8.0+(-2.0)(2.0)\\
&=+4.0\text{ m s}^{-1}
\end{aligned}
\]

The trolley is travelling east.

At \(t=6.0\text{ s}\):

\[
\begin{aligned}
v&=8.0+(-2.0)(6.0)\\
&=-4.0\text{ m s}^{-1}
\end{aligned}
\]

The trolley is travelling west.

**Part C: Explain why a quadratic equation is needed.**

The unknown time appears in both the linear and squared terms:

\[
12=8t-t^2
\]

You cannot isolate \(t^2\) without still having an unknown \(t\) elsewhere in the expression.

Taking a square root at this stage would not correctly isolate time.

The two solutions also have a physical explanation.

The trolley reaches its maximum eastward displacement when its velocity becomes zero:

\[
\begin{aligned}
0&=8.0-2.0t\\
t&=4.0\text{ s}
\end{aligned}
\]

Its displacement at that instant is:

\[
\begin{aligned}
s&=8.0(4.0)-\frac{1}{2}(2.0)(4.0)^2\\
&=32-16\\
&=16\text{ m}
\end{aligned}
\]

The trolley travels through the \(12.0\text{ m}\) position on its way towards the \(16.0\text{ m}\) turning point, then passes through it again while travelling west.

The two mathematical roots represent two different moments in the trolley's actual motion.

### Question 5

A small bead is released from rest on a smooth track leading into a vertical circular loop. The bead travels along the inside of the loop.

The loop has a radius of \(0.80\text{ m}\). The bead starts at a height of \(2.4r\) above the lowest point of the loop, where \(r\) is the loop's radius.

Assume:

- The track is frictionless.
- Air resistance is negligible.
- The bead can be treated as a particle.
- Gravitational acceleration is \(9.8\text{ m s}^{-2}\).
- The bead starts from rest.

A student uses conservation of mechanical energy to calculate the speed at the top of the loop:

\[
v_{\text{top}}^2=2g(h-2r)
\]

Since this produces a positive value for \(v_{\text{top}}^2\), the student concludes that the bead reaches the top while remaining in contact with the track.

Evaluate this conclusion.

Then calculate:

(a) The angle from the upward vertical at which the bead first loses contact with the track.

(b) The bead's speed at the instant it loses contact.

Take the angle \(\theta\) to be measured from the upward vertical through the centre of the loop, with \(\theta=0^\circ\) at the top.

[IMAGE: A bead released from rest on a smooth approach ramp entering the bottom of a vertical circular loop. Show the loop radius r = 0.80 m, the initial release height h = 2.4r above the bottom, and a bead at a general position on the upper right-hand side of the loop. Mark the centre, the upward vertical reference line, and angle theta between that line and the radius to the bead. Do not include force arrows or the separation point.]

### Solution 5

The student is incorrect: **the bead loses contact with the track before reaching the top**.

It first loses contact at approximately:

\[
\boxed{\theta=21.0^\circ}
\]

from the upward vertical.

Its speed at separation is:

\[
\boxed{v=2.7\text{ m s}^{-1}}
\]

The important issue is that conservation of energy alone does not guarantee that an object can maintain contact with a curved track.

We need to combine energy conservation with Newton's second law in the radial direction.

**Step 1: Check the student's energy calculation at the top.**

The initial height is:

\[
h=2.4r
\]

At the top of the loop, the bead is at height:

\[
h_{\text{top}}=2r
\]

Conservation of mechanical energy gives:

\[
mgh=\frac{1}{2}mv_{\text{top}}^2+mg(2r)
\]

Rearrange:

\[
v_{\text{top}}^2=2g(h-2r)
\]

Substitute \(h=2.4r\):

\[
\begin{aligned}
v_{\text{top}}^2
&=2g(2.4r-2r)\\
&=0.8gr
\end{aligned}
\]

This is positive, so the energy equation predicts a real speed at the top.

Using \(g=9.8\text{ m s}^{-2}\) and \(r=0.80\text{ m}\):

\[
\begin{aligned}
v_{\text{top}}^2
&=0.8(9.8)(0.80)\\
&=6.272
\end{aligned}
\]

Therefore:

\[
v_{\text{top}}=2.5\text{ m s}^{-1}
\]

But this calculation assumes the bead actually follows the circular track all the way to the top.

We have not checked whether the track can provide the force needed to keep it moving in a circle.

**Step 2: Establish the condition for maintaining contact.**

At the top of the loop, both gravity and the normal force from the track point towards the centre.

The centripetal force requirement is:

\[
mg+N=\frac{mv_{\text{top}}^2}{r}
\]

where \(N\) is the normal contact force.

Rearrange:

\[
N=\frac{mv_{\text{top}}^2}{r}-mg
\]

Substitute the energy result \(v_{\text{top}}^2=0.8gr\):

\[
\begin{aligned}
N&=\frac{m(0.8gr)}{r}-mg\\
&=0.8mg-mg\\
&=-0.2mg
\end{aligned}
\]

This is impossible.

The normal force would have to be negative, meaning the track would need to pull the bead towards itself. A normal contact force from the track can push the bead, but it cannot pull it back.

Therefore, the bead must lose contact before reaching the top.

[IMAGE: A force diagram of a bead on the inside of a vertical circular loop at a general angle theta from the top. Show gravity mg vertically downwards, normal force N radially towards the centre, and the inward radial direction. Include the inward component of gravity mg cos theta, clearly distinguishing it from the full weight.]

**Step 3: Use conservation of energy at a general position on the loop.**

We now need to find where the bead actually loses contact.

At an angle \(\theta\) from the upward vertical, the bead's height above the bottom of the loop is:

\[
y=r+r\cos\theta
\]

or:

\[
y=r(1+\cos\theta)
\]

Apply conservation of mechanical energy between the release point and this position:

\[
mgh=\frac{1}{2}mv^2+mg r(1+\cos\theta)
\]

Rearrange to isolate \(v^2\):

\[
v^2=2g[h-r(1+\cos\theta)]
\]

Substitute \(h=2.4r\):

\[
\begin{aligned}
v^2&=2g[2.4r-r(1+\cos\theta)]\\
&=2gr(1.4-\cos\theta)
\end{aligned}
\]

This gives the bead's speed at any point where it remains in contact with the track.

**Step 4: Apply Newton's second law towards the centre.**

The normal force \(N\) points towards the centre.

Gravity also has a component towards the centre, equal to \(mg\cos\theta\).

Therefore:

\[
N+mg\cos\theta=\frac{mv^2}{r}
\]

At the instant the bead loses contact:

\[
N=0
\]

So:

\[
mg\cos\theta=\frac{mv^2}{r}
\]

Cancel the mass:

\[
g\cos\theta=\frac{v^2}{r}
\]

**Step 5: Combine the energy and force equations.**

From Step 3:

\[
v^2=2gr(1.4-\cos\theta)
\]

Substitute into the radial force equation:

\[
g\cos\theta=2g(1.4-\cos\theta)
\]

Cancel \(g\):

\[
\cos\theta=2.8-2\cos\theta
\]

Collect like terms:

\[
3\cos\theta=2.8
\]

Therefore:

\[
\cos\theta=\frac{2.8}{3}
=\frac{14}{15}
\]

Take the inverse cosine:

\[
\begin{aligned}
\theta&=\cos^{-1}\left(\frac{14}{15}\right)\\
&\approx21.0^\circ
\end{aligned}
\]

The bead loses contact approximately \(21.0^\circ\) before reaching the top.

**Step 6: Calculate the speed at separation.**

At the instant of separation, the normal force is zero, so:

\[
v^2=gr\cos\theta
\]

Substitute:

\[
v^2=gr\left(\frac{14}{15}\right)
\]

Now use the numerical values:

\[
\begin{aligned}
v&=\sqrt{\frac{14}{15}(9.8)(0.80)}\\
&=\sqrt{7.317}\\
&=2.71\text{ m s}^{-1}
\end{aligned}
\]

Therefore:

\[
\boxed{v=2.7\text{ m s}^{-1}}
\]

The bead leaves the track with this speed and then follows a projectile trajectory, rather than continuing around the circular path.

The key misconception was treating a positive value of \(v^2\) as proof that the entire motion was physically possible. Energy conservation tells us the speed an object would have at a particular height if it reached that height. Newton's second law tells us whether the required forces can actually produce the motion.

That is the extra check that changes the answer.

## 7. The method to remember

When an unknown appears squared in a physics equation, use this sequence:

1. Choose an equation that matches the known quantities and the physical situation.
2. Check the assumptions behind that equation.
3. Rearrange until the squared unknown is isolated.
4. Take the square root of both sides.
5. Consider both mathematical roots.
6. Use the physical meaning of the variable to select the appropriate answer.
7. Check the units and whether the result makes sense.

The biggest distinction is between an isolated squared unknown, such as \(v^2=u^2+2as\), and an unknown that appears in both linear and squared terms, such as \(s=ut+\frac{1}{2}at^2\).

The first can be solved by isolating the square and taking a square root. The second requires a quadratic method.

Once you can recognise that difference, the next step is combining rearranged equations with force diagrams, energy conservation, and multi-stage motion. That is where these algebra skills become particularly useful in HSC Physics.