1. Scientific notation

Why Scientists Use Scientific Notation

Science often deals with numbers that are extremely large or extremely small.

For example:

• The speed of light is approximately 300,000,000 m/s.
• The mass of an electron is approximately 0.000000000000000000000000000000911 kg.
• The distance from Earth to the Sun is approximately 150,000,000,000 m.

Writing numbers like these repeatedly is difficult and increases the chance of making mistakes.

Scientific notation provides a shorter and clearer way to represent them.

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The Basic Form

A number written in scientific notation has the form:

a × 10ⁿ

where:

• a is a number from 1 up to, but not including, 10.
• 10 is the base.
• n is an integer called the exponent.

For example:

4.5 × 10⁶

The coefficient is:

4.5

The exponent is:

6

Understanding Powers of Ten

Powers of ten tell us how many times 10 is multiplied or divided.

10¹ = 10

10² = 100

10³ = 1,000

10⁴ = 10,000

10⁵ = 100,000

10⁶ = 1,000,000

Negative exponents represent values smaller than 1:

10⁻¹ = 0.1

10⁻² = 0.01

10⁻³ = 0.001

10⁻⁴ = 0.0001

10⁻⁵ = 0.00001

10⁻⁶ = 0.000001

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Converting Large Numbers to Scientific Notation

Consider:

5,600,000

Move the decimal point until there is only one non-zero digit before it:

5.6

The decimal point moved 6 places to the left.

Therefore:

5,600,000 = 5.6 × 10⁶

For numbers greater than 10, the exponent will normally be positive.

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Worked Example: 72,000

Start with:

72,000

Move the decimal point:

72,000 → 7.2

It moves 4 places left.

Therefore:

72,000 = 7.2 × 10⁴

Check:

10⁴ = 10,000

so:

7.2 × 10,000 = 72,000

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Worked Example: 458,000,000

Move the decimal point until the coefficient is between 1 and 10:

458,000,000 → 4.58

The decimal point moves 8 places left.

Therefore:

458,000,000 = 4.58 × 10⁸

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Converting Small Numbers to Scientific Notation

Consider:

0.00034

Move the decimal point until the coefficient is between 1 and 10:

0.00034 → 3.4

The decimal point moves 4 places to the right.

Because the original number was smaller than 1, the exponent is negative.

Therefore:

0.00034 = 3.4 × 10⁻⁴

Worked Example: 0.0062

Move the decimal:

0.0062 → 6.2

The decimal moves 3 places right.

Therefore:

0.0062 = 6.2 × 10⁻³

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Worked Example: 0.00000081

Move the decimal:

0.00000081 → 8.1

The decimal moves 7 places right.

Therefore:

0.00000081 = 8.1 × 10⁻⁷

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A Useful Rule

When converting decimal notation to scientific notation:

Large Numbers

Move the decimal left.

The exponent is positive.

Example:

83,000 = 8.3 × 10⁴

Small Numbers

Move the decimal right.

The exponent is negative.

Example:

0.000083 = 8.3 × 10⁻⁵

Notice that the two numbers are very different even though their coefficients are both 8.3.

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Converting Scientific Notation Back to Decimal Form

The exponent tells you how to move the decimal point.

Positive Exponent

Move the decimal point right.

Example:

3.2 × 10⁵

Move 5 places:

3.2 → 320,000

Therefore:

3.2 × 10⁵ = 320,000

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Negative Exponent

Move the decimal point left.

Example:

4.7 × 10⁻⁴

Move 4 places:

4.7 → 0.00047

Therefore:

4.7 × 10⁻⁴ = 0.00047

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Zero Exponent

Remember:

10⁰ = 1

Therefore:

6.3 × 10⁰ = 6.3

Any non-zero number raised to the power zero equals 1.

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Is It Proper Scientific Notation?

The coefficient must satisfy:

1 ≤ coefficient < 10

Therefore:

4.7 × 10⁵

is correctly written in scientific notation.

But:

47 × 10⁴

is not in standard scientific notation because 47 is greater than 10.

We can rewrite it:

47 × 10⁴ = 4.7 × 10⁵

Similarly:

0.52 × 10⁸

should be written:

5.2 × 10⁷

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Comparing Numbers in Scientific Notation

Scientific notation makes very large and very small numbers easier to compare.

Consider:

3.2 × 10⁷

and:

8.9 × 10⁵

Compare the exponents first:

10⁷ > 10⁵

Therefore:

3.2 × 10⁷ > 8.9 × 10⁵

The coefficient does not need to be compared because the exponents are different.

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Comparing Numbers with the Same Exponent

Consider:

4.8 × 10⁶

and:

7.1 × 10⁶

The exponents are the same.

Now compare the coefficients:

7.1 > 4.8

Therefore:

7.1 × 10⁶ > 4.8 × 10⁶

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Comparing Numbers with Negative Exponents

Consider:

3.0 × 10⁻³

and:

3.0 × 10⁻⁶

Convert mentally:

3.0 × 10⁻³ = 0.003

3.0 × 10⁻⁶ = 0.000003

Therefore:

3.0 × 10⁻³ > 3.0 × 10⁻⁶

A more negative exponent represents a smaller positive number when the coefficients are comparable.

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Multiplying in Scientific Notation

When multiplying numbers in scientific notation:

multiply the coefficients

and:

add the exponents

For example:

(2 × 10³)(4 × 10⁵)

Multiply the coefficients:

2 × 4 = 8

Add the exponents:

3 + 5 = 8

Therefore:

(2 × 10³)(4 × 10⁵) = 8 × 10⁸

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Worked Example: Multiplication

Calculate:

(3.0 × 10⁴)(2.0 × 10³)

Multiply coefficients:

3.0 × 2.0 = 6.0

Add exponents:

4 + 3 = 7

Answer:

6.0 × 10⁷

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When the Answer Needs Adjusting

Calculate:

(6 × 10⁴)(3 × 10⁵)

Multiply coefficients:

6 × 3 = 18

Add exponents:

4 + 5 = 9

This gives:

18 × 10⁹

But 18 is not between 1 and 10.

Rewrite:

18 × 10⁹ = 1.8 × 10¹⁰

Therefore:

(6 × 10⁴)(3 × 10⁵) = 1.8 × 10¹⁰

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Dividing in Scientific Notation

When dividing:

divide the coefficients

and:

subtract the exponents

For example:

(8 × 10⁷) ÷ (2 × 10³)

Divide coefficients:

8 ÷ 2 = 4

Subtract exponents:

7 − 3 = 4

Therefore:

4 × 10⁴

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Worked Example: Division

Calculate:

(9.0 × 10⁸) ÷ (3.0 × 10²)

Divide coefficients:

9.0 ÷ 3.0 = 3.0

Subtract exponents:

8 − 2 = 6

Therefore:

3.0 × 10⁶

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Dividing with Negative Exponents

Calculate:

(6 × 10⁻³) ÷ (2 × 10⁻⁵)

Divide coefficients:

6 ÷ 2 = 3

Subtract exponents:

−3 − (−5)

= −3 + 5

= 2

Therefore:

3 × 10² = 300

Be especially careful when subtracting negative exponents.

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Adding and Subtracting

Addition and subtraction require a different approach.

The powers of ten must be the same before the coefficients can be directly added or subtracted.

Consider:

3.2 × 10⁵ + 4.1 × 10⁵

The exponents already match.

Add the coefficients:

3.2 + 4.1 = 7.3

Therefore:

7.3 × 10⁵

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Adding Different Powers of Ten

Calculate:

3.2 × 10⁵ + 4.0 × 10⁴

The exponents are different.

Rewrite:

4.0 × 10⁴ = 0.40 × 10⁵

Now:

3.2 × 10⁵ + 0.40 × 10⁵

Add:

3.60 × 10⁵

Therefore:

3.6 × 10⁵

depending on the required significant figures.

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Subtraction Example

Calculate:

7.5 × 10⁶ − 2.0 × 10⁶

The exponents match.

Subtract coefficients:

7.5 − 2.0 = 5.5

Therefore:

5.5 × 10⁶

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Scientific Notation on a Calculator

Many scientific calculators display scientific notation using:

EXP

or:

EE

or a form such as:

×10ˣ

For example, to enter:

6.02 × 10²³

you may enter:

6.02 EXP 23

depending on the calculator.

Do not enter:

6.02 × 10 × 10²³

That would represent a different number.

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Calculator Displays

A calculator might display:

3.45E8

This means:

3.45 × 10⁸

Similarly:

7.2E−5

means:

7.2 × 10⁻⁵

The letter E in this context means:

× 10 raised to the power of

It does not mean the mathematical constant e.

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Scientific Notation in Physics

Scientific notation is especially important in physics because physical quantities can cover enormous ranges.

For example:

speed of light ≈ 3.00 × 10⁸ m/s

elementary charge ≈ 1.60 × 10⁻¹⁹ C

gravitational constant ≈ 6.67 × 10⁻¹¹ N·m²/kg²

Earth's mass ≈ 5.97 × 10²⁴ kg

proton mass ≈ 1.67 × 10⁻²⁷ kg

Scientific notation makes these values much easier to use in calculations.

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Worked Physics Example: Distance

Light travels at approximately:

3.00 × 10⁸ m/s

How far does it travel in:

2.00 s?

Use:

d = vt

Substitute:

d = (3.00 × 10⁸)(2.00)

Multiply:

d = 6.00 × 10⁸ m

Therefore, light travels approximately:

6.00 × 10⁸ m

in 2.00 seconds.

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Worked Physics Example: Force

Suppose:

m = 2.0 × 10³ kg

and:

a = 3.0 × 10² m/s²

Use:

F = ma

Substitute:

F = (2.0 × 10³)(3.0 × 10²)

Multiply coefficients:

2.0 × 3.0 = 6.0

Add exponents:

3 + 2 = 5

Therefore:

F = 6.0 × 10⁵ N

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Worked Physics Example: Density

Suppose an object has:

m = 6.0 × 10⁻³ kg

and:

V = 2.0 × 10⁻⁶ m³

Use:

ρ = m/V

Substitute:

ρ = (6.0 × 10⁻³) / (2.0 × 10⁻⁶)

Divide coefficients:

6.0 / 2.0 = 3.0

Subtract exponents:

−3 − (−6) = 3

Therefore:

ρ = 3.0 × 10³ kg/m³

or:

ρ = 3000 kg/m³

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Worked Physics Example: Electric Charge

Suppose an object has:

5.0 × 10¹² electrons

and the magnitude of the charge of one electron is approximately:

1.60 × 10⁻¹⁹ C

Magnitude of total charge:

Q = (5.0 × 10¹²)(1.60 × 10⁻¹⁹)

Multiply coefficients:

5.0 × 1.60 = 8.0

Add exponents:

12 + (−19) = −7

Therefore:

Q = 8.0 × 10⁻⁷ C

If the particles are electrons and the question asks for signed charge, the charge would be negative:

Q = −8.0 × 10⁻⁷ C

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Orders of Magnitude

Scientific notation can also help us compare the order of magnitude of quantities.

Consider:

2.0 × 10³ m

and:

2.0 × 10⁶ m

The exponents differ by:

6 − 3 = 3

The second distance is:

10³ = 1000

times larger.

Powers of ten allow scientists to compare scales very quickly.

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From Atoms to Space

Scientific notation allows the same mathematical system to describe extremely different scales.

For example:

An atomic-scale distance might be approximately:

1 × 10⁻¹⁰ m

Earth's radius is approximately:

6.4 × 10⁶ m

Large astronomical distances can reach:

10²⁰ m and beyond.

Scientific notation lets physicists move between these scales without writing long strings of zeros.

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Scientific Notation and SI Prefixes

Scientific notation is closely connected to SI prefixes.

Some common examples are:

For example:

1 km = 1 × 10³ m

1 mm = 1 × 10⁻³ m

1 µm = 1 × 10⁻⁶ m

1 nm = 1 × 10⁻⁹ m

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Worked Example: Nanometres

Convert:

450 nm

to metres.

Since:

1 nm = 1 × 10⁻⁹ m

then:

450 nm = 450 × 10⁻⁹ m

Convert to proper scientific notation:

450 = 4.50 × 10²

Therefore:

450 nm = 4.50 × 10⁻⁷ m

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Worked Example: Kilometres

Convert:

72 km

to metres.

Since:

1 km = 10³ m

then:

72 km = 72 × 10³ m

Convert to proper scientific notation:

72 × 10³ = 7.2 × 10⁴

Therefore:

72 km = 7.2 × 10⁴ m

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Significant Figures

Scientific notation makes significant figures easier to identify.

Consider:

4.50 × 10⁶

The coefficient:

4.50

contains 3 significant figures.

The zeros contained in the power of ten do not determine the number of significant figures.

Compare:

4 × 10⁶ → 1 significant figure

4.5 × 10⁶ → 2 significant figures

4.50 × 10⁶ → 3 significant figures

These values communicate different levels of measurement precision.

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Interpreting Scientific Notation

Consider:

6.2 × 10⁻⁸ m

The negative exponent tells us immediately that this is a very small distance.

Consider:

6.2 × 10⁸ m

The positive exponent tells us that this is a very large distance.

The sign of the exponent is therefore extremely important.

Changing:

10⁸

to:

10⁻⁸

does not make a small change.

It changes the scale enormously.

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How Enormous Is the Difference?

Compare:

1 × 10⁸

with:

1 × 10⁻⁸

Their ratio is:

10⁸ / 10⁻⁸ = 10¹⁶

The first value is:

10,000,000,000,000,000

times larger than the second.

A missing negative sign can therefore completely change a scientific answer.

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Estimating Answers

Scientific notation makes estimation easier.

Suppose:

(4.8 × 10⁶)(2.1 × 10³)

Before calculating exactly, estimate:

5 × 10⁶ × 2 × 10³

Approximately:

10 × 10⁹

which is:

1 × 10¹⁰

So the final answer should be around:

10¹⁰

If your calculator gives:

1.008 × 10⁴

you immediately know something has gone wrong.

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Common Mistakes

Coefficient Is Too Large

Incorrect:

45 × 10⁶

Correct:

4.5 × 10⁷

The coefficient should be at least 1 but less than 10.

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Using the Wrong Sign

Incorrect:

0.00052 = 5.2 × 10⁴

Correct:

0.00052 = 5.2 × 10⁻⁴

Small positive numbers require negative exponents.

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Counting Zeros Instead of Decimal Moves

Focus on how many places the decimal point moves.

Example:

0.0036 → 3.6

The decimal moves 3 places.

Therefore:

3.6 × 10⁻³

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Adding Exponents During Addition

Incorrect:

2 × 10⁵ + 3 × 10⁵ = 5 × 10¹⁰

For addition with matching exponents:

2 × 10⁵ + 3 × 10⁵ = 5 × 10⁵

Exponents are added when multiplying, not when adding.

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Forgetting to Normalize the Answer

If multiplication gives:

24 × 10⁷

rewrite it as:

2.4 × 10⁸

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Forgetting Units

In physics:

3.0 × 10⁸

is incomplete if the quantity requires units.

For the speed of light:

3.0 × 10⁸ m/s

is meaningful.

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Calculator Entry Errors

Entering scientific notation incorrectly can change an answer by many orders of magnitude.

Always estimate the expected size of the answer before trusting the calculator.

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Check Your Understanding

1. Write 65,000 in scientific notation.

2. Write 7,200,000 in scientific notation.

3. Write 0.0048 in scientific notation.

4. Write 0.00000092 in scientific notation.

5. Convert 3.4 × 10⁵ to decimal notation.

6. Convert 7.2 × 10⁻⁴ to decimal notation.

7. Which is larger: 5.2 × 10⁷ or 9.8 × 10⁶?

8. Which is smaller: 4.0 × 10⁻³ or 4.0 × 10⁻⁶?

9. Calculate: (2 × 10³)(4 × 10⁵).

10. Calculate: (3 × 10⁻²)(2 × 10⁴).

11. Calculate: (8 × 10⁷) / (2 × 10³).

12. Calculate: (6 × 10⁻⁴) / (3 × 10⁻⁶).

13. Calculate: 4.2 × 10⁵ + 3.1 × 10⁵.

14. Calculate: 7.8 × 10⁶ − 2.3 × 10⁶.

15. Rewrite 38 × 10⁷ in proper scientific notation.

16. Rewrite 0.72 × 10⁵ in proper scientific notation.

17. Write 350 nm in metres using scientific notation.

18. A vehicle has a mass of 2.5 × 10³ kg and accelerates at 4.0 × 10¹ m/s². Calculate the resultant force.

19. Light travels at 3.0 × 10⁸ m/s. Calculate the distance travelled in 5.0 s.

20. Explain why scientific notation is particularly useful in physics.

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Key Terms

• Scientific notation – a method of writing numbers using a coefficient multiplied by a power of ten.
• Coefficient – the number placed before the power of ten; in standard scientific notation it is at least 1 but less than 10.
• Base – the number being raised to a power; scientific notation uses base 10.
• Exponent – number showing the power to which the base is raised.
• Positive exponent – commonly represents large values when scientific notation is normalized.
• Negative exponent – commonly represents values between 0 and 1 when scientific notation is normalized.
• Power of ten – number such as 10³, 10⁶ or 10⁻⁹.
• Order of magnitude – scale of a quantity expressed in terms of powers of ten.
• Decimal notation – ordinary representation of a number using digits and a decimal point.
• Significant figures – digits used to communicate the precision of a measured or calculated value.
• SI prefix – prefix such as kilo, milli, micro or nano representing a power-of-ten multiplier.

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Key Takeaways

• Scientific notation is used to represent very large and very small numbers efficiently.
• Scientific notation has the form a × 10ⁿ.
• The coefficient must be at least 1 and less than 10.
• Large numbers generally have positive exponents.
• Small positive numbers between 0 and 1 have negative exponents.
• A positive exponent moves the decimal right when converting back to decimal notation.
• A negative exponent moves the decimal left.
• When multiplying scientific notation, multiply the coefficients and add the exponents.
• When dividing, divide the coefficients and subtract the exponents.
• When adding or subtracting, first express the numbers using the same power of ten.
• Answers should be rewritten into proper scientific notation when necessary.
• Scientific notation is especially important in physics because physical quantities range from subatomic scales to astronomical scales.
• SI prefixes such as kilo, milli, micro and nano are closely related to powers of ten.
• Scientific notation makes significant figures easier to communicate.
• Estimating the expected order of magnitude is an excellent way to catch calculator errors.

A useful summary is:

large number → positive exponent

small number → negative exponent

multiply → add exponents

divide → subtract exponents

add/subtract → match exponents first

2. Unit conversions

3. Dimensional analysis

What Is Dimensional Analysis?

Dimensional analysis is a method of examining the units or dimensions of physical quantities to check equations, perform conversions, and identify possible errors.

Physics equations describe relationships between quantities. For an equation to be physically meaningful, the quantities on both sides must have compatible dimensions.

For example:

v = d/t

Distance has the dimension of length and time has the dimension of time.

Therefore:

velocity = length/time

In SI units:

m/s

Dimensional analysis gives us a powerful way to check whether equations and calculations make physical sense.

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Units and Dimensions Are Related but Different

A unit tells us how a quantity is measured.

Examples:

• metre (m)
• second (s)
• kilogram (kg)
• newton (N)

A dimension describes the physical type of the quantity.

Three important fundamental dimensions are:

For example, a distance of 20 m has:

unit → m

dimension → L

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Dimension Symbols

Dimensions are commonly written using square brackets.

For example:

[length] = L

[mass] = M

[time] = T

We can then describe more complicated quantities using combinations of these fundamental dimensions.

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Dimensions of Speed

Speed is:

speed = distance/time

Distance has dimension:

L

Time has dimension:

T

Therefore:

[speed] = L/T

This can also be written:

[speed] = LT⁻¹

Its SI unit is:

m/s

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Dimensions of Acceleration

Acceleration is:

acceleration = change in velocity/time

Velocity has dimensions:

LT⁻¹

Divide by time:

[acceleration] = LT⁻¹/T

Therefore:

[acceleration] = LT⁻²

Its SI unit is:

m/s²

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Dimensions of Force

Newton's second law states:

F = ma

Mass has dimension:

M

Acceleration has dimension:

LT⁻²

Therefore:

[F] = MLT⁻²

The SI unit is the newton:

N = kg·m/s²

This shows that a newton is a derived unit.

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Derived Quantities

Many physics quantities are combinations of fundamental dimensions.

Understanding these relationships makes complicated equations much easier to check.

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Dimensional Consistency

An equation is dimensionally consistent when the dimensions on one side of the equation match the dimensions on the other side.

Consider:

d = vt

Left side:

[d] = L

Right side:

[v][t]

Velocity:

[v] = LT⁻¹

Time:

[t] = T

Therefore:

[vt] = LT⁻¹ × T

The T terms cancel:

[vt] = L

So:

L = L

The equation is dimensionally consistent.

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Checking an Equation

Suppose someone proposes:

d = v/t

Is this dimensionally consistent?

Left side:

[d] = L

Right side:

[v/t] = LT⁻¹/T

Therefore:

[v/t] = LT⁻²

So we obtain:

L ≠ LT⁻²

The dimensions do not match.

Therefore:

d = v/t

cannot be a dimensionally correct equation for distance.

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A Dimensional Check Does Not Prove an Equation Is Correct

This is an important limitation.

Suppose someone proposes:

d = 5vt

Dimensions of the right side:

[5vt] = LT⁻¹ × T

= L

So the equation is dimensionally consistent.

However, dimensional analysis cannot tell us whether the numerical factor 5 is correct.

Therefore:

dimensionally consistent does not necessarily mean physically correct

But:

dimensionally inconsistent means the equation cannot be correct as written

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Dimensionless Numbers

Ordinary numerical constants have no physical dimensions.

Examples include:

2

½

π

For example:

Eₖ = ½mv²

The factor ½ does not affect the dimensional analysis.

We only need to examine:

mv²

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Checking the Kinetic Energy Equation

Consider:

Eₖ = ½mv²

Ignore the dimensionless factor ½.

Mass:

[m] = M

Velocity:

[v] = LT⁻¹

Therefore:

[v²] = L²T⁻²

Multiply by mass:

[mv²] = ML²T⁻²

Energy has dimensions:

[E] = ML²T⁻²

Therefore:

ML²T⁻² = ML²T⁻²

The equation is dimensionally consistent.

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Checking Gravitational Potential Energy

Consider:

Eₚ = mgh

Mass:

[m] = M

Acceleration:

[g] = LT⁻²

Height:

[h] = L

Therefore:

[mgh] = M × LT⁻² × L

= ML²T⁻²

This matches the dimensions of energy.

Therefore:

Eₚ = mgh

is dimensionally consistent.

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Checking Momentum

Momentum is:

p = mv

Mass:

M

Velocity:

LT⁻¹

Therefore:

[p] = MLT⁻¹

The SI unit is:

kg·m/s

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Checking Pressure

Pressure is:

p = F/A

Force:

[F] = MLT⁻²

Area:

[A] = L²

Therefore:

[p] = MLT⁻²/L²

Simplify:

[p] = ML⁻¹T⁻²

The SI unit is:

kg/(m·s²)

This derived unit is called the:

pascal (Pa)

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Checking Density

Density is:

ρ = m/V

Mass:

[m] = M

Volume:

[V] = L³

Therefore:

[ρ] = M/L³

or:

[ρ] = ML⁻³

Its SI unit is:

kg/m³

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Addition and Subtraction Require Matching Dimensions

Quantities can only be meaningfully added or subtracted if they have compatible dimensions.

For example:

5 m + 3 m

makes sense.

But:

5 m + 3 s

does not.

Length and time represent different physical quantities.

The same rule applies inside equations.

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Checking a Motion Equation

Consider:

v = u + at

Check each term.

Velocity:

[v] = LT⁻¹

Initial velocity:

[u] = LT⁻¹

Acceleration × time:

[at] = LT⁻² × T

= LT⁻¹

Therefore:

LT⁻¹ = LT⁻¹ + LT⁻¹

Every term has the dimensions of velocity.

The equation is dimensionally consistent.

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Checking Another Motion Equation

Consider:

s = ut + ½at²

Left side:

[s] = L

First term:

[ut] = LT⁻¹ × T

= L

Second term:

[at²] = LT⁻² × T²

= L

Therefore:

L = L + L

The equation is dimensionally consistent.

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Spotting an Error

Suppose someone writes:

s = ut + ½at

Check the second term:

[at] = LT⁻² × T

= LT⁻¹

The equation would therefore attempt to add:

length + velocity

That is not dimensionally valid.

So the equation contains an error.

The missing factor is related to time:

s = ut + ½at²

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Dimensions Can Reveal Missing Powers

Suppose an equation contains:

distance ∝ timeⁿ

For motion from rest under constant acceleration, suppose we know acceleration must also appear:

d ∝ atⁿ

Determine n using dimensions.

Distance:

L

Acceleration:

LT⁻²

Time:

Tⁿ

Therefore:

L = LT⁻² × Tⁿ

L = LTⁿ⁻²

For the time dimensions to cancel:

n − 2 = 0

Therefore:

n = 2

So:

d ∝ at²

Dimensional reasoning has revealed the required power of time.

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Conversion Factors

Dimensional analysis is also extremely useful for unit conversions.

A conversion factor is a ratio of equivalent quantities.

For example:

1 km = 1000 m

So either of these ratios equals 1:

1000 m / 1 km

or:

1 km / 1000 m

We choose the version that causes the unwanted unit to cancel.

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Example: Kilometres to Metres

Convert:

6.2 km → m

Use:

6.2 km × (1000 m / 1 km)

The kilometres cancel:

6.2 × 1000 m

Therefore:

6.2 km = 6200 m

The cancellation of units provides a built-in check.

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Example: Centimetres to Metres

Convert:

45 cm → m

Use:

45 cm × (1 m / 100 cm)

Cancel cm:

45/100 m

Therefore:

45 cm = 0.45 m

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Multi-Step Dimensional Analysis

Suppose we want to convert:

90 km/h → m/s

Use two conversion factors:

90 km/h × (1000 m / 1 km) × (1 h / 3600 s)

Cancel:

km

and:

h

This leaves:

m/s

Calculate:

90 × 1000 / 3600

= 25

Therefore:

90 km/h = 25 m/s

The units tell us exactly how the conversion factors should be arranged.

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Why Unit Cancellation Is Powerful

Suppose you accidentally write:

90 km/h × (1 km / 1000 m)

The kilometres do not cancel correctly.

You would end up with an inappropriate combination of units.

That immediately warns you that the conversion factor has been arranged incorrectly.

This is one reason dimensional analysis is more reliable than simply memorizing whether to multiply or divide.

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Multi-Step Example

Convert:

72 km/h → cm/s

Start with:

72 km/h

Use:

72 km/h × (1000 m / 1 km) × (100 cm / 1 m) × (1 h / 3600 s)

Cancel:

km

then:

m

then:

h

The remaining unit is:

cm/s

Calculate:

72 × 1000 × 100 / 3600

= 2000

Therefore:

72 km/h = 2000 cm/s

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Squared Units

Dimensional analysis becomes especially useful when converting areas.

Remember:

1 m = 100 cm

For area:

1 m² = (100 cm)²

Therefore:

1 m² = 10,000 cm²

The conversion factor must be squared because the unit itself is squared.

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Example: Area Conversion

Convert:

3.5 m² → cm²

Use:

3.5 m² × (100 cm / 1 m)²

Square the conversion factor:

3.5 × 10,000 cm²

Therefore:

3.5 m² = 35,000 cm²

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Cubed Units

Volume requires the conversion factor to be cubed.

Since:

1 m = 100 cm

then:

1 m³ = (100 cm)³

Therefore:

1 m³ = 1,000,000 cm³

or:

1 m³ = 10⁶ cm³

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Example: Volume Conversion

Convert:

250 cm³ → m³

Use:

250 cm³ × (1 m / 100 cm)³

Therefore:

250 × 10⁻⁶ m³

= 2.50 × 10⁻⁴ m³

The cubic units cancel correctly.

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Dimensional Analysis in Physics Problems

Dimensional analysis can be used before, during, and after solving a physics problem.

Before

Check that the equation is appropriate.

During

Check that units are compatible.

After

Check that the final units match the quantity being calculated.

This can catch many mistakes before they become final answers.

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Worked Example: Force

A student calculates force using:

F = ma

Suppose:

m = 4.0 kg

a = 3.0 m/s²

Look at the units:

kg × m/s²

This gives:

kg·m/s²

which is:

N

Now calculate:

F = 4.0 × 3.0

F = 12 N

The dimensional check confirms that the final unit is appropriate for force.

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Worked Example: Speed

A student travels:

600 m

in:

50 s

Use:

v = d/t

Units:

m/s

Calculation:

v = 600/50

v = 12 m/s

The units match the expected dimensions of speed:

LT⁻¹

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Worked Example: Energy

Suppose:

m = 5.0 kg

v = 4.0 m/s

Use:

Eₖ = ½mv²

Check units:

kg × (m/s)²

= kg·m²/s²

This is equivalent to:

J

Now calculate:

Eₖ = ½(5.0)(4.0)²

Eₖ = 40 J

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Identifying an Incorrect Answer Unit

Suppose a student calculates acceleration and reports:

a = 7.5 m/s

Something is wrong.

Acceleration has dimensions:

LT⁻²

Its SI unit should therefore be:

m/s²

The reported unit:

m/s

belongs to velocity.

Even before checking the arithmetic, dimensional reasoning reveals a problem.

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Identifying an Incorrect Equation

Suppose a student proposes:

F = mv

Check dimensions.

Right side:

[mv] = M × LT⁻¹

= MLT⁻¹

But force has dimensions:

MLT⁻²

Therefore:

MLT⁻¹ ≠ MLT⁻²

So:

F = mv

is not dimensionally consistent.

The quantity mv represents momentum, not force.

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Rearranging Equations and Dimensions

Dimensional analysis can also check rearranged equations.

Suppose:

F = ma

Rearrange for mass:

m = F/a

Check dimensions:

[m] = MLT⁻² / LT⁻²

Cancel:

[m] = M

The result has the dimensions of mass, so the rearrangement is dimensionally consistent.

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Example: Rearranging Density

Start with:

ρ = m/V

Rearrange:

m = ρV

Check:

Density:

ML⁻³

Volume:

L³

Therefore:

[ρV] = ML⁻³ × L³

= M

The result has the dimensions of mass.

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Dimensional Analysis and Constants

Some physical constants have dimensions.

For example, Newton's law of gravitation is:

F = Gm₁m₂/r²

The gravitational constant G must have dimensions that make the entire equation produce force.

We can determine them.

Rearrange:

G = Fr²/(m₁m₂)

Force:

MLT⁻²

Distance squared:

L²

Mass × mass:

M²

Therefore:

[G] = (MLT⁻²)(L²)/M²

Simplify:

[G] = M⁻¹L³T⁻²

This corresponds to the SI units:

N·m²/kg²

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Deriving the Dimensions of Energy

Work is:

W = Fd

Force:

MLT⁻²

Distance:

L

Therefore:

[W] = MLT⁻² × L

[W] = ML²T⁻²

Since work is an energy transfer, energy has the same dimensions:

[E] = ML²T⁻²

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Deriving the Dimensions of Power

Power is:

P = E/t

Energy:

ML²T⁻²

Divide by time:

T

Therefore:

[P] = ML²T⁻³

Its SI unit is the watt:

W = J/s

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Derived Units Can Be Expanded

A named SI unit can often be rewritten using base units.

For example:

Newton

N = kg·m/s²

Joule

J = N·m

Substitute the newton:

J = kg·m²/s²

Pascal

Pa = N/m²

Substitute the newton:

Pa = kg/(m·s²)

Watt

W = J/s

Therefore:

W = kg·m²/s³

Expanding derived units can make dimensional checking easier.

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Comparing Two Sides of an Equation

A systematic method is:

Step 1: Write the dimensions of the left side.

Step 2: Write the dimensions of every quantity on the right side.

Step 3: Combine the dimensions using the mathematical operations.

Step 4: Simplify.

Step 5: Compare both sides.

If they match:

dimensionally consistent

If they do not:

dimensionally inconsistent

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Worked Check

Is this equation dimensionally consistent?

v² = u² + 2as

Left side:

[v²] = (LT⁻¹)²

= L²T⁻²

First term on right:

[u²] = L²T⁻²

Second term:

[as] = LT⁻² × L

= L²T⁻²

The factor 2 has no dimensions.

Therefore:

L²T⁻² = L²T⁻² + L²T⁻²

The equation is dimensionally consistent.

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Worked Check: Finding an Error

Consider:

v² = u² + 2at

Left side:

L²T⁻²

First right-hand term:

L²T⁻²

Second right-hand term:

[at] = LT⁻¹

The equation tries to add:

L²T⁻² + LT⁻¹

These dimensions are different.

Therefore, the equation is dimensionally inconsistent.

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Dimensional Reasoning Without Knowing the Formula

Suppose you forget whether the period of a simple pendulum depends on:

L/g

or:

g/L

We can examine dimensions.

Length:

[L] = L

Acceleration:

[g] = LT⁻²

For:

L/g

the dimensions are:

L / (LT⁻²)

= T²

Taking the square root gives:

T

which is the dimension required for a period.

Therefore, dimensional reasoning suggests a relationship of the form:

period ∝ √(L/g)

Dimensional analysis cannot determine a dimensionless numerical factor, but it can reveal the required dimensional form.

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Limits of Dimensional Analysis

Dimensional analysis is powerful, but it cannot tell us everything.

It cannot usually determine:

• Numerical constants such as ½ or 2π.
• Whether addition or subtraction should be used when terms have the same dimensions.
• The exact physical mechanism involved.
• Whether an equation is experimentally correct.
• Whether two quantities with the same dimensions represent the same physical concept.

For example:

work and torque can both be expressed using:

N·m

but they represent different physical quantities.

---

Dimensionally Correct but Physically Wrong

Consider:

d = vt

This is dimensionally consistent.

Now consider:

d = 17vt

This is also dimensionally consistent because 17 has no dimensions.

But that does not mean both equations correctly describe the same situation.

Dimensional analysis tests consistency, not complete physical correctness.

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Real-World Importance

Dimensional analysis is used throughout science and engineering.

It helps scientists and engineers:

• Check equations.
• Convert measurement units.
• Detect calculation mistakes.
• Design mathematical models.
• Interpret unfamiliar formulas.
• Compare different measurement systems.
• Determine possible relationships between variables.
• Check computer calculations and simulations.

A simple unit check can sometimes prevent a very large mistake.

---

The Mars Climate Orbiter

A famous example of the importance of consistent units occurred with NASA's Mars Climate Orbiter in 1999.

One engineering team supplied data using U.S. customary units while spacecraft navigation software expected metric units. The mismatch contributed to an incorrect spacecraft trajectory and the mission was lost.

The incident demonstrates an important lesson:

A calculation can be numerically performed correctly and still produce a disastrous result if the units are wrong.

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A Practical Problem-Solving Routine

When solving a physics problem:

Identify

Write down the known quantities and their units.

Convert

Convert quantities into compatible units where necessary.

Select

Choose an appropriate equation.

Check Dimensions

Make sure the equation produces the dimensions of the quantity you are trying to calculate.

Substitute

Insert the numerical values with their units.

Calculate

Perform the arithmetic.

Check

Ask:

• Are the final units correct?
• Is the magnitude reasonable?
• Does the sign make sense?
• Does the answer fit the physical situation?

This routine catches many common errors.

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Common Mistakes

Checking Only the Numbers

A calculator can perform arithmetic without understanding units.

Always check both:

number + unit

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Assuming Matching Units Prove an Equation

Matching dimensions show that an equation could be valid.

They do not prove that it is physically correct.

---

Ignoring Powers on Units

If:

1 cm = 10⁻² m

then:

1 cm² = 10⁻⁴ m²

and:

1 cm³ = 10⁻⁶ m³

---

Adding Different Dimensions

You cannot meaningfully add:

5 m + 3 s

Similarly, terms added within a physics equation must have compatible dimensions.

---

Treating Numerical Constants as Physical Dimensions

Numbers such as:

2, ½ and π

are dimensionless.

They do not change the dimensional form of an equation.

---

Confusing Units and Dimensions

metre is a unit.

length is a dimension.

second is a unit.

time is a dimension.

---

Forgetting Compound Units

When converting something such as:

km/h → m/s

both the numerator and denominator must be considered.

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Check Your Understanding

1. What is dimensional analysis?

2. Explain the difference between a unit and a dimension.

3. What are the dimensions of length, mass, and time?

4. Determine the dimensions of speed.

5. Determine the dimensions of acceleration.

6. Determine the dimensions of force.

7. Show that d = vt is dimensionally consistent.

8. Explain why d = v/t is dimensionally inconsistent.

9. Show that F = ma is dimensionally consistent.

10. Determine the dimensions of density.

11. Determine the dimensions of pressure.

12. Determine the dimensions of energy.

13. Determine the dimensions of power.

14. Check whether v = u + at is dimensionally consistent.

15. Check whether s = ut + ½at² is dimensionally consistent.

16. Explain why the equation s = ut + ½at cannot be correct.

17. Convert 4.5 km to metres using a conversion factor.

18. Convert 108 km/h to m/s using dimensional analysis.

19. Convert 2.5 m² to cm² using conversion factors.

20. Convert 500 cm³ to m³.

21. A student reports an acceleration of 8.0 m/s. Use dimensional reasoning to identify the problem.

22. Expand the unit N into SI base units.

23. Expand the unit J into SI base units.

24. Why can dimensional analysis identify an incorrect equation but not prove that a dimensionally consistent equation is physically correct?

25. Explain how dimensional analysis can help detect mistakes when solving physics problems.

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Key Terms

• Dimensional analysis – method of analyzing physical dimensions to check equations, conversions, and calculations.
• Dimension – physical nature of a quantity, such as length, mass, or time.
• Unit – agreed standard used to measure a physical quantity.
• Fundamental dimension – basic dimension such as mass, length, or time.
• Derived quantity – quantity constructed from combinations of fundamental quantities.
• Dimensional consistency – condition in which the dimensions on both sides of an equation agree.
• Conversion factor – ratio of equivalent measurements used to change units.
• Dimensionless quantity – quantity with no physical dimensions.
• Derived unit – unit constructed from combinations of base SI units.
• Unit cancellation – cancellation of identical units in the numerator and denominator during a calculation.

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Key Takeaways

• Dimensional analysis uses the dimensions and units of physical quantities to examine equations and calculations.
• Length, mass, and time are represented by L, M, and T.
• Speed has dimensions LT⁻¹.
• Acceleration has dimensions LT⁻².
• Force has dimensions MLT⁻².
• Energy has dimensions ML²T⁻².
• Power has dimensions ML²T⁻³.
• Density has dimensions ML⁻³.
• Pressure has dimensions ML⁻¹T⁻².
• Both sides of a valid physical equation must be dimensionally compatible.
• Terms being added or subtracted must have compatible dimensions.
• Dimensional analysis can reveal missing or incorrect powers in equations.
• Conversion factors can be arranged so unwanted units cancel.
• Squared units require squared conversion factors.
• Cubed units require cubed conversion factors.
• Dimensional analysis can detect many equation, calculation, and unit-conversion errors.
• A dimensionally inconsistent equation cannot be physically correct as written.
• A dimensionally consistent equation is not automatically physically correct.
• Dimensionless constants cannot normally be determined through dimensional analysis alone.
• Checking units is one of the fastest ways to catch mistakes in physics.

A useful habit for every physics calculation is:

Check the equation → follow the units → calculate → check the final dimensions.

4. Rearranging equations

Why Rearrange Equations?

Physics equations describe relationships between physical quantities. Often, the equation you know is not arranged for the variable you need to calculate.

For example:

v = d/t

This equation is arranged to calculate velocity.

But what if you know velocity and time and need to calculate distance?

The equation must be rearranged:

d = vt

Or, if you need time:

t = d/v

The relationship between the quantities has not changed. We have simply changed which variable is isolated.

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What Does "Isolate a Variable" Mean?

To isolate a variable means to get that variable by itself on one side of the equation.

For example:

F = ma

Here, F is already isolated.

To isolate m:

m = F/a

To isolate a:

a = F/m

The goal of rearranging is usually:

wanted variable = everything else

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The Golden Rule of Algebra

The most important rule is:

Whatever you do to one side of an equation, you must do to the other side.

Think of an equation as a balanced scale.

If:

x + 5 = 12

subtract 5 from both sides:

x + 5 − 5 = 12 − 5

Therefore:

x = 7

The equation remains balanced.

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Inverse Operations

Rearranging equations depends on inverse operations.

Inverse operations undo each other.

For example, if a variable is multiplied by 4:

4x = 20

divide both sides by 4:

x = 20/4

x = 5

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Undo Operations in Reverse Order

Suppose:

3x + 5 = 20

The operations performed on x are:

x → multiply by 3 → add 5

To isolate x, undo them in reverse order:

subtract 5 → divide by 3

Start:

3x + 5 = 20

Subtract 5:

3x = 15

Divide by 3:

x = 5

This idea becomes extremely useful in multi-step physics equations.

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Rearranging Multiplication Equations

Consider Newton's second law:

F = ma

Suppose we want m.

Since m is multiplied by a, divide both sides by a:

F/a = ma/a

The a terms cancel:

m = F/a

Similarly, to find acceleration:

a = F/m

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Worked Example: Force

A 6.0 kg object accelerates at 4.0 m/s².

Use:

F = ma

Substitute:

F = 6.0 × 4.0

F = 24 N

No rearrangement was necessary because F was already isolated.

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Worked Example: Finding Mass

A force of 45 N produces an acceleration of 3.0 m/s².

Start with:

F = ma

We need m.

Divide by a:

m = F/a

Substitute:

m = 45/3.0

m = 15 kg

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Worked Example: Finding Acceleration

A resultant force of 120 N acts on a 30 kg object.

Start with:

F = ma

Rearrange:

a = F/m

Substitute:

a = 120/30

a = 4.0 m/s²

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Rearranging Division Equations

Consider:

v = d/t

Suppose we want distance.

Multiply both sides by t:

vt = d

Usually we write the wanted variable first:

d = vt

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Finding the Denominator

Suppose we want time from:

v = d/t

Multiply by t:

vt = d

Now divide by v:

t = d/v

So the three useful forms are:

v = d/t

d = vt

t = d/v

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Worked Example: Finding Time

A cyclist travels 1500 m at an average speed of 6.0 m/s.

Start with:

v = d/t

Rearrange:

t = d/v

Substitute:

t = 1500/6.0

t = 250 s

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Rearranging Density

Density is:

ρ = m/V

To find mass:

m = ρV

To find volume:

V = m/ρ

So:

ρ = m/V

m = ρV

V = m/ρ

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Worked Example: Density

An object has a density of 2700 kg/m³ and a volume of 0.0020 m³.

Find its mass.

Start with:

ρ = m/V

Rearrange:

m = ρV

Substitute:

m = 2700 × 0.0020

m = 5.4 kg

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Rearranging Pressure

Pressure is:

p = F/A

Possible forms are:

p = F/A

F = pA

A = F/p

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Worked Example: Area

A force of 500 N produces a pressure of 2500 Pa.

Find the area.

Start with:

p = F/A

Rearrange:

A = F/p

Substitute:

A = 500/2500

A = 0.20 m²

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Rearranging Weight

Weight is calculated using:

W = mg

To find mass:

m = W/g

To find gravitational field strength:

g = W/m

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Worked Example: Mass from Weight

An object weighs 196 N where:

g = 9.8 N/kg

Use:

W = mg

Rearrange:

m = W/g

Substitute:

m = 196/9.8

m = 20 kg

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Rearrange Before Substituting

A very useful habit is:

equation → rearrange → substitute → calculate

For example, suppose:

F = ma

and you need mass.

Do this:

F = ma

m = F/a

Then insert the numbers.

This is usually clearer than inserting numbers first and then trying to manipulate them.

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Rearranging Equations with Addition

Consider:

v = u + at

Suppose we want u.

Subtract at from both sides:

v − at = u

Therefore:

u = v − at

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Finding Acceleration

Start with:

v = u + at

We want a.

Subtract u:

v − u = at

Divide by t:

a = (v − u)/t

---

Finding Time

Start with:

v = u + at

Subtract u:

v − u = at

Divide by a:

t = (v − u)/a

So the equation can be rearranged into several useful forms:

v = u + at

u = v − at

a = (v − u)/t

t = (v − u)/a

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Worked Example: Acceleration

A car changes velocity from 8 m/s to 26 m/s in 6.0 s.

Use:

v = u + at

Rearrange:

a = (v − u)/t

Substitute:

a = (26 − 8)/6.0

a = 18/6.0

a = 3.0 m/s²

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Brackets Matter

Consider:

a = (v − u)/t

The entire difference:

v − u

must be divided by t.

If:

v = 30 m/s

u = 10 m/s

t = 5 s

then:

a = (30 − 10)/5

a = 20/5

a = 4 m/s²

Writing:

30 − 10/5

would give a completely different answer because division occurs before subtraction.

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Rearranging Equations Containing Squares

Consider kinetic energy:

Eₖ = ½mv²

Suppose we want mass.

Multiply by 2:

2Eₖ = mv²

Divide by v²:

m = 2Eₖ/v²

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Finding Velocity from Kinetic Energy

Start with:

Eₖ = ½mv²

Multiply by 2:

2Eₖ = mv²

Divide by m:

2Eₖ/m = v²

Now take the square root:

v = √(2Eₖ/m)

This is a multi-step rearrangement.

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Worked Example: Velocity from Kinetic Energy

An object has:

Eₖ = 400 J

m = 8.0 kg

Use:

v = √(2Eₖ/m)

Substitute:

v = √((2 × 400)/8.0)

v = √100

v = 10 m/s

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Squaring and Square Roots Are Inverse Operations

If:

x² = 49

then:

x = √49

For a physical quantity such as speed, we normally use the physically appropriate positive value:

x = 7

This principle appears frequently in physics equations involving:

v²

r²

t²

and other squared quantities.

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Rearranging Gravitational Potential Energy

Consider:

Eₚ = mgh

To find mass:

m = Eₚ/(gh)

To find gravitational field strength:

g = Eₚ/(mh)

To find height:

h = Eₚ/(mg)

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Worked Example: Finding Height

An object has:

Eₚ = 980 J

m = 20 kg

g = 9.8 N/kg

Start with:

Eₚ = mgh

Rearrange:

h = Eₚ/(mg)

Substitute:

h = 980/(20 × 9.8)

h = 980/196

h = 5.0 m

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Rearranging Work Done

For a constant force acting in the direction of motion:

W = Fd

To find force:

F = W/d

To find distance:

d = W/F

Example

If 600 J of work is done by a 150 N force:

d = W/F

d = 600/150

d = 4.0 m

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Rearranging Power

Power is:

P = E/t

To find energy:

E = Pt

To find time:

t = E/P

Example

A 500 W device transfers 15,000 J of energy.

Find the time.

t = E/P

t = 15,000/500

t = 30 s

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Rearranging Momentum

Momentum is:

p = mv

To find mass:

m = p/v

To find velocity:

v = p/m

Example

An object has momentum:

p = 48 kg·m/s

and mass:

m = 6.0 kg

Therefore:

v = p/m

v = 48/6.0

v = 8.0 m/s

---

Rearranging Ohm's Law

Ohm's law is:

V = IR

To find current:

I = V/R

To find resistance:

R = V/I

The same algebraic rules work regardless of the physics topic.

---

Equations with Fractions on Both Sides

Consider:

ρ₁ = m₁/V₁

If you need V₁:

ρ₁V₁ = m₁

Then:

V₁ = m₁/ρ₁

Work systematically rather than trying to move symbols mentally.

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Equations with a Constant

Consider:

F = kx

To find x:

x = F/k

To find k:

k = F/x

The same principle applies to Hooke's law:

F = kx

where k is the spring constant and x is extension.

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Multi-Step Rearranging

Consider:

s = ut + ½at²

Suppose we need acceleration.

Start:

s = ut + ½at²

Subtract ut:

s − ut = ½at²

Multiply by 2:

2(s − ut) = at²

Divide by t²:

a = 2(s − ut)/t²

This is a good example of why rearranging should be done one operation at a time.

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Worked Multi-Step Example

A vehicle travels 100 m in 5.0 s with an initial velocity of 10 m/s.

Find its acceleration.

Use:

s = ut + ½at²

Rearrange:

a = 2(s − ut)/t²

Substitute:

a = 2(100 − (10 × 5.0))/(5.0)²

Calculate inside the brackets:

a = 2(100 − 50)/25

a = 100/25

a = 4.0 m/s²

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Rearranging an Equation with the Variable in a Denominator

Suppose:

P = E/t

and we need t.

Start:

P = E/t

Multiply by t:

Pt = E

Divide by P:

t = E/P

A useful rule is:

If the wanted variable is in the denominator, first multiply to bring it out of the denominator.

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Rearranging an Inverse-Square Equation

Consider:

F = k/r²

Suppose we need r.

Multiply by r²:

Fr² = k

Divide by F:

r² = k/F

Take the square root:

r = √(k/F)

This type of rearrangement occurs in gravitational, electric, and other inverse-square relationships.

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A More Complex Example: Gravitation

Newton's law of gravitation is:

F = Gm₁m₂/r²

Suppose we want r.

Multiply by r²:

Fr² = Gm₁m₂

Divide by F:

r² = Gm₁m₂/F

Take the square root:

r = √(Gm₁m₂/F)

---

Rearranging for One of the Masses

Start with:

F = Gm₁m₂/r²

Suppose we need m₂.

Multiply by r²:

Fr² = Gm₁m₂

Divide by Gm₁:

m₂ = Fr²/(Gm₁)

This demonstrates how even complicated physics equations can be rearranged using the same basic algebraic rules.

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Substituting Known Values

Once the equation is rearranged, substitute the known values.

For example:

a = F/m

If:

F = 75 N

m = 15 kg

then:

a = 75/15

a = 5.0 m/s²

Include units when recording the known quantities and always include the correct unit in the final answer.

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Use Brackets During Substitution

Suppose:

m = 2Eₖ/v²

and:

Eₖ = 900 J

v = 15 m/s

Write:

m = (2 × 900)/(15)²

rather than entering a long expression without brackets.

Calculate:

m = 1800/225

m = 8.0 kg

Brackets reduce calculator-entry errors.

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Check Using the Original Equation

One of the best ways to check a rearrangement is to substitute the rearranged expression back into the original equation.

Suppose:

F = ma

and you rearranged to:

m = F/a

Substitute this into the original equation:

F = (F/a)a

The a terms cancel:

F = F

So the rearrangement is correct.

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Check Using Units

Units can also test a rearrangement.

Suppose:

m = F/a

Force has units:

N = kg·m/s²

Acceleration has units:

m/s²

Therefore:

m = (kg·m/s²)/(m/s²)

The m/s² terms cancel:

m = kg

The result has the correct unit for mass.

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Checking a Speed Rearrangement

Start with:

v = d/t

Suppose:

t = d/v

Check the units:

t = m/(m/s)

Dividing by m/s is equivalent to multiplying by s/m:

t = m × s/m

The metres cancel:

t = s

The result has the correct unit for time.

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Checking a Kinetic Energy Rearrangement

Suppose:

v = √(2Eₖ/m)

Energy has units:

J = kg·m²/s²

Therefore:

Eₖ/m = (kg·m²/s²)/kg

= m²/s²

Taking the square root gives:

m/s

That is the correct unit for velocity.

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Equation Triangles

For simple equations, some students use equation triangles.

For:

v = d/t

a triangle might place:

d

above:

v t

This can help remember:

d = vt

v = d/t

t = d/v

However, equation triangles have limitations.

They work well for simple multiplication and division relationships but are much less useful for equations involving:

• Addition or subtraction.
• Squares.
• Square roots.
• Several terms.
• Brackets.
• Powers.
• Variables appearing more than once.

Learning algebraic rearrangement is therefore much more powerful.

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A General Rearranging Strategy

When rearranging a physics equation:

Identify the variable you need.

↓

Find what operations are being performed on it.

↓

Undo those operations using inverse operations.

↓

Perform the same operation on both sides.

↓

Continue until the variable is isolated.

↓

Simplify the equation.

↓

Check the rearrangement.

↓

Substitute the known values.

↓

Calculate and include units.

This method works for equations across physics.

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Worked Example: A Full Problem

A car increases its velocity from 12 m/s to 30 m/s with an acceleration of 3.0 m/s².

How long does the acceleration take?

Choose the equation

v = u + at

Identify the wanted variable

We need:

t

Rearrange

Subtract u:

v − u = at

Divide by a:

t = (v − u)/a

Substitute

t = (30 − 12)/3.0

Calculate

t = 18/3.0

t = 6.0 s

Check

Use the original equation:

v = u + at

v = 12 + (3.0 × 6.0)

v = 30 m/s

The answer is consistent.

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Worked Example: Force and Area

A pressure of 25,000 Pa is produced by a force acting over an area of 0.040 m².

Find the force.

Start:

p = F/A

We need F.

Multiply by A:

F = pA

Substitute:

F = 25,000 × 0.040

F = 1000 N

Check units:

Pa × m²

Since:

Pa = N/m²

then:

N/m² × m² = N

The units confirm that the rearrangement makes sense.

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Worked Example: Finding Volume

A material has:

mass = 540 kg

density = 2700 kg/m³

Find its volume.

Start:

ρ = m/V

Rearrange:

V = m/ρ

Substitute:

V = 540/2700

V = 0.20 m³

Check the units:

kg ÷ kg/m³

= m³

Correct.

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Worked Example: Finding a Spring Constant

A spring extends 0.080 m when a force of 20 N is applied.

Use:

F = kx

We need k.

Rearrange:

k = F/x

Substitute:

k = 20/0.080

k = 250 N/m

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Worked Example: Finding Distance from Kinetic Energy

Suppose a constant braking force of 4000 N removes 120,000 J of kinetic energy from a car.

Using the magnitude of the work-energy relationship:

W = Fd

Find the braking distance.

Rearrange:

d = W/F

Substitute:

d = 120,000/4000

d = 30 m

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Common Rearrangements Worth Recognizing

Recognizing common patterns can make rearranging faster, but understanding the algebra is more important than memorizing the table.

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Common Mistakes

Moving a Term Without Changing the Operation

Students sometimes think terms simply "move across the equals sign."

For example:

F = ma

It is better to think:

divide both sides by a

rather than:

move a underneath F

Understanding the operation prevents mistakes in more complicated equations.

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Performing an Operation on Only One Side

Incorrect:

3x + 5 = 20

3x = 20

Correct:

3x + 5 = 20

Subtract 5 from both sides:

3x = 15

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Forgetting Brackets

Suppose:

a = (v − u)/t

Writing:

a = v − u/t

changes the equation.

Use brackets whenever several terms must be treated together.

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Forgetting to Square a Quantity

If:

Eₖ = ½mv²

then velocity is squared.

If:

v = 10 m/s

then:

v² = 100 m²/s²

not 20.

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Forgetting the Square Root

From:

v² = 100

you cannot conclude:

v = 100

You need:

v = √100

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Substituting Too Early

Putting numbers into a complicated equation before rearranging can make the algebra harder to follow.

A clearer method is:

formula → rearrange → substitute → calculate

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Ignoring Units

A numerical answer should include an appropriate unit.

If you calculate acceleration:

a = 5.0

the answer is incomplete.

Write:

a = 5.0 m/s²

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Using the Wrong Variable

Physics sometimes uses the same letter for different quantities in different contexts.

For example, p may represent pressure or momentum depending on the equation.

Always identify what each symbol represents.

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Check Your Understanding

1. Rearrange F = ma to make m the subject.

2. Rearrange F = ma to make a the subject.

3. Rearrange v = d/t to make d the subject.

4. Rearrange v = d/t to make t the subject.

5. Rearrange ρ = m/V to make V the subject.

6. Rearrange p = F/A to make F the subject.

7. Rearrange W = mg to make g the subject.

8. Rearrange P = E/t to make t the subject.

9. Rearrange V = IR to make R the subject.

10. Rearrange F = kx to make x the subject.

11. Rearrange v = u + at to make u the subject.

12. Rearrange v = u + at to make a the subject.

13. Rearrange v = u + at to make t the subject.

14. Rearrange Eₚ = mgh to make h the subject.

15. Rearrange Eₖ = ½mv² to make m the subject.

16. Rearrange Eₖ = ½mv² to make v the subject.

17. Rearrange F = k/r² to make r the subject.

18. Rearrange s = ut + ½at² to make a the subject.

19. A 12 kg object experiences a resultant force of 72 N. Calculate its acceleration.

20. An object travels 450 m at 15 m/s. Calculate the travel time.

21. A material has a mass of 780 kg and density of 2600 kg/m³. Calculate its volume.

22. A 6.0 kg object has 300 J of kinetic energy. Calculate its speed.

23. A vehicle changes velocity from 5.0 m/s to 29 m/s in 8.0 s. Calculate its acceleration.

24. A 15 kg object has 1470 J of gravitational potential energy where g = 9.8 N/kg. Calculate its height.

25. Explain two methods that can be used to check whether an equation has been rearranged correctly.

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Key Terms

• Equation – mathematical statement showing that two expressions are equal.
• Variable – symbol representing a quantity that can have different values.
• Subject – variable isolated on one side of an equation.
• Rearrange – change the form of an equation while preserving the relationship between its variables.
• Isolate – get a variable by itself on one side of an equation.
• Inverse operation – operation that reverses another operation.
• Substitute – replace a variable with a known value.
• Square – multiply a quantity by itself.
• Square root – inverse operation of squaring.
• Formula – equation describing a mathematical or physical relationship.

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Key Takeaways

• Rearranging an equation allows you to calculate different variables using the same physical relationship.
• The equation must remain balanced.
• Whatever operation is performed on one side must also be performed on the other.
• Use inverse operations to isolate the wanted variable.
• Addition is undone by subtraction.
• Multiplication is undone by division.
• Squaring is undone by taking a square root.
• Work one step at a time with complicated equations.
• Brackets are important when several terms must be treated together.
• Rearrange the equation before substituting numerical values whenever practical.
• Include units when substituting and in the final answer.
• Check rearrangements by substituting them back into the original equation.
• Dimensional analysis provides another powerful check.
• Multi-step equations use exactly the same algebraic principles as simple equations.

A reliable physics workflow is:

Identify the unknown → choose the equation → rearrange → check → substitute → calculate → check the answer.

5. Estimation techniques