Physics 9702/22 — May/June 2018

Background Concept

A force is defined operationally by Newton’s second law. Momentum is

$$p = mv$$

and Newton’s second law states that the (resultant) force on a body equals the rate of change of its momentum.

$$F = \frac{\mathrm{d}p}{\mathrm{d}t}$$

This definition works in general (including when mass is not constant) and links force to how motion changes.

Understanding the Question

You are asked for a definition (not a description). So you should give a precise physics statement, ideally as an equation, for what force means.

Approach

Use the standard Cambridge definition from Newton’s second law: force equals rate of change of momentum. Write it clearly and, if using finite changes, use (\Delta) symbols.

Step-by-Step Reasoning

1. Start from Newton’s second law definition:

$$F = \frac{\mathrm{d}p}{\mathrm{d}t}$$

1. For exam answers, it is also acceptable to write the finite-difference form:

$$F = \frac{\Delta p}{\Delta t}$$

This communicates the same definition at AS level.

Key Takeaways

• The most fundamental definition at AS is: force is the rate of change of momentum.
• Writing the equation gains the mark clearly.

Common Mistakes

• Writing only “a push or pull” (too vague for an A-level physics definition).
• Writing (F = ma) as the definition; it is a special case that assumes constant mass.

Things to Be Careful About

• Force in Newton’s second law is the resultant (net) force.
• Use correct symbols: momentum (p), time (t), and rate of change (\Delta/\Delta t) or (\mathrm{d}/\mathrm{d}t).

Background Concept

Many physical quantities use derived SI units. Force is measured in newtons (N), defined using Newton’s second law:

$$F = ma$$

Mass (m) has base unit (\text{kg}) and acceleration (a) has base unit (\text{m s}^{-2}). Therefore the newton can be written in base units.

Understanding the Question

The question asks specifically for SI base units (not the named unit newton). So you must express force using only (\text{kg}), (\text{m}), (\text{s}) (and not N).

Approach

Use (F = ma). Replace (m) and (a) by their base units and multiply.

Step-by-Step Reasoning

From

$$F = ma$$

the units are

$$[F] = [m][a] = (\text{kg})(\text{m s}^{-2}) = \text{kg m s}^{-2}.$$

So

$$1\ \text{N} = 1\ \text{kg m s}^{-2}.$$

Key Takeaways

• Use definitions like (F=ma) to convert derived units into base units.
• The SI base units of force are (\text{kg m s}^{-2}).

Common Mistakes

• Writing (\text{kg m s}^{-1}) (missing one power of (s)).
• Writing “N” only (the question asks for base units).

Things to Be Careful About

• Acceleration is (\text{m s}^{-2}), not (\text{m s}^{-1}).
• Use negative indices as Cambridge expects: (\text{s}^{-2}) rather than “per second squared”.

Background Concept

This is a units (dimensional) analysis question. The key idea is that any correct physical equation must be homogeneous: both sides have the same units.

Given

$$F = \frac{Q_1Q_2}{4\pi r^2\varepsilon}$$

• (F) is force (units N).
• (Q) is charge (units C).
• (r) is distance (units m).
• (4\pi) is a pure number (no units).

So the units of (\varepsilon) can be found by rearranging and substituting base units.

Useful unit conversions:

$$1\ \text{N} = 1\ \text{kg m s}^{-2}, \qquad 1\ \text{C} = 1\ \text{A s}.$$

Understanding the Question

You are told the functional form of Coulomb’s law in a medium and asked to find the SI base units of (\varepsilon). That means the final answer must be written only in terms of (\text{kg}), (\text{m}), (\text{s}), (\text{A}) (and powers of them).

Approach

1. Rearrange the equation to make (\varepsilon) the subject.
2. Replace each quantity with its SI units (C, m, N).
3. Convert derived units (C and N) into base units (A, s, kg, m).
4. Simplify indices carefully.

Step-by-Step Reasoning

Start with

$$F = \frac{Q_1Q_2}{4\pi r^2\varepsilon}$$

Rearrange:

$$\varepsilon = \frac{Q_1Q_2}{4\pi r^2 F}$$

Now do units (ignore (4\pi) because it is dimensionless):

$$[\varepsilon] = \frac{[Q]^2}{[r]^2[F]} = \frac{\text{C}^2}{\text{m}^2\,\text{N}}$$

Convert (\text{N}) and (\text{C}) to base units:

$$\text{C}^2 = (\text{A s})^2 = \text{A}^2\text{s}^2$$

and

$$\text{m}^2\text{N} = \text{m}^2 (\text{kg m s}^{-2}) = \text{kg m}^3\text{s}^{-2}$$

So

$$[\varepsilon] = \frac{\text{A}^2\text{s}^2}{\text{kg m}^3\text{s}^{-2}} = \text{A}^2\text{s}^{4}\text{kg}^{-1}\text{m}^{-3}$$

Hence the base units are

$$\text{kg}^{-1}\,\text{m}^{-3}\,\text{s}^4\,\text{A}^2.$$

Key Takeaways

• Rearranging first helps you see what must be in the numerator and denominator.
• Treat constants like (4\pi) as dimensionless.
• Convert everything to base units at the end ((\text{C} = \text{A s}), (\text{N} = \text{kg m s}^{-2})).

Common Mistakes

• Forgetting to square the coulomb when using (Q_1Q_2).
• Using (\text{C} = \text{A}/\text{s}) (wrong; it is (\text{A s})).
• Handling the negative powers of (s) in newtons incorrectly, leading to (\text{s}^0) or (\text{s}^2) instead of (\text{s}^4).

Things to Be Careful About

• Write the final answer in base units only (kg, m, s, A), not in N or C.
• Be systematic with indices: dividing by (\text{s}^{-2}) is multiplying by (\text{s}^2), which is where the (\text{s}^4) comes from overall.
• Any equivalent ordering of the base units is fine, but the powers must be correct.