Momentum Revision Notes — 9709 Paper 4

topic	marks/paper
Kinematics of Motion in a Straight Line	20.9
Forces and Equilibrium	13.8
Newton's Laws of Motion	13.2
Energy, Work and Power	12.9
Momentum	4.6

Linear momentum: definition and vector nature

Syllabus requirement · §4.3

“

use the definition of linear momentum and show understanding of its vector nature (for motion in one dimension only).

”

A lorry and a bicycle rolling at the same speed are not equally hard to stop, and neither are a bicycle at walking pace and the same bicycle at racing speed. The quantity that measures "how much motion" a moving object carries — combining how heavy it is with how fast it is going — is its linear momentum:

p = mv

where $m$ is the mass in kilograms and $v$ is the velocity in metres per second. ("Linear" just means motion along a line, as opposed to spinning; on Paper 4 every momentum is linear, so the word is usually dropped.)

Units. Multiplying $\text{kg}$ by $\text{m s}^{-1}$ gives $\text{kg m s}^{-1}$ . You will also see momentum written in newton seconds, $\text{N s}$ — a question may say "its momentum is $4\text{ N s}$ ". The two units are the same thing: a newton is $1\text{ kg m s}^{-2}$ , so a newton second is $1\text{ kg m s}^{-2} \times 1\text{ s} = 1\text{ kg m s}^{-1}$ .

Why momentum is a vector. Mass is always positive, but velocity has a direction, and on a straight line that direction is recorded as a sign (§4.2): choose one way along the line as positive, and a particle moving the other way has a negative velocity. Momentum is mass times velocity, so it takes the sign of the velocity — it points the way the particle is moving. A $2\text{ kg}$ particle moving at $3\text{ m s}^{-1}$ has momentum $+6\text{ kg m s}^{-1}$ or $-6\text{ kg m s}^{-1}$ depending on which way it is going, and those are genuinely different momenta, not the same one written two ways.

p = mv

Linear momentum

m in kg (always positive) × v in m s⁻¹ (signed) — units kg m s⁻¹, equivalently N s

With right as positive: A (2 kg, moving right at 3 m s⁻¹) has momentum +6; B (1 kg, moving left at 5 m s⁻¹) has momentum −5. The thick momentum arrows are drawn to one scale, so A's is longer even though B is moving faster — mass counts as much as speed.

One positive direction, stated once, held for every particle

Almost every momentum question involves two or more particles at once, so the positive direction has to be the same for all of them, in the same equation — otherwise the signs stop meaning anything relative to each other. Start every solution with a sentence such as "taking the direction of $P$ 's initial motion as positive", or mark an arrow labelled "+" on your diagram, and keep it to the end. Examiners accept either direction as positive; what they do not accept is a direction that changes halfway through.

The momentum of a system. When several particles are involved, the total momentum of the system is the sum of their individual momenta — added with their signs. Two particles moving towards each other partly (or completely) cancel, because one of them contributes a negative amount. This total is the quantity the rest of the topic is about: in §02 you will see that a collision cannot change it.

Momentum of each particle, and of the system

Particle $A$ , of mass $3\text{ kg}$ , moves at $4\text{ m s}^{-1}$ . Particle $B$ , of mass $2\text{ kg}$ , moves at $6\text{ m s}^{-1}$ in the opposite direction. Taking $A$ 's direction of motion as positive, find the momentum of each particle and the total momentum of the system.

Show full working

1
Write $A$ 's velocity with its sign. $A$ moves in the positive direction: $v_A = +4\text{ m s}^{-1}$
Writing the sign explicitly, even when it is +, is the habit that stops the minus sign being forgotten for the next particle.
2
Write $B$ 's velocity with its sign. $B$ moves the opposite way, so its velocity is negative: $v_B = -6\text{ m s}^{-1}$
'Opposite direction' in words has to become a minus sign in numbers before anything is substituted — this conversion is where most sign errors in the topic begin.
3
State the rule for $A$ 's momentum. $p_A = m_A v_A$
Mass times velocity — not mass times weight, not mass times g. The rule is stated before any numbers so each piece can be seen going in.
4
Substitute $m_A = 3$ and $v_A = 4$ : $p_A = 3 \times 4 = 12\text{ kg m s}^{-1}$
Positive, because A moves in the positive direction.
5
Substitute $m_B = 2$ and $v_B = -6$ into $p_B = m_B v_B$ : $p_B = 2 \times (-6) = -12\text{ kg m s}^{-1}$
The mass is positive; the minus sign comes from the velocity and carries straight through into the momentum.
6
Add the two signed momenta for the total: $p_A + p_B = 12 + (-12)$
The total is a signed sum. Adding the sizes, 12 + 12 = 24, would describe two particles moving the same way — a different situation altogether.
7
Evaluate: $p_A + p_B = 0\text{ kg m s}^{-1}$
Zero total momentum does not mean nothing is moving: it means the motion in the two directions exactly balances.

Answer

$p_A = 12\text{ kg m s}^{-1}$ , $p_B = -12\text{ kg m s}^{-1}$ , total $= 0\text{ kg m s}^{-1}$ .

A total momentum of zero is exactly the situation in which two colliding particles can both end up at rest — keep this in mind for §03, where a real exam question uses it.

Momentum and kinetic energy — the same two ingredients, combined differently. A moving particle also has kinetic energy, the energy it has because it is moving (it is taught properly in §4.5, but its formula is needed throughout this topic):

\text{KE} = \tfrac12 mv^2

Compare it with $p = mv$ . Both are built from the mass and the speed, but:

momentum uses $v$ once, so it keeps the sign — it is a vector;
kinetic energy uses $v^2$ , and a square is never negative, so kinetic energy is always positive whichever way the particle moves — it is a scalar. Its unit is the joule, $\text{J}$ .

So a $2\text{ kg}$ particle moving left at $3\text{ m s}^{-1}$ (with right positive) has $p = 2 \times (-3) = -6\text{ kg m s}^{-1}$ but $\text{KE} = \tfrac12 \times 2 \times (-3)^2 = 9\text{ J}$ . Because the two formulae combine $m$ and $v$ differently, knowing both the momentum and the kinetic energy of a particle is enough to find its mass and its speed — which is exactly what the next question asks.

Finding mass and speed from momentum and kinetic energy

9709/41 M/J 2025 Q5(a)3 marks

When a particle $P$ of mass $m\text{ kg}$ has speed $u\text{ m s}^{-1}$ , its momentum is $4\text{ N s}$ and its kinetic energy is $16\text{ J}$ .

Find the value of $m$ and the value of $u$ .

Show full working

1
Write the momentum fact as an equation. Momentum is $mu$ , and it equals $4$ : $mu = 4$
4 N s is the same as 4 kg m s⁻¹, so no conversion is needed — the unit just tells you this number is a momentum.
2
Write the kinetic-energy fact as an equation. Kinetic energy is $\tfrac12 mu^2$ , and it equals $16$ : $\tfrac12 mu^2 = 16$
Two facts, two unknowns (m and u): two equations are exactly what is needed.
3
Notice that the second equation contains the first. $\tfrac12 mu^2$ is $\tfrac12 u \times (mu)$ , so: $\tfrac12 u \times (mu) = 16$
Splitting mu² as u × mu shows the momentum hiding inside the kinetic energy — the quickest way to eliminate m.
4
Substitute $mu = 4$ : $\tfrac12 u \times 4 = 16$
This is the same as dividing the KE equation by the momentum equation; either description is fine.
5
Simplify the left-hand side: $2u = 16$
½ × 4 = 2.
6
Solve for $u$ : $u = 8$
A speed, so positive — which it is.
7
Substitute $u = 8$ back into $mu = 4$ to find $m$ : $8m = 4 \implies m = 0.5$
Use the simpler (momentum) equation to finish.
8
Check in the kinetic-energy equation: $\tfrac12 \times 0.5 \times 8^2 = \tfrac12 \times 0.5 \times 64 = 16 \ \checkmark$
One line of checking catches an arithmetic slip before it spoils part (b), which uses m = 0.5 (see §06).

Answer

$m = 0.5$ , $u = 8$ .

Momentum and kinetic energy together always pin down m and v: divide ½mv² by mv to get ½v, then back-substitute.

Common mistakes

Treating momentum as always positive, because "mass times speed" sounds like a size
Momentum takes the sign of the velocity — write it as a signed quantity from the first line
A momentum with the wrong sign makes every later equation false; the sign is not decoration.
Writing $mgv$ (weight times velocity) for momentum
Momentum is $mv$ — $g$ never appears in a momentum term
Mark schemes repeatedly say 'M1 A0 if g included with the masses': the method mark survives, the answer mark does not.
Giving momentum in joules, or kinetic energy in $\text{kg m s}^{-1}$
Momentum: $\text{kg m s}^{-1}$ or $\text{N s}$ . Kinetic energy: $\text{J}$
They are different quantities built from the same ingredients; the unit shows which one you mean.

Your turn

State a positive direction before writing any velocity.

19709/42 O/N 2020 Q1(a)1 mark
Two particles $P$ and $Q$ , of masses $0.2\text{ kg}$ and $0.5\text{ kg}$ respectively, are at rest on a smooth horizontal plane. $P$ is projected towards $Q$ with speed $2\text{ m s}^{-1}$ .

Write down the momentum of $P$ .
Stuck? Show hint
Mass times velocity — and give the units.
Show solution
1. 1
  Take $P$ 's direction of motion as positive, so $v_P = +2\text{ m s}^{-1}$ .
  With a single moving particle, its own direction is the natural positive direction.
2. 2
  Apply $p = mv$ with $m = 0.2$ and $v = 2$ : $p = 0.2 \times 2 = 0.4\text{ kg m s}^{-1}$
  'Write down' means one line — but the units are part of the answer.
Answer
$0.4\text{ kg m s}^{-1}$ (in the direction of motion of $P$ ).
2
Three particles move along the same straight line. $A$ has mass $1\text{ kg}$ and velocity $5\text{ m s}^{-1}$ ; $B$ has mass $2\text{ kg}$ and velocity $-2\text{ m s}^{-1}$ ; $C$ has mass $0.5\text{ kg}$ and velocity $-4\text{ m s}^{-1}$ , all measured with the same positive direction. Find the total momentum of the system, and state which way it points.
Stuck? Show hint
Find each signed momentum on its own line, then add.
Show solution
1. 1
  $A$ : $p_A = 1 \times 5 = 5\text{ kg m s}^{-1}$ .
  Positive velocity, positive momentum.
2. 2
  $B$ : $p_B = 2 \times (-2) = -4\text{ kg m s}^{-1}$ .
  The minus sign comes from the velocity.
3. 3
  $C$ : $p_C = 0.5 \times (-4) = -2\text{ kg m s}^{-1}$ .
  Same rule — mass positive, velocity negative, momentum negative.
4. 4
  Add the signed values: $5 + (-4) + (-2) = -1\text{ kg m s}^{-1}$
  The two negative momenta together outweigh A's, so the total is negative.
Answer
Total momentum $= -1\text{ kg m s}^{-1}$ , i.e. $1\text{ kg m s}^{-1}$ in the negative direction.
3
A particle of mass $2.5\text{ kg}$ has momentum of magnitude $15\text{ N s}$ . Find its speed.
Stuck? Show hint
Rearrange $p = mv$ for $v$ .
Show solution
1. 1
  Start from the definition: $p = mv$
  Write the rule before rearranging it.
2. 2
  Divide both sides by $m$ : $v = \frac{p}{m}$
  Mass is never zero, so dividing by it is always allowed.
3. 3
  Substitute $p = 15$ and $m = 2.5$ : $v = \frac{15}{2.5} = 6\text{ m s}^{-1}$
  1 N s = 1 kg m s⁻¹, so the units come out as m s⁻¹ directly.
Answer
$6\text{ m s}^{-1}$ .
4
A particle has momentum $6\text{ N s}$ and kinetic energy $9\text{ J}$ . Find its mass and its speed.
Stuck? Show hint
Write $mv = 6$ and $\tfrac12 mv^2 = 9$ , then use the first inside the second.
Show solution
1. 1
  Momentum: $mv = 6$
  First fact, first equation.
2. 2
  Kinetic energy: $\tfrac12 mv^2 = 9$
  Second fact, second equation.
3. 3
  Write $\tfrac12 mv^2$ as $\tfrac12 v \times (mv)$ and substitute $mv = 6$ : $\tfrac12 v \times 6 = 9$
  The momentum is sitting inside the kinetic energy.
4. 4
  Simplify and solve: $3v = 9 \implies v = 3\text{ m s}^{-1}$
  ½ × 6 = 3.
5. 5
  Back-substitute into $mv = 6$ : $3m = 6 \implies m = 2\text{ kg}$
  Check: ½ × 2 × 3² = 9 ✓.
Answer
Mass $2\text{ kg}$ , speed $3\text{ m s}^{-1}$ .

Practise the definition of momentumReal past-paper questions · Definition of linear momentum and its vector nature

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Can you do all of these?

State one positive direction at the start and keep it for every particle and every stage
Write every velocity as a signed number before substituting: towards each other means opposite signs; rebounds means the velocity after is negative
Momentum is mv — never mgv; mark schemes remove the answer mark when g appears
Write momentum before and momentum after as separate lines, including a zero term for a particle at rest
Leave an unknown velocity as a plain letter and let its sign give its direction
Answer 'speed' with a positive number; answer 'velocity' with a size and a direction in words
Coalesce / stick together / remain in contact: one combined mass, one common velocity
With two unknown velocities after an impact, use the question's link (a ratio, w and 2w, a difference, equal speeds) to write both with one letter
With a difference of speeds, the particle in front is the faster one
Find every velocity from momentum first, then KE before, KE after, and loss = before − after (never negative)
Use one KE term per particle before the impact — never the combined mass unless they already move together
A given loss or percentage loss is an equation; eliminate one unknown with the momentum equation first, and reject any quadratic root that contradicts the question
A bare speed after impact means two cases (+s and −s); check each: positive speeds/masses, no particle passing through another, no gain in KE, any stated condition
In a chain, one momentum equation per collision with only the two particles that touch; carry each velocity forward unchanged on a smooth surface
After each collision compare neighbours: left particle faster to the right than its right-hand neighbour ⇒ another collision
'No further collision' is an inequality; greatest/least values come from its boundary
A fixed wall or the ground is not a momentum collision: use the stated new speed and reverse the direction
Between collisions on a smooth horizontal surface speeds are constant: distance = speed × time, time = gap ÷ closing speed
In longer questions, find the velocity immediately before impact by suvat — never reuse the launch speed — then use the velocity after as the new u

Now do the questions

47 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes

Kinematics of Motion in a Straight Line

Newton's Laws of Motion

Momentum

Linear momentum: definition and vector nature

Conservation of momentum: the principle and the standard method

Both particles moving: signs, rebounds, and speed versus velocity

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Linear momentum: definition and vector nature

The rest of this note

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