Momentum

Syllabus coverage

Use the definition of linear momentum and show understanding of its vector nature
Restrict the motion considered to one dimension only
Use conservation of linear momentum to solve problems that may be modelled as the direct impact of two bodies
Include the direct impact of two bodies where the bodies coalesce on impact
Note that knowledge of impulse and of the coefficient of restitution is not required

Part of the AS & A Level Maths syllabus 9709.

A direct impact on a straight line, before and after. Taking the positive direction to the right, the total momentum before the collision equals the total momentum after it, whether the two bodies coalesce or separate.

Linear momentum and its vector nature

The linear momentum of a body is its mass multiplied by its velocity, written $p = m v$ . Mass is a scalar and velocity is a vector, so momentum is a vector: it has both a size and a direction. The SI unit is the kilogram metre per second, kg m/s, which is also the newton second, N s. A body of mass $4$ kg moving at $6$ m/s has momentum $24$ kg m/s in the direction of its motion.

Because this topic is restricted to motion in one dimension, the direction is carried entirely by a sign. Choose a positive direction at the start of the question, exactly as you would in kinematics, and then every velocity you write - before or after an impact - has a sign attached. Taking to the right as positive, a $3$ kg particle moving to the left at $5$ m/s has momentum $3 \times (- 5) = - 15$ kg m/s, and its momentum has magnitude $15$ kg m/s. Cambridge states that vector notation is not used in the mechanics papers, so you describe the direction in words and let the signs do the rest.

A body that is stationary has momentum zero, because its velocity is zero. That sounds trivial but it is the single most common place to drop a mark: in a question where one body is at rest before the impact, its momentum before is $0$ , not $m$ and not $m u$ .

Momentum is not the same quantity as kinetic energy, and questions mix the two up on purpose. A $2$ kg particle moving at $3$ m/s has momentum $2 \times 3 = 6$ kg m/s and kinetic energy $\frac{1}{2} (2) (3)^{2} = 9$ J. Momentum is a vector, so the momenta of two bodies moving in opposite directions partly cancel; kinetic energy is a scalar, so the energies of those same bodies always add. That difference is what makes momentum useful for impacts and energy useful for work and power.

Body	Mass	Velocity (right positive)	Momentum
A	$5$ kg	$+ 4$ m/s	$+ 20$ kg m/s
B	$2$ kg	$- 6$ m/s	$- 12$ kg m/s
C	$0.5$ kg	0	0
A and B together	-	-	$20 - 12 = + 8$ kg m/s

Momentum is a vector, so a body moving the other way has negative momentum. If your total momentum has the wrong sign, the direction of motion of the combined body will be wrong too.

Conservation of linear momentum

When two bodies collide along a straight line and no external force acts on either of them along that line, the total momentum of the two bodies is the same after the impact as it was before. Written out with signs, for bodies of mass $m_{1}$ and $m_{2}$ with velocities $u_{1}$ and $u_{2}$ before the impact and $v_{1}$ and $v_{2}$ after it:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

The reason it works is Newton's third law. During the impact the two bodies push on each other with equal and opposite forces for exactly the same short interval, so the momentum one loses the other gains, and the total is unchanged. Because the interval is so short, any external force such as friction or air resistance has a negligible effect while the impact lasts, which is why the collision may be treated as if the two bodies were alone.

Here is the equation in use. Particle $A$ of mass $3$ kg moves to the right at $6$ m/s and particle $B$ of mass $2$ kg moves to the left at $4$ m/s. Take to the right as positive, so $u_{A} = + 6$ and $u_{B} = - 4$ . The total momentum before the impact is $3 (6) + 2 (- 4) = 18 - 8 = 10$ kg m/s. Suppose that after the impact $B$ moves to the right at $2$ m/s. Then $3 v_{A} + 2 (2) = 10$ , so $3 v_{A} = 6$ and $v_{A} = 2$ m/s to the right. The positive answer confirms that $A$ is still moving in its original direction.

Notice how much of that working is bookkeeping rather than physics. The equation is one line; the marks come from substituting the right masses, writing the correct sign against each velocity and then solving. A question that gives you a velocity 'to the left' and a positive direction 'to the right' has told you to write a minus sign, and the examiner is checking whether you did.

Total momentum before the impact = total momentum after the impact
Every velocity carries the sign that matches your chosen positive direction
A body at rest before the impact contributes zero momentum
A negative answer for a velocity means the body is moving in the negative direction, so it has rebounded

Conservation of momentum is a statement about vectors, so it is the momenta that cancel, not the speeds. Two bodies of equal mass and equal speed moving towards each other have a total momentum of zero.

Direct impact where the bodies coalesce

When two bodies stick together on impact they move on as a single body with one common velocity. Cambridge calls this coalescing, and the same conservation equation applies - it is only the right-hand side that simplifies, because both bodies now share the velocity $V$ :

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$

The masses add on the right-hand side. That is the whole difference between this case and a general impact, and it is where the marks are won or lost: writing $m_{1} V$ alone on the right, or forgetting to include the mass of the body that was at rest, both give the wrong answer.

A ball of mass $0.4$ kg moving at $6$ m/s hits a stationary ball of mass $0.2$ kg and the two coalesce. Taking the direction of the first ball as positive, the momentum before the impact is $0.4 (6) + 0.2 (0) = 2.4$ kg m/s. After the impact the combined mass is $0.6$ kg, so $0.6 V = 2.4$ and $V = 4$ m/s in the original direction of motion. A heavier example behaves in exactly the same way: a $1400$ kg van travelling at $12$ m/s hits a stationary $600$ kg car and they coalesce, so $1400 (12) = (1400 + 600) V$ , giving $V = \frac{16 800}{2000} = 8.4$ m/s.

Kinetic energy is not conserved when bodies coalesce - the total kinetic energy afterwards is always less than before - but you are not asked to work with energy here. Momentum is conserved and that is the only conservation law this topic uses. Cambridge states plainly that the coefficient of restitution is not required at AS, so a coalescence question is always a single conservation equation.

Situation	Momentum before	Momentum after
General direct impact, separate afterwards	$m_{1} u_{1} + m_{2} u_{2}$	$m_{1} v_{1} + m_{2} v_{2}$
The bodies coalesce	$m_{1} u_{1} + m_{2} u_{2}$	$(m_{1} + m_{2}) V$
Second body at rest before the impact	$m_{1} u_{1}$	$m_{1} v_{1} + m_{2} v_{2}$
Second body at rest and the bodies coalesce	$m_{1} u_{1}$	$(m_{1} + m_{2}) V$

Coalescing means one velocity, so the masses add. If the two bodies separate after the impact, each keeps its own unknown velocity and the equation has two unknowns to sort out.

The before-and-after method

Every impact question yields to the same five steps, and writing them out is faster than trying to hold the signs in your head.

Draw a before-and-after diagram with the two bodies on a single straight line. Choose a positive direction and mark it on the diagram with an arrow. Write each velocity with its sign, above or beside the body it belongs to. Write the conservation equation with the masses in place. Solve it, then state the direction of the answer in words.

A particle $A$ of mass $5$ kg moves to the right at $4$ m/s and collides directly with a particle $B$ of mass $2$ kg moving to the left at $3$ m/s. After the impact $A$ moves to the right at $2$ m/s. Taking to the right as positive: the momentum before is $5 (4) + 2 (- 3) = 20 - 6 = 14$ kg m/s, and the momentum after is $5 (2) + 2 v_{B} = 10 + 2 v_{B}$ . Setting the two equal gives $2 v_{B} = 4$ , so $v_{B} = 2$ m/s. The answer is positive, so $B$ is moving to the right, which means it has reversed its direction of travel.

The diagram is not decoration. Drawing the arrows in the two panels forces you to decide the sign of each velocity before you write the equation, and it makes an answer like ' $B$ moves to the right' visible before the algebra confirms it. In an examination the diagram also carries the first method mark in most mark schemes.

Draw the before panel and the after panel on the same line
Choose and mark a positive direction
Attach a sign to every velocity, including the ones you are about to find
Write total momentum before = total momentum after, with the masses
Solve, then state the direction and check it against your diagram

If your answer comes out negative, do not change the sign. A negative velocity is the information that the body has reversed direction, and the question often asks for that conclusion.

Finding an unknown mass, and reading a negative velocity

Not every question asks for a velocity. A common variant gives you the common velocity after a coalescence and asks for the mass of one of the bodies. The method is unchanged: the unknown mass appears in the bracket on the right-hand side, and the equation is linear, so one rearrangement finishes it.

A particle of mass $6$ kg moving at $5$ m/s collides with a stationary particle and the two coalesce, moving on together at $3$ m/s. The momentum before the impact is $6 (5) = 30$ kg m/s. After the impact it is $(6 + m) (3)$ , so $6 + m = 10$ and the stationary particle has mass $4$ kg. Substituting back is a cheap check: $(6 + 4) (3) = 30$ , which is what the momentum was before.

A negative velocity after an impact is a real physical result, not an error. Take a $2$ kg particle moving at $10$ m/s which hits a stationary $3$ kg particle, and suppose the $3$ kg particle moves off at $8$ m/s. Conservation gives $20 = 2 v + 3 (8) = 2 v + 24$ , so $v = - 2$ m/s: the $2$ kg particle is now moving at $2$ m/s in the direction opposite to its original motion. It has rebounded. In a question that asks for the velocity you must give the minus sign or state the direction; in a question that asks for the speed you give $2$ m/s as a positive number.

It is worth checking that an answer is physically sensible before you move on. The common velocity after a coalescence must lie between the velocities the two bodies had before, and it must point the way the total momentum points. If the total momentum before the impact is positive, the combined body must move in the positive direction.

An unknown mass goes inside the bracket $(m_{1} + m_{2})$ and is found by one rearrangement
Substitute your answer back into both sides to check the momentum totals agree
A negative velocity means the body moves in the negative direction, so it has rebounded
Velocity answers carry a sign or a direction; speed answers are positive

Divide the total momentum by the total mass to get the common velocity after a coalescence. That single division also tells you the direction, from the sign of the momentum.

Modelling the impact, and what is not required

Momentum questions are modelling questions, and the assumptions are worth stating because a written part of a question sometimes asks about them. The two bodies are modelled as particles, so their size and any rotation are ignored. The impact is direct, which means the velocities before and after all lie along the same straight line - the line of the centres of the two bodies. The time of the impact is short enough that any external force along the line, such as friction or air resistance, may be ignored during it.

Two points about what the syllabus does not require. The coefficient of restitution is not needed at AS, so you are never asked to relate the relative speeds before and after an impact, and you should not introduce it. Impulse is not needed either, so a question will not ask you to find the force multiplied by the time of the impact. Both belong to the A Level year and to later mechanics, and a solution that reaches for them here is doing more work than the question wants.

The distinction to hold on to is that momentum is conserved in every direct impact covered by this topic, whether the bodies separate or coalesce, while kinetic energy is conserved only in the ideal case of a perfectly elastic impact, which is not part of this syllabus. If a question mentions energy at all, it will be in the energy, work and power topic rather than here.

Finally, remember that conservation applies to the total momentum of the two bodies, not to each one separately. Either body's momentum changes during the impact - that is what the impact does - and the sum is what stays fixed.

One equation, one unknown. If a direct impact question seems to need two equations, check whether you have used the coalescence condition

v_{1} = v_{2} = V

; that condition is what makes a two-unknown problem solvable.

Key terms you must be able to define

Linear momentum: The product of the mass of a body and its velocity, $p = m v$ . A vector, measured in kg m/s.

Vector quantity: A quantity with both a magnitude and a direction. Momentum, velocity and acceleration are vectors; mass and speed are not.

Conservation of linear momentum: When no external force acts along the line of motion, the total momentum of two colliding bodies is the same before and after the impact.

Direct impact: A collision in which the velocities of both bodies lie along the same straight line, so the motion stays one-dimensional.

Coalesce: To stick together on impact. Bodies that coalesce move on as one body with a single common velocity, and their masses add.

Common velocity: The single velocity $V$ of two bodies that have coalesced, equal to the total momentum before the impact divided by the total mass.

External force: A force on a body from outside the system of colliding bodies, such as friction. During a short impact its effect is negligible, which is why momentum is conserved.

Internal force: The force each body exerts on the other during the impact. By Newton's third law the two are equal and opposite, so they cancel in the total momentum.

Positive direction: The direction you declare positive at the start of a question, so that every velocity and momentum carries a sign.

Magnitude of momentum: The size of the momentum without its sign. A momentum of $- 15$ kg m/s has magnitude $15$ kg m/s.

Stationary: At rest, so with velocity zero. A stationary body has zero momentum, whatever its mass.

Rebound: To move off after an impact in the direction opposite to the approach. A rebound shows up as a negative velocity.

Particle model: The model in which a body has mass but no size, so rotation during an impact is ignored and only the linear momentum matters.

Unit of momentum: The kilogram metre per second, kg m/s. It is the same as the newton second, N s.

Kinetic energy: The scalar $\frac{1}{2} m v^{2}$ . Unlike momentum it cannot cancel, so it is not conserved when bodies coalesce.

Coefficient of restitution: A measure of how bouncy an impact is. Cambridge states that knowledge of it is not required at AS, so it is never needed here.

Impulse: The product of a force and the time for which it acts, equal to the change in momentum. Cambridge states that knowledge of it is not required at AS.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Two particles approaching each other

A of mass 3 kg moves right at 6 m/s and B of mass 2 kg moves left at 4 m/s. Taking right as positive, the total momentum before the impact is 3(6) + 2(-4) = 10 kg m/s. If B moves right at 2 m/s afterwards, then 3v + 2(2) = 10 gives v = 2 m/s, so A continues to the right at 2 m/s. The minus sign on B's velocity before the impact is the only difficult part of the question.

A body at rest before the impact

A 0.4 kg ball moving at 6 m/s strikes a stationary 0.2 kg ball and the two coalesce. The momentum before the impact is 0.4(6) = 2.4 kg m/s, because the stationary ball contributes nothing. After the impact the mass is 0.6 kg, so 0.6V = 2.4 and V = 4 m/s in the direction the first ball was moving. Writing 0.4(6) + 0.2(6) is the classic error: the stationary ball has velocity zero, not the velocity of the moving one.

Coalescing after a collision

A 1400 kg van travelling at 12 m/s collides with a stationary 600 kg car and they coalesce. The momentum before is 1400 times 12 = 16 800 kg m/s. After the impact the combined mass is 2000 kg, so V = 16 800 divided by 2000 = 8.4 m/s. The common velocity is lower than the van's original 12 m/s because the same momentum now has to carry more mass.

Finding an unknown mass

A 6 kg particle moving at 5 m/s collides with a stationary particle and the two move on together at 3 m/s. Before the impact the momentum is 30 kg m/s. After it is (6 + m)(3), so 6 + m = 10 and m = 4 kg. Checking: (6 + 4)(3) = 30 kg m/s, which matches the momentum before the impact, so the answer is consistent.

A rebound gives a negative velocity

A 2 kg particle moving at 10 m/s hits a stationary 3 kg particle, and afterwards the 3 kg particle moves at 8 m/s. Conservation gives 20 = 2v + 24, so v = -2 m/s. The minus sign says the 2 kg particle now moves at 2 m/s in the opposite direction: it has rebounded. If the question asked for the speed the answer would be 2 m/s, and if it asked for the velocity the answer must keep the minus sign or name the direction.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1Define linear momentum.

Mass times velocity, p = mv. It is a vector, measured in kg m/s or N s.

Q2Why is momentum a vector?

Because velocity is a vector and mass is a scalar. Multiplying a vector by a positive scalar leaves the direction unchanged.

Q3What is the momentum of a stationary body of mass 5 kg?

Zero. Its velocity is zero, so 5 times 0 = 0.

Q4State the principle of conservation of linear momentum for two colliding bodies.

If no external force acts along the line of motion, the total momentum before the impact equals the total momentum after: m1u1 + m2u2 = m1v1 + m2v2.

Q5How does the equation change when the two bodies coalesce?

They share one velocity V, so the right-hand side becomes (m1 + m2)V. The masses add.

Q6A 2 kg particle moves right at 3 m/s and a 3 kg particle moves left at 2 m/s. What is the total momentum, taking right as positive?

2(3) + 3(-2) = 0 kg m/s. The two momenta cancel exactly, so a coalescence would leave the pair stationary.

Q7How do you find the common velocity after a coalescence?

Divide the total momentum before the impact by the total mass: V = (m1u1 + m2u2) / (m1 + m2).

Q8Why is momentum conserved in an impact?

Newton's third law makes the forces the bodies exert on each other equal and opposite, so the momentum one loses the other gains.

Q9What does a negative velocity after an impact mean?

The body is moving in the direction opposite to the positive direction you chose, so it has rebounded. It is not a sign error.

Q10Is kinetic energy conserved when two bodies coalesce?

No. The kinetic energy afterwards is always less than before. Only momentum is conserved in this topic.

Q11What are the units of momentum, and what is the alternative unit?

Kilogram metres per second, kg m/s. It is also the newton second, N s.

Q12What is a direct impact?

One in which the velocities of both bodies lie along the same straight line, so the motion stays one-dimensional.

Q13Is the coefficient of restitution required at AS Level?

No. Cambridge states that knowledge of the coefficient of restitution is not required, and impulse is not required either.

Q14Why is the time of an impact ignored in these questions?

It is short enough that any external force along the line, such as friction, changes the momentum by a negligible amount, so the two bodies may be treated as alone.

0 of 14 marked as known

Exam technique for this topic

Write 'taking the direction of motion of A as positive' on your first line. It costs six words and it makes every subsequent sign defensible, whichever direction the examiner would have chosen.
Draw the two-panel before-and-after diagram with an arrow on each body. The diagram is the method mark in most mark schemes and it prevents the commonest error, which is treating a body moving in the negative direction as if it had positive momentum.
Put the sign in at the moment you substitute. '2(-5)' is unmistakable; '2 x 5' followed later by a remembered minus is where the mark disappears.
When the bodies coalesce, write the combined mass in brackets before you do any arithmetic: $(m_{1} + m_{2}) V$ . That is the condition the question is really testing, and writing it down proves you have used it.
Give the direction as part of your answer. A velocity of $- 2$ m/s is a complete answer only if it is clear what the positive direction was, and an examiner reading '2 m/s' alone cannot tell whether the particle has rebounded.

Why students lose marks here

A student wroteA is moving to the left at 5 m/s, so its momentum is 5m.

Why it loses marksMomentum is mass times velocity, and the velocity is negative when the body moves in the negative direction. Using the speed instead of the velocity gives a momentum with the wrong sign, and every total built on it is wrong.

The answer that scoresTaking to the right as positive, a 3 kg particle moving to the left at 5 m/s has momentum 3(-5) = -15 kg m/s, with magnitude 15 kg m/s.

A student wroteThe second ball is at rest, so its momentum before the impact is 0.2.

Why it loses marksThe mass has been used in place of the momentum. A stationary body has velocity zero, so its momentum is mass times zero.

The answer that scoresThe momentum of a stationary 0.2 kg ball is 0.2(0) = 0. In the coalescence of a 0.4 kg ball at 6 m/s with it, the equation is 2.4 = 0.6V, giving V = 4 m/s.

A student wroteThe two bodies coalesce, so 0.4(6) = 0.4V and V = 6 m/s.

Why it loses marksThe mass of the second body has been left out of the right-hand side. Coalescing means one body of the combined mass moving with one velocity, so both masses must appear.

The answer that scores0.4(6) = (0.4 + 0.2)V, so 0.6V = 2.4 and V = 4 m/s. The combined mass is 0.6 kg, not 0.4 kg.

A student wroteThe velocity comes out as -2 m/s, so I have made a sign error somewhere.

Why it loses marksA negative velocity after an impact is a genuine result: it means the body is now moving in the direction opposite to the positive direction you chose, so it has rebounded. Changing the sign to make it look tidier loses the mark.

The answer that scoresKeep the minus sign and interpret it. A 2 kg particle that hits a stationary 3 kg particle and comes away at -2 m/s has rebounded and is travelling at 2 m/s in the opposite direction to its original motion.

A student wroteThe kinetic energy before the impact equals the kinetic energy after it, so I can use energy as well as momentum.

Why it loses marksKinetic energy is only conserved in a perfectly elastic impact, which is not part of this syllabus. When bodies coalesce the kinetic energy always decreases, so an energy equation gives the wrong answer.

The answer that scoresUse conservation of momentum only: total momentum before equals total momentum after. For a 0.4 kg ball at 6 m/s coalescing with a stationary 0.2 kg ball, the energy falls from 7.2 J to 4.8 J, while the momentum stays at 2.4 kg m/s.

Syllabus coverage

Linear momentum and its vector nature

Conservation of linear momentum

Direct impact where the bodies coalesce

The before-and-after method

Finding an unknown mass, and reading a negative velocity

Modelling the impact, and what is not required

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here