Physics · 1 Motion, forces and energy

1.6 Momentum

Before you start

Mass in kg and velocity in m/s, and that velocity has a direction while speed does not.
Newton's first law: with no resultant force, an object keeps moving at constant velocity.
Rearranging a two-variable equation, e.g. getting $v$ out of $p = m v$ .

Syllabus objectives covered (4)

Define momentum as mass × velocity; recall and use the equation p = mv
Define impulse as force × time for which force acts; recall and use the equation impulse = FΔt = Δ(mv)
Apply the principle of the conservation of momentum to solve simple problems in one dimension
Define resultant force as the change in momentum per unit time; recall and use the equation F = Δp / Δt

Part 1 of 5

Momentum is how hard something is to stop

A tennis ball and a lorry can travel at the same speed, but you would happily catch one and not the other. What separates them is mass. Momentum is the quantity that combines the two.
Momentum is defined as mass × velocity. That definition is what you write if you are asked to $𝐷𝑒𝑓𝑖𝑛𝑒 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚$ — not the symbols, the words.
$p = m v$
The unit follows from the equation: kg for mass, m/s for velocity, so momentum is in kg m/s. There is no special name for it. If you write 'N' or 'J' you have lost the mark.
Momentum is a vector, because velocity is a vector. It has a direction. This is the single most important thing about momentum, and everything awkward in this topic comes from it.
In one dimension you handle direction with a sign. Choose one direction as positive — say, right — and stick to it for the whole question. Anything moving left then has a negative velocity, and therefore negative momentum.
So a $0.50$ kg ball moving right at $4.0$ m/s has momentum $+ 2.0$ kg m/s, and the same ball moving left at $4.0$ m/s has momentum $- 2.0$ kg m/s. Same size, opposite sign, genuinely different momentum.

Part 2 of 5

Impulse: what a force does over time

A force applied for a long time changes motion more than the same force applied for an instant. Impulse is the quantity that captures 'force multiplied by how long it acted'.
Impulse is defined as force × time for which the force acts.
$impulse = F Δ t$
Here is the part worth understanding rather than memorising: that impulse is exactly equal to the change in momentum it produces.
$F Δ t = Δ (m v)$
Read the equation both ways. Left to right: apply a force for a time, and you change the momentum. Right to left: if you know how much the momentum changed, you know the impulse that did it — without knowing the force or the time separately.
The unit of impulse is N s, and because impulse equals a change in momentum, N s and kg m/s are the same unit. Either is accepted; do not mix them inside one answer.
$Δ (m v)$ means final momentum minus initial momentum, in that order. Getting this backwards flips the sign of your answer, which is the most common error in this whole subtopic.

Part 3 of 5

Resultant force is the rate of change of momentum

Take $F Δ t = Δ p$ and divide both sides by $Δ t$ . You get an equation that defines force in a more general way than $F = m a$ does.
$F = \frac{Δ p}{Δ t}$
In words: resultant force is the change in momentum per unit time. 'Per unit time' is the phrase the mark scheme wants — 'rate of change of momentum' also earns it.
This is not a new law competing with $F = m a$ . For constant mass, $Δ p = m Δ v$ , so $F = m Δ v / Δ t = m a$ . The momentum form is simply the more honest one, because it still works when mass changes — a rocket burning fuel, for instance.
The practical payoff is the one every exam question is about: for a fixed change in momentum, a longer collision time means a smaller force. That is the whole physics of crumple zones, airbags, crash mats and bending your knees when you land.
So when a question asks why a safety feature reduces injury, the structure of a full-mark answer is always the same: the change in momentum is fixed, the feature increases the time of the collision, and since $F = Δ p / Δ t$ a larger $Δ t$ gives a smaller $F$ .

Part 4 of 5

Conservation of momentum

The principle: in a closed system, total momentum before an interaction equals total momentum after it. 'Closed system' means no external resultant force acts — so no friction, no external push.
Written out for two objects colliding in a line, with $u$ for before and $v$ for after:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
This equation is not on your formula sheet, because there is no formula sheet in Paper 4. You have to be able to build it from the principle. Practise writing it from scratch rather than copying it.
The method is always the same four moves: choose a positive direction; write total momentum before; write total momentum after; set them equal and solve.
If two objects stick together after the collision, they share one mass and one velocity afterwards, so the right-hand side becomes $(m_{1} + m_{2}) v$ . Watch for the words 'coalesce', 'couple together' or 'embeds itself'.
The same total momentum, redistributed — 6.0 kg m/s before, 6.0 kg m/s after.
Energy is a separate question. Momentum is conserved in every collision; kinetic energy is not conserved unless the collision is elastic. Never argue from one to the other.

Part 5 of 5

Putting the four ideas in order

When you meet a momentum question, decide first which of the two situations you are in: one object whose momentum changes, or two objects interacting.
One object changing momentum — reach for $p = m v$ to get the before and after momenta, then $Δ p = F Δ t$ if force or time appears.
Two objects interacting — reach for conservation of momentum, total before equals total after.
In both cases, write your positive direction down on the paper before you write any numbers. Almost every lost mark in this subtopic is a sign error, not a physics error.

Worked examples

Example 1. A trolley of mass $0.80$ kg moves along a straight track at $2.5$ m/s. Calculate its momentum, and state its momentum after it rebounds from a wall and travels at $1.5$ m/s in the opposite direction.

Show the working

Take the initial direction of travel as positive. Everything else follows from this choice.
Use $p = m v$ with $m = 0.80$ kg and $v = + 2.5$ m/s.
$p = 0.80 \times 2.5 = 2.0 kg m/s$
After the rebound the trolley moves the other way, so its velocity is $- 1.5$ m/s, not $+ 1.5$ m/s.
$p = 0.80 \times (- 1.5) = - 1.2 kg m/s$
Say what the sign means rather than leaving a bare negative number — it tells the examiner you understand momentum is a vector.

Answer: $2.0$ kg m/s in the original direction; after rebounding, $1.2$ kg m/s in the opposite direction (i.e. $- 1.2$ kg m/s).

Example 2. A ball of mass $0.15$ kg travelling at $12$ m/s hits a wall and rebounds at $8.0$ m/s. It is in contact with the wall for $0.030$ s. Calculate the impulse on the ball and the average force the wall exerts on it.

Show the working

Take the initial direction of the ball as positive, so the rebound velocity is negative.
Momentum before: $p_{1} = 0.15 \times 12 = + 1.8$ kg m/s. Momentum after: $p_{2} = 0.15 \times (- 8.0) = - 1.2$ kg m/s.
Impulse equals the change in momentum, final minus initial. This is where the sign work pays off.
$Δ p = (- 1.2) - (+ 1.8) = - 3.0 kg m/s$
Notice the change is larger than either momentum on its own. That is because the ball reversed — it had to be stopped and then sent back.
Now use $F = Δ p / Δ t$ to get the average force.
$F = \frac{- 3.0}{0.030} = - 100 N$
The negative sign says the force acts opposite to the ball's original motion, which is exactly what a wall does. Quote the magnitude and describe the direction.

Answer: Impulse $= 3.0$ N s directed away from the wall; average force $= 100$ N, directed away from the wall.

Example 3. A railway truck of mass $2000$ kg moving at $3.0$ m/s collides with a stationary truck of mass $4000$ kg. The trucks couple together. Calculate their common velocity after the collision.

Show the working

Take the direction of the moving truck as positive. The stationary truck contributes zero momentum, not zero mass.
Total momentum before:
$p_{before} = 2000 \times 3.0 + 4000 \times 0 = 6000 kg m/s$
After coupling they move as one object of mass $2000 + 4000 = 6000$ kg with a single velocity $v$ .
$p_{after} = 6000 v$
Momentum is conserved, so set the two equal and solve.
$6000 v = 6000 \Rightarrow v = 1.0 m/s$
Sanity-check the answer: the same momentum is now carried by three times the mass, so the velocity should be a third of $3.0$ m/s. It is.

Answer: $1.0$ m/s in the direction the first truck was originally moving.

Where people go wrong

Mistake: Treating momentum as a scalar and adding $2.0$ kg m/s and $1.2$ kg m/s to get $3.2$ when one object moves the other way.

Instead: Assign signs from your chosen positive direction first, then add. Opposite directions subtract.

Momentum is mass × velocity, and velocity is a vector. Nearly every dropped mark in this subtopic starts here.

Mistake: Calculating $Δ p$ as initial minus final.

Instead: Always final minus initial: $Δ p = p_{2} - p_{1}$ .

$Δ$ means 'change in', and a change is always what you ended with minus what you started with. Reversing it flips the direction of your force.

Mistake: Saying an airbag 'reduces the momentum change' or 'absorbs the force'.

Instead: It increases the time over which the momentum changes, and so reduces the force.

The change in momentum is fixed by the crash — you were moving, now you are not. Only $Δ t$ is available to change, and $F = Δ p / Δ t$ then gives a smaller $F$ .

Mistake: Assuming kinetic energy is conserved because momentum is.

Instead: Momentum is conserved in all collisions in a closed system; kinetic energy is only conserved in elastic ones.

They are independent conservation statements. In a coupling collision, energy is transferred to internal energy and sound, but momentum still balances.

Mistake: Forgetting the unit, or writing momentum in N.

Instead: kg m/s for momentum, N s for impulse.

The final answer mark in a science mark scheme is an A mark, and A marks require the unit. A correct number with no unit scores the C mark only.

Check yourself

A $1200$ kg car travels at $15$ m/s. What is its momentum, and in what unit?
Answer
$p = m v = 1200 \times 15 = 18 000$ kg m/s, in the direction of travel.
Why can two objects have momentum of the same size but not the same momentum?
Answer
Momentum is a vector. If they travel in opposite directions their momenta have opposite signs, so they are different despite having equal magnitude.
A force of $25$ N acts on a box for $0.40$ s. What is the impulse, and what does it tell you about the box?
Answer
Impulse $= F Δ t = 25 \times 0.40 = 10$ N s. The box's momentum changes by $10$ kg m/s in the direction of the force.
State the definition of resultant force in terms of momentum, and give the equation.
Answer
Resultant force is the change in momentum per unit time: $F = Δ p / Δ t$ .
A gymnast bends her knees on landing. Explain, in terms of momentum, why this reduces the force on her.
Answer
Her change in momentum on landing is fixed. Bending her knees increases the time taken to stop, and since $F = Δ p / Δ t$ , a larger $Δ t$ gives a smaller force.
State the principle of the conservation of momentum, including the condition under which it applies.
Answer
In a closed system — one with no external resultant force — the total momentum before an interaction equals the total momentum after it.
A $0.20$ kg ball at $6.0$ m/s strikes a stationary $0.40$ kg ball and stops dead. How fast does the second ball move?
Answer
Before: $0.20 \times 6.0 = 1.2$ kg m/s. After: $0.40 v = 1.2$ , so $v = 3.0$ m/s in the original direction.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A ball of mass $0.42$ kg is kicked. It leaves the ground at $18$ m/s. The foot is in contact with the ball for $0.012$ s.

(a) Define momentum. 1 mark No calculator

Mark scheme

B1 mass × velocity (accept the equation p = mv with symbols identified)

(b) Calculate the momentum of the ball as it leaves the ground. 2 marks

Mark scheme

C1 p = mv or 0.42 × 18 seen
A1 7.6 kg m/s (accept 7.56); unit required

(c) Calculate the average force exerted by the foot on the ball. 3 marks

Mark scheme

C1 F = Δp / Δt or impulse = FΔt used
C1 Δp = 7.56 (kg m/s) used, ecf from (b)
A1 630 N (accept 6.3 × 10² N); unit required

(d) The ball is later kicked with the same force but the foot stays in contact for longer. Explain the effect on the ball's momentum. 2 marks No calculator

Mark scheme

B1 impulse is larger because impulse = FΔt and Δt is larger
B1 so the change in momentum is larger / the ball leaves faster, ORA

Question 2. Two ice skaters stand at rest, facing each other on frictionless ice. Skater P has mass $60$ kg and skater Q has mass $45$ kg. They push apart. Skater P moves off at $1.5$ m/s.

(a) State the total momentum of the two skaters before they push apart. 1 mark No calculator

Mark scheme

B1 zero (accept 0 kg m/s)

(b) Calculate the speed of skater Q immediately after they push apart. 3 marks

Mark scheme

C1 total momentum after = total momentum before = 0, or 60 × 1.5 = 45 × v
C1 90 (kg m/s) seen for P's momentum
A1 2.0 m/s; unit required

(c) Explain why the two skaters move in opposite directions. 2 marks No calculator

Mark scheme

B1 total momentum must remain zero / momentum is conserved
B1 so their momenta must be equal in size and opposite in direction, owtte

Question 3. A car is fitted with a crumple zone at the front.

(a) Explain, in terms of momentum, how a crumple zone reduces the force on the driver during a collision. 3 marks No calculator

Mark scheme

B1 the change in momentum of the driver is the same / fixed
B1 the crumple zone increases the time taken to stop
B1 F = Δp / Δt, so a larger Δt gives a smaller force, ORA

(b) Suggest one other feature of a car designed to reduce the force on a person in a collision. 1 mark No calculator

Mark scheme

B1 any one from: airbag / seat belt that stretches / padded dashboard / collapsible steering column

Words to know

Momentum: Mass × velocity. A vector quantity, measured in kg m/s.
Impulse: Force × the time for which the force acts, equal to the change in momentum produced. Measured in N s.
Closed system: A system on which no external resultant force acts, so its total momentum cannot change.
Elastic collision: A collision in which kinetic energy as well as momentum is conserved.
Resultant force: The change in momentum per unit time: $F = Δ p / Δ t$ .

All Physics lessons Next: 1.7.1 Energy