Momentum & Impulse

Momentum and impulse are the entry point to AQA Further Mechanics. They convert Newton's second law from a statement about acceleration at an instant into a statement about change accumulated over time — and that accumulated form is exactly what you need for collisions, where the force is enormous, unknown, and lasts only a few milliseconds. Master the vector bookkeeping here and every later collision topic falls out as algebra.

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1 Where this sits in AQA 7367

This is the opening topic of the Mechanics option on Paper 3 (7367/3M), one of the two optional applications you choose alongside the compulsory pure of Papers 1 and 2. Paper 3 is weighted AO1 40% / AO2 25% / AO3 35% — markedly more problem-solving (AO3) than the pure papers, so expect questions that combine impulse with vectors, restitution or energy rather than testing a single formula. The prerequisite is the constant-acceleration mechanics and the vector work $\vec r = x\,\vec i + y\,\vec j$r=xi+yj​ from A-Level Maths; we build directly on $\vec F = m\vec a$F=ma.

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2 Core theory

Momentum

The linear momentum of a particle of mass $m$m moving with velocity $\vec v$v is

$\vec p = m\vec v.$p​=mv.

Property	Detail
Type	Vector — same direction as $\vec v$v
SI unit	$\mathrm{kg\,m\,s^{-1}}$kgms−1, identically $\mathrm{N\,s}$Ns
Depends on	mass and velocity

Because momentum is a vector, in one dimension its sign carries the direction and in two dimensions you must split it into components. Always declare a positive direction before you write a single number — most lost marks in this topic are sign errors, not algebra.

Impulse

The impulse of a force is the time-integral of that force. For a constant force $\vec F$F acting for time $\Delta t$Δt,

$\vec J = \vec F\,\Delta t,$J=FΔt,

and for a force that varies with time,

$\vec J = \int_{t_1}^{t_2} \vec F(t)\,\mathrm{d}t.$J=∫t1​t2​​F(t)dt.

Its unit is the newton-second, $\mathrm{N\,s}$Ns, which is dimensionally identical to $\mathrm{kg\,m\,s^{-1}}$kgms−1 — the same unit as momentum, which is the first clue that the two are linked.

The impulse–momentum principle

Newton's second law in its fundamental form is $\vec F = \dfrac{\mathrm{d}\vec p}{\mathrm{d}t}$F=dtdp​​. Integrating both sides over the contact time,

$\int_{t_1}^{t_2}\vec F\,\mathrm{d}t = \int_{t_1}^{t_2}\frac{\mathrm{d}\vec p}{\mathrm{d}t}\,\mathrm{d}t = \vec p(t_2) - \vec p(t_1),$∫t1​t2​​Fdt=∫t1​t2​​dtdp​​dt=p​(t2​)−p​(t1​),

so

$\boxed{\;\vec J = \Delta \vec p = m\vec v - m\vec u\;}$J=Δp​=mv−mu​

where $\vec u$u is the initial and $\vec v$v the final velocity. Impulse equals change in momentum. This is the engine of the whole topic: it lets you sidestep the unknown peak force during a collision and work only with the velocities before and after.

Why the time-integral form matters

During a real impact — a bat on a ball, a hammer on a nail — the contact force $F(t)$F(t) is a sharp, complicated spike lasting milliseconds. You could never write it down explicitly, let alone solve $F = ma$F=ma instant by instant. The genius of the impulse form is that the detail of $F(t)$F(t) is irrelevant: only its integral over the contact, $\int F\,\mathrm{d}t$∫Fdt, affects the outcome, and that integral is fixed by the velocities before and after. Two completely different force profiles delivering the same area produce the same change in momentum. This is why the impulse–momentum principle, not Newton's second law in its raw form, is the right tool for collisions.

Impulse and force–time area

Because $\vec J = \int_{t_1}^{t_2} \vec F\,\mathrm{d}t$J=∫t1​t2​​Fdt, the impulse of a one-dimensional force is the signed area under the force–time graph between $t_1$t1​ and $t_2$t2​. For a rectangle (constant force) the area is $F\,\Delta t$FΔt; for a triangle (force ramping linearly from zero) it is $\tfrac12 F_{\max}\,\Delta t$21​Fmax​Δt; for a trapezium it is the average height times the width. This geometric reading is examined directly and saves you from integrating when the graph is piecewise-linear.

Momentum of a system

For a collection of particles the total momentum is the vector sum of the individual momenta,

$\vec p_{\text{total}} = \sum_i m_i \vec v_i.$p​total​=∑i​mi​vi​.

Each particle obeys its own impulse–momentum equation, and summing them shows that the total momentum of the system changes only in response to the external resultant force (the internal forces, by Newton's third law, cancel in pairs). This is the seed of conservation of momentum, the topic of the next lesson: when the external resultant is zero, $\vec p_{\text{total}}$p​total​ is constant. For now the key skill is the bookkeeping — sum momenta as vectors, component by component, never as bare magnitudes.

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3 Worked examples with mark scheme

Worked Example 1 — variable force

A force $F = 4t\ \mathrm{N}$F=4t N acts on a particle of mass $2\ \mathrm{kg}$2 kg, initially at rest, for $0 \le t \le 3\ \mathrm{s}$0≤t≤3 s. Find the impulse and the final speed.

$$\begin{aligned} J &= \int_0^3 4t\,\mathrm{d}t = \Big[\,2t^2\,\Big]_0^3 = 18\ \mathrm{N\,s} \\ J &= mv - mu \;\Rightarrow\; 18 = 2v - 2(0) \\ v &= 9\ \mathrm{m\,s^{-1}}. \end{aligned}$$JJv​=∫03​4tdt=[2t2]03​=18 Ns=mv−mu⇒18=2v−2(0)=9 ms−1.​

Mark scheme: M1 integrate $F$F with respect to $t$t; A1 $J = 18\ \mathrm{N\,s}$J=18 Ns; M1 apply $J = mv - mu$J=mv−mu; A1 $v = 9\ \mathrm{m\,s^{-1}}$v=9 ms−1. (4 marks.)

Worked Example 2 — a rebound (sign discipline)

A ball of mass $0.1\ \mathrm{kg}$0.1 kg moving at $20\ \mathrm{m\,s^{-1}}$20 ms−1 to the right is struck and returns at $15\ \mathrm{m\,s^{-1}}$15 ms−1 to the left. Find the impulse on the ball.

Take right as positive, so $u = +20$u=+20 and $v = -15$v=−15:

$$J = mv - mu = 0.1(-15) - 0.1(20) = -1.5 - 2 = -3.5\ \mathrm{N\,s}.$$J=mv−mu=0.1(−15)−0.1(20)=−1.5−2=−3.5 Ns.

The impulse is $3.5\ \mathrm{N\,s}$3.5 Ns directed to the left.

Mark scheme: B1 consistent positive direction with $v=-15$v=−15; M1 $J = mv - mu$J=mv−mu with signed values; A1 $3.5\ \mathrm{N\,s}$3.5 Ns leftward. The single most common error is writing $0.1(15)-0.1(20)$0.1(15)−0.1(20) and losing the reversal.

Worked Example 3 — average force in a collision

A cricket ball of mass $0.15\ \mathrm{kg}$0.15 kg is bowled at $30\ \mathrm{m\,s^{-1}}$30 ms−1 and driven straight back at $40\ \mathrm{m\,s^{-1}}$40 ms−1. The bat is in contact for $0.01\ \mathrm{s}$0.01 s. Find the average force on the ball.

Take the direction of the return as positive, so $u = -30,\ v = +40$u=−30, v=+40. The average force is the impulse divided by the contact time, $F_{\text{avg}} = \dfrac{\Delta p}{\Delta t}$Favg​=ΔtΔp​:

$$\begin{aligned} J &= 0.15(40) - 0.15(-30) = 6 + 4.5 = 10.5\ \mathrm{N\,s} \\ F_{\text{avg}} &= \frac{J}{\Delta t} = \frac{10.5}{0.01} = 1050\ \mathrm{N}. \end{aligned}$$JFavg​​=0.15(40)−0.15(−30)=6+4.5=10.5 Ns=ΔtJ​=0.0110.5​=1050 N.​

Mark scheme: M1 signed $\Delta p$Δp; A1 $10.5\ \mathrm{N\,s}$10.5 Ns; M1 divide by $\Delta t$Δt; A1 $1050\ \mathrm{N}$1050 N. Note the force is over a thousand times the ball's weight $(\approx 1.5\ \mathrm{N})$(≈1.5 N) — that is why the impulse method, which never needs that peak force explicitly, is so useful.

Worked Example 4 — simple momentum, then a 2-D impulse

(a) A particle of mass $3\ \mathrm{kg}$3 kg moves at $8\ \mathrm{m\,s^{-1}}$8 ms−1. Find its momentum. (b) A particle of mass $4\ \mathrm{kg}$4 kg has velocity $(2\vec i + 3\vec j)\ \mathrm{m\,s^{-1}}$(2i+3j​) ms−1. Find its momentum and the magnitude of that momentum.

(a) $p = mv = 3 \times 8 = 24\ \mathrm{kg\,m\,s^{-1}}$p=mv=3×8=24 kgms−1, in the direction of motion.

(b) $\vec p = m\vec v = 4(2\vec i + 3\vec j) = (8\vec i + 12\vec j)\ \mathrm{kg\,m\,s^{-1}}$p​=mv=4(2i+3j​)=(8i+12j​) kgms−1; magnitude $|\vec p| = \sqrt{8^2 + 12^2} = \sqrt{208} = 4\sqrt{13} \approx 14.4\ \mathrm{kg\,m\,s^{-1}}$∣p​∣=82+122​=208​=413​≈14.4 kgms−1.

Mark scheme: B1 $24\ \mathrm{kg\,m\,s^{-1}}$24 kgms−1; M1 scalar-multiply the velocity vector; A1 $(8\vec i + 12\vec j)$(8i+12j​); A1 $4\sqrt{13}$413​. In two dimensions you scale each component, then take the magnitude of the resultant — never add the component magnitudes.

Worked Example 5 — impulse from a force–time graph

A constant force of $50\ \mathrm{N}$50 N acts for $0.02\ \mathrm{s}$0.02 s on a stationary $0.25\ \mathrm{kg}$0.25 kg ball. Find (a) the impulse, (b) the speed the ball acquires.

(a) With a constant force the force–time graph is a rectangle: $J = F\,\Delta t = 50 \times 0.02 = 1\ \mathrm{N\,s}$J=FΔt=50×0.02=1 Ns.

(b) $J = mv - mu$J=mv−mu: $1 = 0.25v - 0 \Rightarrow v = 4\ \mathrm{m\,s^{-1}}$1=0.25v−0⇒v=4 ms−1.

Mark scheme: M1 $J = F\,\Delta t$J=FΔt (or area of rectangle); A1 $1\ \mathrm{N\,s}$1 Ns; M1 $J = mv - mu$J=mv−mu; A1 $4\ \mathrm{m\,s^{-1}}$4 ms−1. A useful sanity check on every impulse problem: the units of $F\,\Delta t$FΔt $(\mathrm{N\,s})$(Ns) and of $mv$mv $(\mathrm{kg\,m\,s^{-1}})$(kgms−1) are identical, confirming the principle is dimensionally consistent.

Worked Example 6 — trapezium force–time graph

A $0.5\ \mathrm{kg}$0.5 kg trolley starts at rest. A force rises linearly from $0$0 to $8\ \mathrm{N}$8 N over $0\le t\le 1\ \mathrm{s}$0≤t≤1 s, stays at $8\ \mathrm{N}$8 N for $1\le t\le 3\ \mathrm{s}$1≤t≤3 s, then falls linearly back to $0$0 over $3\le t\le 4\ \mathrm{s}$3≤t≤4 s. Find the final speed.

The impulse is the total area under the graph — a trapezium with parallel sides (the two "plateau" durations) handled most easily as triangle + rectangle + triangle:

$$\begin{aligned} J &= \underbrace{\tfrac12(1)(8)}_{\text{rise}} + \underbrace{(2)(8)}_{\text{plateau}} + \underbrace{\tfrac12(1)(8)}_{\text{fall}} = 4 + 16 + 4 = 24\ \mathrm{N\,s}\\ 24 &= 0.5\,v - 0 \;\Rightarrow\; v = 48\ \mathrm{m\,s^{-1}}. \end{aligned}$$J24​=rise21​(1)(8)​​+plateau(2)(8)​​+fall21​(1)(8)​​=4+16+4=24 Ns=0.5v−0⇒v=48 ms−1.​

Mark scheme: M1 decompose the area (or use the trapezium rule $\tfrac12(a+b)h$21​(a+b)h with $a=4,\ b=2,\ h=8$a=4, b=2, h=8); A1 $24\ \mathrm{N\,s}$24 Ns; M1 $J = mv - mu$J=mv−mu; A1 $48\ \mathrm{m\,s^{-1}}$48 ms−1. The trapezium formula $\tfrac12(\text{sum of parallel sides})(\text{height}) = \tfrac12(4+2)(8) = 24$21​(sum of parallel sides)(height)=21​(4+2)(8)=24 gives the same area in one line — a neat check.

Worked Example 7 — two impulses in succession

A particle of mass $2\ \mathrm{kg}$2 kg moving at $5\ \mathrm{m\,s^{-1}}$5 ms−1 receives an impulse of $6\ \mathrm{N\,s}$6 Ns in the direction of motion, and then a second impulse of $14\ \mathrm{N\,s}$14 Ns directly opposing the motion. Find the final velocity.

Apply the impulse–momentum principle to each stage in turn, carrying the velocity forward. Take the original direction as positive.

$$\begin{aligned} \text{After 1st impulse:}\quad & 6 = 2v_1 - 2(5) \;\Rightarrow\; 2v_1 = 16 \;\Rightarrow\; v_1 = 8\ \mathrm{m\,s^{-1}}\\ \text{After 2nd impulse:}\quad & -14 = 2v_2 - 2(8) \;\Rightarrow\; 2v_2 = 2 \;\Rightarrow\; v_2 = 1\ \mathrm{m\,s^{-1}}. \end{aligned}$$After 1st impulse:After 2nd impulse:​6=2v1​−2(5)⇒2v1​=16⇒v1​=8 ms−1−14=2v2​−2(8)⇒2v2​=2⇒v2​=1 ms−1.​

The particle ends moving at $1\ \mathrm{m\,s^{-1}}$1 ms−1 in the original direction. Mark scheme: M1 apply $J = mv - mu$J=mv−mu to the first stage; A1 $v_1 = 8$v1​=8; M1 apply it again with the second (negative) impulse, using $v_1$v1​ as the new initial velocity; A1 $v_2 = 1$v2​=1.

Alternatively — and this is worth seeing — because impulses simply add as vectors, the net impulse is $6 - 14 = -8\ \mathrm{N\,s}$6−14=−8 Ns, so $-8 = 2v_2 - 2(5) \Rightarrow v_2 = 1\ \mathrm{m\,s^{-1}}$−8=2v2​−2(5)⇒v2​=1 ms−1 in one step. The two impulses could be combined because impulse is a vector quantity that obeys ordinary addition.

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4 Specimen-style exam question

(Specimen-style.) A particle of mass $2\ \mathrm{kg}$2 kg has velocity $(5\vec i - 3\vec j)\ \mathrm{m\,s^{-1}}$(5i−3j​) ms−1. It receives an impulse $(-4\vec i + 6\vec j)\ \mathrm{N\,s}$(−4i+6j​) Ns. (a) Find the new velocity. (b) Find the speed before and after, and comment.

(a) Using $\vec J = m\vec v - m\vec u \Rightarrow m\vec v = m\vec u + \vec J$J=mv−mu⇒mv=mu+J:

$$\begin{aligned} m\vec v &= 2(5\vec i - 3\vec j) + (-4\vec i + 6\vec j) = (10\vec i - 6\vec j) + (-4\vec i + 6\vec j) = 6\vec i \\ \vec v &= \tfrac{1}{2}(6\vec i) = 3\vec i\ \mathrm{m\,s^{-1}}. \end{aligned}$$mvv​=2(5i−3j​)+(−4i+6j​)=(10i−6j​)+(−4i+6j​)=6i=21​(6i)=3i ms−1.​

(b) Before: $|\vec u| = \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83\ \mathrm{m\,s^{-1}}$∣u∣=52+32​=34​≈5.83 ms−1. After: $|\vec v| = 3\ \mathrm{m\,s^{-1}}$∣v∣=3 ms−1. The impulse cancelled the $\vec j$j​-component entirely and reduced the speed, so it acted partly to oppose the motion.

Extension — magnitude and direction of the impulse. The impulse itself is $\vec J = (-4\vec i + 6\vec j)\ \mathrm{N\,s}$J=(−4i+6j​) Ns, with magnitude $|\vec J| = \sqrt{(-4)^2 + 6^2} = \sqrt{52} = 2\sqrt{13} \approx 7.21\ \mathrm{N\,s}$∣J∣=(−4)2+62​=52​=213​≈7.21 Ns, at angle $\arctan\!\dfrac{6}{-4}$arctan−46​ (second quadrant) $\approx 123.7^\circ$≈123.7∘ to $\vec i$i. Notice we never needed the contact time or the peak force to answer any part — exactly the economy the impulse–momentum principle buys.

Choosing your method

This question could in principle be done with $\vec F = m\vec a$F=ma, but only if you knew the force and the time over which it acted — which you do not. Whenever a question gives you an impulse, a force–time graph, or velocities before and after an interaction, reach for $\vec J = m\vec v - m\vec u$J=mv−mu. Save $\vec F = m\vec a$F=ma for problems where the force and acceleration are the unknowns of interest at a single instant.

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5 Synoptic links

• Vectors (A-Level Maths): $\vec p$p​ and $\vec J$J obey the same component algebra as displacement; magnitude is always $\sqrt{p_x^2 + p_y^2}$px2​+py2​​.
• Calculus (A-Level Maths): impulse is $\int F\,\mathrm{d}t$∫Fdt and equals the area under a force–time graph — connect this to definite integration as area.
• Conservation of momentum (next lesson): summing impulse–momentum over an interacting pair, with Newton's third law making the internal impulses cancel, gives total-momentum conservation.
• Work–energy (later): the space-integral $\int F\,\mathrm{d}x$∫Fdx gives work and energy; the time-integral $\int F\,\mathrm{d}t$∫Fdt gives impulse and momentum. Knowing which integral the question wants is a key Further-Mechanics skill.

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6 Mark-scheme literacy

• "Find the impulse" — examiners want $J = mv - mu$J=mv−mu written down with signed velocities before substitution; the method mark hangs on the signed form.
• "State your positive direction" or an unstated convention — declare it; markers award follow-through only if your signs are internally consistent.
• A vector impulse must be given as a vector (or magnitude and direction); a bare magnitude when direction is asked drops the final A1.
• Units $\mathrm{N\,s}$Ns (impulse) versus $\mathrm{kg\,m\,s^{-1}}$kgms−1 (momentum) are interchangeable and never penalised, but a missing unit can cost the accuracy mark.
• "Find the magnitude of the impulse" — give a positive number with units; the sign you used during working tells you the direction, which you should then state in words ("to the left", "back the way it came").
• Follow-through (ft) marks: if you make an early arithmetic slip but apply the correct method thereafter, examiners award the later method marks on your (wrong) value — so always show the method explicitly even when unsure of a number.
• "Hence" signals that you must use the previous result; a fresh independent method, even if correct, may not earn the "hence" marks.

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7 Grade-band model answers

Question. A force–time graph for a $0.2\ \mathrm{kg}$0.2 kg particle (initially at rest) rises linearly from $0$0 to $10\ \mathrm{N}$10 N over $0 \le t \le 4\ \mathrm{s}$0≤t≤4 s. Find the final speed.

Mid-band. "Area $= \tfrac12 \times 4 \times 10 = 20$=21​×4×10=20. $20 = 0.2 v$20=0.2v, so $v = 100$v=100." (Correct value, but no statement that area = impulse, no $J=mv-mu$J=mv−mu, no units.)

Stronger. "Impulse = area under the graph $= \tfrac12(4)(10) = 20\ \mathrm{N\,s}$=21​(4)(10)=20 Ns. Then $J = mv - mu$J=mv−mu: $20 = 0.2v - 0$20=0.2v−0, so $v = 100\ \mathrm{m\,s^{-1}}$v=100 ms−1."

Top-band. "The impulse is the area under the force–time graph (a triangle): $J = \tfrac12 \times 4 \times 10 = 20\ \mathrm{N\,s}$J=21​×4×10=20 Ns. By the impulse–momentum principle $J = mv - mu$J=mv−mu; the particle starts from rest so $u=0$u=0: $20 = 0.2v \Rightarrow v = 100\ \mathrm{m\,s^{-1}}$20=0.2v⇒v=100 ms−1 in the direction of the force."

Examiner-style commentary. All three reach $100\ \mathrm{m\,s^{-1}}$100 ms−1, but only the top answer earns every communication mark: it names the principle, justifies area = impulse, carries units throughout, and states direction. In AO2-weighted Paper 3 marking, the reasoning is assessed as much as the number.

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8 Common misconceptions

1. "Impulse is a force." No — impulse is force accumulated over time, with units $\mathrm{N\,s}$Ns, not $\mathrm{N}$N. Dimensionally $[\,\mathrm{N\,s}\,] = [\,\mathrm{kg\,m\,s^{-1}}\,]$[Ns]=[kgms−1], which is momentum, not force.
2. "Momentum is a scalar like energy." Momentum is a vector; direction (sign) is part of the answer. Two equal masses moving at the same speed in opposite directions have zero total momentum but plenty of total kinetic energy — the contrast is worth holding in mind.
3. Forgetting to reverse the sign on a rebound. A ball returning has $v$v and $u$u of opposite sign — this is the largest single source of error. If a $0.2\ \mathrm{kg}$0.2 kg ball arrives at $10$10 and leaves at $8$8 the other way, $J = 0.2(-8) - 0.2(10) = -3.6\ \mathrm{N\,s}$J=0.2(−8)−0.2(10)=−3.6 Ns, not $0.2(8) - 0.2(10) = -0.4$0.2(8)−0.2(10)=−0.4.
4. "Heavier means more momentum." Only if the velocities match; a light fast object can out-momentum a heavy slow one — a $0.05\ \mathrm{kg}$0.05 kg bullet at $400\ \mathrm{m\,s^{-1}}$400 ms−1 ($20\ \mathrm{kg\,m\,s^{-1}}$20 kgms−1) beats a $2\ \mathrm{kg}$2 kg ball at $5\ \mathrm{m\,s^{-1}}$5 ms−1 ($10\ \mathrm{kg\,m\,s^{-1}}$10 kgms−1).
5. Confusing $\int F\,\mathrm{d}t$∫Fdt with $\int F\,\mathrm{d}x$∫Fdx. The first is impulse (momentum); the second is work (energy). The differential — $\mathrm{d}t$dt versus $\mathrm{d}x$dx — tells you which.
6. Treating average force as the peak force. $F_{\text{avg}} = J/\Delta t$Favg​=J/Δt is a mean over the contact; the instantaneous peak is larger, often much larger.
7. Adding momentum magnitudes in 2-D. You must add components, then take the magnitude of the resultant, not add magnitudes. $|\,(3\vec i) + (4\vec j)\,| = 5$∣(3i)+(4j​)∣=5, not $7$7.

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9 Common errors

• Sign on a reversed velocity. Dropping the negative when a velocity points in the negative direction — the single commonest slip.
• Unit mismatch. A contact time given in milliseconds left as a bare number; convert $20\ \mathrm{ms} = 0.02\ \mathrm{s}$20 ms=0.02 s first.
• Form of the answer. Quoting a vector answer as a magnitude only (or vice versa) when the question specifies which it wants.
• Graph misread. Reading a force–time graph as if the gradient were the impulse — it is the area.
• Integration limits. In a variable-force integral, forgetting to substitute the lower limit (a particularly easy slip when the lower limit is $0$0 but the integrand has a constant term).
• Mass omitted. Writing $J = v - u$J=v−u instead of $J = mv - mu$J=mv−mu — dimensionally wrong and a guaranteed lost mark.

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10 Going further (stretch)

For a force $\vec F(t)$F(t) with no closed-form antiderivative, the impulse is still $\int \vec F\,\mathrm{d}t$∫Fdt — exactly the setting where numerical integration (trapezium rule, Simpson's rule) earns its keep in applied problems. A STEP-flavoured extension: a particle of mass $m$m is subject to a resistive force $F = -kv$F=−kv. Then $m\dfrac{\mathrm{d}v}{\mathrm{d}t} = -kv$mdtdv​=−kv, a separable first-order ODE giving $v = u\,e^{-kt/m}$v=ue−kt/m. The total impulse delivered as the particle decays from $u$u to rest is

$J = \int_0^{\infty} -kv\,\mathrm{d}t = -k\int_0^{\infty} u\,e^{-kt/m}\,\mathrm{d}t = -k \cdot u \cdot \frac{m}{k} = -mu,$J=∫0∞​−kvdt=−k∫0∞​ue−kt/mdt=−k⋅u⋅km​=−mu,

which is precisely $\Delta p = 0 - mu$Δp=0−mu. The impulse–momentum principle and the differential equation give the same answer — a satisfying consistency check that links Mechanics to the Further-Pure improper-integral toolkit.

A second stretch — variable mass. The truly fundamental form of Newton's second law is $\vec F = \dfrac{\mathrm{d}\vec p}{\mathrm{d}t} = \dfrac{\mathrm{d}(m\vec v)}{\mathrm{d}t}$F=dtdp​​=dtd(mv)​. When the mass is constant this is the familiar $m\vec a$ma; but for a rocket, which ejects mass, the product rule gives

$\vec F = m\frac{\mathrm{d}\vec v}{\mathrm{d}t} + \vec v\frac{\mathrm{d}m}{\mathrm{d}t},$F=mdtdv​+vdtdm​,

and analysing this leads to the celebrated Tsiolkovsky rocket equation $\Delta v = v_e \ln\!\dfrac{m_0}{m_f}$Δv=ve​lnmf​m0​​, where $v_e$ve​ is the exhaust speed. The whole of rocketry rests on the momentum principle you have just met — the engine throws mass backwards, and conservation of momentum throws the rocket forwards. Deriving the rocket equation from $\dfrac{\mathrm{d}p}{\mathrm{d}t}$dtdp​ and a separable integral is a classic STEP-level synthesis of Mechanics and Further-Pure calculus.

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11 Additional A* practice

1. A pile-driver of mass $500\ \mathrm{kg}$500 kg hits a pile at $6\ \mathrm{m\,s^{-1}}$6 ms−1 and is brought to rest in $0.04\ \mathrm{s}$0.04 s. Find the average force.
2. A $0.06\ \mathrm{kg}$0.06 kg tennis ball arrives horizontally at $25\ \mathrm{m\,s^{-1}}$25 ms−1 and leaves at $30\ \mathrm{m\,s^{-1}}$30 ms−1 in the opposite direction. Find the impulse.
3. A force $\vec F = (6t\,\vec i + 2\,\vec j)\ \mathrm{N}$F=(6ti+2j​) N acts on a $3\ \mathrm{kg}$3 kg particle for $0\le t\le 2\ \mathrm{s}$0≤t≤2 s. Find the impulse vector and its magnitude.
4. A particle of mass $4\ \mathrm{kg}$4 kg changes velocity from $(2\vec i + \vec j)$(2i+j​) to $(-\vec i + 4\vec j)\ \mathrm{m\,s^{-1}}$(−i+4j​) ms−1. Find the impulse and the angle it makes with $\vec i$i.
5. A force $F = (12 - 6t)\ \mathrm{N}$F=(12−6t) N acts on a $2\ \mathrm{kg}$2 kg particle, initially at rest, for $0 \le t \le 2\ \mathrm{s}$0≤t≤2 s. Find the impulse over the interval and the velocity at $t = 2$t=2. Comment on the sign of $F$F near $t = 2$t=2.

Worked answers.

1. $F_{\text{avg}} = \dfrac{\Delta p}{\Delta t} = \dfrac{500(0)-500(6)}{0.04} = \dfrac{-3000}{0.04} = -75000\ \mathrm{N}$Favg​=ΔtΔp​=0.04500(0)−500(6)​=0.04−3000​=−75000 N; magnitude $75\ \mathrm{kN}$75 kN, directed to oppose the motion.
2. Take arrival positive: $J = 0.06(-30) - 0.06(25) = -1.8 - 1.5 = -3.3\ \mathrm{N\,s}$J=0.06(−30)−0.06(25)=−1.8−1.5=−3.3 Ns; i.e. $3.3\ \mathrm{N\,s}$3.3 Ns back the way it came.
3. $\vec J = \displaystyle\int_0^2 (6t\,\vec i + 2\,\vec j)\,\mathrm{d}t = \big[3t^2\,\vec i + 2t\,\vec j\big]_0^2 = 12\vec i + 4\vec j\ \mathrm{N\,s}$J=∫02​(6ti+2j​)dt=[3t2i+2tj​]02​=12i+4j​ Ns; magnitude $\sqrt{12^2 + 4^2} = \sqrt{160} = 4\sqrt{10} \approx 12.6\ \mathrm{N\,s}$122+42​=160​=410​≈12.6 Ns.
4. $\vec J = 4\big[(-\vec i + 4\vec j) - (2\vec i + \vec j)\big] = 4(-3\vec i + 3\vec j) = -12\vec i + 12\vec j\ \mathrm{N\,s}$J=4[(−i+4j​)−(2i+j​)]=4(−3i+3j​)=−12i+12j​ Ns; the angle to $\vec i$i is $\arctan\!\dfrac{12}{-12}$arctan−1212​, in the second quadrant, $= 135^{\circ}$=135∘.
5. $J = \displaystyle\int_0^2 (12 - 6t)\,\mathrm{d}t = \big[12t - 3t^2\big]_0^2 = 24 - 12 = 12\ \mathrm{N\,s}$J=∫02​(12−6t)dt=[12t−3t2]02​=24−12=12 Ns. Then $12 = 2v - 0 \Rightarrow v = 6\ \mathrm{m\,s^{-1}}$12=2v−0⇒v=6 ms−1. For $t > 2$t>2 the force $12 - 6t$12−6t is negative (it changes sign at $t = 2$t=2), so beyond this interval it would decelerate the particle — the impulse over $0\le t\le 2$0≤t≤2 captures only the accelerating phase.

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12 Board alignment

Aligned to AQA 7367 Paper 3 Mechanics (7367/3M), momentum and impulse strand. The treatment is essentially identical on Edexcel (Further Mechanics 1) and OCR / OCR (MEI) Mechanics options; only notation for the modulus and the choice of $g$g differ between boards.

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13 Visual summary

$$\boxed{\;\vec p = m\vec v \qquad \vec J = \int \vec F\,\mathrm{d}t = \vec F\,\Delta t \qquad \vec J = m\vec v - m\vec u \qquad F_{\text{avg}} = \frac{\Delta p}{\Delta t}\;}$$p​=mvJ=∫Fdt=FΔtJ=mv−muFavg​=ΔtΔp​​

The impulse is the shaded area under the force–time graph:

xychart-beta
    title "Impulse = area under F–t graph"
    x-axis "time t (s)" 0 --> 4
    y-axis "force F (N)" 0 --> 12
    bar [2.5, 5, 7.5, 10]
    line [2.5, 5, 7.5, 10]

Concept	Formula	Unit
Momentum	$\vec p = m\vec v$p​=mv	$\mathrm{kg\,m\,s^{-1}}$kgms−1
Impulse (constant)	$\vec J = \vec F\,\Delta t$J=FΔt	$\mathrm{N\,s}$Ns
Impulse (variable)	$\vec J = \int F\,\mathrm{d}t$J=∫Fdt	$\mathrm{N\,s}$Ns
Impulse–momentum	$\vec J = m\vec v - m\vec u$J=mv−mu	—
Average force	$F_{\text{avg}} = \Delta p/\Delta t$Favg​=Δp/Δt	$\mathrm{N}$N

Recap. Momentum is mass × velocity (a vector). Impulse is the time-integral of force, equal to the change in momentum. Declare a positive direction, keep signs honest, and you can solve any collision without ever knowing the peak force.