Momentum — p = mv

Spec mapping: OCR H556 Module 3.3 — Work, energy and power; momentum (definition of linear momentum $\vec{p} = m\vec{v}$p​=mv as a vector; units; the relationship to kinetic energy; foundations for conservation of momentum in collisions and explosions). Refer to the official OCR H556 specification document for exact wording.

Linear momentum is arguably the single most important quantity in A-Level mechanics. It sits at the heart of Newton's Second Law ($\vec{F} = d\vec{p}/dt$F=dp​/dt), underpins the conservation principle that dominates every collision and explosion problem (lessons 6-10), and reappears later in circular motion, orbital mechanics, and quantum physics (where $\vec{p}$p​ becomes the canonical conjugate of position and forms the basis of the de Broglie wavelength $\lambda = h/p$λ=h/p).

This lesson defines momentum carefully as a vector, works through its calculation habits in 1-D and 2-D, derives the relationship $p^2 = 2mE_k$p2=2mEk​ between momentum and kinetic energy (a frequent OCR A* discriminator), and establishes the sign-convention discipline you'll need for collisions. By the end you should be able to compute momentum, change of momentum, and KE-from-momentum without thinking.

---

Definition

Linear momentum is the product of an object's mass and its velocity:

$\vec{p} = m\vec{v}$p​=mv

• $\vec{p}$p​ is momentum, a vector quantity.
• $m$m is mass in kilograms (kg).
• $\vec{v}$v is velocity in metres per second (m s$^{-1}$−1).
• SI units: kg m s$^{-1}$−1 (equivalently N s — see units section below).

Because velocity is a vector, momentum is a vector: it has both magnitude and direction, and the direction of $\vec{p}$p​ matches the direction of $\vec{v}$v. A lorry and a bicycle can have momenta of equal magnitude if the bicycle is moving very fast and the lorry very slowly — but the comparison is only sensible if we specify the direction. Two equal-magnitude momenta in opposite directions are not the same momentum; their vector sum is zero, not double.

The vector nature is the single most-tested feature of momentum at A-Level. The OCR mark schemes routinely award a mark for the wording "vector quantity" or for explicit treatment of direction in numerical answers.

---

Why Momentum is a Vector — The Model Answer

A common OCR question runs: "State what is meant by linear momentum and explain why it is a vector quantity."

Model A-level answer: "Linear momentum is the product of an object's mass and its velocity, $\vec{p} = m\vec{v}$p​=mv. It is a vector because velocity is a vector, and multiplying a vector ($\vec{v}$v) by a scalar (mass $m$m) yields another vector with the same direction. Momentum therefore has both magnitude $|\vec{p}| = mv$∣p​∣=mv and direction — that of $\vec{v}$v."

Omitting either "vector" or "direction" loses marks. Stating $p = mv$p=mv without explicit reference to the vector nature is incomplete.

---

Worked Example 1 — Bullet vs Lorry

Compare the magnitudes of momentum of:

(a) a $0.010$0.010 kg bullet travelling at $400$400 m s$^{-1}$−1;

(b) a $20\,000$20000 kg lorry moving at $0.50$0.50 m s$^{-1}$−1.

• (a) $p_{\text{bullet}} = 0.010 \times 400 = 4.0$pbullet​=0.010×400=4.0 kg m s$^{-1}$−1.
• (b) $p_{\text{lorry}} = 20\,000 \times 0.50 = 10\,000$plorry​=20000×0.50=10000 kg m s$^{-1}$−1.

The lorry has 2500 times the bullet's momentum even though the bullet travels 800 times faster. The "stopping power" of a slow lorry far exceeds that of a fast bullet — a sobering reminder of why heavy goods vehicles need very long braking distances and dominate the fatality statistics in traffic accidents.

This worked example illustrates the linear nature of momentum in mass and velocity: doubling mass doubles momentum, doubling speed also doubles momentum. Kinetic energy, by contrast, scales quadratically with speed ($E_k = \tfrac{1}{2}mv^2$Ek​=21​mv2) — see the "Momentum vs Kinetic Energy" section below.

---

Worked Example 2 — Car on a Bend (Change of Direction)

A $1200$1200 kg car travels at $25$25 m s$^{-1}$−1 due East. After $10$10 s of driving around a bend, it is travelling at $25$25 m s$^{-1}$−1 due North (same speed, different direction). Find the magnitude and direction of the change in momentum.

Beginners look at "same speed" and answer "no change". But momentum is a vector, so direction matters.

• $\vec{p}_1 = 1200 \times 25 = 30\,000$p​1​=1200×25=30000 kg m s$^{-1}$−1 due East.
• $\vec{p}_2 = 1200 \times 25 = 30\,000$p​2​=1200×25=30000 kg m s$^{-1}$−1 due North.
• $\Delta\vec{p} = \vec{p}_2 - \vec{p}_1$Δp​=p​2​−p​1​ (vector subtraction).

Resolving into components (East as $+x$+x, North as $+y$+y):

• $\Delta p_x = 0 - 30\,000 = -30\,000$Δpx​=0−30000=−30000 kg m s$^{-1}$−1.
• $\Delta p_y = 30\,000 - 0 = +30\,000$Δpy​=30000−0=+30000 kg m s$^{-1}$−1.

Magnitude: $|\Delta\vec{p}| = \sqrt{30\,000^2 + 30\,000^2} = 30\,000\sqrt{2} \approx 4.24 \times 10^4$∣Δp​∣=300002+300002​=300002​≈4.24×104 kg m s$^{-1}$−1.

Direction: $\arctan(30\,000/30\,000) = 45°$arctan(30000/30000)=45° measured from $-x$−x toward $+y$+y — i.e. north-west.

By Newton 2, a resultant force acted on the car during the bend (friction from tyres providing the centripetal force). The average force: $\vec{F}_{\text{avg}} = \Delta\vec{p}/\Delta t = 4.24 \times 10^4 / 10 = 4240$Favg​=Δp​/Δt=4.24×104/10=4240 N north-west — roughly the weight of a small car, applied sideways through the tyres.

This example is critical: a change of direction is a change of momentum, even at constant speed. Circular motion at constant speed is non-zero momentum change. Centripetal force exists precisely because circular motion demands a continuous change of momentum direction.

---

Worked Example 3 — Ball Bouncing Off a Wall (Sign Convention)

A $0.20$0.20 kg ball travels horizontally at $5.0$5.0 m s$^{-1}$−1, hits a wall, and rebounds at $4.0$4.0 m s$^{-1}$−1 in the opposite direction. Find the change in momentum.

Take the initial direction (toward the wall) as positive.

• $p_{\text{initial}} = 0.20 \times (+5.0) = +1.0$pinitial​=0.20×(+5.0)=+1.0 kg m s$^{-1}$−1.
• $p_{\text{final}} = 0.20 \times (-4.0) = -0.80$pfinal​=0.20×(−4.0)=−0.80 kg m s$^{-1}$−1 (negative because rebound is in the opposite direction).
• $\Delta p = p_{\text{final}} - p_{\text{initial}} = -0.80 - 1.0 = -1.80$Δp=pfinal​−pinitial​=−0.80−1.0=−1.80 kg m s$^{-1}$−1.

The negative sign tells you the change is in the opposite direction to the initial motion — i.e. away from the wall, back toward the source. Magnitude: $1.80$1.80 kg m s$^{-1}$−1.

Note that the magnitude of the change ($1.80$1.80 kg m s$^{-1}$−1) is greater than either the initial or final momentum alone — because the direction reverses. A classic OCR error is to compute $\Delta p = 0.80 - 1.0 = -0.20$Δp=0.80−1.0=−0.20 kg m s$^{-1}$−1 by forgetting the negative sign on the final velocity. Always define and stick to a positive direction; substitute signed values.

The change is twice as large as you'd get from a "ball that stops dead" scenario ($p_{\text{final}} = 0$pfinal​=0, giving $\Delta p = -1.0$Δp=−1.0 kg m s$^{-1}$−1): the rebound contribution adds another $0.80$0.80 kg m s$^{-1}$−1 magnitude of momentum change. This is why rebounding from a hard wall delivers more impulse — and feels more violent — than landing on a soft mat.

---

Momentum and Newton's Second Law Revisited

From lesson 2:

$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$Fnet​=dtdp​​

So the resultant force on a body equals the rate of change of its momentum. Rearranging (and writing for constant force):

$\Delta \vec{p} = \vec{F}\,\Delta t$Δp​=FΔt

This product is called the impulse, the subject of lesson 5. For variable forces, integrating:

$\Delta\vec{p} = \int_{t_1}^{t_2} \vec{F}(t)\,dt$Δp​=∫t1​t2​​F(t)dt

which is the area under the force-time graph. A-Level candidates are expected to interpret this graphically (counting squares, computing triangle/rectangle areas) even if they do not formally evaluate integrals.

The deep connection: force is the rate of momentum change. This single statement is the general Newton 2, valid for variable-mass systems too. The familiar $F = ma$F=ma is the constant-mass special case.

---

Units — kg m s$^{-1}$−1 vs N s

Momentum has two equivalent unit forms:

$\text{kg m s}^{-1} = (\text{kg m s}^{-2}) \times \text{s} = \text{N} \times \text{s} = \text{N s}$kg m s−1=(kg m s−2)×s=N×s=N s

Both are correct and mark schemes accept either. The "N s" form makes the connection to impulse $J = F\Delta t$J=FΔt explicit; the "kg m s$^{-1}$−1" form makes the connection to $\vec{p} = m\vec{v}$p​=mv explicit. Use whichever fits the question's framing.

---

Momentum as a Vector — Component Form

In two dimensions, resolve velocity into components and apply $\vec{p} = m\vec{v}$p​=mv to each:

$p_x = mv_x \qquad p_y = mv_y$px​=mvx​py​=mvy​

Total magnitude: $|\vec{p}| = \sqrt{p_x^2 + p_y^2}$∣p​∣=px2​+py2​​. Direction: $\theta = \arctan(p_y/p_x)$θ=arctan(py​/px​).

This component decomposition is the workhorse for 2-D collision problems (lesson 9) and projectile-motion momentum analyses.

Worked Example 4 — Projectile Momentum

A $0.50$0.50 kg ball is projected with initial velocity $20$20 m s$^{-1}$−1 at $60°$60° above the horizontal.

• $v_x = 20 \cos 60° = 10.0$vx​=20cos60°=10.0 m s$^{-1}$−1.
• $v_y = 20 \sin 60° = 17.32$vy​=20sin60°=17.32 m s$^{-1}$−1.
• $p_x = 0.50 \times 10.0 = 5.0$px​=0.50×10.0=5.0 kg m s$^{-1}$−1.
• $p_y = 0.50 \times 17.32 = 8.66$py​=0.50×17.32=8.66 kg m s$^{-1}$−1.
• $|\vec{p}| = \sqrt{5.0^2 + 8.66^2} = \sqrt{25 + 75} = \sqrt{100} = 10$∣p​∣=5.02+8.662​=25+75​=100​=10 kg m s$^{-1}$−1.

which equals $m|\vec{v}| = 0.50 \times 20 = 10$m∣v∣=0.50×20=10 kg m s$^{-1}$−1 — exactly as expected (vector magnitude unchanged by decomposition).

During the flight (neglecting drag), $p_x$px​ remains constant (no horizontal force) while $p_y$py​ changes with time according to $p_y(t) = p_{y,0} - mgt$py​(t)=py,0​−mgt (gravity provides the only force, downward). At the peak of the trajectory $p_y = 0$py​=0; at the landing moment (same height as launch) $p_y$py​ is the negative of the initial value, i.e. $-8.66$−8.66 kg m s$^{-1}$−1.

So the total momentum change over the flight is:

$\Delta\vec{p}_{\text{flight}} = (\Delta p_x, \Delta p_y) = (0, -17.32 \text{ kg m s}^{-1})$Δp​flight​=(Δpx​,Δpy​)=(0,−17.32 kg m s−1)

— purely vertical, magnitude $17.32$17.32 kg m s$^{-1}$−1 downward. This equals $-mgT$−mgT where $T$T is the flight time, an expression of Newton 2 + projectile kinematics.

---

Momentum and Kinetic Energy — The $p^2 = 2mE_k$p2=2mEk​ Relationship

Momentum and kinetic energy are closely related but distinct:

$E_k = \tfrac{1}{2}mv^2 = \frac{(mv)^2}{2m} = \frac{p^2}{2m}$Ek​=21​mv2=2m(mv)2​=2mp2​

Equivalently:

$p = \sqrt{2m E_k}$p=2mEk​​

This is an OCR Top-band discriminator. The two key insights:

1. They are not proportional. Doubling speed doubles $p$p but quadruples $E_k$Ek​.
2. At equal momentum, lighter bodies have more KE: $E_k = p^2/(2m)$Ek​=p2/(2m), so for fixed $p$p, smaller $m$m gives larger $E_k$Ek​.
3. At equal KE, heavier bodies have more momentum: $p = \sqrt{2mE_k}$p=2mEk​​, so for fixed $E_k$Ek​, larger $m$m gives larger $p$p.

In lessons 7-8 the distinction becomes critical when we classify collisions as elastic (KE conserved) or inelastic (KE lost), with momentum always conserved.

Quantity	Formula	Type	Units
Momentum	$\vec{p} = m\vec{v}$p​=mv	Vector	kg m s$^{-1}$−1 or N s
Kinetic energy	$E_k = \tfrac{1}{2}mv^2 = p^2/(2m)$Ek​=21​mv2=p2/(2m)	Scalar	J
Bridge	$p = \sqrt{2mE_k}$p=2mEk​​	—	—

---

Worked Example 5 — Same KE, Different Mass, Different Momentum

A $2.0$2.0 kg object and a $0.50$0.50 kg object each have a kinetic energy of $100$100 J. Find their momenta and compare.

Using $p = \sqrt{2m E_k}$p=2mEk​​:

• $2.0$2.0 kg object: $p = \sqrt{2 \times 2.0 \times 100} = \sqrt{400} = 20$p=2×2.0×100​=400​=20 kg m s$^{-1}$−1.
• $0.50$0.50 kg object: $p = \sqrt{2 \times 0.50 \times 100} = \sqrt{100} = 10$p=2×0.50×100​=100​=10 kg m s$^{-1}$−1.

Equal kinetic energies give different momenta. The heavier object has twice the momentum. Conversely, at equal momentum the heavier object would have less KE (a lorry and a bullet with the same $p$p have wildly different $E_k$Ek​: lorry is small, bullet is large).

Numerical check: velocities are $v = p/m$v=p/m. For the heavier: $v = 20/2.0 = 10$v=20/2.0=10 m s$^{-1}$−1; KE check $\tfrac{1}{2} \times 2.0 \times 10^2 = 100$21​×2.0×102=100 J. ✓. For the lighter: $v = 10/0.50 = 20$v=10/0.50=20 m s$^{-1}$−1; KE check $\tfrac{1}{2} \times 0.50 \times 400 = 100$21​×0.50×400=100 J. ✓.

The lighter object goes twice as fast for the same KE — exactly because KE scales as $v^2$v2, so doubling speed at quarter mass keeps KE constant.

---

Worked Example 6 — Hydrogen vs Helium Atoms at Equal Temperature

A hydrogen molecule ($m \approx 3.3 \times 10^{-27}$m≈3.3×10−27 kg) and a helium atom ($m \approx 6.6 \times 10^{-27}$m≈6.6×10−27 kg) are in thermal equilibrium at $300$300 K. From the equipartition theorem, both have the same average translational KE: $\bar{E}_k = \tfrac{3}{2}k_B T \approx 6.2 \times 10^{-21}$Eˉk​=23​kB​T≈6.2×10−21 J. Compare their average momenta.

$\bar{p} = \sqrt{2m\bar{E}_k}$pˉ​=2mEˉk​​:

• H$_2$2​: $\bar{p} = \sqrt{2 \times 3.3 \times 10^{-27} \times 6.2 \times 10^{-21}} \approx 2.0 \times 10^{-24}$pˉ​=2×3.3×10−27×6.2×10−21​≈2.0×10−24 kg m s$^{-1}$−1.
• He: $\bar{p} = \sqrt{2 \times 6.6 \times 10^{-27} \times 6.2 \times 10^{-21}} \approx 2.9 \times 10^{-24}$pˉ​=2×6.6×10−27×6.2×10−21​≈2.9×10−24 kg m s$^{-1}$−1.

He has $\sqrt{2}$2​ times the momentum of H$_2$2​ at the same KE — about $40\%$40% more. This is a preview of A2 thermodynamics (Module 5.1), where the rms speed is $\sqrt{3k_BT/m}$3kB​T/m​ and lighter molecules diffuse faster (which is why hydrogen escapes Earth's atmosphere over geological time while heavier gases don't). The $p^2 = 2mE_k$p2=2mEk​ identity is the kinematic backbone of statistical mechanics.

---

Graphical Interpretation

Momentum-time graph: plot $p$p on $y$y-axis against $t$t on $x$x-axis. The gradient is the resultant force, by Newton 2 ($F = dp/dt$F=dp/dt). A constant force gives a straight line; a varying force gives a curve from which $F(t)$F(t) can be read off the slope.

Force-time graph: plot $F$F on $y$y-axis against $t$t on $x$x-axis. The area under the curve is the change in momentum (impulse), by $\Delta p = \int F\,dt$Δp=∫Fdt. Counting squares or computing geometric areas is the practical technique — lesson 5 elaborates with worked examples.

These two graphical perspectives are dual: the $p$p-$t$t graph encodes force as its gradient; the $F$F-$t$t graph encodes momentum change as its area. They are two faces of the same physics ($F = dp/dt$F=dp/dt, $\Delta p = \int F\,dt$Δp=∫Fdt).

---

Sign Conventions — Always Define Positive First

Before any $\Delta p$Δp calculation, write on your page something like:

"Take rightward as positive."

— or "take initial direction as positive", "take upward as positive", or whatever fits the geometry. Then substitute signed values throughout. This single discipline eliminates 90% of sign errors in 1-D collision and rebound questions.

For 2-D problems, define two perpendicular positive axes (typically $+x$+x and $+y$+y), then resolve every velocity and force into signed components.

Common pitfall: changing the sign convention mid-problem. If "$+x$+x = east" at the start, keep it that way throughout — don't switch to "$+x$+x = direction of motion" after a collision flips the direction of motion. The convention is fixed to space, not to the moving body.

---

Synoptic Links