Linear momentum p = mv: Worked examples — Physics deep dive

This page is the practice arena for Linear momentum $p=mv$. The parent note told you what momentum is; here we drill every kind of question it can ask you — every sign, both directions, zero and infinite limits, a real-world word problem, and an exam-style trap. Nothing new is assumed: if a symbol shows up, we re-earn it right here.

Recall The one formula everything rests on

$p = m v$ $m$ is mass (always a positive number, measured in kilograms, $kg$ ). $v$ is velocity — speed with a direction. The little arrow on $p$ means momentum is a vector: it has a size and points somewhere. Its unit is $kg\cdotm/s$ .

The scenario matrix

Before solving anything, let's list every distinct kind of situation this topic can throw at you. A worked example below hits each cell.

#	Cell (case class)	What makes it different	Example that covers it
A	Plain 1D, one object, positive velocity	The bare formula, no signs to worry about	Ex 1
B	1D negative velocity	Momentum can be negative — direction encoded as a sign	Ex 2
C	1D, two objects, opposite directions	Vector sum with cancellation	Ex 3
D	Zero / degenerate input	Object at rest, or massless idealisation — what happens to $p$ ?	Ex 4
E	2D, perpendicular components	Adding at right angles → Pythagoras + angle	Ex 5
F	2D, general angle (both components signed)	A component points the "wrong" way	Ex 6
G	Limiting behaviour	Same $p$ , mass $\to$ huge / speed $\to$ huge	Ex 7
H	Real-world word problem	Translate messy English into $m$ and $v$	Ex 8
I	Exam-style twist (units + $K E$ link)	Mixed units, and $K E = p^{2} /2 m$	Ex 9

We cover cells A–I with nine examples. Let's go.

Setting up our picture of direction

Everything below uses a sign convention: pick one direction as positive. We'll always draw right = positive $(+)$ and left = negative $(-)$ . This single choice turns "direction" into "a plus or minus sign," which is the whole trick that makes 1D momentum easy.

Cell A — Plain 1D, positive velocity

Worked example Ex 1 · A rolling bowling ball

A $7.0 kg$ bowling ball rolls to the right at $4.0 m/s$ . Find its momentum.

Forecast: Guess the number and its sign before reading on. Bigger than 10? Positive or negative?

Identify the pieces. $m = 7.0 kg$ , $v = + 4.0 m/s$ (right = positive). Why this step? Momentum needs exactly two inputs, mass and velocity — pin them down first.
Apply $p = m v$ . $p = (7.0) (+ 4.0) = + 28 kg\cdotm/s$ Why this step? Direct use of the definition; nothing else is needed for one object in 1D.
Read the sign. $+ 28$ means "28 units of oomph, pointing right." Why this step? The sign IS the direction — it's part of the answer, not decoration.

Verify: Units: $kg \times m/s = kg\cdotm/s$ ✓. Magnitude $28$ is heavier×faster than a $10 kg\cdotm/s$ football, which matches "bowling ball = big oomph." ✓

Cell B — 1D negative velocity

Worked example Ex 2 · Ball fired leftward

A $0.30 kg$ ball moves left at $5.0 m/s$ . Find its momentum.

Forecast: Same size as before? What's the sign this time?

Translate "left" into a sign. Right is $+$ , so left is $-$ : $v = - 5.0 m/s$ . Why this step? Our convention (Fig s01) converts a direction word into a signed number.
Apply $p = m v$ . $p = (0.30) (- 5.0) = - 1.5 kg\cdotm/s$ Why this step? Mass is always positive, so the velocity's sign carries straight through to $p$ .
Interpret. The minus sign says the momentum points left. Its magnitude (its size, bars defined just above) is $∣ p ∣ = 1.5 kg\cdotm/s$ . Why this step? Magnitude drops the sign; the sign is separate information about direction.

Verify: A negative $p$ never means "less than nothing of oomph" — it means oomph aimed the other way. If it hit Ex 1's ball head-on, this $- 1.5$ would subtract from that ball's $+ 28$ . That's exactly why we keep signs. ✓

Cell C — 1D, opposite directions, total momentum

Worked example Ex 3 · Two carts approaching

Cart A: $2.0 kg$ at $+ 4.0 m/s$ . Cart B: $3.0 kg$ at $- 2.0 m/s$ (moving left). Find the total momentum of the system.

Forecast: Will the total be positive (net rightward), negative, or zero?

Momentum of each cart. $p_{A} = (2.0) (+ 4.0) = + 8.0 kg\cdotm/s, p_{B} = (3.0) (- 2.0) = - 6.0 kg\cdotm/s$ Why this step? Total momentum of a system is the vector sum of the parts, so we need each part first.
Add with signs. $p_{total} = (+ 8.0) + (- 6.0) = + 2.0 kg\cdotm/s$ Why this step? Adding a positive and a negative lets the opposite directions partly cancel — that's a vector sum on a line.

Verify: The green net arrow in Fig s02 points right and is short, matching $+ 2.0$ . Sanity: A's rightward oomph ( $8$ ) beats B's leftward oomph ( $6$ ), so the leftover leans right — consistent. ✓

Cell D — Zero and degenerate inputs

Worked example Ex 4 · At rest, and the massless limit

(a) A $12 kg$ box sits still. (b) An idealised "massless" tracer speck drifts at $10 m/s$ . Momentum of each?

Forecast: Can something moving have zero momentum? Can something at rest?

Box at rest: $v = 0$ , so $p = (12) (0) = 0 kg\cdotm/s .$ Why this step? Zero velocity kills the product regardless of how heavy the box is — no motion, no oomph.
Massless speck: in the idealisation $m \to 0$ , so $p = (0) (10) = 0 kg\cdotm/s .$ Why this step? Zero mass also kills the product, no matter how fast it goes.

Verify: Both degenerate inputs ( $v = 0$ or $m = 0$ ) give $p = 0$ . This checks the structure: $p = m v$ is a product, so it vanishes the instant either factor is zero. A truck parked at a light and a hypothetical massless dust mote both carry zero momentum — for opposite reasons. ✓

Cell E — 2D, perpendicular components

Now motion leaves the line, so a single $\pm$ sign is no longer enough. We need components: split the velocity into how much goes east ( $x$ ) and how much goes north ( $y$ ), handle each separately, then rebuild.

$p_{x} = m v_{x}, p_{y} = m v_{y}, ∣ p ∣ = p_{x}^{2} + p_{y}^{2}$

Worked example Ex 5 · A puck on an air table

A $0.20 kg$ puck has $v_{x} = 3.0 m/s$ (east) and $v_{y} = 4.0 m/s$ (north). Find $∣ p ∣$ and its direction.

Forecast: A tempting-but-wrong guess is to just add the two component speeds, $v_{x} + v_{y} = 3.0 + 4.0 = 7.0$ , and multiply by mass to get $0.20 \times 7 = 1.4$ . Do you think the true magnitude is bigger or smaller than that $1.4$ ? (Perpendicular things don't add straight — keep that in mind.)

Component momenta. $p_{x} = (0.20) (3.0) = 0.60, p_{y} = (0.20) (4.0) = 0.80 (kg\cdotm/s)$ Why this step? Mass scales each direction independently — momentum inherits velocity's components.
Combine with Pythagoras. $∣ p ∣ = 0.6 0^{2} + 0.8 0^{2} = 0.36 + 0.64 = 1.00 = 1.0 kg\cdotm/s$ Why this step? The legs $0.6$ and $0.8$ are perpendicular, so the diagonal is the hypotenuse.
Direction (angle north-of-east). The angle $θ$ satisfies $tan θ = p_{y} / p_{x}$ . We use $tan$ = opposite/adjacent because it's the ratio of the two legs, which pins the tilt. $θ = arctan (\frac{0.80}{0.60}) = arctan (1.333 \dots) \approx 53. 1^{\circ}$ Why this step? $arctan$ answers "which angle has this leg-ratio?" — it undoes the tan to hand back the direction.

Verify: $0.6, 0.8, 1.0$ is a scaled $3$ - $4$ - $5$ triangle — a classic exact right triangle. ✓ And $1.0 < 1.4$ (the naive straight-add guess), confirming perpendicular parts add smaller than a straight sum. ✓

A consistent rule for angles in all four quadrants

Before Cell F mixes signs, let's fix one way to name a direction so we never get lost. We measure every angle as the standard angle $ϕ$ : start at the positive $x$ -axis (east) and sweep counter-clockwise. East $= 0^{\circ}$ , north $= 9 0^{\circ}$ , west $= 18 0^{\circ}$ , south $= 27 0^{\circ}$ .

Formula The quadrant recipe (do this every time)

Compute the reference angle $α = arctan (\frac{∣ p _{y} ∣}{∣ p _{x} ∣})$ — always feed $arctan$ the positive leg-ratio, so $α$ is a clean acute angle between $0^{\circ}$ and $9 0^{\circ}$ .
Read the signs of $p_{x}, p_{y}$ to find the quadrant, then convert:

Quadrant	sign of $p_{x}$	sign of $p_{y}$	Standard angle $ϕ$
I (NE)	$+$	$+$	$ϕ = α$
II (NW)	$-$	$+$	$ϕ = 18 0^{\circ} - α$
III (SW)	$-$	$-$	$ϕ = 18 0^{\circ} + α$
IV (SE)	$+$	$-$	$ϕ = 36 0^{\circ} - α$

Why a recipe and not raw $arctan$ ? The plain $arctan$ button only returns answers between $- 9 0^{\circ}$ and $+ 9 0^{\circ}$ (it can't tell NW from SE — both give the same ratio). The sign table is what disambiguates all four quadrants. In Cell E the puck was in Quadrant I, so $ϕ = α = 53. 1^{\circ}$ directly.

Cell F — 2D, a component pointing the "wrong" way

Worked example Ex 6 · Puck with a westward drift

A $0.50 kg$ puck has $v_{x} = - 6.0 m/s$ (that's west) and $v_{y} = + 8.0 m/s$ (north). Find $∣ p ∣$ and its standard-angle direction.

Forecast: Does the minus sign shrink the total momentum? Which quadrant does the arrow point into?

Component momenta — keep the sign. $p_{x} = (0.50) (- 6.0) = - 3.0, p_{y} = (0.50) (+ 8.0) = + 4.0 (kg\cdotm/s)$ Why this step? In 2D each axis keeps its own sign; the $x$ -component points west, so it stays negative.
Magnitude — squares erase the sign. $∣ p ∣ = (- 3.0)^{2} + (4.0)^{2} = 9 + 16 = 25 = 5.0 kg\cdotm/s$ Why this step? Squaring $(- 3.0)$ gives $+ 9$ : size doesn't care about direction, so a negative component adds to the length just like a positive one.
Direction — use the quadrant recipe. Reference angle first: $α = arctan (\frac{∣ p _{y} ∣}{∣ p _{x} ∣}) = arctan (\frac{4.0}{3.0}) \approx 53. 1^{\circ} .$ Signs are $p_{x} < 0$ , $p_{y} > 0$ ⇒ Quadrant II (north-west), so from the table $ϕ = 18 0^{\circ} - α \approx 18 0^{\circ} - 53. 1^{\circ} = 126. 9^{\circ} .$ Why this step? Magnitude alone loses the direction; the signs pick the quadrant, and the recipe converts the acute reference angle into the true standard angle.

Verify: $3$ - $4$ - $5$ again ⇒ length $5.0$ . ✓ The standard angle $126. 9^{\circ}$ lies between $9 0^{\circ}$ (north) and $18 0^{\circ}$ (west), i.e. the upper-left — matching Fig s04 and the negative $p_{x}$ . ✓

Cell G — Limiting behaviour (same momentum, extreme mass/speed)

Worked example Ex 7 · The mosquito and the glacier

Two objects each carry exactly $p = 12 kg\cdotm/s$ . Object 1 is a light dart, $m_{1} = 0.010 kg$ . Object 2 is a slow slab, $v_{2} = 0.0020 m/s$ . Find $v_{1}$ and $m_{2}$ , and comment on the limits.

Forecast: As mass shrinks toward zero for fixed $p$ , what must speed do?

Solve $p = m v$ for the missing factor. Rearranging, $v = p / m$ and $m = p / v$ . Why this step? Whichever factor is unknown, divide the fixed $p$ by the known one.
Dart's speed. $v_{1} = \frac{p}{m _{1}} = \frac{12}{0.010} = 1200 m/s .$ Why this step? Same oomph packed into tiny mass forces a huge speed.
Slab's mass. $m_{2} = \frac{p}{v _{2}} = \frac{12}{0.0020} = 6000 kg .$ Why this step? Same oomph at a crawling speed forces an enormous mass.
State the limits. For fixed $p$ : as $m \to 0$ , $v = p / m \to \infty$ ; as $v \to 0$ , $m = p / v \to \infty$ . Why this step? $p$ is a product, so shrinking one factor blows up the other to hold the product constant.

Verify: $0.010 \times 1200 = 12$ ✓ and $6000 \times 0.0020 = 12$ ✓. Same momentum, wildly different mass and speed — the exact "equal oomph, unequal everything else" idea from the parent note. ✓

Cell H — Real-world word problem

Worked example Ex 8 · A recoiling cannon

A $500 kg$ cannon, initially at rest, fires a $5.0 kg$ ball forward at $60 m/s$ . Using momentum conservation, find the cannon's recoil velocity. (Total momentum before = total after, because the firing forces are internal — see Conservation of Linear Momentum.)

Forecast: Before you compute — will the cannon move fast or slow, and which way?

Total momentum before. Everything is at rest: $p_{before} = 0.$ Why this step? No motion ⇒ zero oomph (Cell D again).
Total momentum after must also be $0$ . Let the cannon's velocity be $V$ and the ball's direction be $+$ . Writing $m_{cannon} = 500 kg$ and $m_{ball} = 5.0 kg$ : $0 = m_{ball} v_{ball} + m_{cannon} V = (5.0) (+ 60) + (500) V .$ Why this step? Internal forces can't change the system's total momentum, so "after" equals "before" $= 0$ .
Solve for $V$ . $500 V = - 300 \Rightarrow V = - 0.60 m/s .$ Why this step? Rearranging the balance; the minus sign means the cannon moves backward, opposite the ball.

Verify: Ball momentum $+ 300$ , cannon momentum $500 \times (- 0.60) = - 300$ ; sum $= 0$ ✓, matching the initial $0$ . The heavy cannon recoils slowly ( $0.60 m/s$ ) while the light ball flies fast — heavy ⇒ small $v$ for equal $∣ p ∣$ , consistent with Ex 7. ✓

Cell I — Exam twist: mixed units + kinetic-energy link

Worked example Ex 9 · Grams, km/h, and energy

A $200 g$ ball flies at $36 km/h$ . (a) Find $p$ in SI units. (b) Find its kinetic energy using $K E = p^{2} /2 m$ .

Forecast: Which trips students up more — the grams or the km/h? Convert both first.

Convert to SI. $200 g = 0.200 kg$ (divide by $1000$ ). $36 km/h = 36 \times \frac{1000}{3600} = 10 m/s$ . Why this step? $p = m v$ only gives $kg\cdotm/s$ if $m$ is in $kg$ and $v$ in $m/s$ ; mixed units are the classic trap.
Momentum. $p = (0.200) (10) = 2.0 kg\cdotm/s .$ Why this step? Plain formula once units are clean.
Kinetic energy via $K E = p^{2} /2 m$ . We use this form (instead of $\frac{1}{2} m v^{2}$ ) to practise the momentum–energy bridge from the parent note. $K E = \frac{p ^{2}}{2 m} = \frac{( 2.0 ) ^{2}}{2 ( 0.200 )} = \frac{4.0}{0.40} = 10 J .$ Why this step? $K E = p^{2} /2 m$ answers "how much work-capacity for this much oomph and mass?" — it links the two quantities directly.

Verify: Cross-check with $\frac{1}{2} m v^{2} = \frac{1}{2} (0.200) (10)^{2} = \frac{1}{2} (0.200) (100) = 10 J$ ✓ — both routes agree. Units: $\frac{( kg\cdotm/s ) ^{2}}{kg} = \frac{kg ^{2} m ^{2} / s ^{2}}{kg} = kg\cdotm^{2} / s^{2} = J$ ✓. See Kinetic Energy. ✓

Recall

Recall Which cell is which?

Positive 1D single object ::: Ex 1 (Cell A) Where does the direction "live" in 1D? ::: In the $\pm$ sign of $v$ , hence of $p$ . Both $m = 0$ and $v = 0$ give ::: $p = 0$ , because $p = m v$ is a product. Perpendicular components combine by ::: Pythagoras, $∣ p ∣ = p_{x}^{2} + p_{y}^{2}$ . A negative component changes the magnitude how? ::: Not at all — squaring erases its sign; it only sets the quadrant. How do you get the true direction in any quadrant? ::: Take $α = arctan (∣ p_{y} ∣/∣ p_{x} ∣)$ , then use the sign table (I: $α$ , II: $180 - α$ , III: $180 + α$ , IV: $360 - α$ ). For fixed $p$ , as $m \to 0$ the speed ::: $\to \infty$ (and as $v \to 0$ , $m \to \infty$ ). Cannon recoil direction ::: Opposite the ball, so total $p$ stays $0$ . Convert $36 km/h$ to m/s ::: $36 \times 1000/3600 = 10 m/s$ . $K E$ from momentum ::: $K E = p^{2} /2 m$ .

Connections

Parent: Linear momentum $p=mv$ — the definitions these examples exercise.
Conservation of Linear Momentum — the rule behind Ex 8 (cannon recoil).
Impulse–Momentum Theorem — how a stationary wall (Ex 4) still stops a car.
Kinetic Energy — the $K E = p^{2} /2 m$ bridge used in Ex 9.
Elastic and Inelastic Collisions — where signed 1D sums (Ex 3) get used constantly.
Centre of Mass — total momentum $= M v_{c m}$ , the system view of Ex 3 and Ex 8.