Momentum and Impulse — Grade 12 Mechanics

00 Topic Overview Mechanics — Paper 1

Momentum and Impulse falls under Mechanics in Paper 1. It is examined in both multiple-choice and structured questions, and can be integrated with vertical projectile motion and work, energy & power.

CAPS Examination Requirements — You Must Be Able To

Define momentum and describe its vector nature
Calculate momentum using \(p = mv\)
Draw vector diagrams relating \(p_i\), \(p_f\), and \(\Delta p\)
State and apply Newton's second law in terms of momentum
Define impulse and apply the impulse-momentum theorem
Explain how impulse applies to safety devices (airbags, seatbelts, arrestor beds, crumple zones)
State and apply the principle of conservation of linear momentum in isolated systems
Distinguish between elastic and inelastic collisions by calculation

Concept	Key Formula	Key Idea
Momentum	\(p = mv\)	Vector — same direction as \(v\)
Change in Momentum	\(\Delta p = mv_f - mv_i\)	Newton's 2nd law: \(F_{NET} = \Delta p / \Delta t\)
Impulse	\(F_{NET}\Delta t = \Delta p\)	Area under the F-t graph; basis for safety devices
Conservation of Momentum	\(\Sigma p_i = \Sigma p_f\)	Isolated system — bounce apart, stick together, or start joined and separate
Elastic Collisions	\(\Sigma E_{k,i} = \Sigma E_{k,f}\)	Momentum and kinetic energy both conserved; objects do not combine
Inelastic Collisions	\(\Sigma E_{k,i} \neq \Sigma E_{k,f}\)	Momentum conserved, kinetic energy is not; energy → heat, sound, deformation

01 Momentum — Definition & Vector Nature p = mv

East (+)

p = 1 350 kg·m·s⁻¹

West (−)

p = 500 kg·m·s⁻¹

Direction of p is always the same as the direction of v

CAPS Definition — Learn this exactly

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

\[p = mv\]

Symbol	Quantity	SI Unit
\(p\)	momentum	\(\text{kg}\cdot\text{m}\cdot\text{s}^{-1}\)
\(m\)	mass	kg
\(v\)	velocity	\(\text{m}\cdot\text{s}^{-1}\)

1

Momentum is a vector

Momentum has both magnitude and direction. Its direction is always the same as the direction of the object's velocity. Momentum can be zero only when the velocity is zero (the object is at rest).

2

Why this matters for calculations

You must always state a direction in your answer (unless only magnitude is asked), and you must use a sign convention in every calculation. A negative result is not an error — it means the object moves in the direction you chose as negative.

Sign Convention — Do This First, Every Time

Before any vector calculation, declare your positive direction and write it at the top of your working, e.g. "Take East as positive (+)". Maintain this convention throughout the entire question — the same convention chosen in sub-question (a) must be used in (b), (c), etc.

02 Unit Conversions & Worked Examples p = mv

Given unit	Convert to	Method
grams (g)	kilograms (kg)	÷ 1 000
km·h⁻¹	m·s⁻¹	÷ 3,6
tons (t)	kilograms (kg)	× 1 000

Example 1 — Basic calculation

A dog with mass 30 kg runs at 45 m·s⁻¹. Calculate the dog's momentum. Take direction of motion as positive (+)

\[p = mv = (30)(45) = \color{#E8FF47}{1\,350 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} \text{ in the direction of motion}}\]

Example 2 — Unit conversions required

A bowler bowls a 150 g cricket ball towards the batsman at 135 km·h⁻¹. Calculate the momentum of the ball. Take towards the batsman as positive (+)

\[m = \frac{150}{1\,000} = 0{,}15 \text{ kg} \qquad v = \frac{135}{3{,}6} = 37{,}5 \text{ m}\cdot\text{s}^{-1}\]

\[p = mv = (0{,}15)(37{,}5) = \color{#E8FF47}{5{,}63 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} \text{ towards the batsman}}\]

Example 3 — Solving for velocity

A truck (mass 1,5 t) has a momentum of 19 590 kg·m·s⁻¹. At what velocity is the truck travelling? Take direction of motion as positive (+)

\[m = 1{,}5 \times 1\,000 = 1\,500 \text{ kg}\]

\[p = mv \implies 19\,590 = (1\,500)(v) \implies v = \frac{19\,590}{1\,500} = \color{#E8FF47}{13{,}06 \text{ m}\cdot\text{s}^{-1} \text{ in the direction of motion}}\]

Exam Technique: Momentum Calculations

1. State the positive direction.
2. List all given values with SI units converted.
3. Write the formula as given on the formula sheet — do not manipulate it first.
4. Substitute, then solve.
5. Final answer must include magnitude, unit, and direction.
6. Round to a minimum of two decimal places.

03 Δp & Newton's Second Law Fnet = Δp / Δt

\[\Delta p = p_f - p_i = mv_f - mv_i = m(v_f - v_i)\]

Symbol	Meaning	Unit
\(\Delta p\)	change in momentum	kg·m·s⁻¹
\(p_f\)	final momentum	kg·m·s⁻¹
\(p_i\)	initial momentum	kg·m·s⁻¹
\(v_f\)	final velocity	m·s⁻¹
\(v_i\)	initial velocity	m·s⁻¹

Common Exam Language Trap

"Momentum changes by x kg·m·s⁻¹" → refers to \(\Delta p\)
"Momentum changes to x kg·m·s⁻¹" → refers to \(p_f\)
Confusing these leads to substitution errors.

CAPS Definition — Newton's Second Law (momentum form)

Newton's second law of motion states that the resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force.

\[F_{NET} = \frac{\Delta p}{\Delta t} = \frac{mv_f - mv_i}{\Delta t}\]

Formula Sheet Note

The formula sheet gives \(F_{NET}\Delta t = \Delta p\) under FORCE. Newton's second law and the impulse equation are the same relationship rearranged.

04 Vector Diagrams for Δp — Three Cases examination skill

Case A — reverses direction

pi East

pf West

Δp West — largest

Case B — decelerates

pi East (large)

pf East (smaller)

Δp West — opposes motion

Case C — accelerates

pi East (small)

pf East (larger)

Δp East — same direction

Drawing the \(\Delta p\) vector diagram is an examination skill. In every case, \(\Delta p = p_f - p_i\). Three cases appear in exams:

Case	What happens	Direction of Δp
A — Reverses direction	e.g. a ball is returned the opposite way it came (\(p_i\) East, \(p_f\) West)	Same direction as \(p_f\) — \(\Delta p\) is larger than either \(p_i\) or \(p_f\) alone
B — Decelerates	object slows down but keeps moving the same way (\(p_f < p_i\), same direction)	Opposes the motion — \(\Delta p\) points backward even though the object still moves forward
C — Accelerates	object speeds up in the same direction (\(p_f > p_i\), same direction)	Same direction as the motion

Key Insight

The direction of \(\Delta p\) is always the direction of the net force that caused the change — not necessarily the direction the object is still moving in. This is why Case B is the one students get wrong most often.

05 Worked Examples — Change in Momentum Δp = mvf − mvi

Example 1 — Ball reverses direction (classic exam question)

A tennis ball (100 g) is served East at 220 km·h⁻¹. The opponent returns it West at 154,8 km·h⁻¹. Calculate the change in momentum of the ball. Take East as positive (+)

\[m = 0{,}1 \text{ kg}, \quad v_i = +\frac{220}{3{,}6} = +61{,}11 \text{ m}\cdot\text{s}^{-1}, \quad v_f = -\frac{154{,}8}{3{,}6} = -43 \text{ m}\cdot\text{s}^{-1}\]

\[\Delta p = mv_f - mv_i = (0{,}1)(-43) - (0{,}1)(61{,}11) = -10{,}41\]

\[\color{#E8FF47}{\Delta p = 10{,}41 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} \textbf{ West}}\]

Example 2 — Ball slows in the same direction

Same ball (100 g) served East at 220 km·h⁻¹ hits the net and continues East at 120 km·h⁻¹. Take East as positive (+)

\[\Delta p = m(v_f - v_i) = (0{,}1)\!\left(\frac{120}{3{,}6} - \frac{220}{3{,}6}\right) = (0{,}1)(33{,}33 - 61{,}11) = -2{,}78\]

\[\color{#E8FF47}{\Delta p = 2{,}78 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} \textbf{ West}}\]

Even though the ball still moves East, \(\Delta p\) is West — because the ball lost momentum. The direction of \(\Delta p\) is always the direction of the net force that caused the change.

Example 3 — Finding final momentum from Δp

An arrow has an initial momentum of 20 kg·m·s⁻¹ towards a target. Its momentum changes by 4 kg·m·s⁻¹ in the opposite direction. Find the final momentum. Take towards the target as positive (+)

\[\Delta p = p_f - p_i \implies -4 = p_f - 20 \implies \color{#E8FF47}{p_f = 16 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} \textbf{ towards the target}}\]

06 Impulse & the Impulse-Momentum Theorem FnetΔt = Δp

CAPS Definition — Learn this exactly

Impulse is the product of the resultant/net force acting on an object and the time the resultant/net force acts on the object.

\[\text{Impulse} = F_{NET} \cdot \Delta t\]

Impulse is a vector quantity. Its direction is the same as the direction of the net force (and of \(\Delta p\)). SI units: N·s, equivalently kg·m·s⁻¹.

From Newton's second law, \(F_{NET} = \dfrac{\Delta p}{\Delta t}\), therefore:

\[F_{NET}\Delta t = \Delta p = mv_f - mv_i\]

Symbol	Quantity	SI Unit
\(F_{NET}\)	net/resultant force	N
\(\Delta t\)	contact time	s
\(\Delta p\)	change in momentum	N·s = kg·m·s⁻¹

07 The Force-Time (F–t) Graph area = impulse

\(t\text{ (s)}\)

\(F\text{ (N)}\)

\(F_{max}\)

Maximum force

Area = Impulse

\(\Delta t\) (contact time)

During a collision, the net force rises from zero to a maximum and falls back to zero. The area under the F-t graph equals the impulse (\(= \Delta p\)).

Newton's Third Law and Impulse

When objects A and B collide, \((F_{NET}\Delta t)_{A \text{ on } B} = -(F_{NET}\Delta t)_{B \text{ on } A}\). The impulses are equal in magnitude and opposite in direction. The contact time \(\Delta t\) is identical for both objects.

Critical Rule

When calculating impulse or net force during a collision, use the data of only ONE object. Never mix data from both objects in a single \(\Delta p\) calculation. Choose the object for which you have the most complete data (or the one the question specifies).

08 Impulse and Safety Applications F inversely ∝ Δt

Same Δp — different Δt and F

No safety device

Short Δt, large F

t

F

small Δt

With safety device

Long Δt, small F

t

F

large Δt

Both shaded areas are equal — same Δp in both cases

Since \(F_{NET} = \dfrac{\Delta p}{\Delta t}\), and \(\Delta p\) is fixed (the person must decelerate from the car's speed to rest), the only way to reduce \(F_{NET}\) is to increase \(\Delta t\). \(F_{NET}\) is inversely proportional to \(\Delta t\) when \(\Delta p\) is constant.

Safety Device	How it increases Δt
Airbags	Inflated cushion extends the time over which the head decelerates
Seatbelts	Stretchy webbing spreads the stopping over a longer time than a rigid dashboard
Crumple zones	The vehicle front deforms progressively, extending the collision duration
Arrestor beds	Sand/gravel pit stops runaway trucks gradually over a longer distance
Padded surfaces	Helmets, crash mats compress on impact, increasing contact time

How to Answer "Explain How [Device] Reduces Injury"

1. Physical action: "The airbag increases the contact time (\(\Delta t\)) during the collision."
2. Physics law: "From \(F_{NET} = \dfrac{\Delta p}{\Delta t}\), since the change in momentum (\(\Delta p\)) remains constant, an increase in \(\Delta t\) causes a decrease in \(F_{NET}\)."
3. Safety link: "The reduced net force on the driver/passenger decreases the risk of injury."

09 Worked Examples — Impulse FnetΔt = Δp

Example 1 — Calculate impulse

A tennis ball (100 g) is served West at 220 km·h⁻¹. It is returned East at 154,8 km·h⁻¹. Calculate the impulse on the ball. Take West as positive (+)

\[m = 0{,}1 \text{ kg}, \quad v_i = +61{,}11 \text{ m}\cdot\text{s}^{-1}, \quad v_f = -43 \text{ m}\cdot\text{s}^{-1}\]

\[\text{Impulse} = mv_f - mv_i = (0{,}1)(-43) - (0{,}1)(61{,}11) = -10{,}41 \text{ N}\cdot\text{s}\]

\[\color{#E8FF47}{\text{Impulse} = 10{,}41 \text{ N}\cdot\text{s} \textbf{ East}}\]

Exam Hint: Replace "\(F_{NET}\Delta t\)" with the word "Impulse" in your working. This prevents accidentally solving for \(F_{NET}\) when the question asks for impulse.

Example 2 — Calculate average net force

A cricket ball (156 g) is bowled at 37 m·s⁻¹. Hit back in the opposite direction at 45 m·s⁻¹. Contact time is 0,04 s. Take initial direction as positive (+)

\[F_{NET}\Delta t = mv_f - mv_i\]

\[F_{NET}(0{,}04) = (0{,}156)(-45) - (0{,}156)(37) = -7{,}02 - 5{,}772 = -12{,}792\]

\[F_{NET} = \frac{-12{,}792}{0{,}04} = \color{#E8FF47}{319{,}80 \text{ N} \textbf{ away from the bowler}}\]

By Newton's 3rd law, the force the ball exerts on the bat = 319,80 N in the direction of the bowler.

Example 3 — Calculate contact time

A 300 g soccer ball hits a goalpost at 15 m·s⁻¹ and bounces back at 10 m·s⁻¹. A net force of 37,5 N acts on the ball away from the post. Take towards the goalpost as positive (+)

\[F_{NET}\Delta t = mv_f - mv_i\]

\[(-37{,}5)\,\Delta t = (0{,}3)(-10) - (0{,}3)(15) = -3 - 4{,}5 = -7{,}5\]

\[\Delta t = \frac{-7{,}5}{-37{,}5} = \color{#E8FF47}{0{,}20 \text{ s}}\]

Example 4 — F-t graph: triangular area

A triangular F-t graph has a maximum force of 60 N and a contact time of 0,3 s. Calculate the impulse.

\[\text{Impulse} = \text{Area} = \tfrac{1}{2} \times \Delta t \times F_{max} = \tfrac{1}{2} \times 0{,}3 \times 60 = \color{#E8FF47}{9 \text{ N}\cdot\text{s}}\]

10 Closed/Isolated Systems & the Principle Σpi = Σpf

CAPS Definitions — Learn these exactly

Closed/isolated system (in Physics): A system on which the resultant/net external force is zero.

Internal forces: forces that objects within the system exert on each other (e.g. the contact force between two colliding objects). These do not change the total momentum of the system.

External forces: forces from outside the system (e.g. friction with the floor). If non-zero, the system is not isolated and momentum is not conserved.

In exam questions, "ignore the effects of friction" or "smooth horizontal surface" signals you can treat the system as isolated.

CAPS Definition — Learn this exactly

The principle of conservation of linear momentum: The total linear momentum of a closed/isolated system remains constant (is conserved).

\[\Sigma p_i = \Sigma p_f\]

Not on the Formula Sheet

\(\Sigma p_i = \Sigma p_f\) is NOT on the formula sheet. You must write it at the start of every conservation problem before expanding it.

11 Three Collision Scenarios how objects end up

Scenario 1 — Separate → Separate

BEFORE

AFTER

e.g. snooker balls bouncing apart

Scenario 2 — Separate → Joined (perfectly inelastic)

BEFORE

AFTER

e.g. vehicles locking together; person jumping onto a skateboard

Scenario 3 — Joined → Separate (explosion)

BEFORE — at rest

AFTER

e.g. spring-loaded trolleys; a gun firing a bullet; rocket exhaust

If \(v_i = 0\), then \(m_1v_{1f} = -m_2v_{2f}\)

Scenario	Equation	Example
1 — Separate → Separate objects bounce apart	\(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\)	Snooker balls colliding and bouncing apart
2 — Separate → Joined objects stick together (perfectly inelastic)	\(m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f\)	Vehicles locking together; person jumping onto a skateboard
3 — Joined → Separate objects start joined and separate (explosion)	\((m_1 + m_2)v_i = m_1v_{1f} + m_2v_{2f}\)	Spring-loaded trolleys; a gun firing a bullet; rocket exhaust

Explosion Shortcut

If \(v_i = 0\) in Scenario 3 (objects start at rest), then \(\Sigma p_i = 0\), so \(m_1v_{1f} = -m_2v_{2f}\) — the two final momenta are equal in magnitude and opposite in direction.

Sign Conventions in Conservation Problems

Direction is critical. Establish your positive direction first and substitute all velocities with the correct sign. A negative result simply means the object moves opposite to your chosen positive direction. Always re-state the final answer as a positive value with the direction written out.

12 Worked Examples — Conservation of Momentum Σpi = Σpf

Example 1 — Scenario 2: Person jumping onto a skateboard

A girl (30 kg) runs at 5 m·s⁻¹ and jumps onto a 2 kg skateboard at rest. Find their velocity after she jumps on. Take direction of motion as positive (+)

\[\Sigma p_i = \Sigma p_f \implies m_g v_{gi} + m_b v_{bi} = (m_g + m_b)\,v_f\]

\[(30)(5) + (2)(0) = (30 + 2)\,v_f \implies 150 = 32\,v_f\]

\[\color{#E8FF47}{v_f = 4{,}69 \text{ m}\cdot\text{s}^{-1} \textbf{ in the direction of motion}}\]

Example 2 — Scenario 3: Spring-loaded trolleys (explosion)

Trolleys P (400 g) and Q (600 g) are held together by a compressed spring on a frictionless surface, both initially at rest. When released, Q moves right at 4 m·s⁻¹. Find the velocity of P. Take right as positive (+)

\[\Sigma p_i = \Sigma p_f \implies (m_P + m_Q)\,v_i = m_P v_{Pf} + m_Q v_{Qf}\]

\[(0{,}4 + 0{,}6)(0) = (0{,}4)\,v_{Pf} + (0{,}6)(4) \implies 0 = 0{,}4\,v_{Pf} + 2{,}4\]

\[v_{Pf} = \frac{-2{,}4}{0{,}4} = -6 \text{ m}\cdot\text{s}^{-1} \implies \color{#E8FF47}{v_{Pf} = 6 \text{ m}\cdot\text{s}^{-1} \textbf{ towards the left}}\]

The negative confirms P moves opposite to Q — as expected from Newton's 3rd law.

Example 3 — Scenario 1: Equal-mass balls, head-on collision

Ball A and Ball B have the same mass. A moves right at 15 m·s⁻¹, B moves left at 8 m·s⁻¹. After the collision, A moves right at 5 m·s⁻¹. Find the final velocity of B. Take right as positive (+)

\[\Sigma p_i = \Sigma p_f \implies m(v_{Ai} + v_{Bi}) = m(v_{Af} + v_{Bf})\]

Mass cancels (equal on both sides):

\[(15) + (-8) = (5) + v_{Bf} \implies 7 = 5 + v_{Bf} \implies \color{#E8FF47}{v_{Bf} = 2 \text{ m}\cdot\text{s}^{-1} \textbf{ to the right}}\]

When both objects have the same mass, it cancels — you don't need a numerical value for mass.

Example 4 — Conservation + Impulse: find contact time

A car (1 000 kg, rightward at 40 m·s⁻¹) collides head-on with a truck (5 000 kg, leftward at 20 m·s⁻¹). They combine and move together. The car experiences a net force of 100 000 N to the left. Find the contact time. Take right as positive (+)

Step 1 — Find \(v_f\):

\[(1\,000)(40) + (5\,000)(-20) = (6\,000)\,v_f \implies v_f = \frac{-60\,000}{6\,000} = -10 \text{ m}\cdot\text{s}^{-1} \text{ (leftward)}\]

Step 2 — Find \(\Delta t\) using the car's data:

\[F_{NET}\Delta t = mv_f - mv_i \implies (-100\,000)\,\Delta t = (1\,000)(-10) - (1\,000)(40) = -50\,000\]

\[\color{#E8FF47}{\Delta t = 0{,}50 \text{ s}}\]

13 How to Solve Any Momentum Problem method

1

Read & identify

Read the question and identify the given values and the unknown.

2

Declare a positive direction

e.g. "Take East as positive (+)" — then convert every value to SI units (g ÷ 1000 → kg; km·h⁻¹ ÷ 3,6 → m·s⁻¹) and apply the sign to each vector.

3

Is it a collision or explosion?

If yes — decide how the objects end up: bounce apart (Scenario 1), stick together (Scenario 2), or start joined and separate (Scenario 3) — then apply the matching equation from Card 11.

4

If not a collision/explosion

Is force or time given? If yes, use the impulse theorem \(F_{NET}\Delta t = mv_f - mv_i\). If no, use \(p = mv\) or \(\Delta p = mv_f - mv_i\) directly.

5

Substitute, then solve

Do not rearrange the formula before substituting. Give your final answer with magnitude, unit, and direction.

14 Definitions & How to Determine Collision Type Ek = ½mv²

CAPS Definitions — Learn both exactly

Elastic collision: A collision in which both the total linear momentum and the total kinetic energy are conserved.

Inelastic collision: A collision during which the total linear momentum is conserved, but the total kinetic energy is not conserved.

In both types of collision, total linear momentum is always conserved (in an isolated system). The only difference is whether kinetic energy is also conserved. In an inelastic collision, kinetic energy is converted to heat, sound, and deformation.

1

Calculate total KE before

\(\Sigma E_{k,i} = \tfrac{1}{2}m_1v_{1i}^2 + \tfrac{1}{2}m_2v_{2i}^2\) — use magnitudes only, no signs.

2

Calculate total KE after

\(\Sigma E_{k,f} = \tfrac{1}{2}m_1v_{1f}^2 + \tfrac{1}{2}m_2v_{2f}^2\) — use magnitudes only, no signs.

3

Compare the two totals

\(\Sigma E_{k,i} = \Sigma E_{k,f}\) → elastic (momentum and KE both conserved, objects do not combine, e.g. snooker balls). Otherwise → inelastic (momentum conserved, KE not conserved, energy → heat/sound/deformation, e.g. a car crash).

\[E_k = \frac{1}{2}mv^2 \qquad \text{Units: Joules (J) — scalar, no direction}\]

Property	Elastic	Inelastic
Momentum conserved?	Yes	Yes
Kinetic energy conserved?	Yes	No
Objects combine?	No	Sometimes
Real-world example	Snooker balls	Car crash

Kinetic Energy — Exam Tips

\(E_k\) is a scalar — no direction required and no negative signs on velocities.

Calculate \(\Sigma E_{k,i}\) and \(\Sigma E_{k,f}\) separately — never in one step.

Find the kinetic energy formula on the formula sheet under Work, Energy and Power.

15 Worked Example — Elastic or Inelastic? ΣEk,i vs ΣEk,f

A car collides with a truck and they combine

A car (1 000 kg) at 40 m·s⁻¹ rightward collides with a truck (5 000 kg) at 20 m·s⁻¹ leftward. They combine and move at 10 m·s⁻¹ leftward. Elastic or inelastic? Use magnitudes only for \(E_k\).

\[\Sigma E_{k,i} = \tfrac{1}{2}(1\,000)(40)^2 + \tfrac{1}{2}(5\,000)(20)^2 = 800\,000 + 1\,000\,000 = 1{,}8 \times 10^6 \text{ J}\]

\[\Sigma E_{k,f} = \tfrac{1}{2}(6\,000)(10)^2 = 3{,}0 \times 10^5 \text{ J}\]

\[1{,}8 \times 10^6 \neq 3{,}0 \times 10^5 \implies \Sigma E_{k,i} \neq \Sigma E_{k,f}\]

\[\color{#E8FF47}{\therefore \text{ The collision is \textbf{inelastic.}}}\]

16 Formula Summary all formulas

Concept	Formula	On formula sheet?
Momentum	\(p = mv\)	Yes — FORCE
Change in momentum	\(\Delta p = mv_f - mv_i\)	Yes — FORCE
Impulse-momentum theorem	\(F_{NET}\Delta t = \Delta p\)	Yes — FORCE
Newton's 2nd law (momentum)	\(F_{NET} = \dfrac{\Delta p}{\Delta t}\)	Derived from above
Conservation of momentum	\(\Sigma p_i = \Sigma p_f\)	No — memorise
Kinetic energy	\(E_k = \frac{1}{2}mv^2\)	Yes — WORK, ENERGY & POWER

17 Glossary of Key Terms CAPS definitions

Term	CAPS Definition
Linear momentum	The product of an object's mass and its velocity; a vector quantity with the same direction as the velocity of the object
Newton's second law (momentum form)	The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force
Impulse	The product of the resultant/net force acting on an object and the time the resultant/net force acts on the object
Closed/isolated system	A system on which the resultant/net external force is zero
Principle of conservation of linear momentum	The total linear momentum of a closed/isolated system remains constant (is conserved)
Elastic collision	A collision in which both total linear momentum and total kinetic energy are conserved
Inelastic collision	A collision during which total linear momentum is conserved but total kinetic energy is not conserved
Internal forces	Forces that objects within the system exert on each other (e.g. contact force during a collision)
External forces	Forces from outside the system that act on objects within it (e.g. friction with the floor)
Contact forces	Forces that arise from the physical contact between two objects
Non-contact forces	Forces between objects not in physical contact (e.g. gravitational force)
System	A part of space chosen for studying the changes that take place within it
Environment	Everything outside the system