Momentum and Impulse
Orientation
Lesson goal: build accurate physics fluency for momentum and impulse and use that fluency to support clear HSC-style scientific writing.
This page is materialised into the MentorMind course shell from existing teaching, textbook, and eduKG material. Use it as the main lesson surface; use the tutor for targeted repair, worked examples, and concise writing feedback.
Source Lesson Material
Syllabus inquiry question
- How can momentum models predict outcomes of interactions?
From The Feynman Lectures on Physics, Vol I, Chapter 10:
Momentum is a conserved quantity that often reveals the final motion even when the detailed forces are unknown.
Learning Objectives
- Define momentum and impulse with units.
- Apply the impulse-momentum theorem.
- Use conservation of momentum in one dimension.
- Distinguish elastic and inelastic collisions.
Content
Momentum
Momentum is the product of mass and velocity:
$$\vec{p} = m\vec{v}$$
- Momentum is a vector (has direction)
- SI units: kg·m/s (or N·s)
- A fast, heavy object has large momentum
- A stationary object has zero momentum
Momentum is conserved in collisions, making it a powerful tool for predicting outcomes without knowing the detailed forces.
Interactive: Comparing Momentum
Different objects with the same momentum:
Impulse
Impulse is the change in momentum caused by a force acting over time:
$$\vec{J} = \vec{F}\Delta t = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i$$
This is the impulse-momentum theorem: the impulse equals the change in momentum.
$$\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}$$
A large force for a short time, or a small force for a long time, can produce the same impulse.
Units of impulse: N·s (equivalent to kg·m/s)
Why Impulse Matters
Impulse explains why:
- Airbags reduce injury: same impulse over longer time → smaller force
- Following through in sport: contact time increases → greater impulse
- Crumple zones save lives: extending collision time reduces peak force
Conservation of Momentum
In an isolated system (no external forces), total momentum is conserved:
$$\vec{p}{before} = \vec{p}{after}$$
For two objects: $$m_1\vec{v}{1i} + m_2\vec{v}{2i} = m_1\vec{v}{1f} + m_2\vec{v}{2f}$$
Interactive: Collision Before and After
Two objects collide and exchange momentum:
Before: Total momentum = $3 \times 4 + 2 \times 0 = 12$ kg·m/s
After (if they stick): $(3 + 2) \times v_f = 12$ → $v_f = 2.4$ m/s
Types of Collisions
| Type | Momentum | Kinetic Energy | Example |
|---|---|---|---|
| Elastic | Conserved | Conserved | Billiard balls, atomic collisions |
| Inelastic | Conserved | NOT conserved | Car crash, ball catches |
| Perfectly inelastic | Conserved | Maximum KE lost | Objects stick together |
Momentum is ALWAYS conserved in collisions (if the system is isolated). Kinetic energy is only conserved in elastic collisions.
Interactive: Elastic vs Inelastic Collision
Compare the outcomes of different collision types:
In an elastic collision between equal masses where one is at rest, the moving object stops and the stationary object moves with the original velocity.
Worked Examples
Example 1: Calculate momentum
A 0.25 kg ball moves at 18 m/s.
Solution:
-
Use $p = mv$
-
$p = 0.25 \times 18 = 4.5$ kg·m/s
-
Momentum is in the direction of motion
Example 2: Impulse and velocity change
A 1.5 kg cart experiences a 12 N force for 0.50 s.
Solution:
-
Calculate impulse: $J = F\Delta t = 12 \times 0.50 = 6.0$ N·s
-
Impulse equals change in momentum: $J = \Delta p = m\Delta v$
-
Velocity change: $\Delta v = \frac{J}{m} = \frac{6.0}{1.5} = 4.0$ m/s
The cart's velocity increases by 4.0 m/s in the direction of the force.
Example 3: Perfectly inelastic collision
A 2.0 kg cart moving at 3.0 m/s collides and sticks to a 1.0 kg cart at rest.
Solution:
-
Initial momentum: $p_i = m_1v_1 + m_2v_2 = 2.0 \times 3.0 + 1.0 \times 0 = 6.0$ kg·m/s
-
Final mass (stuck together): $m_f = 2.0 + 1.0 = 3.0$ kg
-
Conservation: $p_f = p_i$
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Final velocity: $v_f = \frac{p_i}{m_f} = \frac{6.0}{3.0} = 2.0$ m/s
Example 4: Force from impulse
A 0.40 kg ball changes velocity from 12 m/s (right) to 8 m/s (left) in 0.020 s. Find the average force.
Solution:
-
Taking right as positive:
- Initial velocity: $v_i = +12$ m/s
- Final velocity: $v_f = -8$ m/s
-
Change in momentum: $$\Delta p = m(v_f - v_i) = 0.40 \times (-8 - 12) = 0.40 \times (-20) = -8.0 \text{ kg·m/s}$$
-
Average force: $$F = \frac{\Delta p}{\Delta t} = \frac{-8.0}{0.020} = -400 \text{ N}$$
-
The force is 400 N to the left (negative direction)
Example 5: Elastic collision
A 1.0 kg cart moving at 4.0 m/s collides elastically with a 2.0 kg cart at rest. Find the final velocities.
Solution:
For elastic collisions between two objects (object 2 initially at rest):
$$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{1.0 - 2.0}{1.0 + 2.0} \times 4.0 = \frac{-1}{3} \times 4.0 = -1.33 \text{ m/s}$$
$$v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} = \frac{2 \times 1.0}{1.0 + 2.0} \times 4.0 = \frac{2}{3} \times 4.0 = 2.67 \text{ m/s}$$
Cart 1 bounces back at 1.33 m/s; Cart 2 moves forward at 2.67 m/s.
Verification: Check momentum is conserved:
- Before: $1.0 \times 4.0 = 4.0$ kg·m/s
- After: $1.0 \times (-1.33) + 2.0 \times 2.67 = -1.33 + 5.33 = 4.0$ kg·m/s ✓
Common Misconceptions
-
Misconception: Momentum depends only on speed. Correction: Momentum is $p = mv$. Both mass AND velocity matter. A slow truck can have more momentum than a fast bicycle.
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Misconception: Impulse equals force. Correction: Impulse is $J = F\Delta t$. The same force applied for different times produces different impulses.
-
Misconception: Momentum is always conserved. Correction: Only in isolated systems with no external forces. External forces change total momentum.
-
Misconception: Kinetic energy is always conserved in collisions. Correction: Only in elastic collisions. In inelastic collisions, some KE is converted to other forms (sound, heat, deformation).
-
Misconception: Heavier objects always have more momentum. Correction: A light, fast object can have more momentum than a heavy, slow object.
Practice Questions
Easy (2 marks)
Find the momentum of a 3.0 kg object moving at 2.5 m/s.
- Use $p = mv$ (1)
- Correct value: $p = 3.0 \times 2.5 = 7.5$ kg·m/s with units (1)
Answer: 7.5 kg·m/s
Medium (4 marks)
A 0.40 kg ball changes velocity from 12 m/s (east) to 8 m/s (west) in 0.020 s. Find the average force.
- Correct change in velocity: $\Delta v = -8 - (+12) = -20$ m/s (1)
- Change in momentum: $\Delta p = 0.40 \times (-20) = -8.0$ kg·m/s (1)
- Force calculation: $F = \Delta p / \Delta t = -8.0 / 0.020 = -400$ N (1)
- Direction: 400 N west (1)
Answer: 400 N west
Hard (5 marks)
A 1.0 kg cart moving at 4.0 m/s collides elastically with a 2.0 kg cart at rest. Find the final velocities.
- State conservation of momentum (1)
- State conservation of kinetic energy for elastic collision (1)
- Apply elastic collision formulas or solve simultaneous equations (1)
- Correct final velocity of cart 1: $v_{1f} = -1.33$ m/s (1)
- Correct final velocity of cart 2: $v_{2f} = 2.67$ m/s (1)
Solution:
Using elastic collision formulas:
- $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{-1}{3} \times 4.0 = -1.33$ m/s
- $v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} = \frac{2}{3} \times 4.0 = 2.67$ m/s
Answer: Cart 1: 1.33 m/s backward; Cart 2: 2.67 m/s forward
Multiple Choice Questions
Test your understanding with these interactive questions:
Summary
- Momentum: $\vec{p} = m\vec{v}$ (units: kg·m/s)
- Impulse: $\vec{J} = \vec{F}\Delta t = \Delta\vec{p}$ (units: N·s)
- Conservation: In isolated systems, $\vec{p}{before} = \vec{p}{after}$
- Elastic collisions: Both momentum AND kinetic energy conserved
- Inelastic collisions: Only momentum conserved; KE is lost
Self-Assessment
Check your understanding:
After studying this section, you should be able to:
- Calculate momentum using $p = mv$
- Apply the impulse-momentum theorem
- Use conservation of momentum in collisions
- Distinguish elastic from inelastic collisions
- Explain why airbags reduce injury forces
Scientific Writing And Exam Support
When answering questions from this lesson, separate:
- the physical quantity being discussed,
- the model or law being applied,
- the mathematical relationship, including units,
- the conclusion in words.
For explanation questions, write in the pattern: claim -> physics reason -> consequence. For calculation questions, state the formula, substitute with units, calculate, then interpret the answer.
Tutor Context
Use this lesson context when the student asks about momentum and impulse, related calculations, representations, or scientific writing. Prefer a short diagnostic before re-teaching. Check whether the student is confusing closely related categories such as force, velocity, acceleration, field, energy, momentum, model evidence, or mathematical representation.
Useful tutor diagnostic:
Which quantity is changing here, what causes that change, and what unit should the final answer use?
Maintenance Loop
One-minute retrieval:
- State the key law, model, or relationship used in this lesson.
- Identify one common misconception that would lead to a wrong answer.
- Write one sentence that links the calculation or evidence back to the physical meaning.
Source Trace
This content record was materialised from:
edu/physics-prep/hsc/physics/textbook/module-2/sections/m2-4-momentum-impulse.qmdedu/physics-prep/hsc/physics/eduKG/lessons/program/year-11/module-2/T2W2L2-m2-4-momentum-impulse.qmd/Users/philiphaynes/devel/teaching/teaching/cromer/2026/classes/11PHY5/lessons/mod-2/lesson-3-momentum-impulse.qmd