Module 2 · Week 4 · Lesson

Momentum and Impulse

PH11-9

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.

Syllabus inquiry question

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

Content

Momentum

Momentum is the product of mass and velocity:

$$\vec{p} = m\vec{v}$$

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:

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

TypeMomentumKinetic EnergyExample
ElasticConservedConservedBilliard balls, atomic collisions
InelasticConservedNOT conservedCar crash, ball catches
Perfectly inelasticConservedMaximum KE lostObjects 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:

  1. Use $p = mv$

  2. $p = 0.25 \times 18 = 4.5$ kg·m/s

  3. 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:

  1. Calculate impulse: $J = F\Delta t = 12 \times 0.50 = 6.0$ N·s

  2. Impulse equals change in momentum: $J = \Delta p = m\Delta v$

  3. 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:

  1. Initial momentum: $p_i = m_1v_1 + m_2v_2 = 2.0 \times 3.0 + 1.0 \times 0 = 6.0$ kg·m/s

  2. Final mass (stuck together): $m_f = 2.0 + 1.0 = 3.0$ kg

  3. Conservation: $p_f = p_i$

  4. 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:

  1. Taking right as positive:

    • Initial velocity: $v_i = +12$ m/s
    • Final velocity: $v_f = -8$ m/s

  2. Change in momentum: $$\Delta p = m(v_f - v_i) = 0.40 \times (-8 - 12) = 0.40 \times (-20) = -8.0 \text{ kg·m/s}$$

  3. Average force: $$F = \frac{\Delta p}{\Delta t} = \frac{-8.0}{0.020} = -400 \text{ N}$$

  4. 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:

Common Misconceptions

Practice Questions

Easy (2 marks)

Find the momentum of a 3.0 kg object moving at 2.5 m/s.

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.

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.

Solution:

Using elastic collision formulas:

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

Self-Assessment

Check your understanding:

After studying this section, you should be able to:

Scientific Writing And Exam Support

When answering questions from this lesson, separate:

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.

Maintenance Loop

One-minute retrieval:

  1. State the key law, model, or relationship used in this lesson.
  2. Identify one common misconception that would lead to a wrong answer.
  3. Write one sentence that links the calculation or evidence back to the physical meaning.

Student Working