12  Momentum and Impulse

12.1 Syllabus inquiry question

  • How can momentum models predict outcomes of interactions?
Feynman Insight

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.

12.2 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.

12.3 Content

12.3.1 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
Why Momentum Matters

Momentum is conserved in collisions, making it a powerful tool for predicting outcomes without knowing the detailed forces.

12.3.2 Interactive: Comparing Momentum

Different objects with the same momentum:

12.3.3 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.

Impulse-Momentum Theorem

\[\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}\]

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)

12.3.4 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

12.3.5 Conservation of Momentum

In an isolated system (no external forces), total momentum is conserved:

\[\vec{p}_{before} = \vec{p}_{after}\]

For two objects: \[m_1\vec{v}_{1i} + m_2\vec{v}_{2i} = m_1\vec{v}_{1f} + m_2\vec{v}_{2f}\]

12.3.6 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

12.3.7 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
Key Point

Momentum is ALWAYS conserved in collisions (if the system is isolated). Kinetic energy is only conserved in elastic collisions.

12.3.8 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.

12.4 Worked Examples

12.4.1 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

12.4.2 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.

12.4.3 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

12.4.4 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)

12.4.5 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 ✓

12.5 Common Misconceptions

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.

  • 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.

12.6 Practice Questions

12.6.1 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

12.6.2 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

12.6.3 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

12.7 Multiple Choice Questions

Test your understanding with these interactive questions:

12.8 Summary

Key Takeaways
  • 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

12.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: