Momentum Revision Notes for HSC SSCE Physics

Momentum

What is momentum?

Momentum is a fundamental concept in physics that helps us understand how objects move and interact with each other. Think about the difference between being hit by a tennis ball and being hit by a watermelon, both moving at the same speed. The watermelon would have a much greater effect, even though both are moving at the same velocity. This difference can be explained using momentum.

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Real-World Analogy:

The tennis ball versus watermelon example illustrates why momentum depends on both mass and velocity. Even at identical speeds, the watermelon's greater mass gives it significantly more momentum, resulting in a much more forceful impact.

Understanding momentum is particularly important when analysing collisions, such as car crashes. During a collision, there are large forces acting on vehicles and passengers, and energy is transformed as the state of motion changes. Momentum helps us predict and understand these interactions.

Definition and formula

Momentum ( $\vec{p}$ ) is defined as the product of an object's mass ( $m$ ) and its velocity ( $\vec{v}$ ):

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

Key characteristics of momentum:

Units: kilogram metres per second ( $\text{kg} \cdot \text{m} \cdot \text{s}^{-1}$ )
Type: Vector quantity (has both magnitude and direction)
Direction: Same direction as the object's velocity

chatImportant

Because momentum is a vector, we must consider its direction when performing calculations. An object moving north has different momentum from the same object moving south at the same speed.

Calculating momentum

Let's look at a practical example of calculating momentum.

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Worked Example: Momentum of a tomato

During a food fight, a child is hit by a $150 \text{ g}$ tomato travelling at $10 \text{ m} \cdot \text{s}^{-1}$ . What is the magnitude of the momentum of the tomato?

Solution:

Answer	Logic
$m = 150 \text{ g}$ ; $v = 10 \text{ m} \cdot \text{s}^{-1}$	Identify the relevant data
$m = 0.15 \text{ kg}$	Convert to SI units (grams to kilograms)
$p = mv$	Use the momentum formula
$p = (0.15 \text{ kg}) \times (10 \text{ m} \cdot \text{s}^{-1})$	Substitute the known values
$p = 1.5 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1}$	Calculate the final answer

The momentum of the tomato is 1.5 kg·m·s⁻¹.

chatImportant

Exam tip: Always convert masses to kilograms before calculating momentum. This is a common source of errors in exams.

Momentum and Newton's laws

Sir Isaac Newton originally formulated his second law using the concept of momentum rather than acceleration. This connection is important for understanding how forces relate to changes in motion.

Newton's second law in terms of momentum

Newton's second law can be written as:

$\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t}$

This tells us that the net force acting on an object equals the rate of change of its momentum.

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We can show this is equivalent to the more familiar form ( $\vec{F}_{\text{net}} = m\vec{a}$ ). If we substitute the definition of momentum and assume mass is constant:

$\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = \frac{\Delta(m\vec{v})}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$

Connection to Newton's third law

Newton's third law states that forces come in equal and opposite pairs: $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$ .

Using the momentum form of Newton's second law:

$\vec{F}_{\text{A on B}} = \frac{\Delta \vec{p}_B}{\Delta t} = -\frac{\Delta \vec{p}_A}{\Delta t} = -\vec{F}_{\text{B on A}}$

This leads to an important result:

$\Delta \vec{p}_A = -\Delta \vec{p}_B$

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Key Connection to Conservation:

In any interaction between two objects, the change in momentum of one object is equal in magnitude but opposite in direction to the change in momentum of the other object. This is essentially a statement of conservation of momentum.

The law of conservation of momentum

For any conserved quantity, the total amount of that quantity in the universe, or in any isolated system, remains constant. Momentum is one such conserved quantity.

chatImportant

The law of conservation of momentum states:

The sum of the momentum of all objects in an isolated system remains constant.

Mathematically:

$\vec{p}_{\text{net}} = \sum \vec{p} = \text{constant}$

or equivalently:

$\sum \vec{p}_{\text{initial}} = \sum \vec{p}_{\text{final}}$

Important considerations when applying conservation of momentum:

Vector nature: Momentum is a vector, so we must add momenta using vector addition
Sign convention: In one dimension, define positive and negative directions carefully
Components: In two dimensions, use vector decomposition and add components separately
Isolated system: The law applies to systems with no external forces

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Worked Example: Falling textbook and Earth

A $1.5 \text{ kg}$ maths textbook falls through a height of $2.0 \text{ m}$ when dropped from a window. The book falls due to its interaction with Earth through Earth's gravitational field. Assume the book starts at rest and ignore air resistance.

Part 1: What is the change in momentum of the book due to the fall?

Answer	Logic
$m_b = 1.5 \text{ kg}$ ; $\Delta h = 2.0 \text{ m}$ ; $u_b = 0.0 \text{ m} \cdot \text{s}^{-1}$ ; $v_b = ?$	Identify relevant data
$\sum E_{\text{initial}} = \sum E_{\text{final}}$	Use conservation of energy
$m_b g \Delta h = \frac{1}{2}m_b v_b^2$	Expand (noting $u_b = 0$ )
$v_b = \sqrt{2g\Delta h}$	Rearrange for $v_b$
$\Delta p_b = m_b v_b = m_b \sqrt{2g\Delta h}$	Expression for momentum change
$= (1.5 \text{ kg})\sqrt{2(9.8 \text{ m} \cdot \text{s}^{-2})(2.0 \text{ m})}$	Substitute values
$= 9.39 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1}$	Calculate
$\Delta p_b = 9.4 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1}$	Final answer with correct significant figures

Part 2: What is the change in momentum of Earth?

Momentum is conserved in any interaction:

$\Delta \vec{p}_b = -\Delta \vec{p}_{\text{Earth}}$

Therefore: Δp_Earth = -9.4 kg·m·s⁻¹

Part 3: What is the change in speed of Earth?

Answer	Logic
$\Delta p_{\text{Earth}} = m_{\text{Earth}} \Delta v_{\text{Earth}}$	Relate momentum to velocity
$\Delta v_{\text{Earth}} = \frac{\Delta p_{\text{Earth}}}{m_{\text{Earth}}}$	Rearrange
$m_{\text{Earth}} = 5.97 \times 10^{24} \text{ kg}$	Look up Earth's mass
$\Delta v_{\text{Earth}} = \frac{-9.4 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1}}{5.97 \times 10^{24} \text{ kg}}$	Substitute values
$= -1.57 \times 10^{-24} \text{ m} \cdot \text{s}^{-1}$	Calculate
$\Delta v_{\text{Earth}} = -1.6 \times 10^{-24} \text{ m} \cdot \text{s}^{-1}$	Final answer

The negative sign indicates that Earth's velocity change is opposite to that of the book. This value is so incredibly small that we cannot observe changes in Earth's velocity due to its interactions with everyday objects.

Conservation of momentum in one dimension

When applying conservation of momentum in one dimension, we must be careful with signs. Define a positive direction (e.g., to the right, upward, north) and ensure all velocity and momentum values have the correct sign.

Investigation 6.1: Conservation of momentum in one dimension

Aim: To verify conservation of momentum in one-dimensional collisions.

Materials:

2 dynamics trolley cars
Weighing scale
Motion-sensor and data-logger
Spring with loops at each end
String
Scissors

Risk assessment:

What are the risks?	How to stay safe
The spring may flick back or flick an object into a person's eye	Wear safety glasses when working with springs

Method:

Weigh the two dynamics trolley cars
Tie the trolleys together using string with a compressed spring between them (see Figure 6.2)
Position motion-sensors so you can measure the velocity of each car
With the cars at rest, cut the string holding them together
Measure and record the velocities of each trolley after cutting the string
Repeat at least three times using the same spring compression each time

Results:

Record the mass of each car and your measured velocities in a table:

	$v$ of Car A ( $\text{m} \cdot \text{s}^{-1}$ )	$p$ of Car A ( $\text{kg} \cdot \text{m} \cdot \text{s}^{-1}$ )	$v$ of Car B ( $\text{m} \cdot \text{s}^{-1}$ )	$p$ of Car B ( $\text{kg} \cdot \text{m} \cdot \text{s}^{-1}$ )	$\Delta p$ of system ( $\text{kg} \cdot \text{m} \cdot \text{s}^{-1}$ )
Trial 1
Trial 2
...
Average

Analysis:

Calculate the momentum of each trolley car after the string is cut for each trial
Define positive and negative directions clearly
Since the cars started at rest (initial momentum = zero), find the change in momentum of the system
Calculate average values and uncertainties

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Expected Results:

The results should show that the total change in momentum of the system is zero (within experimental uncertainty), confirming conservation of momentum. Any deviations can be attributed to friction, air resistance, or variations in initial conditions.

Conservation of momentum in two dimensions

When dealing with collisions in two dimensions, we apply conservation of momentum separately in each direction:

$\sum p_{\text{initial}, x} = \sum p_{\text{final}, x}$

$\sum p_{\text{initial}, y} = \sum p_{\text{final}, y}$

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This approach uses vector decomposition: we break each momentum vector into its $x$ and $y$ components, apply conservation in each direction separately, then combine the results to find the final velocity vector.

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Worked Example: Pool ball collision

In a game of pool, the cue ball moves with initial velocity $3.0 \text{ m} \cdot \text{s}^{-1}$ N30°W and hits a stationary red ball. The red ball moves off with speed $2.5 \text{ m} \cdot \text{s}^{-1}$ N45°W. Calculate the speed and direction of the cue ball's recoil. The balls have the same mass.

Given information:

$m_c = m_r$ (same mass)
Before collision: $u_c = 3.0 \text{ m} \cdot \text{s}^{-1}$ N30°W; $u_r = 0.0 \text{ m} \cdot \text{s}^{-1}$
After collision: $v_r = 2.5 \text{ m} \cdot \text{s}^{-1}$ N45°W; $v_c = ?$

Step 1: Apply conservation in the x-direction

$\sum p_{\text{initial}, x} = \sum p_{\text{final}, x}$

$m_c u_{c,x} + m_r u_{r,x} = m_c v_{c,x} + m_r v_{r,x}$

Since $m_c = m_r$ and $u_r = 0$ :

$u_{c,x} = v_{c,x} + v_{r,x}$

$v_{c,x} = u_{c,x} - v_{r,x}$

Taking west as negative $x$ -direction:

$v_{c,x} = (-3.0 \text{ m} \cdot \text{s}^{-1} \sin 30°) - (-2.5 \text{ m} \cdot \text{s}^{-1} \sin 45°)$

$v_{c,x} = 0.268 \text{ m} \cdot \text{s}^{-1}$

Step 2: Apply conservation in the y-direction

$u_{c,y} = v_{c,y} + v_{r,y}$

$v_{c,y} = u_{c,y} - v_{r,y}$

Taking north as positive $y$ -direction:

$v_{c,y} = (3.0 \text{ m} \cdot \text{s}^{-1} \cos 30°) - (2.5 \text{ m} \cdot \text{s}^{-1} \cos 45°)$

$v_{c,y} = 0.830 \text{ m} \cdot \text{s}^{-1}$

Step 3: Calculate magnitude and direction

$v_c = \sqrt{v_{c,x}^2 + v_{c,y}^2} = \sqrt{(0.268)^2 + (0.830)^2}$

$v_c = 0.87 \text{ m} \cdot \text{s}^{-1}$

For direction:

$\tan \theta = \frac{v_{c,y}}{v_{c,x}} = \frac{0.830}{0.268}$

$\theta = 72°$

Final answer: v_c = 0.87 m·s⁻¹, N18°E

chatImportant

Exam tip: Always draw a clear diagram showing the velocity components before and after the collision. This helps prevent sign errors.

Investigation 6.2: Collisions in two dimensions

Aim: To describe and analyse interactions in two dimensions using video analysis.

Materials:

2 billiard balls of different colours
Large sheet of paper (butcher's paper) with grid
Tape, pencil
Webcam with stand (or phone with selfie-stick)
Stopwatch
Weighing scale

Risk assessment:

What are the risks?	How to stay safe
Billiard balls can hurt your foot if they fall on it	Wear enclosed shoes

Method:

Weigh the two billiard balls
Draw a grid on the paper with lines spaced $1 \text{ cm}$ apart
Tape the paper to the bench and mark a starting position for the first ball
Place the stopwatch face-up on the paper
Mount the camera looking down at the paper to capture the whole area
Place the first ball on the mark
Start the stopwatch and camera recording
Gently roll the second ball towards the first to create a collision
Repeat several times

Analysis:

Select frames before the collision to find the initial velocity of the incoming ball
Use the grid to measure displacement ( $\Delta x$ and $\Delta y$ )
Use the stopwatch reading to find time intervals ( $\Delta t$ )
Calculate velocities: $v_x = \Delta x / \Delta t$ and $v_y = \Delta y / \Delta t$
Calculate momenta: $p_x = m v_x$ and $p_y = m v_y$
Verify that $\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}$ and $\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}$

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Key Finding:

This investigation confirms that momentum is conserved in each direction independently during two-dimensional collisions.

bookmarkSummary

Key Points to Remember:

Momentum is the product of mass and velocity: $\vec{p} = m\vec{v}$
Momentum is a vector quantity with units of $\text{kg} \cdot \text{m} \cdot \text{s}^{-1}$
Conservation of momentum: In any isolated system, total momentum remains constant ( $\sum \vec{p} = \text{constant}$ )
Newton's third law is a statement of conservation of momentum: in any interaction, $\Delta \vec{p}_A = -\Delta \vec{p}_B$
In two dimensions, apply conservation separately in $x$ and $y$ directions using vector components

Momentum (HSC SSCE Physics): Revision Notes

Momentum

What is momentum?

Definition and formula

Calculating momentum

Momentum and Newton's laws

Newton's second law in terms of momentum

Connection to Newton's third law

The law of conservation of momentum

Conservation of momentum in one dimension

Investigation 6.1: Conservation of momentum in one dimension

Conservation of momentum in two dimensions

Investigation 6.2: Collisions in two dimensions

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