Kinematics Graphs Interactive

Understanding motion in physics often involves analyzing displacement, velocity, and acceleration graphs. With the interactive GeoGebra graph at this link, you can dynamically explore how these concepts are connected.

How It Works

This interactive simulation lets you visualize an object’s motion and its corresponding displacement-time, velocity-time, and acceleration-time graphs. You can interact with the model in two key ways:

  1. Adjust Initial Conditions:
    • Move the dots on the graph to change the starting displacement, velocity, or acceleration.
    • Observe how these changes influence the overall motion of the object.
  2. Use the Slider to Animate Motion:
    • Slide through time to see how the object moves along its path.
    • Watch the displacement vector, velocity vector, and acceleration vector update in real time.

Key Observations

  • When displacement changes, the velocity and acceleration graphs adjust accordingly.
  • A constant acceleration results in a straight-line velocity graph and a quadratic displacement graph.
  • Negative acceleration (deceleration) slows the object down and can cause direction reversals.
  • If velocity is constant, the displacement graph is linear, and acceleration remains at zero.

Why This is Useful

This GeoGebra tool is perfect for students and educators looking to build intuition about kinematics. Instead of just solving equations, you get a visual and hands-on way to see the relationships between these key motion variables.

Try it out yourself and experiment with different conditions to deepen your understanding of motion!

Exploring Motion with a Ticker Tape Timer

This week, I conducted a lesson on motion for my IP3 class using a simple yet effective tool: a simulated ticker tape timer. The objective was to help students develop an intuitive understanding of uniform and non-uniform motion by actively engaging in an experiment.

Introduction to the Ticker Tape Timer

To kickstart the lesson, I showed my students a YouTube video that explains how a ticker tape timer works:

This video provided a visual demonstration of how a ticker tape timer marks regular intervals on a moving strip of paper, allowing us to analyze motion quantitatively.

Hands-On Experiment: Simulating a Ticker Tape Timer

After the video, I had students pair up for a hands-on activity. Instead of using an actual ticker tape timer, we simulated the process using paper strips cut from A3-sized sheets. Each pair had one student act as the “moving arm,” responsible for placing dots on the strip, while the other played the role of the “puller,” responsible for pulling the paper strip at different speeds.

To ensure a consistent time interval between each dot, I used a Metronome App that I created:

This app produces a steady rhythm at 120 beats per minute, meaning that the interval between each beep (and consequently each dot) is 0.5 seconds. To improve accuracy, the student acting as the moving arm was instructed to close their eyes and focus solely on the beep.

Step 1: Recording Uniform Motion

In the first trial, the puller was asked to pull the paper at a constant rate. As the paper moved steadily, the moving arm marked dots at regular intervals based on the metronome beat. After completing the trial, students used a ruler to measure the distances between successive dots. Since the time interval was fixed, they could easily calculate the speed of the paper by using:

Step 2: Recording Accelerated Motion

Next, the students switched roles. This time, the new puller was asked to gradually increase the speed of the paper. As expected, the spacing between dots increased progressively, providing a clear visual representation of acceleration. This led to discussions on how motion can be analyzed using dot patterns and how acceleration differs from uniform motion.

Reflections and Key Takeaways

This activity was highly effective in reinforcing key motion concepts. Since we do not have an actual ticker tape machine, it allowed students to engage in a hands-on simulation while visually and physically experience motion rather than just reading about it.

Next Steps

To extend this lesson, I plan to introduce velocity-time graphs and have students plot their measured speeds to analyze changes in motion further. Additionally, incorporating digital tools like video analysis with Tracker software could help reinforce these concepts further.

If you have any feedback or suggestions, feel free to share them in the comments below!

Symposium on Tech for Engagement at SISTC 2024

This deck of slides are the ones I will be using for the Symposium on “Leveraging Technology for Engaging and Effective Learning” at the Singapore International Science Teachers’ Conference (SISTC) 2024 on Day 2 of the Conference (20 November). Feel free to download for your reference.

A new AI-enabled workflow for generating diagrams for questions

In recent years, AI tools like DALL-E and MidJourney have gained popularity for generating stunning, artistic images. However, when it comes to creating scientific illustrations, particularly for exams, we need accuracy, clarity, and precision. This is where LaTeX, a document preparation system widely used in academia, shines. Combined with the intelligence of ChatGPT, LaTeX can produce exactly the images you need without the artistic flair or ambiguity that tools like DALL-E might introduce.

The process of generating science exam images using ChatGPT and LaTeX is straightforward:

  1. Describe the Diagram: You begin by providing a detailed description of the diagram or scientific illustration you need. For example, “Generate a diagram using Latex showing a free-body diagram of a block on an inclined plane with friction.”
  2. ChatGPT Generates LaTeX Code: Based on your description, ChatGPT can generate the appropriate LaTeX code, using packages like TikZ or PGFPlots to create the required scientific diagram. The output might look like this:
    \documentclass{standalone}
    \usepackage{tikz}
    \begin{document}
    \begin{tikzpicture}
    % Inclined plane
    \draw[thick] (0,0) -- (4,2);
    \draw[thick] (0,0) -- (4,0);
    % Block
    \draw[fill=gray] (2,1) rectangle (2.5,1.5);
    % Forces
    \draw[->,thick] (2.25,1.25) -- (2.25,2.25) node[above] {Normal Force};
    \draw[->,thick] (2.25,1.25) -- (2.25,0.25) node[below] {Weight};
    \draw[->,thick] (2.25,1.25) -- (3.25,1.25) node[right] {Friction};
    \end{tikzpicture}
    \end{document}
  3. Compile the LaTeX Code: Once the code is generated, you can compile it using any LaTeX editor (such as Overleaf or a local LaTeX distribution).
  4. Check for Errors: The diagram is unlikely to look perfect in the first iteration. For example, the above code gives the following:
  5. Edit: You can either instruct ChatGPT to modify specific sections of the diagram or make the changes yourself. After 3 more iterations, for example, ChatGPT produced the following codes:
    \begin{tikzpicture}
    % Inclined plane
    \draw[thick] (0,0) -- (4,2);
    \draw[thick] (0,0) -- (4,0);
    % Block (rotated to match the slope)
    \draw[fill=gray, rotate around={26.565:(2.25,1.25)}] (2,1) rectangle (2.5,1.5);
    % Forces (adjusted for friction up the slope)
    % Normal Force (perpendicular to the slope)
    \draw[->,thick] (2.25,1.25) -- ++(-0.447,0.894) node[above left] {Normal Force};
    % Weight (straight down)
    \draw[->,thick] (2.25,1.25) -- (2.25,0.25) node[below] {Weight};
    % Friction (along the slope, now pointing up the incline)
    \draw[->,thick] (2.25,1.25) -- ++(0.894,0.447) node[above right] {Friction};
    \end{tikzpicture}
  6. This is the output image:
  7. Integrate into Teaching Materials: Once you are satisfied with the output, the compiled image can then be saved as a PDF, PNG, or any other image format and directly embedded into your exam materials.

The following are similar images made using the same workflow and their corresponding codes, which I made changes to manually instead as it was faster for me once I became familiar with the coordinate-system based drawing method.

      \begin{tikzpicture}
        
        % Draw the base container (mercury reservoir)
        \draw[draw=none, fill=gray!30] (-2,-0.2) rectangle (2,-1);
        \draw[thick] (-2,-1) -- (-2,0.4);
        \draw[thick] (2,-1) -- (2,0.4);
        \draw[thick] (-2,-1) -- (2,-1);
        \draw[thick] (-2,-0.2) -- (-0.1,-0.2);
        \draw[thick] (0.1,-0.2) -- (2,-0.2);
                
        
        % Mercury inside the tube
        \draw[draw=none, fill=gray!30] (-0.1,3) -- (-0.1,-0.3) -- (0.1,-0.3) -- (0.1,3) -- cycle;
        \draw[thick] (-0.1,3) -- (0.1,3);
        
        % Draw the glass tube
        \draw[thick] (0.1,4) -- (0.1,-0.4); % Tube
        \draw[thick] (-0.1,4) -- (-0.1,-0.4); % Tube
        % Label for the mercury level
        \draw[<-] (0,2) -- (-1.1,2) node[left]{mercury column};
        \draw[<-] (0,3.7) -- (-1.1,3.7) node[left]{vacuum};
        \draw[thick] (0,4) ++(0:0.1cm) arc (0:180:0.1cm);
        
        % Labels for height
        \draw[<->] (0.5,3) -- (0.5,-0.2) node[midway,right]{h};
        
        % Add some text labels
        \node at (0,-1.5) {mercury reservoir};
        
        \end{tikzpicture} 
\begin{tikzpicture}

        % Draw principal axis
        \draw[thick] (-5,0) -- (2,0) node at (-4.5,0.2) {principal axis};
    
        % Draw the lens
        \draw[thick] (-2,-1.5) arc[start angle=270, end angle=90, x radius=0.15cm, y radius=1.5cm];
        \draw[thick] (-2,1.5) arc[start angle=90, end angle=270, x radius=-0.15cm, y radius=1.5cm];
    
        % Draw the focal points
        \node at (1, 0) [above] {focal point, $F$};
        \draw[fill] (1, 0) circle [radius=0.05];
        \node at (-2, 0) [below] {optical centre};
        \draw[fill] (-2, 0) circle [radius=0.05];    
        % Draw the 3 rays parallel to the principal axis (before hitting the lens)
        \draw[thick] (-5,1) -- (-2,1);
        \draw[thick] (-5,0.5) -- (-2,0.5);
        \draw[thick] (-5,-0.5) -- (-2,-0.5);
        \draw[thick] (-5,-1) -- (-2,-1);
        \draw[thick, ->] (-4,1) -- (-3,1);
        \draw[thick, ->] (-4,0.5) -- (-3,0.5);
        \draw[thick, ->] (-4,-0.5) -- (-3,-0.5);
        \draw[thick, ->] (-4,-1) -- (-3,-1);
        \draw[thick, ->] (-4,0) -- (-3,0);
    
        % Rays converging to the focal point F
        \draw[thick] (-2,1) -- (1,0);
        \draw[thick] (-2,0.5) -- (1,0);
        \draw[thick] (-2,-0.5) -- (1,0);
        \draw[thick] (-2,-1) -- (1,0);
        \draw[thick, ->] (-2,1) -- (-0.5,0.5);
        \draw[thick, ->] (-2,0.5) -- (-0.5,0.25);
        \draw[thick, ->] (-2,0) -- (-0.5,0);
        \draw[thick, ->] (-2,-0.5) -- (-0.5,-0.25);
        \draw[thick, ->] (-2,-1) -- (-0.5,-0.5);
    
        \node at (-0.5, 1.3) {converging lens};
        % Focal length
        \draw[thick, <->] (-2,-1.2) -- (1,-1.2);
        \node at (-0.5, -1.5) {focal length, $f$};        
    \end{tikzpicture}
 \begin{tikzpicture}
        % Draw the wave
        \draw[thick, domain=0:6.28, smooth, variable=\x] plot ({\x}, {sin(2*\x r)});
        
        % Draw the x-axis
        \draw[->] (0, 0) -- (6.5, 0) node at (6.4,0.3) {distance};
        
        % Draw the y-axis
        \draw[->] (0, -1.5) -- (0, 1.5) node[above] {displacement};
        
        % Draw the amplitude arrow
        \draw[<->, thick] (0.785, 0) -- (0.785, 1) node[midway, right] {amplitude};
    
        % Draw the wavelength arrow
        \draw[<->, thick] (0.785, 1.1) -- (3.925, 1.1) node at (2.355,1.4) {wavelength};
    
        % Label the points
        \node at (-0.2, 1) {$A$};
        \node at (-0.3, -1) {$-A$};
        \node at (3.14, -0.2) {$\lambda$};
        \node at (6.28, -0.2) {2$\lambda$};
        \end{tikzpicture}

Turning Effect due to Non-Perpendicular Forces

When applying the principle of moments to calculate the magnitude of a force creating a turning effect, where the force is not perpendicular to the length of the object, there are two approaches.

Take the following problem:

A uniform rectangular beam has negligible thickness and weight 850 N. Its length is 5.0 m and it is in contact with the top of a support at point P. P is 0.80 m from one end of the beam.

The beam is held stationary, at an angle of 30° to the horizontal, by a rope that is attached to the bottom corner of the other end of the beam.

Calculate the magnitude of the force T on the beam due to the tension in the rope.

Approach 1: Identifying the perpendicular distance between line of action of the force and pivot

In the applet above, check the box that says “Show perp dist” to see the lines representing the perpendicular distances between each line of action of the force and the pivot P.

Taking moments about P,

Clockwise moment due to T = Anti-clockwise moment due to W

$$T \times (5.0\text{ m} – 0.8\text{ m}) \sin 30^{\circ} = W \times (\dfrac{5.0\text{ m}}{2} – 0.8\text{ m}) \cos 30^{\circ}$$

$$T \times 2.10 \text{ m} = 850 \text{ N} \times 1.472 \text{ m}$$

$$T = 596 \approx 600 \text{ N}$$

Approach 2: Resolving the forces to obtain components of forces perpendicular to the beam

In the same applet, check the box that says “Resolve forces”

Taking moments about P,

Clockwise moment due to $T_{\perp}$ = Anti-clockwise moment due to $W_{\perp}$

$$T \sin 30^{\circ} \times (5.0\text{ m} – 0.8\text{ m}) = 850 \text{ N} \cos 30^{\circ} \times (\dfrac{5.0\text{ m}}{2} – 0.8\text{ m})$$

$$T \times 0.500 \times 4.2 \text{ m} = 850 \text{ N} \times 0.866 \times 1.7 \text{ m}$$

$$T = 596 \approx 600 \text{ N}$$

DC Circuits Practice

The simulation below allows students to practise calculating potential differences and currents of a slightly complex circuit, involving three different modes that can be toggled by clicking on the switch.

Link: https://www.geogebra.org/m/jkckp9pr

Mode 1: Two Resistors in Series

When resistors \( R_1 \) and \( R_2 \) are connected in series, the total resistance is simply the sum of the individual resistances:

\[ R_{\text{total}} = R_1 + R_2 \]

The current \( I \) through the circuit is given by Ohm’s Law:

\[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{V_{\text{total}}}{R_1 + R_2} \]

where \( V_{\text{total}} \) is the total potential difference supplied by the source.

The potential difference across each resistor can be calculated using:

\[ V_1 = I \cdot R_1, \quad V_2 = I \cdot R_2 \]

Mode 2: \( R_1 \) and \( R_3 \) in Parallel, \( R_2 \) in Series

In this mode, resistors \( R_1 \) and \( R_3 \) are in parallel, and \( R_2 \) is in series with the combination. First, calculate the equivalent resistance of the parallel combination:

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_3} \]

Thus, the total resistance is:

\[ R_{\text{total}} = R_{\text{parallel}} + R_2 \]

The current through the circuit is:

\[ I = \frac{V_{\text{total}}}{R_{\text{total}}} \]

The potential difference across \( R_2 \) is:

\[ V_2 = I \cdot R_2 \]

Since \( R_1 \) and \( R_3 \) are in parallel, they share the same potential difference:

\[ V_1 = V_3 = V_{\text{total}} – V_2 \]

The current through each parallel resistor can be found using Ohm’s Law:

\[ I_1 = \frac{V_1}{R_1}, \quad I_3 = \frac{V_3}{R_3} \]

Mode 3: \( R_1 \) and \( R_2 \) in Series, \( R_3 \) in Parallel

Here, resistors \( R_1 \) and \( R_2 \) are connected in series, and the combination is in parallel with \( R_3 \). First, calculate the resistance of the series combination:

\[ R_{\text{series}} = R_1 + R_2 \]

Then, find the total resistance of the parallel combination:

\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3} \]

The total current is:

\[ I = \frac{V_{\text{total}}}{R_{\text{total}}} \]

The voltage across the parallel combination is the same for both branches:

\[ V_1 + V_2 = V_3 = V_{\text{total}} \]

The current through \( R_3 \) is:

\[ I_3 = \frac{V_3}{R_3} \]

The current through \( R_1 \) and \( R_2 \), which are in series, is the same:

\[ I_{\text{series}} = \frac{V_{\text{total}}}{R_1 + R_2} \]

The voltage across each series resistor is:

\[ V_1 = I_{\text{series}} \cdot R_1, \quad V_2 = I_{\text{series}} \cdot R_2 \]