I added a check for the text input so that users have to key in the correct number of decimal places according to the precision of the instrument. For instance, a reading of 1 V should be recorded as 1.00 V and 1.5 V recorded as 1.50 V. Users need to read to half the smallest division, e.g. if the needle is between 2.4 and 2.5, they should input 2.45 V.
It took a while due to the need to adjust the equations used based on the position of the graphs, but here it is: https://www.geogebra.org/m/dfb53dps
The kinematics of a bouncing ball can be explained by considering the dynamics and forces involved in its motion. In this simulation, air resistance is assumed negligible. When a ball is dropped from a certain height and bounces off the ground, several key principles of physics come into play. Let’s break down the process step by step:
Free Fall: When the ball is released, it enters a state of free fall. During free fall, the only force acting on the ball is gravity. This force is directed downward and can be described by W = mg
W is the gravitational force. m is the mass of the ball. g is the acceleration due to gravity (approximately 9.81 m/s² near the surface of the Earth).
Impact with the Ground and Bounce: When the ball reaches the ground, it experiences a force due to the collision with the surface. This force is an example of a contact force and much larger than the gravitational force. This force depends on the elasticity of the ball and the surface it bounces off.
During the collision with the ground, the ball’s momentum changes rapidly. If the ball and the ground are both ideal elastic materials, the ball will bounce back with the same speed it had just before impact. In reality, some energy is lost during the collision, causing the bounce to be less than perfectly elastic. This simulation assumes elastic collisions.
Post-Bounce Motion: After the bounce, the ball starts moving upward. Gravity acts on it as it ascends, decelerating its motion until it reaches its peak height.
Second Descent: The ball then starts descending again, experiencing the force of gravity pulling it back down towards the ground.
This process continues with each bounce. In practice, with each bounce, some energy is lost due to the non-ideal nature of the collision and other dissipative forces like air resistance. As a result, each bounce is typically lower than the previous one until the ball eventually comes to rest. However, for simplicity, the simulation assumes no energy is lost during the collision and to dissipative forces.
An animated gif file is included here for use in powerpoint slides:
Did this simple interactive upon request by a colleague who is teaching the JC1 topic of Oscillations.
Based on the following question, this is used as a quick visual to demonstrate why there must be a minimum depth before the boat approaches harbour.
The rise and fall of water in a harbour is simple harmonic. The depth varies between 1.0 m at low tide and 3.0 m at high tide. The time between successive low tides is 12 hours. A boat, which requires a minimum depth of water of 1.5 m, approaches the harbour at low tide. How long will the boat have to wait before entering?
The equation of the depth of water H based on the amplitude of the tide a can be given by $H = H_o + a \cos \omega t$ where $H_o$ is the average depth of the water.
In preparing for blended learning lessons for my JC2 students, I tweaked the Gravitational Potential applet made last year for a similar display of the electric potential between two point charges. This is a testament to the similarities between the two concepts as well as the ease of adapting a GeoGebra applet for education.
We can scaffold students’ learning using this interactive applet by asking questions such as:
By observing the electric potential graph, are you able to find a point when the net field / force acting on a test charge is zero? What are the necessary conditions?
The slope of the sum of the electric potentials is analogous to that of a physical slope where a ball will roll downhill in the same way that a positive test charge will accelerate based on the potential gradient. However, this analogy will work differently for a test charge that is negative. Why?
Given that $E = -\dfrac{dV}{dx}$, where x is the distance from the point charge, is the direction of the E-field vector consistent with the negative of slope?
To paste this applet into SLS, use the following embed code. In SLS, create a new component within an Activity within a Lesson using the “+” button. Choose Text/Media and select the button that shows “</>” or reads Embed Website/App”. Copy and paste the following codes to the box.
<iframe scrolling="no" title="Electric Potential of Two Point Charges" src="https://www.geogebra.org/material/iframe/id/z8cr66wb/width/640/height/480/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/true/ctl/false" width="640px" height="480px" style="border:0px;"> </iframe>
By embedding the gravitational potential distance graph for two masses, a comparison can be made between the two. This will help students draw connections between the two concepts based on the fact that the forces both follow an inverse-square law.
This is the embed code for the applet on gravitational potential.
<iframe scrolling="no" title="Gravitational Potential between Two Planets" src="https://www.geogebra.org/material/iframe/id/ff55x6vr/width/638/height/478/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/true/rc/false/ld/false/sdz/false/ctl/false" width="638px" height="478px" style="border:0px;"> </iframe>
Using a chain of rubber bands, I swung a ball around in a vertical loop. This demonstration shows how the tension in an elastic band changes according to the position of the ball, by referring to the length of the elastic band.
Securing the elastic band to the ball with a shoelace
When the ball of mass $m$ is at the bottom of the loop, the centripetal force is given by the difference between tension $T_{bottom}$ and weight $mg$, where $T_{bottom}$ varies depending on the speed of the ball $v_{bottom}$ and the radius of the curvature $r_{bottom}$.
When the ball is at the top of the path, it is given by
$T_{top} + mg = \dfrac{mv_{top}^2}{r_{top}}$
As the weight is acting in the same direction to tension when the ball is at the top, a smaller tension is exerted by the elastic band to maintain a centripetal force. Therefore , $T_{bottom} > T_{top}$.
The GeoGebra app below shows a simpler version of a vertical loop – a circular path with a fixed radius $r$. Consider a ball sliding around a smooth circular loop. The normal contact force varies such that
$N_{bottom} = \dfrac{mv_{bottom}^2}{r} + mg$
$N_{top} = \dfrac{mv_{top}^2}{r} – mg$
It can be shown that the minimum height at which the ball must be released in order for it to complete the loop without losing contact with the track is 2.5 times the radius of the frictionless circular track.
If we were to consider the rotational kinetic energy required for the ball to roll, the required initial height will have to be 2.7 times the radius, as shown in the video below:
Many thanks to Dr Darren Tan for his input. Do check out his EJSS simulation of a mass-spring motion in a vertical plane, which comes with many more features such as the ability to vary the initial velocity of the mass, graphs showing the variation of energy and velocity, as well as an option for a mass-string motion as well.
Using the GeoGebra app above, I intend to demonstrate the relationship between total energy, kinetic energy and gravitational potential energy in a rocket trying to escape a planet’s gravitational field.
By changing the total energy of the rocket, you will increase the initial kinetic energy, thus allowing it to fly further from the surface of the planet. The furthest point to which the rocket can fly can be observed by moving the slider for “distance”. You will notice that the furthest point is where kinetic energy would have depleted.
Gravitational potential energy of an object is taken as zero at an infinite distance away from the source of the gravitational field. This means gravitational potential energy anywhere else takes on a negative value of $\dfrac{-GMm}{r}$. Therefore, the total energy of the object may be negative, even after taking into account its positive kinetic energy as total energy = kinetic energy + gravitational potential energy.
The minimum total energy needed for the rocket to leave the planet’s gravitational field is zero, as that will mean that the minimum initial kinetic energy will be equal to the increase in gravitational potential energy needed, according to the equation $\Delta U = 0 – (-\dfrac{GMm}{R_P})$, where $R_P$ is the radius of the planet.
Since $\dfrac{1}{2}mv^2 = \dfrac{GMm}{R_P}$, escape velocity, $v = \sqrt{\dfrac{2GM}{R_P}}$.
