22 September

Misconception: Skydiver goes up when parachute opens

Seng Kwang Tan GeoGebra, IP3 03 Dynamics 0 Comments

When a parachute opens, many people think the parachutist suddenly shoots upward. This is not what really happens. The parachutist is always moving downward, but the parachute causes a very sharp deceleration. The large canopy produces a big upward drag force that slows the fall dramatically.

When people see videos of parachutists opening their parachutes, the camera angle can create a powerful illusion that the parachutist suddenly shoots upward. What really happens is that the parachutist decelerates sharply while the camera, usually attached to another skydiver, continues falling at almost the same high speed.

From the perspective of the camera, the parachutist with the open parachute is no longer keeping pace in the fall. The camera-holder is still dropping rapidly, but the parachutist has slowed down. In the video, this relative motion makes it look like the parachutist has bounced upward, when in fact they are still moving downward—just not as quickly.

The physics of air resistance shows clearly why the velocity cannot turn upward. Air resistance always acts opposite to the velocity and its size depends on speed. Before and after the parachute opens, the parachutist’s velocity is downward, so the drag force must be upward. If the parachutist’s velocity really were upward, then the drag would have to point downward. In that case, the resultant force would also point downward, making the acceleration greater than gravity—something we never observe. Instead, the drag force remains upward, which proves the parachutist is still moving downward the whole time. The chute simply reduces that downward speed to a safe value.

For a simulation on how the forces and velocity change with time, refer to this GeoGebra app.

19 September

Moving charge between two charged spheres

Seng Kwang Tan 14 Electric Fields 1 Comment

This simulation is made upon request by a colleague teaching JC2 this year.

The motion of a mobile charge between two source charges is governed by Coulomb’s law ($F = \dfrac{Q_1Q_2}{4\pi\epsilon_0r^2}$) and the electric field. Each source charge produces a field in space, exerting a force on the test charge according to $F = qE$. The total field is the vector sum of all source charges, with positive charges moving along the field and negative charges moving opposite to it.

The test charge’s acceleration depends on the net force, changing its velocity and trajectory according to Newton’s second law. Its motion shows how attractive and repulsive forces combine, providing an intuitive view of electrostatic interactions and field lines.

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16 September

A dynamics problem – apparent weight

Seng Kwang Tan IP3 03 Dynamics 0 Comments

This is a problem posed by a student today:

A boy of mass 40 kg is standing on a weighing machine inside a lift which is moving upwards. At a certain moment, the speed of the lift is $3.0 \text{ m s}^{-1}$ and it is decelerating at $2.0 \text{ m s}^{-2}$. What is the reading (in kg) shown on the weighing scale?

My advice for any dynamics problem is to take the direction of acceleration as positive. This way, you can apply Newton’s second law directly without worrying about inserting extra negative signs for deceleration. After all, a negative value of acceleration simply comes from the chosen sign convention—it reflects direction. For simplicity, putting the direction of acceleration as positive (in this case, downward is positive), we have

$$F_{net} = ma$$

The net force should therefore, be $W – N$, where $N$ is smaller than $W$.

$$W – N = ma$$

Substituting, we have

$$mg – N = ma$$

$$(40 \text{ kg} \times 9.81 \text{ N kg}^{-1}) – N = 40 \text{ kg} \times 2.0 \text{ m s}^{-2}$$

$$N = 312 \text{ N}$$

Note that $N$ is also known as apparent weight.

The mass reading due to this normal force = $\dfrac{312\text{ N}}{9.81 \text{ N kg}^{-1}} = 32 \text{ kg}$

(Notice here that the speed of the lift is irrelevant.)

The GeoGebra simulation allows you to modify the acceleration and observe the change in the vector representing the normal contact force, which is force acting on the weighing scale.

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For a real-life experiment along with a visualisation of the changes in the vectors based on this scenario, check out my video and simulation.

14 September

Kinematics Equations – which to use?

Seng Kwang Tan IP3 02 Kinematics 0 Comments
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When tackling kinematics problems, a quick way to decide which equation to use is to look carefully at which variables are given and which one you need to find. The four equations involve only five quantities — $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration), $t$ (time), and $s$(displacement). Each equation links four of them, leaving one out. So if, for example, time $t$ is not mentioned anywhere in the problem, the best choice is usually the equation $v^2 = u^2 + 2as$, since that one does not involve $t$. By matching the known and unknown quantities against the “missing variable” in each equation, you can quickly narrow down the correct equation to apply.

13 September

Electrostatic Precipitator Simulation

Seng Kwang Tan IP4 10 Static Electricity 0 Comments
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Interact with the simulation of an electrostatic precipitator (ESP) above. The ESP is a big air cleaner for factories and power plants. It uses electricity to pull tiny dust and mist out of smoky gas before the gas goes up the stack. Inside the box are thin wires kept at a very high voltage next to large metal plates that are grounded. The high voltage makes a faint “corona” around the wires, which is a cloud of charged ions. As the dirty gas flows past, those ions bump into the dust particles and give them an electric charge, kind of like when a balloon rubs on your sweater and becomes “static.”

Once the particles are charged, the electric field pushes them sideways toward the plates. Positive particles are pulled one way, negative the other, but either way they end up sticking to a plate instead of staying in the air stream. The cleaned gas keeps moving forward and exits the ESP. Every so often, the plates are cleaned: in a dry ESP, mechanical “rappers” gently shake the plates so the dust falls into hoppers; in a wet ESP, water rinses the plates so sticky mist and very fine droplets wash off.

ESP performance depends on keeping the electric field strong and the gas flow smooth, and on the dust being “just right” electrically—if it holds charge too well or not well enough, efficiency drops. When everything is tuned properly, an ESP can remove well over 99% of particles with very little resistance to airflow. That’s why they’re widely used to protect the air around power plants, cement kilns, and other heavy industries.

11 September

RC Circuit

Seng Kwang Tan 16 Circuits, GeoGebra 0 Comments
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This is meant for the A-level topic of Circuits, for which students have to describe and represent the variation with time, of quantities like current, charge and potential difference, for a capacitor that is charging or discharging through a resistor, using equations of the form $x = x_0e^{-\frac{t}{\tau}}$ or $x = x_0(1 – e^{-\frac{t}{\tau}})$, where $\tau = RC$ is the time constant.

This GeoGebra interactive by Dave Nero is well-designed. It illustrates how the charge, voltage, and current in an RC circuit change over time. You can adjust the resistance, capacitance, and supply voltage using the sliders provided. The two circuit switches can be opened or closed by selecting the check boxes. A drop-down menu allows you to choose which quantity to display on the graph, and pressing the play button in the lower left corner starts the time-dependent plot.

When a capacitor is connected in series with a resistor, the changes in current, charge and potential difference follow an exponential pattern, controlled by the time constant $\tau = RC$.

During charging, the capacitor begins with no charge, so the battery’s full potential difference appears across the resistor, giving a maximum initial current. As charge accumulates on the plates, the potential difference across the capacitor rises. This reduces the potential difference across the resistor, causing the current to decrease. The charge on the capacitor and its potential difference both increase with time according to the equation $x = x_0 \left(1 – e^{-t/\tau}\right)$, approaching their maximum values asymptotically. Meanwhile, the current decreases exponentially with time, following $x = x_0 e^{-t/\tau}$.

During discharging, the capacitor starts with an initial charge and potential difference. Once connected across the resistor, this stored energy drives a current in the circuit. As the charge leaves the plates, the potential difference across the capacitor falls. Both charge and potential difference decrease exponentially with time according to $x = x_0 e^{-t/\tau}$, and the current also decays exponentially to zero, reversing direction compared to charging.

The time constant $\tau = RC$ sets the rate of change. After one time constant, a charging capacitor reaches about 63% of its final charge, or a discharging capacitor falls to about 37% of its initial charge. After about five time constants, the process is practically complete.