The cartridge pushes gas backwards, and the gas pushes the car forwards. At the end of the lane the car pushes the stop, and the stop pushes the car. It is the same law at two moments in one race. This pack is built around the one drawing mistake almost everyone makes with it.
State Newton’s third law: forces come in pairs that are equal in size, opposite in direction, and act on two different objects.
02
Identify the launch pair (gas and car) and the impact pair (car and stop) in a STEM Racing run.
03
Explain why the two forces in a pair never cancel each other out.
04
Estimate the average stopping force from the car’s kinetic energy and the stopping distance.
02From the Nightmare car
Our launch is a burst of CO₂. Gas leaves the cartridge backwards through the nozzle, and the car gains speed forwards. About one second later the car has to be caught safely at the end of the 20 m lane. People treat these as two different topics, but they are the same law at two different times. We learned the hard way that the catch matters. A car arriving at 20 m/s carries about 11 J, and a badly padded stop sends that energy straight back into the nose.
Fig. 2 The launch pair and the impact pair. Each arrow is drawn on the object the force acts on. Two arrows on two different objects can never cancel each other.
03Core idea
FA on B = − FB on A Same size. Opposite direction. Always on two different objects. Always the same type of force.
If you draw both arrows on the car, the diagram is wrong. At launch, the forward arrow on the car is “gas pushes car”. The backward arrow belongs on the gas, not on the car. A free-body diagram shows one object and only the forces acting on that object.
Why the car still slows down at the stop
At impact the pair is “stop pushes car” (on the car, backwards) and “car pushes stop” (on the stop, forwards). Only the first one is on the car. So the net force on the car points backwards and the car slows down. The stop feels the other force, which is why the foam squashes and why a solid wall would crack the nose cone.
Favg × d = ½ m v² so Favg = ½mv² / d. This is the average force over the stopping distance d.
!
The classic mistake: “The forces are equal and opposite, so they cancel and nothing moves.” They would only cancel if they acted on the same object. They never do.
04Force-pair animator
Switch between the launch and the impact. Move the thrust slider and watch both arrows change together. They can never be different sizes. Then press “Show the mistake” to see what happens if you draw both arrows on the car.
Both arrows always have the same length. The readouts show what each object does with its half of the pair.
—NForce on the car
—NForce on the gas
—m/s²Car acceleration
—NNet force on the car
05Worked example
Claim to test: “The forward and backward forces cancel, so the car should not move.”
Fix: during the launch, the forward force on the car comes from the gas. The backward force from the car acts on the gas. They are equal and opposite, but they are on different objects, so they cannot cancel on the car’s diagram.
Result: the car has a net forward force and speeds up. The gas is pushed backwards. Both things happen. That is the third law working, not failing.
Stopping force: m = 0.055 kg, v = 20 m/s at the stop, and the foam squashes by d = 0.05 m.
KE = ½mv² = 0.5 × 0.055 × 20² = 11 J
Favg = KE / d = 11 / 0.05 = 220 N
220 N is about the weight of a 22 kg child, pushing on a 55 g nose. The stop feels the same 220 N pointing forwards at the same moment. That is why the foam is designed to take it, not the car.
06Student questions Teacher mode is off. Answers are hidden.
Q1 · Identify[2]
Name the two objects in the launch pair, then write the two forces in the form “X pushes Y”.
Answer key
Objects: the gas (CO₂) and the car. Forces: “gas pushes car” (forwards) and “car pushes gas” (backwards). One mark for the objects, one for both forces with directions.
Q2 · Explain[3]
The two forces at the stop are equal and opposite. So why does the car still slow down?
Answer key
The two forces act on different objects (1). “Stop pushes car” acts on the car (1) and it is the only sideways force on the car, so the car has a net backward force and slows down. “Car pushes stop” acts on the stop and does not appear on the car’s diagram (1).
Q3 · Diagram[3]
Sketch the car on its own halfway down the track, coasting. Which force pairs still matter, and which one has already finished?
Answer key
On the car: weight down, the track pushing up, drag backwards, rolling resistance backwards. Partners that still matter: the car pushing down on the track, the car pushing the air forwards, the car pushing on its own axles. Finished: the gas pushing the car, because the cartridge is empty (1). No thrust arrow (1). No backward “reaction to thrust” arrow (1).
Q4 · Calculate[3]
A 0.055 kg car reaches the stop at 20 m/s and the foam squashes by 8 cm. Find the kinetic energy and the average stopping force. What force does the stop feel?
Answer key
KE = ½ × 0.055 × 20² = 11 J
Favg = 11 / 0.08 = 137.5 N ≈ 140 N, backwards, on the car
The stop feels 137.5 N forwards. Same size, opposite direction. One mark each.
07Printable worksheet The answer key prints as well when teacher mode is on
Team Nightmare · STEM Subject Lab
NM-LAB-02 · Newton’s Third Law
Name ______________________ Class __________ Date __________
Physics 35 minutes Launch and impact
The law in one line
Forces come in pairs. Same size, opposite direction, different objects, same type of force.
FA on B = − FB on A
Stopping
Favg = ½mv² / d
pairfree-body diagramnet forcedecelerateimpulse
[4]1. Launch table.Fill in both sides with the name of the force, its direction, and the object it acts on.
Force on the car
Force on the gas
[3]2.A student says the stop force and the car force cancel out, so nothing can slow down. Correct the student in three sentences.
[3]3.Draw and label the impact pair. Put each arrow on the correct object. Draw the stop as a block.
[3]4.m = 0.060 kg, v = 18 m/s at the stop, and the foam squashes by d = 0.06 m. Find the kinetic energy, then the average stopping force. Show three lines of working.
[2]5.The foam is replaced by a wooden block (d ≈ 0.005 m). Without calculating, say what happens to the stopping force and why teams use foam.
Accept any equivalent working. The marks in brackets match the student sheet.
Teacher copy Do not hand out
1. [4] On the car: “gas pushes car”, forwards, acts on the car (2). On the gas: “car pushes gas”, backwards, acts on the gas (2). Same size.
2. [3] The two forces act on different objects (1). Only “stop pushes car” acts on the car, which gives it a net backward force (1). So the car slows down while the stop feels the equal forward push (1).
3. [3] The car with one arrow pointing backwards labelled “stop pushes car” (1). The block with one arrow pointing forwards labelled “car pushes stop” (1). Both arrows the same length (1). Take a mark off if both arrows are on the car.
4. [3] KE = ½ × 0.060 × 18² = 9.72 JFavg = KE/d = 9.72 / 0.06 = 162 N One mark each for the formula, the KE, and the force with a unit.
5. [2] The stopping distance is about 12 times smaller, so the force is about 12 times bigger, roughly 1900 N (1). Foam spreads the same energy over a longer distance and time, which lowers the peak force on the nose (1).