Welcome! Every new car is smashed into a wall before it goes on sale – on purpose. In this lesson you’ll learn how forces act in a crash, then become a crash test engineer and decide which car design keeps its passengers safest. It takes about one hour. Work through the steps in order.
Before you start
- You need: your exercise book or paper, a pen and a calculator.
- Write the title and date: Forces and car safety – Crash Test Engineer.
- The golden rule: write your answer first, then tap Show answer. Correct mistakes in a different colour.
- Look out for the blue 📝 boxes. Copy each one into your book neatly – these are your revision notes.
📝 Copy into your book
By the end of this lesson I will be able to:
1. use F = ma to calculate the force in a crash
2. explain how crumple zones, seatbelts and airbags reduce injuries
3. (Higher) calculate momentum and use conservation of momentum.
Step 1: Starter (5 minutes)
If you catch a raw egg with a stiff hand, it breaks. If you move your hand backwards as you catch it, it often survives. Why do you think this works? Write your best explanation.
Show answer
Moving your hand back makes the egg take longer to stop. A longer stopping time means a smaller deceleration, so a smaller force acts on the egg. Car safety features work in exactly the same way!
📝 Copy into your book
Making a collision last LONGER makes the force SMALLER.
Step 2: Read and write (8 minutes)
First, copy these three questions into your book. Then read the text below carefully and answer them in full sentences.
- What equation links resultant force, mass and acceleration?
- Name three car safety features.
- How do these safety features reduce the force on a passenger?
📖 Read: What happens in a crash?
Newton’s second law says that the acceleration of an object depends on the resultant force acting on it and on its mass: resultant force = mass × acceleration (F = ma). In a crash, a car and the people inside it slow down very suddenly. Slowing down is a negative acceleration, called a deceleration. The bigger the deceleration, the bigger the force needed – and large forces on the body cause serious injuries.
Cars are designed to make the people inside take longer to stop. Crumple zones at the front and back of the car squash in a controlled way. Seatbelts stretch slightly. Airbags inflate and give a soft surface to hit. All three increase the time it takes for a passenger to stop, which reduces their deceleration, so the force on them is smaller.
Every moving object also has momentum: momentum = mass × velocity (p = mv), measured in kg m/s. In a collision, the total momentum before equals the total momentum after, as long as no external forces act. This is called conservation of momentum.
Show answers
(a) Resultant force = mass × acceleration, F = ma.
(b) Crumple zones, seatbelts and airbags.
(c) They increase the time taken for the passenger to stop, so the deceleration is smaller and the force on the passenger is smaller.
📝 Copy and complete
Crumple zones, seatbelts and airbags all __________ the time it takes a passenger to stop. This makes the __________ smaller, so the __________ on the passenger is smaller.
Check your gaps
increase · deceleration · force
Step 3: Key facts (5 minutes)
Copy these equations and the chain carefully – you’ll use them for the rest of the lesson.
📝 Copy into your book
acceleration = change in velocity ÷ time a = Δv ÷ t
resultant force = mass × acceleration F = m a
(Higher) momentum = mass × velocity p = m v (kg m/s)
📝 Copy into your book
Safety feature increases stopping TIME
→ so the DECELERATION is smaller
→ so the FORCE on the passenger is smaller (F = ma)
→ so there are fewer serious injuries.
Step 4: Prove it with a calculation (10 minutes)
A 60 kg passenger is travelling at 15 m/s when the car crashes. Show every step.
Q1 With no seatbelt, the passenger hits the dashboard and stops in 0.05 s. Calculate their deceleration.
Show answer
a = Δv ÷ t = 15 ÷ 0.05 = 300 m/s²
Q2 Calculate the force on the passenger.
Show answer
F = m a = 60 × 300 = 18 000 N – about the weight of a small car!
Q3 Now the passenger wears a seatbelt and the airbag inflates, so they stop in 0.30 s. Calculate the new deceleration and force.
Show answer
a = 15 ÷ 0.30 = 50 m/s²
F = 60 × 50 = 3000 N – six times smaller!
Q4 (Higher) Calculate the momentum of a 1200 kg car travelling at 15 m/s.
Show answer
p = m v = 1200 × 15 = 18 000 kg m/s
✅ Score out of 4.
Step 5: Main task – Crash Test Engineer (14 minutes)
You work for a car company’s safety lab. Three prototype cars have been crash-tested at 14 m/s (about 30 mph). Each car carried a 70 kg crash test dummy. Here are the results:
| Design | Safety features | Time for dummy to stop | Extra cost per car |
|---|---|---|---|
| A | Rigid front, no airbag | 0.04 s | £0 |
| B | Crumple zone, seatbelt | 0.10 s | £300 |
| C | Crumple zone, seatbelt and airbag | 0.25 s | £800 |
Your task: write a report for the company’s boss.
- Calculate the deceleration and the force on the dummy for each design.
- Explain the physics: why does design C give the smallest force? Use the chain in your 📝 box.
- Recommend which design the company should sell, and whether the extra cost is worth it.
Stuck? Use a = Δv ÷ t first, then F = ma. The change in velocity is 14 m/s every time.
Show a model report
1. Results:
A: a = 14 ÷ 0.04 = 350 m/s²; F = 70 × 350 = 24 500 N
B: a = 14 ÷ 0.10 = 140 m/s²; F = 70 × 140 = 9800 N
C: a = 14 ÷ 0.25 = 56 m/s²; F = 70 × 56 = 3920 N
2. Physics: in design C, the crumple zone, seatbelt and airbag all increase the time the dummy takes to stop. A longer time means a smaller deceleration, so by F = ma the force on the dummy is smaller. The force in C is over six times smaller than in A.
3. Recommendation: sell design C. The extra £800 is a small part of the cost of a car, and a force of 3920 N instead of 24 500 N could be the difference between minor injuries and death.
✅ Check: did you show your working for all six calculations, and finish with a clear recommendation?
Step 6: Exam practice (13 minutes)
Look at the marks – they tell you how many points to make. Answer in full sentences, then mark yourself.
Q1 (3 marks) A resultant force of 2700 N acts on a car of mass 900 kg. Calculate its acceleration.
Show model answer
F = m a (1)
2700 = 900 × a (1)
a = 3.0 m/s² (1)
Q2 (4 marks) Explain how an airbag reduces the risk of injury to a driver in a crash.
Hint: use the chain in your 📝 box from Step 3.
Show model answer
The airbag increases the time taken for the driver to stop (1). So the driver’s deceleration is smaller (1). Since F = ma, a smaller deceleration means a smaller force on the driver (1). The airbag also spreads the force over a larger area of the body, so injuries are less serious (1).
Q3 (4 marks – Higher) A 1000 kg car travelling at 12 m/s crashes into a stationary 1500 kg car. The two cars stick together. Calculate their velocity just after the crash.
Hint: total momentum before = total momentum after.
Show model answer
Momentum before = 1000 × 12 = 12 000 kg m/s (1)
Momentum after = momentum before = 12 000 kg m/s (1)
Total mass after = 1000 + 1500 = 2500 kg (1)
v = 12 000 ÷ 2500 = 4.8 m/s (1)
✅ Add up your marks out of 11 (or out of 7 if you’re doing Foundation and skipped Q3). Write your score in your book.
Step 7: Exit ticket (5 minutes)
- A seatbelt that doesn’t stretch would be dangerous because…
- If a car has more mass, the force needed to stop it is… because…
- One thing I’m still not sure about is…
🎉 Well done – lesson complete! Your 📝 boxes are your revision notes for this topic. Want more? Try the Forces questions and the Road Safety Campaign lesson.
