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# Solving force and motion problems

Year 10 Physics · Newton's laws

<!-- Speaker note: Students solve multi-step force and motion problems using F = ma and a = Δv ÷ t with correct units, use a graph to support a quantitative argument, and check answers for reasonableness (AC9S10U05, node 7, a practise node). About 20 minutes (Engage 3 min · Explain 12 min · Evaluate 3 min · Typical adds 1 min, Wider about 1 min). -->

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## Today

**We are learning to** a multi-step problem becomes manageable with a routine: list what you know, work out the net force, find the acceleration with F = ma, use a = Δv ÷ t to link it to speed and time, then check the answer makes sense. Keep units (N, kg, m/s², m/s, s) with every number.

- I can solve a problem that needs two or more steps.
- I can keep the units straight.
- I can tell whether an answer is reasonable.

<!-- Speaker note: Before class: A practice lesson with no simulation, following Free body diagrams. Set the width with `?width=minimum`, `typical` (default) or `wider`. The Year 10 progression marks this node as a place for structured practice, so students need pencil and calculator more than a screen: ask them to show each line in their books before entering an answer. Answers with a sign are entered as numbers (−2.5). Round figures are chosen for clean numbers; real braking and engine forces vary with speed. Teacher: In your own words: the routine and the motorbike problem. -->

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## Why it matters

> A car travelling at 25 m/s brakes. How long does it take to stop, and how big is the force? The answer takes three steps, and the hardest part is knowing which one comes first.

<!-- Speaker note: A short practice lesson with no simulation. You know F = ma, a = Δv ÷ t and the three laws. Now use them together. -->

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<span style="background:#dbeafe; color:#1a1a1a; border:1.5px solid #3b82f6; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">💬 Discussion</span>

## Engage: predict

**A 1500 kg car has an engine force of 4000 N forwards and a resistance of 1000 N backwards. You are asked how fast its speed is changing. What do you work out first?**

- The speed, from the mass
- Nothing: you need the time first
- The net force: 4000 − 1000 = 3000 N
- The acceleration, using 4000 N

<!-- Speaker note: What happens: The net force, 3000 N. Acceleration depends on the net (unbalanced) force, not the engine force alone. Then a = F ÷ m = 3000 ÷ 1500 = 2 m/s². Take two or three predictions before anyone runs it. -->

<!-- Speaker note: Teacher: tally the predictions. Ask the 'acceleration' group what force the acceleration depends on. -->

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## What happened?

* The net force, 3000 N. Acceleration depends on the net (unbalanced) force, not the engine force alone. Then a = F ÷ m = 3000 ÷ 1500 = 2 m/s².

<!-- Speaker note: Reveal after running. Ask who predicted it, and what made them think so. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Net force

**A 1500 kg car has an engine force of 4000 N and a resistance of 1000 N. What is its acceleration (m/s²)?**

* Answer: 2

<!-- Speaker note: Net force = 4000 − 1000 = 3000 N. a = F ÷ m = 3000 ÷ 1500 = 2 m/s². -->

<!-- Speaker note: Teacher: the five-step routine is the success criterion; collect a different problem from each group. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Speed after 6 s

**It accelerates at 2 m/s² from rest. What is its speed after 6 s (m/s)?**

* Answer: 12

<!-- Speaker note: Δv = a × t = 2 × 6 = 12 m/s. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Braking

**A 1200 kg car travelling at 20 m/s brakes to a stop in 8 s. What is its acceleration (m/s²)? Use a negative sign for slowing down.**

* Answer: -2.5

<!-- Speaker note: a = Δv ÷ t = (0 − 20) ÷ 8 = −2.5 m/s². -->

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## Braking force

**What braking force does the 1200 kg car need (N)? Give the size only.**

* Answer: 3000

<!-- Speaker note: F = ma = 1200 × 2.5 = 3000 N. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Is it reasonable

**A student calculates the braking force of a car as 3 000 000 N. What should they do?**

- Check the working: a force that large is not reasonable for a family car
- Divide by 10 to make it look sensible
- Report it: the calculation was careful

* Answer: Check the working: a force that large is not reasonable for a family car

<!-- Speaker note: Compare the answer with what you know: a family car's braking force is a few thousand newtons. A huge number suggests a unit slip (grams for kilograms, for example). -->

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## The graph

**A speed–time graph for a trolley is a straight line rising from 0 m/s at 0 s to 6 m/s at 3 s. What is the acceleration, and what does the gradient show?**

- 2 m/s²: the gradient of a speed–time graph is the acceleration
- 6 m/s²
- 3 m/s²: the gradient is the time

* Answer: 2 m/s²: the gradient of a speed–time graph is the acceleration

<!-- Speaker note: Gradient = change in speed ÷ change in time = acceleration. Steeper means a bigger acceleration. -->

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## Elaborate: two-step problem

**A 60 kg sprinter reaches 9 m/s from rest in 3 s. What net force acts on the sprinter (N)?**

* Answer: 180

<!-- Speaker note: a = 9 ÷ 3 = 3 m/s². F = ma = 60 × 3 = 180 N. What this step is for: Connecting a graph to the formulas in a third context checks the transfer. -->

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## Elaborate: school zones

**A car's stopping distance at 60 km/h is about 36 m, and at 40 km/h about 20 m. What is the best argument for a 40 km/h school-zone limit?**

- At the lower speed the car stops in much less distance, and the force on anyone it hits is smaller
- Stopping distance goes down in proportion to speed, so the lower limit saves only a third of the distance
- Cars go slowly so it is quieter

* Answer: At the lower speed the car stops in much less distance, and the force on anyone it hits is smaller

<!-- Speaker note: A quantitative argument uses numbers: going 50% faster (60 instead of 40 km/h) takes nearly twice the distance to stop, and a faster car has more energy to lose. What this step is for: A quantitative argument from stopping distances connects the laws to a real decision. -->

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## Evaluate: look back

Write one sentence: "At first I thought… Now I think… because…"

**Revision summary**

- A routine: **list what you know, find the net force, use F = ma for acceleration, use a = Δv ÷ t for speed and time, then check** the answer is reasonable.
- Keep **units** with every number: N, kg, m/s², m/s, s.
- On a **speed–time graph**, the gradient is the acceleration.
- A negative acceleration means slowing down.

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Exit ticket

**A 50 kg skateboarder is pushed with a net force of 100 N from rest. How fast is she moving after 4 s?**

* a = 100 ÷ 50 = 2 m/s². After 4 s: 2 × 4 = 8 m/s.

<!-- Speaker note: Students answer on paper or a mini whiteboard before the reveal. Anyone who cannot answer needs the main idea again next lesson. Where this leads: Next: Newton's laws in the real world. In class: collect braking data with a toy car and a ramp; use a speed–time graph from a data logger; work through a vehicle's stopping distance table. -->