--- title: "Solving force and motion problems" subtitle: "Year 10 Physics · Newton's laws · Lesson 7" --- # Solving force and motion problems **Name:** {{field-24}} **Date:** {{field-15}} **Class:** {{field-15}} > **Key question:** 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. :::teacher-note **In one sentence:** 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). **Students should be able to say:** 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. **Width:** the unmarked parts are the floor every student does (Engage, Explore, Explain, Evaluate). Typical and Wider sections add contexts, not difficulty. Students work in the online tutorial alongside this sheet. ::: ## Engage :::keep-together **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 My prediction, and why: {{lines-3}} ::: :::teacher-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. ::: ## Explain: practise the routine :::keep-together ### 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: {{field-15}} ::: :::teacher-note **Answer:** 2 Net force = 4000 − 1000 = 3000 N. a = F ÷ m = 3000 ÷ 1500 = 2 m/s². ::: :::keep-together ### Speed after 6 s **It accelerates at 2 m/s² from rest. What is its speed after 6 s (m/s)?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 12 Δv = a × t = 2 × 6 = 12 m/s. ::: :::keep-together ### 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: {{field-15}} ::: :::teacher-note **Answer:** -2.5 a = Δv ÷ t = (0 − 20) ÷ 8 = −2.5 m/s². ::: :::keep-together ### Braking force **What braking force does the 1200 kg car need (N)? Give the size only.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 3000 F = ma = 1200 × 2.5 = 3000 N. ::: :::keep-together ### 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 ::: :::teacher-note **Answer:** Check the working: a force that large is not reasonable for a family car 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). ::: :::keep-together ### 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 ::: :::teacher-note **Answer:** 2 m/s²: the gradient of a speed–time graph is the acceleration Gradient = change in speed ÷ change in time = acceleration. Steeper means a bigger acceleration. ::: :::typical ## Elaborate: two-step problem :::keep-together **A 60 kg sprinter reaches 9 m/s from rest in 3 s. What net force acts on the sprinter (N)?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 180 a = 9 ÷ 3 = 3 m/s². F = ma = 60 × 3 = 180 N. ::: ::: :::wider ## Elaborate: school zones :::keep-together **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 ::: :::teacher-note **Answer:** At the lower speed the car stops in much less distance, and the force on anyone it hits is smaller 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. ::: ::: ## 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. ## Exit ticket :::keep-together **A 50 kg skateboarder is pushed with a net force of 100 N from rest. How fast is she moving after 4 s?** {{lines-3}} ::: :::teacher-note **Answer:** a = 100 ÷ 50 = 2 m/s². After 4 s: 2 × 4 = 8 m/s. :::