The shape of the lesson
By the end of the lesson, learners can
- Height into speed: Release the ball and watch the two energy bars trade: gravitational potential energy falls as kinetic energy rises, while the total stays fixed.
- Two derivations to recall: Both formulas come straight from work done by a force.
- What is conserved, and what cancels: Energy conservation often lets the mass cancel and the path drop out.
- GPE change: ΔEₚ = mgΔh, vertical height only.
- KE: Eₖ = ½mv², depends on v².
- Conservation: mgΔh = ½mv² with no resistance.
One picture every idea returns to
The change in gravitational potential energy is ΔEₚ = mgΔh, from W = Fs with F = mg. Kinetic energy is Eₖ = ½mv², from the work done by a resultant force using v² = u² + 2as. With no resistance, GPE and KE interchange while their total is conserved.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Energy on the move.. Lift an object and you store gravitational potential energy; let it fall and that store pours into kinetic energy. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Height into speed | Release the ball and watch the two energy bars trade: gravitational potential energy falls as kinetic energy rises, while the total stays fixed. Project tab 2 of the lab (Kinetic and potential energy) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two derivations to recall | Both formulas come straight from work done by a force. | Pairs |
| 21 to 30 min | Teach and check: What is conserved, and what cancels | Energy conservation often lets the mass cancel and the path drop out. | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in What is conserved, and what cancels, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Kinetic energy depends on v², not v, so doubling the speed quadruples the kinetic energy (and the braking distance). | The change in gravitational potential energy uses the vertical height only, never the distance along a slope. And in free energy conversion the mass cancels, so a heavy and a light ball dropped from the same height reach the same speed. |
Kinetic and potential energy: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Define kinetic energy and state the formula used to calculate it.
Find the gravitational potential energy gained when a 5.0 kg mass is lifted through 2.0 m (g = 9.8 m s⁻²).
A 0.20 kg ball is dropped from rest from a height of 5.0 m. Using energy conservation (g = 9.8 m s⁻²), find its speed on landing.
A 1200 kg car speeds up from 10 m s⁻¹ to 20 m s⁻¹. Find the increase in its kinetic energy.
A pendulum bob is pulled aside until it is 0.45 m above its lowest point, then released. Find its speed at the lowest point (g = 9.8 m s⁻²).
Explain why, for a constant braking force, doubling a car's speed roughly quadruples its braking distance.
Answer key and marking notes
- Kinetic energy is the energy a body has because of its motion ✓
- Eₖ = ½mv² ✓
- ΔEₚ = mgΔh = 5.0 × 9.8 × 2.0 ✓
- ΔEₚ = 98 J ✓
- mgΔh = ½mv², and the mass cancels ✓
- v = √(2gΔh) = √(2 × 9.8 × 5.0) ✓
- v = 9.9 m s⁻¹ ✓
- KE at 20 = ½ × 1200 × 20² = 240 000 J ✓
- KE at 10 = ½ × 1200 × 10² = 60 000 J ✓
- Increase = 240 000 − 60 000 = 180 000 J = 180 kJ ✓
- GPE lost becomes KE: mgΔh = ½mv² ✓
- v = √(2gΔh) = √(2 × 9.8 × 0.45) ✓
- v = 3.0 m s⁻¹ ✓
- Kinetic energy Eₖ = ½mv² is proportional to v², so doubling v gives four times the KE ✓
- The braking force does work Fd to remove this KE, so d ∝ KE ∝ v², and the distance quadruples ✓
20 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The kinetic energy of a 4.0 kg object moving at 5.0 m s⁻¹ is:
- A50 J
- B20 J
- C100 J
- D10 J
The GPE gained by a 3.0 kg mass raised 4.0 m (g = 9.8) is:
- A12 J
- B1.3 J
- C118 J
- D59 J
Kinetic energy depends on the:
- Aspeed only
- Bsquare of the speed
- Cinverse of the speed
- Dsquare root of the speed
A ball is dropped from 3.2 m. Ignoring air resistance (g = 9.8), its landing speed is:
- A5.6 m s⁻¹
- B31 m s⁻¹
- C63 m s⁻¹
- D7.9 m s⁻¹
As a ball falls freely, the sum of its kinetic and gravitational potential energy:
- Astays constant
- Bincreases
- Cdecreases
- Dbecomes zero
A 1500 kg car slows from 20 m s⁻¹ to rest. The kinetic energy lost is:
- A1.5 × 10⁴ J
- B30 J
- C3.0 × 10⁵ J
- D6.0 × 10⁴ J
Two balls of different mass are dropped from the same height (no air resistance). They land with:
- Athe heavier one faster
- Bthe same speed
- Cthe lighter one faster
- Dspeeds that depend on shape
A 0.40 kg ball is thrown up at 8.0 m s⁻¹. Its maximum height (g = 9.8) is:
- A0.82 m
- B6.5 m
- C1.6 m
- D3.3 m
If a car's speed triples, its kinetic energy becomes:
- Anine times as large
- Bthree times as large
- Csix times as large
- Dunchanged
A pendulum bob rises 0.20 m at the end of a swing. Its speed at the lowest point was (g = 9.8):
- A3.9 m s⁻¹
- B0.4 m s⁻¹
- C2.0 m s⁻¹
- D4.0 m s⁻¹
A 2.0 kg block slides down a frictionless ramp from a height of 1.5 m. Its speed at the bottom is:
- A2.7 m s⁻¹
- B5.4 m s⁻¹
- C29 m s⁻¹
- D3.0 m s⁻¹
The formula for gravitational potential energy change near the Earth is:
- A½mv²
- Bmgh²
- C½mgh
- DmgΔh
For a constant braking force, the braking distance is proportional to:
- Athe square of the speed
- Bthe speed
- Cthe mass only
- Dthe inverse of the speed
A 0.060 kg ball leaves a bat at 30 m s⁻¹. Its kinetic energy is:
- A54 J
- B1.8 J
- C27 J
- D900 J
A roller-coaster car is highest at the start. As it descends, its GPE is mainly transferred to:
- Aelastic energy
- Bkinetic energy
- Cchemical energy
- Dlight
A stone of mass 0.50 kg is thrown down a cliff at 4.0 m s⁻¹ from 20 m. Using energy (g = 9.8), its landing speed is:
- A14 m s⁻¹
- B4.0 m s⁻¹
- C400 m s⁻¹
- D20 m s⁻¹
Kinetic energy is the energy a body has because of its:
- Amotion
- Bposition
- Ctemperature
- Dcharge
On a frictionless track, the speed of a ball at the bottom of a dip depends on:
- Athe length of the track
- Bthe steepness only
- Cthe vertical drop only
- Dthe mass
A 10 kg mass has 490 J of GPE relative to the ground (g = 9.8). Its height is:
- A50 m
- B5.0 m
- C0.5 m
- D49 m
A 1200 kg car accelerates from rest to 15 m s⁻¹. The work done on it is:
- A1.8 × 10⁴ J
- B9000 J
- C2.7 × 10⁵ J
- D1.35 × 10⁵ J
Quiz key
Eₖ = ½mv² = ½ × 4.0 × 25 = 50 J.
ΔEₚ = mgΔh = 3.0 × 9.8 × 4.0 = 118 J.
Eₖ = ½mv², proportional to v².
v = √(2gΔh) = √(2 × 9.8 × 3.2) = 7.9 m s⁻¹.
With no resistance, total mechanical energy is conserved.
Eₖ = ½mv² = ½ × 1500 × 400 = 3.0 × 10⁵ J.
v = √(2gΔh); the mass cancels, so speeds are equal.
Δh = v²/(2g) = 64/19.6 = 3.3 m.
Eₖ ∝ v², so tripling v gives 3² = 9 times.
v = √(2gΔh) = √(2 × 9.8 × 0.20) = 2.0 m s⁻¹.
v = √(2gΔh) = √(2 × 9.8 × 1.5) = 5.4 m s⁻¹.
ΔEₚ = mgΔh.
Work Fd removes ½mv², so d ∝ v².
Eₖ = ½ × 0.060 × 30² = 27 J.
Descending converts GPE into KE.
½v² = ½u² + gΔh gives v = √(4.0² + 2 × 9.8 × 20) = √(16+392) = 20 m s⁻¹.
Kinetic energy is energy due to motion.
Energy conservation gives v = √(2gΔh), set by the vertical drop alone.
h = Eₚ/(mg) = 490/(10 × 9.8) = 5.0 m.
Work = gain in KE = ½ × 1200 × 15² = 1.35 × 10⁵ J.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.