The shape of the lesson
By the end of the lesson, learners can
- Three laws, one equation: In the lab, set the friction coefficient to zero and double the applied force: the acceleration doubles.
- Combining with suvat: Forces give acceleration; acceleration links to motion through the constant acceleration equations.
- Vertical motion and lifts: For vertical motion, resolve vertically with the direction of the acceleration as positive.
- Diagram: all forces, plus an arrow for the acceleration.
- Perpendicular to motion: forces balance; this usually gives \(R\).
- Friction: moving means \(F = \mu R\), opposing the motion.
- Along the motion: resultant force \(= ma\).
- Motion: use suvat with the acceleration you found (or the reverse).
One picture every idea returns to
Newton's second law: resultant force \(= ma\), resolved in the direction of the acceleration. Perpendicular to the motion the forces still balance. Weight is \(W = mg\) with \(g = 10\ \text{m s}^{-2}\), and a sliding particle has friction \(F = \mu R\).
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, Resultant force, ma.. Equilibrium was the special case where the resultant force is zero. 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: Three laws, one equation | In the lab, set the friction coefficient to zero and double the applied force: the acceleration doubles. Project tab 1 of the lab (Dynamics lab, Newton's laws (F = ma)) and let a student drive it while the class predicts. Model the worked example A driving force against resistance on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Combining with suvat | Forces give acceleration; acceleration links to motion through the constant acceleration equations. Model the worked example Motion to force on the board, one line at a time. Your turn: A car of mass 1200 kg brakes uniformly from \(20\ \text{m s}^{-1}\) to rest in 40 m. Find the braking force, assuming no other horizontal forces. | Pairs |
| 21 to 30 min | Teach and check: Vertical motion and lifts | For vertical motion, resolve vertically with the direction of the acceleration as positive. Model the worked example A person in a lift on the board, one line at a time. | Frayer Model |
| 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, 24 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 Vertical motion and lifts, 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
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 |
|---|---|
| \(F = ma\) uses the RESULTANT force. | Writing \(50 = 8a\), or \(50\cos 30^\circ = 8a\), ignores friction and is the most common lost mark. |
| Keep signs consistent. | If the particle is slowing down, \(a\) is negative in your positive direction, and so is the resultant force. |
| "Moving upwards" does not mean "accelerating upwards". | A lift moving up while slowing down has a DOWNWARD acceleration, so the reaction is less than the weight. |
Newton's laws of motion: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
State Newton's first law. A particle of mass 3 kg on a smooth horizontal surface is acted on by a horizontal force of 12 N. Find its acceleration.
A car of mass 1200 kg brakes uniformly from \(20\ \text{m s}^{-1}\) to rest in a distance of 40 m. Find the braking force, assuming no other resistance.
A boy of mass 60 kg stands in a lift. Find the normal reaction from the floor when the lift (a) accelerates upwards at \(1.5\ \text{m s}^{-2}\), (b) moves at constant speed, (c) moves downwards while slowing down at \(1.5\ \text{m s}^{-2}\).
A block of mass 8 kg starts from rest on rough horizontal ground, \(\mu = 0.25\). It is pulled by a force of 50 N at \(30^\circ\) above the horizontal. Find the distance it travels in the first 4 s.
A puck is projected at \(8\ \text{m s}^{-1}\) across rough horizontal ice, \(\mu = 0.32\). Find how far it slides and how long it takes to stop.
A crate of mass 50 kg is raised from rest by a vertical rope. For the first 3 s the tension is 560 N. The tension then changes to a constant value \(T\) N and the crate comes to rest after a further 2 s.
(a) Find the speed of the crate after 3 s.
(b) Find \(T\).
(c) Find the total height through which the crate is raised.
Answer key and marking notes
- A particle stays at rest or moves with constant velocity unless a resultant force acts on it.
- \(12 = 3a\), so \(a = 4\ \text{m s}^{-2}\).
- \(0 = 20^2 + 2a(40)\), so \(a = -5\ \text{m s}^{-2}\).
- Braking force \(= 1200 \times 5 = 6000\) N.
- (a) \(R - 600 = 60(1.5)\), so \(R = 690\) N.
- (b) \(a = 0\), so \(R = 600\) N.
- (c) Moving down and slowing means the acceleration is UPWARDS, \(1.5\ \text{m s}^{-2}\): \(R = 690\) N, the same as (a).
- \(R = 80 - 50\sin 30^\circ = 55\) N; friction \(= 0.25 \times 55 = 13.75\) N.
- \(50\cos 30^\circ - 13.75 = 8a\), so \(a = 3.694\ \text{m s}^{-2}\).
- \(s = \tfrac{1}{2}(3.694)(4^2) = 29.6\) m.
- Friction \(= 0.32 \times 10m = 3.2m\), so the deceleration is \(3.2\ \text{m s}^{-2}\).
- \(0 = 64 - 2(3.2)s\), so \(s = 10\) m.
- \(0 = 8 - 3.2t\), so \(t = 2.5\) s.
- (a) \(560 - 500 = 50a\), \(a = 1.2\); \(v = 1.2 \times 3 = 3.6\ \text{m s}^{-1}\).
- (b) Deceleration \(= 3.6 \div 2 = 1.8\ \text{m s}^{-2}\): \(T - 500 = 50(-1.8)\), so \(T = 410\).
- (c) Heights: \(\tfrac{1}{2}(3.6)(3) = 5.4\) m and \(\tfrac{1}{2}(3.6)(2) = 3.6\) m.
- Total \(= 9\) m.
8 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.
A particle of mass 4 kg on a smooth horizontal surface is pulled by a horizontal force of 30 N against a resistance of 6 N. Find its acceleration.
- A\(6\ \text{m s}^{-2}\)
- B\(7.5\ \text{m s}^{-2}\)
- C\(9\ \text{m s}^{-2}\)
- D\(60\ \text{m s}^{-2}\)
A person stands in a lift that is moving upwards but slowing down. How does the normal reaction from the floor compare with the person's weight?
- AIt is less than the weight
- BIt is greater than the weight
- CIt is equal to the weight
- DIt is zero
A car of mass 1200 kg travelling at \(24\ \text{m s}^{-1}\) brakes uniformly and stops in 48 m. Find the braking force, assuming no other horizontal forces act.
- A7200 N
- B14400 N
- C600 N
- D12000 N
A man of mass 70 kg stands in a lift that is accelerating downwards at \(1.2\ \text{m s}^{-2}\). Find the normal reaction from the floor on the man.
- A616 N
- B784 N
- C700 N
- D84 N
A block of mass 10 kg is pulled along rough horizontal ground by a force of 60 N at \(20^\circ\) above the horizontal, as shown. The coefficient of friction is 0.3. Find the acceleration of the block.
- A\(3.25\ \text{m s}^{-2}\)
- B\(2.64\ \text{m s}^{-2}\)
- C\(5.64\ \text{m s}^{-2}\)
- D\(2.02\ \text{m s}^{-2}\)
A puck slides across rough horizontal ice. Its speed falls from \(8\ \text{m s}^{-1}\) to zero in a distance of 16 m. Find the coefficient of friction between the puck and the ice.
- A0.2
- B0.4
- C0.5
- D2
A crate of mass 40 kg is raised from rest by a vertical rope. For the first 4 s the tension is 480 N. The tension is then reduced to 360 N, and the crate keeps moving upwards until it comes to rest. Find the total height risen.
- A48 m
- B16 m
- C32 m
- D19.6 m
A particle of mass 0.8 kg falls from rest. A constant air resistance of 2 N acts on it. Find its speed after it has fallen 25 m.
- A\(19.4\ \text{m s}^{-1}\)
- B\(20\ \text{m s}^{-1}\)
- C\(22.4\ \text{m s}^{-1}\)
- D\(25\ \text{m s}^{-1}\)
Quiz key
Resultant force \(= 30 - 6 = 24\) N, so \(a = 24 \div 4 = 6\ \text{m s}^{-2}\). The tempting 7.5 uses the 30 N pull alone and ignores the resistance. 9 adds the resistance instead of subtracting it, and 60 multiplies by the mass instead of dividing.
Moving up and slowing down means the acceleration is DOWNWARDS, so the resultant force is downwards: \(mg - R = ma\), giving \(R = m(g - a) < mg\). The tempting "greater than the weight" confuses the direction of motion with the direction of acceleration. The reaction equals the weight only at constant velocity, and is zero only if the lift accelerates down at \(g\).
\(0 = 24^2 + 2a \times 48\) gives \(a = -6\ \text{m s}^{-2}\), so the braking force is \(1200 \times 6 = 7200\) N. The tempting 14400 N drops the 2 in \(2as\) (giving \(a = 12\)). 600 N uses \(a = 24 \div 48\), and 12000 N is the car's weight, which acts vertically and does not stop the car.
\(70 \times 10 - R = 70 \times 1.2\), so \(R = 700 - 84 = 616\) N. The tempting 784 N has the acceleration the wrong way round; that is the reaction when the lift accelerates UPWARDS. 700 N is his weight (true only at constant velocity), and 84 N is the resultant force, not the reaction.
Vertically: \(R = 100 - 60\sin 20^\circ = 79.48\) N, so friction \(= 0.3 \times 79.48 = 23.84\) N. Horizontally: \(60\cos 20^\circ - 23.84 = 10a\), so \(a = \dfrac{56.38 - 23.84}{10} = 3.25\ \text{m s}^{-2}\). The tempting 2.64 uses \(R = 100\) N, ignoring the upward component of the pull. 5.64 leaves out friction, and 2.02 adds the vertical component to the weight.
\(0 = 64 - 2a \times 16\) gives \(a = 2\ \text{m s}^{-2}\). Friction is the only horizontal force: \(\mu mg = ma\), so \(\mu = \dfrac{2}{10} = 0.2\). The tempting 0.4 drops the 2 in \(v^2 = u^2 + 2as\), giving \(a = 4\). 0.5 is \(8 \div 16\), and 2 is the deceleration itself, not \(\mu\).
Stage 1: \(a = \dfrac{480 - 400}{40} = 2\ \text{m s}^{-2}\), so after 4 s, \(v = 8\ \text{m s}^{-1}\) and the height is \(\tfrac{1}{2} \times 2 \times 16 = 16\) m. Stage 2: \(a = \dfrac{360 - 400}{40} = -1\ \text{m s}^{-2}\), so \(0 = 64 - 2s\) gives \(s = 32\) m. Total \(= 48\) m. The tempting 19.6 m uses a deceleration of \(360 \div 40 = 9\) in stage 2, treating the tension as if it opposed the motion. 16 m and 32 m are the separate stages.
\(8 - 2 = 0.8a\), so \(a = 7.5\ \text{m s}^{-2}\). Then \(v^2 = 2 \times 7.5 \times 25 = 375\) and \(v = 19.4\ \text{m s}^{-1}\). The tempting 20 subtracts the 2 N straight from \(g\) (\(a = 8\)), forgetting to divide the force by the mass. 22.4 ignores the resistance, and 25 adds it.
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.