The shape of the lesson
By the end of the lesson, learners can
- Smooth slopes: On a smooth plane the only force along the slope is the weight component \(mg\sin\alpha\), so \(a = g\sin\alpha\) down the plane whichever way the particle is moving.
- Rough slopes: up and down differ: Friction always opposes the motion.
- Extra forces on the slope: With a pulling force, resolve it too.
- Weight: \(mg\sin\alpha\) down the plane, \(mg\cos\alpha\) into it.
- Perpendicular: \(R = mg\cos\alpha \pm\) components of other forces.
- Friction: \(\mu R\), opposite to the motion (it switches direction when the particle turns round).
- Along the plane: resultant \(= ma\), then suvat.
- At rest at the top? It stays if \(\mu \ge \tan\alpha\).
One picture every idea returns to
Along the plane: resultant force \(= ma\). Perpendicular: \(R = mg\cos\alpha\) (plus or minus any other perpendicular components). Moving up a rough plane, friction acts down it, so \(mg\sin\alpha + \mu R\) slows the particle; moving down, friction acts up, so the driving force is \(mg\sin\alpha - \mu R\).
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, Down the slope, then up.. On a slope, resolve along and perpendicular to the plane. 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: Smooth slopes | On a smooth plane the only force along the slope is the weight component \(mg\sin\alpha\), so \(a = g\sin\alpha\) down the plane whichever way the particle is moving. Model the worked example Released on a smooth plane on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Rough slopes: up and down differ | Friction always opposes the motion. Model the worked example Sliding down on the board, one line at a time. Your turn: A particle slides from rest down a rough plane inclined at \(25^\circ\) and reaches \(3\ \text{m s}^{-1}\) after travelling 3 m. Find the coefficient of friction. | Pairs |
| 21 to 30 min | Teach and check: Extra forces on the slope | With a pulling force, resolve it too. Model the worked example Pulled up the plane on the board, one line at a time. | Connect-Extend-Challenge |
| 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, 28 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 Extra forces on the slope, 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.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
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 |
|---|---|
| \(a = g\sin\alpha\), not \(g\cos\alpha\). | Check with a limiting case: on a vertical "slope" (\(\alpha = 90^\circ\)) the particle falls freely with \(a = g\), and \(\sin 90^\circ = 1\). |
| After a particle stops at the top, do not assume it slides back. | Compare the weight component \(mg\sin\alpha\) with the maximum friction \(\mu mg\cos\alpha\): it stays put if \(\mu \ge \tan\alpha\). |
| Using \(R = mg\cos\alpha\) when the applied force has a component perpendicular to the plane. | In the second example that gives friction 5.20 N instead of 9.20 N. |
Motion on an inclined plane: worksheet
6 questions, 28 marks. Written in the style of the papers; no past paper question is reproduced.
A particle is released from rest on a smooth plane inclined at \(30^\circ\). Find its acceleration, its speed after sliding 4 m, and the time taken.
A particle of mass 2 kg is released from rest on a rough plane inclined at \(\alpha\), where \(\sin\alpha = 0.6\). The coefficient of friction is 0.25. Find its speed after 2 s and the distance it has travelled.
A particle slides from rest down a rough plane inclined at \(30^\circ\) and travels 4 m in 2 s. Find the coefficient of friction, giving your answer in exact form.
A particle of mass 0.5 kg is projected up a line of greatest slope of a rough plane inclined at \(20^\circ\), with speed \(6\ \text{m s}^{-1}\). The coefficient of friction is 0.4.
(a) Find the distance it travels up the plane before coming to rest.
(b) Show that it then remains at rest.
A particle of mass 3 kg is pushed up a rough plane inclined at \(30^\circ\) by a horizontal force of 40 N. The coefficient of friction is 0.2. Find the acceleration of the particle.
A skier of mass 70 kg starts from rest and skis straight down a slope inclined at \(15^\circ\) to the horizontal. The coefficient of friction between skis and snow is 0.1, and there is a constant air resistance of 40 N. Find the skier's speed after travelling 100 m.
Answer key and marking notes
- \(a = 10\sin 30^\circ = 5\ \text{m s}^{-2}\).
- \(v = \sqrt{2 \times 5 \times 4} = \sqrt{40} = 6.32\ \text{m s}^{-1}\).
- \(t = 6.325 \div 5 = 1.26\) s.
- \(R = 16\) N, friction 4 N; \(12 - 4 = 2a\), so \(a = 4\ \text{m s}^{-2}\).
- \(v = 4 \times 2 = 8\ \text{m s}^{-1}\).
- \(s = \tfrac{1}{2}(4)(2^2) = 8\) m.
- \(4 = \tfrac{1}{2}a(2^2)\), so \(a = 2\ \text{m s}^{-2}\).
- Per unit mass: \(10\sin 30^\circ - 10\mu\cos 30^\circ = 2\), so \(5 - 5\sqrt{3}\mu = 2\).
- \(\mu = \dfrac{3}{5\sqrt{3}} = \dfrac{\sqrt{3}}{5}\) (0.346).
- (a) \(R = 5\cos 20^\circ = 4.698\) N; friction \(= 1.879\) N down the plane.
- \(-(5\sin 20^\circ + 1.879) = 0.5a\), so \(-(1.710 + 1.879) = 0.5a\) and \(a = -7.18\ \text{m s}^{-2}\).
- \(0 = 36 + 2(-7.179)s\), so \(s = 2.51\) m.
- (b) Weight component \(1.710\) N is less than the maximum friction \(1.879\) N (equivalently \(\mu = 0.4 > \tan 20^\circ = 0.364\)), so it stays at rest.
- \(R = 30\cos 30^\circ + 40\sin 30^\circ = 45.98\) N.
- Friction \(= 0.2 \times 45.98 = 9.196\) N down the plane.
- \(40\cos 30^\circ - 30\sin 30^\circ - 9.196 = 3a\).
- \(10.44 = 3a\), so \(a = 3.48\ \text{m s}^{-2}\).
- \(R = 700\cos 15^\circ = 676.1\) N, so friction \(= 67.61\) N.
- Down the slope: \(700\sin 15^\circ - 67.61 - 40 = 70a\).
- \(181.17 - 67.61 - 40 = 73.56 = 70a\), so \(a = 1.051\ \text{m s}^{-2}\).
- \(v^2 = 2(1.051)(100) = 210.2\), so \(v = 14.5\ \text{m s}^{-1}\).
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 is released from rest on a smooth plane inclined at \(25^\circ\) to the horizontal. Find its speed after it has slid 6 m down the plane.
- A\(7.12\ \text{m s}^{-1}\)
- B\(10.4\ \text{m s}^{-1}\)
- C\(11.0\ \text{m s}^{-1}\)
- D\(2.25\ \text{m s}^{-1}\)
A particle is projected up a line of greatest slope of a SMOOTH plane inclined at \(\alpha\) to the horizontal. Which statement about its acceleration is correct?
- AIt is \(g\sin\alpha\) down the plane, both while moving up and while sliding back down
- BIt is \(g\sin\alpha\) up the plane while moving up, and down the plane while sliding down
- CIt is zero while moving up and \(g\sin\alpha\) down the plane while sliding down
- DIt is \(g\cos\alpha\) down the plane in both directions
A particle of mass 3 kg slides down a rough plane inclined at \(\alpha\) to the horizontal, where \(\sin\alpha = 0.6\). The coefficient of friction is 0.5. Find its acceleration.
- A\(2\ \text{m s}^{-2}\)
- B\(10\ \text{m s}^{-2}\)
- C\(1\ \text{m s}^{-2}\)
- D\(5\ \text{m s}^{-2}\)
A particle is projected up a line of greatest slope of a rough plane inclined at \(\alpha\) to the horizontal, where \(\sin\alpha = 0.6\), with speed \(8\ \text{m s}^{-1}\). The coefficient of friction is 0.5. Find how far up the plane it travels before coming to rest.
- A3.2 m
- B16 m
- C5.33 m
- D6.4 m
A particle is projected up a line of greatest slope of a rough plane, where \(\sin\alpha = 0.6\) and \(\mu = 0.5\). It comes to rest 3.2 m above its starting point A, measured along the plane. Find its speed when it returns to A, or decide that it stays at rest.
- A\(3.58\ \text{m s}^{-1}\)
- B\(8\ \text{m s}^{-1}\)
- C\(6.20\ \text{m s}^{-1}\)
- DIt stays at rest
A particle of mass 4 kg is pushed up a rough plane inclined at \(25^\circ\) to the horizontal by a horizontal force of 50 N, as shown. The coefficient of friction is 0.3. Find the acceleration of the particle.
- A\(2.80\ \text{m s}^{-2}\)
- B\(4.38\ \text{m s}^{-2}\)
- C\(7.10\ \text{m s}^{-2}\)
- D\(11.2\ \text{m s}^{-2}\)
A particle is released from rest on a rough plane inclined at \(30^\circ\) to the horizontal. It slides 9 m down the plane in 3 s. Find the coefficient of friction.
- A0.346
- B0.462
- C0.231
- D0.6
A particle of mass 6 kg is pulled up a line of greatest slope of a rough plane inclined at \(20^\circ\) by a force of \(P\) N parallel to the plane, as shown. The coefficient of friction is 0.25, and the particle accelerates at \(1.5\ \text{m s}^{-2}\). Find \(P\).
- A43.6
- B15.4
- C34.6
- D44.5
Quiz key
\(a = 10\sin 25^\circ = 4.226\ \text{m s}^{-2}\), so \(v^2 = 2 \times 4.226 \times 6 = 50.71\) and \(v = 7.12\ \text{m s}^{-1}\). The tempting 10.4 uses \(g\cos 25^\circ\), the component into the plane. 11.0 uses the full \(g\), as if the particle were falling vertically, and 2.25 forgets the \(g\) in \(g\sin\alpha\).
The only force along a smooth plane is \(mg\sin\alpha\) down the slope, so the acceleration is \(g\sin\alpha\) down the plane whichever way the particle moves; going up, it decelerates. The tempting "up the plane while moving up" confuses the direction of motion with the direction of acceleration. \(g\cos\alpha\) is the wrong component.
\(R = 30 \times 0.8 = 24\) N, so friction \(= 12\) N up the plane. Down the plane: \(30 \times 0.6 - 12 = 3a\), so \(a = 2\ \text{m s}^{-2}\). The tempting 10 adds the friction, as if it acted down the plane; that is the deceleration when the particle moves UP. 1 uses friction \(\mu mg = 15\) N, and 5 swaps sin and cos.
Deceleration \(= g\sin\alpha + \mu g\cos\alpha = 6 + 4 = 10\ \text{m s}^{-2}\). Then \(0 = 64 - 2 \times 10 \times s\), so \(s = 3.2\) m. The tempting 16 m uses the downhill acceleration of \(2\ \text{m s}^{-2}\), with friction pointing the wrong way. 5.33 m ignores friction (\(a = 6\)), and 6.4 m drops the 2 in \(v^2 = u^2 + 2as\).
\(\tan\alpha = 0.75 > 0.5\), so the particle slides back. Down the plane, \(a = 10(0.6 - 0.5 \times 0.8) = 2\ \text{m s}^{-2}\), so \(v^2 = 2 \times 2 \times 3.2 = 12.8\) and \(v = 3.58\ \text{m s}^{-1}\). The tempting \(8\ \text{m s}^{-1}\) assumes it returns with its launch speed, which ignores the energy lost to friction. 6.20 ignores friction on the way down, and "stays at rest" would need \(\mu \ge \tan\alpha\).
\(R = 40\cos 25^\circ + 50\sin 25^\circ = 57.38\) N, so friction \(= 0.3 \times 57.38 = 17.21\) N down the plane. Along the plane: \(50\cos 25^\circ - 40\sin 25^\circ - 17.21 = 4a\), giving \(45.32 - 16.90 - 17.21 = 11.20\) and \(a = 2.80\ \text{m s}^{-2}\). The tempting 4.38 uses \(R = 40\cos 25^\circ\), missing the push into the plane. 7.10 ignores friction, and 11.2 is the resultant force, not the acceleration.
\(9 = \tfrac{1}{2}a \times 9\), so \(a = 2\ \text{m s}^{-2}\). Then \(2 = 10\sin 30^\circ - 10\mu\cos 30^\circ\), so \(\mu = \dfrac{3}{8.660} = 0.346\). The tempting 0.462 drops the \(\tfrac{1}{2}\) in \(\tfrac{1}{2}at^2\) (\(a = 1\)). 0.231 uses \(a = 9 \div 3 = 3\), the average speed rather than the acceleration, and 0.6 uses \(\sin 30^\circ\) for the normal reaction.
\(P = 6 \times 1.5 + 60\sin 20^\circ + 0.25 \times 60\cos 20^\circ = 9 + 20.52 + 14.10 = 43.6\). The tempting 15.4 points friction up the plane; the particle moves up, so friction acts down. 34.6 is the force for constant speed (it leaves out \(ma\)), and 44.5 uses friction \(\mu mg = 15\) N instead of \(\mu mg\cos 20^\circ\).
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.