Newton's laws of motion

• Forces on the left, \(ma\) on the right, and friction on a sliding particle against its velocity.
• The derived results are quick checks, not starting points: a full solution writes the equation of motion from the force diagram every time, and the result falls out of it.
• Resultant = ma, forces only on the left. Friction against the velocity, so it reverses on the way down. One string, one tension; a negative bar force is a thrust.

• 4.4.1 I can — apply Newton's laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod
• 4.4.2 I can use — use the relationship between mass and weight
• 4.4.3 I can — solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration
• 4.4.4 I can — solve simple problems which may be modelled as the motion of connected particles

Accuracy and working are marked. The syllabus states that non-exact numerical answers are to be given correct to three significant figures, or one decimal place for angles in degrees, unless the question specifies otherwise, and that to earn accuracy marks you should avoid rounding until the final answer. It also states that no marks are given for unsupported answers from a calculator, and that graphic calculators and calculators with symbolic algebra or calculus are not permitted. Every worked example in this chapter is therefore written the way an answer must be written: the method line first, the substitution visible, full precision carried, one rounding at the end.

• “\(T - 30 + 3a = 0\)”, with an arrow labelled \(ma\) on the force diagram. Correct \(ma\) is not a force. It is what the resultant force equals. Draw the forces only, draw the acceleration as a separate arrow beside the particle, and write the equation of motion as resultant \(= ma\): \(T - 30 = 3a\) (section A).
• “The particle goes up the rough plane and comes back down with the same acceleration.” Correct Friction on a sliding particle acts against its velocity, so it changes side when the particle turns round. Going up, weight component and friction both act down the slope; coming down, friction acts up the slope. Two stages, two equations, two different accelerations (section C).
• “The lift is going up, so the scale reads more than the weight.” Correct The direction of the acceleration decides. A lift moving up but slowing down accelerates downwards, and \(R < mg\). At constant velocity, up or down, \(R = mg\).
• “\(T_1\) on one side of the pulley, \(T_2\) on the other.” Correct A light string over a smooth pulley has one tension throughout: the same \(T\) on both particles. And \(T\) never appears in the whole-system equation, because it is internal to the system.
• “\(F = \mu R\), so the box accelerates at \((20 - 25)/5 = -1\ \mathrm{m\,s^{-2}}\).” Correct Check that motion starts at all. If the pull on a particle at rest is no more than \(\mu R\), it stays at rest and \(F\) equals the pull. Friction can stop motion; it cannot reverse it.
• “The tension in the tow-bar is \(-450\) N.” Correct A negative tension means the bar is pushing: say “a thrust of 450 N”. A rope cannot push, so under hard braking a rope goes slack.
• “\(T - 30 + 3a = 0\)”, with \(ma\) drawn as a force. Repair \(ma\) is not a force. Write resultant \(= ma\): \(T - 30 = 3a\), and draw the acceleration beside the particle.
• “\(R = mg\) on the rough ground”, when the pull is at an angle. Repair Resolve vertically. A pull at \(30^\circ\) above the horizontal has an upward component, so it reduces \(R\) and with it the friction.
• The same acceleration used up and down a rough plane. Repair Friction opposes the velocity, so it changes side when the particle turns round. Resolve each stage separately.
• “The lift is going up, so \(R > mg\).” Repair The direction of the acceleration decides. A lift going up and slowing down has \(R < mg\).
• Using \(F = \mu R\) for a particle at rest under a small pull. Repair Check first whether it moves. If the pull is less than \(\mu R\), it stays at rest with \(F\) equal to the pull.
• Different tensions \(T_1\) and \(T_2\) in one light string over a smooth pulley. Repair One string, one tension.
• Including \(T\) in the whole-system equation. Repair \(T\) is internal to the system and cancels; only external forces appear.
• “\(a = \dfrac{(m_1 - m_2)g}{m_1 - m_2}\)”. Repair The resultant \((m_1 - m_2)g\) accelerates the total mass: \(a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}\).
• “The force on the pulley is \(T\).” Repair Both parts of the string pull on it: \(2T\) when both are vertical; \(T\sqrt2\) when one is horizontal and one vertical.
• “The tension in the tow-bar is \(-450\) N.” Repair A negative tension is a thrust: say “a thrust of 450 N”.
• The driving force put on the trailer as well as the car. Repair The engine drives the car only; the trailer is moved by the coupling.
• Continuing to use \(T\) after the hanging particle hits the floor. Repair The string is slack: \(T = 0\), and the other particle moves under its remaining forces — friction alone on a rough table, its weight alone if it is hanging.
• “\(W = 70\) N for a 70 kg man.” Repair \(W = mg = 700\) N. Mass in kg on the right of the equation of motion; weight in N among the forces.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Mass in kg, weight in N, and never mix them. “A particle of mass 4 kg” puts \(40\) N on the force diagram and \(4a\) on the right of the equation of motion. Writing \(4\) on the diagram or \(40a\) on the right is the same error in two places.
• Draw it, then read the equation off it. Once the diagram is right, the equation of motion is mechanical: along the arrow \(a\), every force pointing the same way is positive, every force pointing against it is negative, and the right-hand side is \(ma\). If you find yourself wanting to write \(ma\) on the left, look for it on your diagram — and rub it out.
• Interleave with the chapters that use this one. Chapter 22 reuses the equation of motion for a car whose driving force comes from its engine’s power, on the level and on a hill: when you reach it, re-answer worked example 6 and worked example 3 there, the second by energy, and check the answers agree. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Mechanics content, examined in Paper 4. Paper 4 (Mechanics) is 40% of an AS Level that includes it and 20% of an A Level that includes it. It assumes the Paper 1 content. An A Level route that includes Paper 4 takes Papers 1, 3, 4 and 5; Paper 4 cannot be combined with Paper 6. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A question on this section is structured: draw or read the forces, write \(\text{resultant} = ma\) for one particle or for each of two, then use a constant-acceleration formula for a speed, distance or time. Later parts change the situation — the string goes slack, the car brakes, the particle turns round on a rough slope — and each change needs a fresh equation of motion. When a “show that” gives you an acceleration, use the given value in the later parts even if your own working went wrong.
• MF19 prints only \(v = u + at\), \(s = \tfrac12(u + v)t\), \(s = ut + \tfrac12at^2\) and \(v^2 = u^2 + 2as\). Newton’s laws, \(F = ma\), \(W = mg\), \(F = \mu R\) for sliding and the value \(g = 10\ \mathrm{m\,s^{-2}}\) are not printed and must be known (see the MF19 card).
• Each equation of motion states its positive direction. Keep \(R\), \(F\) and the acceleration unrounded (\(17.321\), \(3.4641\), \(6.7321\)) and round once, to 3 significant figures with units, at the end; exact answers such as \(\tfrac53\) or \(\sqrt2\) may be left exact. A tow-bar force that comes out negative is reported as a thrust of that size, never as a negative tension.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.