Edexcel A-level Maths · Topic M8
Practice questions covering forces and newton's laws from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
6 questions on Forces and Newton's laws, each with its mark scheme and worked method. Reveal the answer when you have had a go.
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(3)
(Total for Question 1 is 3 marks)
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(5)
(Total for Question 2 is 5 marks)
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(4)
(Total for Question 3 is 4 marks)
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(5)
(Total for Question 4 is 5 marks)
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(5)
(Total for Question 5 is 5 marks)
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(4)
(Total for Question 6 is 4 marks)
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Printable question pack — M8 Forces and Newton's lawsAll 60 questions across 6 specification points — 54 more than the 6 above — with the full mark scheme. Guest accessible · browser print.13 questions shown as printed, 9 reference-only · newest series first
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Edexcel 9MA0 · June 2024 · Paper 32 · Q1 · 3 marks
3 marks
Examiner report
This proved to be a very friendly starter. (a) The vast majority of candidates were able to score the mark, with answers of 0.5g or 4.9 being the most common. The most common error was to omit the g whilst a few used g = 9.81 or gave a fractional answer, both of which were penalised once for the whole question. A few, seeing the word ‘magnitude’, squared and/or square rooted their answer. (b) Most candidates found the value of the limiting friction by multiplying their answer for (a) by 2 7 which earned the M mark and then realised that X had to be the same value. A significant number lost the A mark after having found the correct value of F for not stating a value for X, as the question required.
A particle P of mass 0.5 kg rests on a rough horizontal plane. The coefficient of friction is 2/7. A horizontal force of magnitude X newtons is applied and P is in limiting equilibrium.
Edexcel 9MA0 · June 2024 · Paper 32 · Q3 · 7 marks
7 marks
Examiner report
A very well answered question with many candidates using a clearly labelled force diagram and gaining full marks for (a) and (b). It is always good exam technique to annotate the diagram to support the answer. (a) Most candidates were able to resolve perpendicular to the plane to obtain R=mgcos and then replacecos to obtain the required answer. A few multiplied by sin instead of cos but this still earned the M mark. However, those who used tan or divided by sin or cos lost both marks. A few responses chose to use a decimal approximation for the angle which could score both marks in this part. (b) This was a “show that” question, so full and accurate working was required. The vast majority wrote down an equation of motion parallel to the plane, sin mg F ma − = and then used their answer from (a) to replace F, cancelled the m’s and factorised to obtain the given answer. Note that when there is a printed answer to show, candidates must give their answer in exactly the same form to guarantee earning all the marks. Common errors were setting sin mg F = , using an incorrect trigonometrical ratio, missing m’s or g’s, using 𝜇= 5 12 . (c) This was worth one mark, so the answer needed to be clear e.g. the particle does not move, it remains at rest. Some candidates went on to try to justify their answer, but this was not required. A considerable number said it would slide down more slowly, or more worryingly, it would slide up the slope. Stating only that the acceleration was zero or that the forces were in equilibrium did not score the mark, as the particle could still be travelling at constant speed. However, stating that the acceleration was zero or that the forces were in equilibrium and that the particle remained at rest did earn the mark.
A particle P of mass m is released from rest on a rough plane inclined at angle alpha, where tan(alpha) = 5/12. The coefficient of friction is mu, initially with mu < 5/12, and air resistance is negligible.
Edexcel 9MA0 · June 2024 · Paper 32 · Q6 · 9 marks
9 marks
Examiner report
The most successful candidates in this question tended to indicate what they were doing at each stage – taking moments about A, resolving horizontally, etc, in words or symbols. It is always good exam technique to annotate the diagram to support the answer. The majority of errors affecting performance in the question as a whole were due to incorrect force diagrams: the normal reaction force at C was sometimes drawn vertically upwards rather than perpendicular to the rod, and an extra reaction force was occasionally put in at G and/or B. Most candidates worked with the angle theta, but a few did convert into terms such as ( ) O sin 90 − and used these successfully. (a) There were a good number of fully correct responses to this ‘Show that’ question. The majority of candidates found and used the trig ratios successfully, possibly because the answer was given in the question, which allowed them to identify whether they had calculated or used the wrong one. If a candidate gained one or two marks, but not all three, it was usually due to sin/cos confusion or making arithmetic errors when rearranging the equation. Some candidates lost all the marks because they omitted the a’s and so did not have a moments equation, but most included these initially and then cancelled appropriately. The final mark was occasionally lost because explicit use of 3 cos 5 = was not seen or the candidate chose to give the force another name (such as RC) and so did not arrive at the printed answer. A common error was using 2a instead of 1.5a as the distance. A few candidates gave the impression of having arrived at the required answer, but on closer inspection were not correct in their work. The most common example of incorrect working was 2 0.5 sin 5 S Mg Mg = = . Some tried to formulate a proof based on the given answer that did not involve taking moments, but instead tried to resolve forces, and received no credit. Most candidates presented their answer in the given form, although S on the RHS was occasionally seen. (b) Many candidates scored all six marks in this part. Resolving vertically and horizontally was the most common and the most successful method. There were few errors apart from sin and cos confusion. Taking moments about C to give an equation in R and F and then resolving was the next most common method, but candidates were more likely to make errors using this approach, either by confusing sin with cos or by omitting a force entirely. Very few candidates resolved parallel and perpendicular to the rod and most of those who used this method failed to arrive at the correct answer due to the more challenging algebra. Occasionally candidates wrote dimensionally inconsistent equations by failing to include lengths in all of their moment terms or, more rarely, erroneously including lengths in their force terms. Errors included missing a g where there should have been one or not having the correct number of terms in a moment or a resolution equation. A surprisingly common error was for students to confuse the reaction force S with the reaction force at A, resulting in the incorrect equation 2 5 F Mg = Some candidates lost method marks by resolving forces that did not need resolving (for example, by writing an equation with both sin Mg and cos S ). Whichever method was used, early substitution for F into their solutions was often seen, with almost all candidates realising that they needed to use F = μR. Some candidates clearly still incorrectly believe that μ cannot be greater than 1 and were ‘dismissive’ when their value was. It was not uncommon to see very messy responses, with multiple attempts and work crossed out. Many candidates made a good attempt at solving for µ, but this was often impeded by the lack of clarity over the previous equations. The mark scheme enabled candidates who used a correct method but who made sign errors, confused sin and cos or used incorrect distances to score half marks (generally, M1A0M1A0M1A0), while those with more fundamental misunderstandings were prevented from accessing the dependent method mark; this seemed to sort out the candidates well. When µ was found, most candidates gave the answer as a fraction or to an appropriate degree of accuracy.
A uniform rod AB has mass M and length 2a. End A rests on rough horizontal ground. The rod is in equilibrium against a small smooth fixed horizontal peg P. The contact point on the rod is C, where AC = 1.5a. The rod makes angle theta with the ground, tan(theta) = 4/3. The peg's normal reaction on the rod at C is S.
Edexcel 9MA0 · June 2023 · Paper 32 · Q2 · 4 marks
4 marks
Examiner report
The vast majority were able to score the mark in part (a), with answers of 5g or 49 being the most common. A few candidates used g = 9.81 but were not penalised here. There were very many correct answers for the second part also, but some ignored the 28N force completely and gave 5 1.4 7 = , as their answer or made a sign error and ended up with F = 35. In the final part, F R = was very well known, and this was a follow through mark on their answers to (a) and (b). However, some left their answer as a fraction or didn’t give it to 2 sf and lost the mark anyway.
A particle P of mass 5 kg is pulled to the right along a rough horizontal plane by a horizontal force of 28 N. The only resistance to motion is a frictional force of magnitude F newtons acting to the left.
Edexcel 9MA0 · June 2023 · Paper 32 · Q5 · 11 marks
11 marks
Examiner report
In part (a), the vast majority of candidates recognised that the horizontal component of velocity of the particle remained constant throughout the motion; this was then used, together with the horizontal distance, to form an equation in T and α and to derive the given result. The second part, however, provided a greater challenge and, although many realised that they needed to consider the vertical motion, a few did not know how to proceed, with some solving the given quadratic in tanα rather deriving it. However, many did make progress, using 2 1 2 s ut at = + with a vertical distance of 20 although there were occasional sign errors. They mostly then substituted the expression for T from part (a) to obtain an equation in α although slips were not uncommon at this stage. Those starting with different suvat equations tended to make little significant progress. To achieve the given result it was necessary to use the trig identity sec2 α = 1 + tan2 α (or equivalent work) to obtain an unsimplified quadratic equation in tan α before reaching the given equation. This was the main stumbling block for many and it was not unusual to see candidates make more than one attempt. Since the final equation was given and the demand of the question was “show that”, it was essential that all steps in the working were correct and the final answer was stated exactly as printed in order to achieve all the marks. In part (c) most candidates realised they should solve the given quadratic in tanα to give two possible angles of projection. Consideration of the situation should have indicated that the larger of the two angles would lead to the greatest possible height. Although many recognised that the greatest height is reached when the vertical component of velocity is zero, some chose to use the smaller 45o angle. Some found the greatest heights for both angles but then failed to select the relevant one. Other errors included using the time to greatest height as half the answer in part (a) or choosing sin α = 1 to give the maximum height in an otherwise correct suvat equation. Rounding errors were not uncommon leaving the final answer incorrect to 3 significant figures. In the final part, most candidates were able to suggest an acceptable limitation of the model such as it not including spin, dimensions of the ball, effect of wind or a more accurate value for g. The most common incorrect answers were that the model did not include weight/mass of the ball and the ground was not perfectly flat which was not specified as a modelling assumption in the question.
A small ball is projected from point O on horizontal ground with speed 28 m s^-1 at angle alpha above the ground. After T seconds it passes through A, which is 40 m horizontally and 20 m vertically from O. The ball is modelled as a particle moving freely under gravity.
Edexcel 9MA0 · June 2023 · Paper 32 · Q6 · 13 marks
13 marks
Examiner report
In the first part, the majority of candidates stated that the frictional force would act towards the right on the diagram and many gave a satisfactory reason. A minority gave the reason that the friction had to balance the reaction force at the wall which acted to the left. Most mentioned that it prevented the rod from slipping to the left. Those who did not achieve the mark often used language that implied the rod was actually moving. A small number thought that the direction was left or gave no reason at all. In part (b), the question directed the candidates to take moments about A and most were able to do so successfully with many achieving the three available marks. The occasional incorrect response had the distance to the reaction at B unresolved as 2a. Also a small number had a’s missing from both sides of their equation. Since this was a “show that” question it was important that all stages of the solution were shown and were correct. It should be noted that the candidate’s final answer had to be written exactly as the printed answer, using the same symbols in the same order, to achieve the final mark. Part (c) required a calculation of the coefficient of friction. There were many fully correct solutions involving vertical and horizontal resolution but many incorrect responses were also seen. The majority who tried other approaches, such as moments about B or resolving parallel and perpendicular to the rod, were usually unsuccessful as they would often miss out a term or there would be a sin/cos error. Some candidates attempted to resolve vertically and horizontally but included an extra force. It was not uncommon for the horizontal resolution F = S to be correct but an extra force would appear in the vertical resolution. Almost all recalled F = µR and tried to use it in some way. Part (d) involved finding the magnitude of the resultant force acting at A. It was not always well answered and sometimes omitted entirely. A significant minority of responses were, however, fully correct and there were many who, having made an error in part (c), were still able to apply Pythagoras’ theorem correctly with their values of F and S. However, some candidates did not understand what was required with R = mg, or an attempt to just combine R and F by adding, seen quite often. A small number failed to deal properly with the square root making an error with the simplification in terms of mg in an otherwise correct solution. In the final part, the majority of candidates were unable to gain the mark often because of lack of precision in the wording of their response. Many stated that S was larger because the weight was closer to B or it exerted more force/pressure at B. Some who did refer to an increased moment about A failed to specify it as the moment of the weight. Some mentioned distance but did not explicitly refer to moments. Candidates who used their moments equation to show that a distance of the weight from A greater than a led to a larger value of S were generally successful. A small number thought S was smaller or did not change.
A rod AB has mass M and length 2a. End A rests on rough horizontal ground and end B rests against a smooth vertical wall to the right of A. The rod makes an angle theta with the ground and is at rest in limiting equilibrium. The wall's normal reaction at B has magnitude S and acts horizontally to the left. In an initial model the rod is uniform, so its weight Mg acts at its midpoint.
Edexcel 9MA0 · June 2022 · Paper 32 · Q2 · 10 marks
10 marks
Examiner report
Almost a quarter of the candidates scored zero whilst 21% scored full marks on this question. In part (a) the most common approach was to resolve perpendicular and parallel to the plane to first find a value for X (49) before finding a value for the friction. Very few used the most efficient method and resolved vertically. The resolving was generally done well with most candidates resolving all the forces that needed to be resolved and most without sin/cos confusion and sign errors. Many who reached the correct value for friction were able to give the correct direction of the force, including those that had assumed that it acted up the plane and had obtained a negative answer. Not all remembered to give the magnitude of the force and numerical calculation errors were seen in some working. µR was seen in some responses which lost the last mark as friction is not limiting in this part of the question. Part (b) was generally done well, even by those that had struggled with the first part, with many candidates writing down a correct F = ma equation. The most common mistake was to take R as 68.6N, the value given in part (a), and not resolve to find the new value of R. Those that resolved perpendicular to the plane generally went on to use F = µR correctly and then accurately calculate the acceleration of the block. A few candidates did not remove the force X. It was noticeable that candidates who produced well labelled diagrams and who presented their solutions in a logical, neat and concise manner often continued to produce a correct solution in both parts.
A rough plane rises to the right at angle alpha to the horizontal, where tan(alpha) = 3/4. A 5 kg block B is held in equilibrium on the plane by a horizontal force of magnitude X newtons acting to the right. The force lies in the vertical plane containing the line of greatest slope. The block is modelled as a particle and the plane's normal reaction has magnitude 68.6 N.
Edexcel 9MA0 · June 2022 · Paper 32 · Q3 · 9 marks
9 marks
Examiner report
In part (a), the majority of candidates were able to achieve the first mark for adding the two forces and collecting the i and j components. Many just equated this resultant force to the vector (3i + j) rather than using (3i + j) to define the direction. Some set up equations λ + 4 = 3 and µ - 1 = 1 and proceeded to solve them simultaneously gaining no further credit. Those who did use ratio usually went on to derive the required equation correctly; however, it should be remembered that, if required to derive a given answer, this must be written as a final conclusion in the form exactly as stated in the question. Another successful approach was to equate the i and j coefficients to 3k and k and solve as simultaneous equations. A small minority of candidates applied F = 4a or used integration to find v at this stage and then used ratio appropriately to obtain the required answer. In the second part, many candidates were able to find the resultant force and acceleration vectors. Some candidates found the magnitude of the force or acceleration at this stage and proceeded correctly with suvat formulae from there, while others continued with vectors. A number of candidates did not identify the initial velocity as zero, reading back to the information given at the start of the question being important here, so failed to use suvat formulae correctly. Use of u = 3i + j was not uncommon. Although most used suvat formulae, some successfully integrated the constant acceleration to find the velocity and displacement vectors. The majority of candidates knew how to find the magnitude of a vector using Pythagoras. The most significant errors seen included finding the resultant force but then using it as a displacement or velocity, showing a lack of understanding of the situation. A small number of candidates found an incorrect value for µ when substituting λ and there was occasional use of only one of the forces.
In this question i and j are horizontal unit vectors. A particle P of mass 4 kg is initially at rest at point A on a smooth horizontal plane. At t = 0, forces F1 = (4i - j) N and F2 = (lambda i + mu j) N are applied. P moves in the direction 3i + j. At t = 4 seconds it passes through B.
Edexcel 9MA0 · June 2022 · Paper 32 · Q4 · 11 marks
11 marks
Examiner report
In part (a), very few candidates gave a fully correct reason for why the frictional force must act to the right on the diagram. Many stated that friction must oppose the direction of motion, saying that the rod would otherwise slip to the left. Some did mention it opposing the horizontal component of the tension but then failed to include that friction was the only other horizontal force in the system which was a key part of the argument. In the second part, it was required to derive a given expression for T (tension). Although some candidates struggled to make a valid start and basically just wrote down the answer and stated ’by resolving ’, it was well done by a fair number who set up a correct equation for moments about A and hence deduced the given result. Alternative approaches by either taking moments about other points or resolving forces were very rare and almost never successful, mostly because of missing terms. A few candidates failed to use perpendicular distances for the moments of the weight terms or failed to include the weight of the rod leading to a missing term. Part (c) was generally well answered by those candidates who resolved vertically, substituted for T and re-arranged to obtain the given answer thereby gaining full marks in a few lines of working. Candidates who used alternative equations as a starting point were generally unsuccessful, usually a result of a missing term in one or both equations used, this often involving friction. In part (d) those candidates who resolved horizontally to give F = T sinƟ usually managed to score all four marks for deriving the exact value of the coefficient of friction. There were a variety of equations, resolving or taking moments, which would have led directly to a value for F since the tension and normal reaction were known at that stage; however, these involved more working and therefore greater opportunity for error. Nevertheless, successful solutions by these methods were seen on occasion. Most of those candidates who found an expression for F were able to achieve the independent method mark for the use of F = μR even if their method for finding F was incorrect. Since all the answers were given in this question, candidates needed to ensure that they showed sufficient detail in their working to warrant being awarded all the available marks. In particular, candidates should be encouraged to state their resolved forces and moments as equations, rather than considering them in isolation, and to ensure the substitution stages are shown.
A uniform rod AB has mass M and length 2a. A particle of mass 2M is attached at C, where AC = 1.5a. End A rests on rough horizontal ground and the rod is held in equilibrium at angle theta above the ground by a light string attached at B. The string is perpendicular to the rod and runs upwards and to the left from B. Its tension is T.
Edexcel 9MA0 · October 2021 · Paper 32 · Q2 · 12 marks
12 marks
Examiner report
Candidates need to make sure that they read and then answer questions carefully. In part (a) an equation of motion for A was required. Most realized that this required the use of Newton’s second law on the 3m mass. However, some ignored this requirement and went straight to a whole system approach to the problem. Errors in part (a) were mainly due to the omission of a relevant force or a misunderstanding of direction. It was essential that the frictional force on A and the tension were in the same direction i.e. had the same sign in the equation. In part (b) either an equation of motion for B was required or an equation of motion for the whole system. This was generally well answered provided that the reaction force from the plane on A had been found and used with F R µ = to give 1 6 F R = . The question was a “Show that” question so full marks were only awarded if there was a fully correct solution, with all the steps included. Errors involved sin/cos confusion, the omission of g, an incorrect direction for the reaction force, omission of a force in an equation of motion or an incorrect reaction such as R = 3mg. Some candidates did make algebraic errors in solving their equations. A few candidates assumed that the tension in the string was mg. Those candidates who attempted a whole system approach were mostly successful. In part (c), most of the graphs given were correct but there are still candidates who add a solid vertical line at the end of the graph and this lost the mark. However, the second mark here was frequently lost. All that was required was a comment that the acceleration was constant. For the final part, candidates needed to answer the question which asked for “ how this would affect the working in part (b)”. Those who referred to how it would affect the answer received no credit.
A small stone A of mass 3m rests on a rough plane inclined at angle alpha, where tan(alpha) = 3/4. It is connected by a light inextensible string, parallel to a line of greatest slope and passing over a small smooth pulley P at the top of the plane, to a freely hanging stone B of mass m. The coefficient of friction between A and the plane is 1/6. A is released from rest and moves down the plane, so B moves upwards, until B reaches the pulley. Both stones are modelled as particles.
Edexcel 9MA0 · October 2021 · Paper 32 · Q3 · 10 marks
10 marks
Examiner report
Part (a) required candidates “ To show that…”. Consequently, full marks were only awarded if there was a fully correct solution, with all the steps included. Thus successful candidates needed one moments equation together with a vertical resolution and/or a horizontal resolution, depending on which point they had used for their moments equation. These equations, combined with F R µ ≤ , could then lead to full marks being earned. Many candidates assumed that the beam was in limiting equilibrium i.e. that F R µ = , produced an equation and then changed it to the required inequality. These candidates lost the final mark. Other errors included omission of g, sin/cos confusion in moments equations and assuming that the reaction at A and the reaction at B were the same. Candidates tended to be more successful in part (b), where a similar strategy was required, together with F R µ = , since the beam was now in limiting equilibrium. A common error was to have the friction force acting towards the wall. This lead to a k value of – 0.1 and the negative sign should have flagged up an error to candidates.
A beam AB has mass m and length 2a. It rests in equilibrium in a vertical plane perpendicular to a wall, with A on rough horizontal ground and B against a smooth vertical wall. The beam is inclined at angle theta to the horizontal and is modelled as a uniform rod. The coefficient of friction at A is mu.
Edexcel 9MA0 · October 2020 · Paper 32 · Q1 · 9 marks
9 marks
Examiner report
The first part was well done by the majority of candidates who resolved perpendicular to the plane to give a correct expression for the normal reaction. In part (b), most resolved parallel to the plane and then used F=µR to derive the given answer for µ. There was occasional sin/cos confusion seen in the relevant equations but this was fairly rare. Horizontal resolution was an alternative valid approach. The third part was found to be much more challenging. Those who realised that m cancels out of the resolution equation parallel to the plane achieved the one mark available very easily but such instances were rare. Most attempted to explain the situation in words. A common response was that as m increases, R increases and so F increases to a larger value than before, meaning that the brick will remain in equilibrium, with no consideration of the component of weight acting down the plane. It was essential to the argument that either the friction and the force pulling it down the plane increase by the same amount or that the friction and weight component increase in the same proportion; just stating that both increase was not deemed sufficient. In part (d), few were able to relate their earlier work to help them answer this question and the majority of candidates failed to realise that the brick would continue to move down the plane with a constant speed since there would be no resultant force. Most thought it would move with acceleration because of weight/gravity or with deceleration because of friction. A few just stated it would move slowly down the plane.
A rough plane is inclined at angle alpha to the horizontal, where tan(alpha) = 3/4. A brick P of mass m is in equilibrium on the point of sliding down the plane. The coefficient of friction is mu and P is modelled as a particle. P is then removed and a much heavier brick Q is placed on the same plane; Q has the same coefficient of friction. Finally Q is projected at 0.5 m s^-1 down a line of greatest slope and is modelled as a particle.
Edexcel 9MA0 · October 2020 · Paper 32 · Q4 · 10 marks
10 marks
Examiner report
A significant number of candidates made little or no progress with this question; a number of responses showed a complete lack of understanding of the nature and directions of the forces acting on the ladder. In part (a) a fair number attempted to take moments about A to derive the given reaction at C although sometimes perpendicular distances were not used and this invalidated the method. The most straightforward approach for part (b) was to resolve horizontally and vertically. Some candidates chose to take moments about various points or to resolve perpendicular and/or parallel to the ladder. These methods often led to errors including missing terms and incorrect distances. Some thought the normal reaction at C was horizontal. Those who attempted to use a reaction perpendicular to the ladder generally used this in F=µR and were unaware that this is only valid for a normal reaction. Confusion between sin/cos in resolving the reaction at C horizontally and vertically was surprisingly common. There were, nevertheless, some well-expressed fully correct solutions seen.
A uniform ladder AB has mass M and length 6a. End A rests on rough horizontal ground. The ladder rests at point C against a fixed smooth horizontal rail, with the vertical plane containing AB perpendicular to the rail. Point C is 4a vertically above the ground. The ladder is inclined at angle alpha to the horizontal, where sin(alpha) = 4/5. The coefficient of friction at A is mu and the ladder is in limiting equilibrium. Since AC sin(alpha) = 4a, AC = 5a; the ladder's centre of mass is 3a from A.
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Everything here stays free. If forces and newton s laws is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.