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In part (b), most responses gave a good description of the motion ‘as decelerating to a stop or slowing down’. However, a significant number thought that the box would stop immediately or continue at a constant speed.
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In part (d), learners needed to plan their method carefully. Most responses gained the mark for finding the value of the friction as 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚= 𝜇𝜇𝜇𝜇= 0.2 x 19.6. Successful responses then applied Newton’s 2nd law, ±𝐹𝐹= 𝑚𝑚𝑚𝑚= 2𝑎𝑎 and then used a suvat approach to arrive at the distance as 4.1 or 4.08(m) to 2 or 3 s.f. Leaving the final answer to more than 3 s.f resulted in the final mark not being awarded. Errors were mainly in the 𝐹𝐹= 𝑚𝑚𝑚𝑚 equation when the force was incorrectly formed. Use of a = 0.4g or a mu = - were common mistakes. Fractional answers, after the use of g = 9.8, were not acceptable.
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In part (e), a variety of acceptable limitations to the model were given. Learners were required to refer to the model, as stated in the question, and most referred to the box being modelled as a particle or that it had size or dimensions. Others correctly said that the coefficient of friction may vary as the box continues in its motion or that the friction may vary. Common incorrect answers were: use a more accurate value of g, the plane might not be horizontal or a reference to the mass of the box. None of these are mentioned in the model in the question and so gained no credit. Note that if more than one answer is given, any incorrect answer in the list will mean the mark is not awarded.
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In the third part although many learners recognised that R (with no friction acting at B) was greater than R1 (with friction acting at B), the explanations were not always clear. Those who 3 R Mg used equations generally performed better and gained the mark e.g. and = 3 so R is greater. Some learners thought that the frictional force would act R Mg F = - 1 downwards concluding that R1 was greater whilst others either just stated ‘because of friction’ or did not state that friction acted upwards (or in the same direction as R1).
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Part (d) was the most challenging part of the question. It was often left blank or multiple half- finished attempts were seen with little clear structure. The most common approach was to take moments about A, to follow from the work in part (b). Responses that did this usually
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This proved to be a very friendly starter. (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.
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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. (b) This was a “show that” question, so full and accurate working was required. The vast - = sin mg F ma majority wrote down an equation of motion parallel to the plane, 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 = mg sin F were setting , using an incorrect trigonometrical ratio, missing m’s or g’s, 5 using 𝜇= 12 .
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A very well answered question with many candidates using a clearly labelled force (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.
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(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 2 = F Mg with the reaction force at A, resulting in the incorrect equation Some 5 candidates lost method marks by resolving forces that did not need resolving (for and ). Whichever method Mg sin example, by writing an equation with both cos S 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 mu , 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 mu was found, most candidates gave the answer as a fraction or to an appropriate degree of accuracy.
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= F = 35. In the final part, F was very well known, and this was a follow through R mark on their answers to (a) and (b). However, some left their answer as a fraction or
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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
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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 = mu R and tried to use it in some way. Part (d) involved finding the magnitude of the 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
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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. mu R 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 = mu R 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.
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Almost a quarter of the candidates scored zero whilst 21% scored full marks on this
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Almost a quarter of the candidates scored zero whilst 21% scored full marks on this 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 =
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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.
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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
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Candidates need to make sure that they read and then answer questions carefully. In part 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 1 reaction force from the plane on A had been found and used with F to give F R = R mu = 6 . 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 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.
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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 , could then lead to full marks being R mu <= earned. Many candidates assumed that the beam was in limiting equilibrium i.e. that , produced an equation and then changed it to the required inequality. These F R mu = 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
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same. Candidates tended to be more successful in part (b), where a similar strategy was required, together with F , since the beam was now in limiting equilibrium. A R = mu 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.
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the plane to give a correct expression for the normal reaction. In part (b), most resolved parallel to the plane and then used F= mu R to derive the given answer for mu . 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
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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
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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.
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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 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= mu R 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.
![3. P =a S A eas Bin] 8 NG ; Figure 1 Two blocks, A and B, of masses 2m and 3m respectively, are attached to the ends of a light string. Initially A is held at rest on a fixed rough plane. The plane is inclined at angle a to the horizontal ground, where tana = D The string passes over a small smooth pulley, P, fixed at the top of the plane. The part of the string from A to P is parallel to a line of greatest slope of the plane. Block B hangs freely below P, as shown in Figure 1. 5 The coefficient of friction between A and the plane is 3 The blocks are released from rest with the string taut and A moves up the plane. The tension in the string immediately after the blocks are released is 7. The blocks are modelled as particles and the string is modelled as being inextensible. 12mg (a) Show that 7 = a (8)](/img/JUN19_9MA032/JUN19_9MA032_03(a).webp)
In part (a), many candidates earned the first mark for correctly finding R, but there were a number who resolved incorrectly, using sin instead of cos . Almost all candidates obtained the second mark, 2 F R very often from simply stating somewhere on their paper. Most candidates who found the 3 equations of motion correctly did proceed to find T. The most successful methods were either adding the two initial equations to eliminate T, finding a and substituting back into one of the equations or setting both equations to 6ma and finding T directly. Candidates who rearranged their equations to give a and then equated the quotients more frequently made errors when solving for T. Common errors were inconsistent use or loss of m, including an extra g in the acceleration term or forgetting to include g in their weight and, on the RHS of the equation, putting 3a and 2a rather than 3ma and 2ma. A significant number of candidates set their equation of motion to zero, finding the tension which would keep A at rest on the slope and received little credit. It was surprising how many candidates simply wrote 12 mg T at the end of their incorrect working, assuming it would not be checked. In the second part, who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.
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12 mg T at the end of their incorrect working, assuming it would not be checked. In the second part, 5 many candidates tried to describe what would happen without any supporting calculations. Many talked about the continued motion of A, ignoring the actual question. Others simply said that it would continue moving as B had more mass than A, ignoring the effects of the plane. Many who showed calculations did successfully compare the component of the weight acting down the slope with the maximum friction, although others continued to include tension or actually compared friction with tension, ignoring the weight altogether. A few candidates showed a deeper understanding of the problem and simply Some candidates did not notice that the terms required were available from the compared and tan . equation of motion for A. Some candidates did not read the question carefully and considered what happened immediately after B reached the ground, rather than when block A came to rest. Some used an equation of motion rather than considering forces separately. A minority showed their lack of understanding by using tension in their explanation. The final part was generally well done, although who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.
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understanding by using tension in their explanation. The final part was generally well done, although candidates still persist in giving more answers than are required, which very often means that they include a wrong one and so lose marks. A few just restated the model – “light string” etc. Candidates who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.
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Part (a) was well done by the majority of students. Virtually all attempted to resolve vertically and then use ‘F = ma’ horizontally to find the acceleration. A few omitted the vertical component of the tension when finding the reaction which simplified the problem and resulted in the loss of 3 of the available 6 marks. ‘F = 0.14R’ was almost invariably used to deduce the value of the acceleration. Full marks were often achieved for this part of the question. In the second part, many clear and concise explanations were seen for why the acceleration would be
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In the second part, many clear and concise explanations were seen for why the acceleration would be less if the brush were pushed rather than pulled. To achieve both marks it was necessary to comment on the pushing down increasing the normal reaction, which in turn causes an increase in the available friction and a consequent decrease in acceleration. Those who failed to refer explicitly to the reaction or, more commonly, failed to draw the final conclusion about the acceleration achieved one of the two available marks. Occasionally, there was some confusion between the ‘reaction’ force and the ‘resultant’ force. Further calculations were not required here although some included them in their explanations.
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