M8.4 Newton’s third law, equilibrium, pulleys, connected particles

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), there were many correct answers seen for the equation of motion of Q. However, a significant number of candidates did not seem to know what constituted an equation of motion, sometimes making an attempt at a suvat equation, despite then using it correctly in part (b).

Common mistakes

• Q: 3g − T = 3a (two terms on the left, weight in g)
• a suvat equation is not an equation of motion
• including m (3mg − T = 3ma) loses the A mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The most common error in part (b) was to include a weight term in the horizontal equation of motion for P i.e. writing T-2g = 2a rather than T = 2a. Sometimes the direction of the acceleration was not treated consistently in the two equations resulting in a sign error in one of them. A relatively small number of candidates used a whole system approach but the requirement to use a combined mass in the ‘ma’ term ruled out some attempts.

Common mistakes

• P on the horizontal table: T = 2a (no weight term: T − 2g = 2a was the most common error)
• or whole system 3g = 5a with combined mass 5
• a = 3g/5 = 5.9 m s⁻² (2 or 3 s.f.)
• inconsistent directions in the two equations

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (d) required finding the time taken by P to move across the table to the pulley once Q had hit the ground. This proved to be the most challenging part on the paper and few valid approaches were seen. A significant number used the acceleration from the first stage of the motion, or even g, in a suvat equation without realising that P would now travel with constant velocity since the table was modelled as smooth and there was no longer a tension acting. Other common errors seen were to use the distance as 1.2 m, 0.4 m or even 1.6 m rather than 0.8 m. This again showed a lack of appreciation of the mechanics of the situation.

Common mistakes

• after Q hits the floor the rope is slack and the table smooth, so P moves at constant speed 2.17 m/s for 1.2 − 0.4 = 0.8 m: K = 0.8/2.17 = 0.37
• common errors: using the earlier acceleration or g, distance 1.2, 0.4 or 1.6 m
• hardest part on the paper

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (e), almost all candidates stated, with a reason, that the acceleration would be less as a result of including a resistance. Those who lost the mark tended to not give a reason or omit this part.

Common mistakes

• acceleration is less, with a reason (resistance opposes the motion, reducing the resultant force)
• no reason scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Again, part (f) was generally answered well. The assumptions of the model were clearly stated as bullet points in the question. To gain credit, any suggested refinement had to relate specifically to one of those, and common valid responses included considering the mass of the ropes or modelling the particles with dimensions. Using a more accurate value for g was not credited as it was not a given modelling assumption. A further refinement was required so including resistance/friction or air resistance also did not gain the mark.

Common mistakes

• refine one of the listed assumptions: packages not particles (dimensions), mass/weight or friction of the pulley, mass or elasticity of the rope
• a more accurate g, "make it 3D", or air resistance/table friction are not credited

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was generally well done with some making good use of the diagram. After identifying the three forces, successful responses resolved vertically to form an equation with the vertical component of 5 N clearly found. This was a ‘Show that…’ question and so those who just wrote ‘3’ down, with no supporting evidence of how it had been obtained, received no reward. Very occasionally cosine was used instead of sine and a few learners used the value of g as 9.81, which meant that the accuracy mark was not awarded.

Common mistakes

• resolve vertically: 5 sin α + R = 2g
• the vertical component 5 × 3/5 = 3 must be shown, just writing 3 scores nothing
• sin/cos confusion
• g = 9.8 not 9.81

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (c) was well answered and the correct result of 2g or 19.6 (N) was stated for the magnitude of the reaction once the force of 5 N had been removed. Common incorrect answers were 2 or 16.6.

Common mistakes

• R = 2g = 19.6 N once the 5 N force is removed
• common wrong answers 2 and 16.6

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a) most wrote down the correct answer of 3Mg for the reaction force using vertical resolution. The most common error was omitting the weight term for the rod and a few included trigonometric functions in their resolutions.

Common mistakes

• resolve vertically: R = Mg + 2Mg = 3Mg
• omitting the weight of the rod
• no trig needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b) the most popular and simplest approach was to take moments about A to derive the given expression for the normal reaction at B. The angle was given between the vertical and the rod, rather than the possibly more familiar case of between the horizontal plane and the rod. Although this was not a problem for the vast majority, a significant number used (90o - alpha ) for their angle which sometimes led to confusion, particularly for those who did not recognise the equivalence with sin alpha /cos alpha and instead tried to use the addition formula to write this in terms of alpha only. In addition to sin/cos confusion, errors seen included total omission or inconsistent inclusion of a, slips in copying from one line to the next and occasional omission of the weight of the rod. Since the requirement was to ‘show that tan S Mg ’ it was essential that working was complete and correct to achieve all the alpha = marks. A very small minority attempted to take moments about B but they often did not complete their solution by resolving forces in order to eliminate F and R. In the third part although many learners recognised that R (with no friction acting at B) was take moments about A, to follow from the work in part (b). Responses that did this usually scored some marks for their horizontal and/or vertical resolutions even if their moments equation was incorrect. Occasionally the reaction at the wall was considered to be perpendicular to the rod rather than to the wall which generally resulted in no marks being achieved. Errors in the moments equation included sin/cos confusion, inconsistent inclusion of a and incorrect perpendicular distances. Some omitted the weight of the rod leading to a missing term in both the moments equation and the vertical resolution. Fragmented respondes, simplifying terms before writing down a full moments equation, sometimes made slips in the process. Those who attempted to take moments about B often achieved the mark for their vertical resolution but often did not gain all the marks for their moments equation. Attempts to resolve parallel or perpendicular to the rod were rare and tended to have a term missing. Nevertheless, there were a fair number of entirely correct solutions seen with clear and concise working.

Common mistakes

• moments about A: S × 2a cos α = Mga sin α + 2Mg × 0.5a sin α
• angle is with the wall, not the ground (using 90° − α causes errors)
• sin/cos confusion
• inconsistent a
• at least one line of working before the given answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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).

Common mistakes

• friction at a rough wall acts upward, so R1 + F = 3Mg and R is greater
• "because of friction" alone or friction acting downward scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• include the 3M particle at B: N = 6Mg (not R from (a))
• moments about A give the wall reaction P = 2Mg = F
• limiting equilibrium so F = μN
• reaction at the wall is perpendicular to the wall, not the rod
• omitting the rod's weight
• incorrect perpendicular distances

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a challenging question for many candidates since forces were given as vectors, but the acceleration was given as a scalar. Although most realised that the two forces needed to be combined to produce a resultant, some subtracted the components rather than adding, thererby failing to achieve the first mark. This also required the i and j terms to be collected to enable an expression for the magnitude to be found. A common starting point was to equate the vector sum of the forces to a scalar ‘ma’ term often leading to no further valid progress. Those who attempted to work in scalars sometimes equated the magnitude of their resultant to sqrt 5 (acceleration) rather than 2 sqrt 5 (ma term) and managed to secure one of the two available method marks. Where a correct unsimplified equation in c was reached, processing errors such as not squaring each side or in expanding brackets sometimes led to incorrect answers. The few who found the acceleration vector and equated the magnitude to sqrt 5 or the square of the magnitude to 5 tended to do so successfully despite the slightly harder algebra.

Common mistakes

• resultant force (2 + c)i + 2j (add, not subtract, the components)
• F = ma with magnitudes: (2 + c)² + 2² = (2√5)² = 20, or acceleration vector with magnitude √5
• c = 2 or c = −6
• equating a vector to the scalar ma
• using √5 instead of 2√5 for ma

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), those candidates who attempted to produce an equation of motion for the whole system were mostly successful in finding the resistance on the car, possibly helped by the fact that it was a given answer. Some used equations for the car and trailer separately, often finding the tension from the trailer equation and using it in the car equation to find R.

Common mistakes

• whole system: 1740 − 400 − R = (800 + 600) × 0.6 gives R = 500
• or car and trailer separately (T = 860)
• all terms needed
• brackets around 600 + 800

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b) required the tension to be found. Either the car or the trailer equation could be used, and some did both just to confirm their answer. However, there was confusion evident in many attempts with either extra or incorrect terms included or the wrong mass in the ma term. Some even included weight terms which were not relevant since the motion was horizontal. Few scored the mark in part (c). Although often a correct equation was used to find a = 5/6 or

Common mistakes

• car: 1740 − T − 400 = 800 × 0.6, or trailer: T − 500 = 600 × 0.6
• T = 860
• wrong mass in the ma term, extra or missing forces, weight terms included

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• resolve perpendicular to the plane R = mg cos α = 12mg/13 with cos α = 12/13 from tan α = 5/12
• using tan or dividing by sin or cos scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(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.

Common mistakes

• resolve horizontally F = S sin θ and vertically R = Mg − S cos θ, then F = μR: μ = 8/19 ≈ 0.42
• confusing S with the reaction at A
• sin/cos confusion
• resolving forces that need no resolving
• μ may be greater than 1 in general

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many candidates made no attempt at part (a). Those who did mostly realised that they needed to use F=ma although many were unable to use the correct combination of forces in their car/trailer/whole system equations. Some never substituted T = 2400 and others omitted the R terms or had extra R’s in their equations of motion. Sign errors in solving their two simultaneous equations were common, especially when solving by substitution. Most who answered the second part were successful although some tied themselves up in knots explaining

Common mistakes

• car: 7400 − 2R − 2400 = 1200a
• trailer: 2400 − R = 400a (or whole system 7400 − 3R = 1600a)
• a = 0.5
• missing or extra R terms, not using T = 2400, sign errors solving simultaneously

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

answered the second part were successful although some tied themselves up in knots explaining their answer even though no reason was asked for. There was a surprisingly large number of candidates who made no attempt at this part. The most common error in the final part was to

Common mistakes

• a₁ is less than a (more mass to accelerate)
• no reason needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• friction acts to the right (towards the wall) to balance the reaction at B / stop the rod slipping
• wording implying the rod is moving scores B0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• resolve vertically R = Mg and horizontally F = S
• F = μR in limiting equilibrium
• μ = 2/3
• extra forces in the vertical resolution
• moments about B or resolving along the rod often lose a term

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although fully correct answers were seen, it was clear that many candidates had little experience of answering this type of question. In particular, there were many equations seen helped some candidates, but these were not often forthcoming. In part (a), many candidates gained all 3 marks. The most common error occurred where the direction of travel had been mixed up. For those who set up an equation of motion for the cage and block together, the most common mistake was ' 50g - T = 50 x 0.2' (as if it was accelerating downwards). Some students attempted 'F = ma' for the cage, but omitted 'R', scoring zero marks as there were an incorrect number of terms. There were instances of numerical mistakes in calculating '50 x 0.2' ('25' being a common one). Most students used the correct value of 9.8 for g. There were a few cases where g was omitted from the weight or added incorrectly in the 'ma' term. In part (b), few students who had made a good attempt in part (a). For students who knew what to do here, there were a few instances of '10g - R = 10 x 0.2'. Some students attempted 'F = ma' for the cage rather than the block, but omitted 'T' or they tried to combine the cage and the block. For those students scoring zero marks for Q4, their responses were either completely blank, or they tried to equate forces, ignoring the fact that the cage and block were accelerating.

Common mistakes

• whole system: T − 50g = 50 × 0.2 gives T = 500 N (g = 9.8)
• direction of motion mixed up (50g − T = 50 × 0.2)
• g omitted from weight or put in the ma term

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although fully correct answers were seen, it was clear that many candidates had little experience of answering this type of question. In particular, there were many equations seen with an incorrect number of terms, particularly in part (b). Clear force diagrams would have helped some candidates, but these were not often forthcoming. In part (a), many candidates where g was omitted from the weight or added incorrectly in the 'ma' term. In part (b), few candidates gained the full 3 marks. Some failed to identify the forces which were acting on the block; others considered the forces acting on the cage rather than the block. A common error was to get the direction of motion mixed up. There were quite a few blanks here, even by students who had made a good attempt in part (a). For students who knew what to do here,

Common mistakes

• block alone: R − 10g = 10 × 0.2 gives 100 N (or cage: T − 40g − R = 40 × 0.2)
• wrong number of terms, e.g. omitting R or including T for the block
• draw a force diagram

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• resolving vertically is quickest (most found X = 49 first)
• sin α = 3/5, cos α = 4/5
• friction is not limiting here so μR is wrong
• F = 9.8 N must be positive
• sin/cos interchanged gives 9.8 in the wrong direction

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Almost a quarter of the candidates scored zero whilst 21% scored full marks on this

Common mistakes

• down the plane (a negative F from assuming up the plane means down)
• must be stated as the answer, not only shown on the diagram

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• horizontal component of T acts to the left and friction is the only other horizontal force
• "friction opposes motion" scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• resolve vertically R + T cos θ = 3Mg with T = 2Mg cos θ and cos θ = 3/5
• other equations need F eliminated and often miss a term

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• resolve horizontally F = T sin θ (sin θ = 4/5)
• F = μR with R = 57Mg/25 must be used, not just quoted
• show substitution as the answer is given

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(a) There were many completely correct answers and those who took the acceleration to be positive in a different direction could also score full marks. Some had no idea where g should go so answers like ‘T – 2m = 2mga’ were sometimes seen. The mass m was occasionally left out also. A few quoted suvat equations, connecting time and distance, instead of equations of motion for P and Q but if they went on to use correct equations of motion in part (b) they scored the marks for part (a). Some had no idea at all and T never appeared.

Common mistakes

• T − 2mg = 2ma for P and 5mg − T = 5ma for Q
• g placed wrongly (e.g. T − 2m = 2mga)
• m omitted
• suvat equations are not equations of motion

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

motion for P and Q but if they went on to use correct equations of motion in part (b) they scored the marks for part (a). Some had no idea at all and T never appeared. (b)Those who had correct equations of motion generally went on to find the acceleration but some missed g in their answer. Not many progressed successfully after this - some found T because that is what is usually required but in this question, it was not necessary. Some calculated the time taken for Q to reach the ground and used it in an equation for P, not realising that they had to find the velocity that P had acquired. Of those who did make progress, some missed the h in their answer for v, and others only added 2h instead of 3h in the final step.

Common mistakes

• a = 3g/7
• P's speed when Q lands from v² = 2ah (P has risen h)
• then P moves freely under gravity: H = 3h/7
• total height 2h + h + H = 24h/7
• T is not needed
• adding only 2h, or using the time Q takes to fall, are common errors

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(c) A very small number answered correctly. Mass was again a common response.

Common mistakes

• the distance Q falls to the ground would not be exactly h (size of the balls)
• "mass" is a common wrong answer
• incorrect extras score B0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(d) Again, there was a very small number of correct responses, some of which were spoilt by saying that the reason for the accelerations being different was that the tensions were different because of friction at the pulley.

Common mistakes

• the accelerations of P and Q would not have equal magnitude
• saying the tensions differ because of friction at the pulley is wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• equation for A only (not the whole system): 3mg sin α − F − T = 3ma
• friction and tension both act up the slope (same sign)
• missing a force or wrong direction

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• R = 3mg cos α (not 3mg)
• F = R/6
• sin α = 3/5, cos α = 4/5 (sin/cos confusion)
• T − mg = ma for B or whole-system equation
• tension is not mg
• full working needed for the given answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates need to make sure that they read and then answer questions carefully. In part 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

Common mistakes

• tensions on A and B would be different in the two equations
• say how the working changes, not the answer
• incorrect extras score B0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• one moments equation (e.g. about A: mga cos θ = 2Sa sin θ) plus resolving R = mg, F = S
• use F ≤ μR to get the inequality (assuming limiting F = μR and then switching to ≥ loses the final mark)
• omitting g
• sin/cos confusion
• reactions at A and B are not equal

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• limiting so F = μR = ½mg
• friction now acts away from the wall (friction towards the wall gives k = −0.1, a negative k should flag the error)
• moments equation plus horizontal resolution N = kmg − F
• k = 0.9

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), the equations of motion were mostly correct when attempted. A significant number of candidates obviously did not understand the model or that T was required and stopped at that point. There were a few with missing m’s but this was quite rare as were sign errors. A very small number of candidates obtained the equations but then completely misunderstood the situation and replaced a by g and lost all the marks. Only a handful realised that 2T was required to find the force exerted on the pulley by the string. The second part was

Common mistakes

• 4mg − T = 4ma and T − 3mg = 3ma give T = 24mg/7
• force on the pulley is 2T = 48mg/7 (few realised)
• omitting m loses the M mark
• do not replace a by g

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

to find the force exerted on the pulley by the string. The second part was reasonably well answered but, as in question 1(f), it should be remembered that, when only one limitation of the model is required, any extra answers given would all have to be correct for the mark to be awarded.

Common mistakes

• string has weight or is extensible, or pulley not smooth
• any incorrect extra loses the mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• resolve perpendicular to the plane: R = mg cos α = 4mg/5 with cos α = 4/5 from tan α = 3/4
• mg cos 36.86° (decimal angle) is A0
• sin/cos confusion

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• resolve parallel to the plane F = mg sin α (or horizontally R sin α = F cos α)
• F = μR must be used, not just stated, to reach the given 3/4
• limiting friction acts up the plane (on the point of sliding down)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• the mass cancels: friction increases in the same proportion as the weight component down the plane
• saying only that R and F increase (ignoring the weight component) scores nothing
• no calculation needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• Q slides down with constant speed because there is no resultant force (Newton's first law)
• most wrongly said it accelerates (gravity) or decelerates (friction)
• must say constant

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• resolve horizontally F = (9Mg/25) sin α and vertically R + (9Mg/25) cos α = Mg
• the reaction at C is perpendicular to the ladder, not horizontal
• F = μR only with the normal reaction at A (not 9Mg/25)
• sin/cos confusion
• μ = 18/49 ≈ 0.37

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although the first part of this question was well done by some candidates, a significant minority failed to write down an equation of motion of the correct structure for each mass. Some included a weight term in the equation for the particle moving horizontally whilst others showed no real understanding of what was required and confused equations of motion with suvat equations or omitted this part completely. Those who managed to produce two valid equations generally had no difficulty in deriving the given answer for the acceleration. Since it was a ‘show that’ question, it was important that the final answer followed from sufficient and entirely correct working; those who went straight from the two simultaneous equations to writing down the answer were not awarded the final mark. Such instances were, however, fairly rare. Some attempted a ‘whole system’ approach which was equivalent to writing down just one equation with the tension eliminated. This was deemed not sufficient for a complete method and scored a maximum of two marks.

Common mistakes

• separate equations of motion: Q: 0.6g − T = 0.6a, P: T = 0.8a (no weight term for P on the horizontal table)
• solve to a = 4.2 showing working
• a whole-system equation alone is incomplete

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The final part required the identification of a possible limitation of the model. The assumptions of this model were listed in the question and so any one of these (such as smooth pulley, light string, inextensible string, modelling the balls as particles) could be used in describing such a limitation. Often other factors were identified which were not part of the specified assumptions for this model and so were not awarded the mark.

Common mistakes

• choose an assumption stated in the question: rope light or inextensible, pulley smooth or small, balls as particles
• factors outside the stated model are not credited

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• R = 2mg cos α (not sin α)
• F = ⅔R
• separate equations for A and B (or whole system) with 2ma and 3ma, not 2a, 3a or = 0
• exact trig ratios from tan α = 5/12 (no decimal angle)
• do not just write the given answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• e.g. extensible or heavy string, rough pulley, blocks not particles, air resistance
• restating the model (light string) scores nothing
• extra wrong answers are penalised

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

second part involved finding the components of the resultant force on the ramp at the point of contact with the ground and hence its magnitude. This proved challenging for many and some made little valid progress. A clearly labelled diagram would have helped to identify forces and distances. Most used a normal reaction and a horizontal frictional force at A; however, the latter was sometimes omitted which significantly reduced the number of marks available for subsequent equations. Some employed components parallel and perpendicular to the ramp; such attempts were rare but generally successful. A variety of resolution and moments equations were attempted but sometimes components were omitted, forces not resolved or perpendicular distances not used. Working was often difficult to decipher. Those who attempted vertical and horizontal resolutions and a ‘moments about A’ equation tended to be the more successful although a significant number stopped when they had found the normal reaction only. Some introduced a coefficient of friction which was not necessary to answer the question (even giving it the value 2 from the previous question). Occasionally those who completed a correct solution by 3 squaring and adding components to find the magnitude of the resultant failed to achieve the final accuracy mark for not rounding to 2 or 3 significant figures following the use of g = 9.8 m s-2. Some who had achieved very few marks for the rest of the question managed to secure the mark in part (c)

Common mistakes

• friction at A is horizontal and must be included
• at least one moments equation (e.g. M(A): 20g × 4cos θ = 5N) with perpendicular distances
• three equations then magnitude √(F² + R²) ≈ 67 N (2 or 3 s.f. after g = 9.8)
• μ is not needed
• stopping at R

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), most students had some idea that it was necessary to write down an equation of motion for P in an attempt to find the value of the tension in this pulley question. Some wrote down equations for both particles at this stage. Sign errors and missing ‘m’ from terms were fairly common. Those who reached the correct expression ‘ T = 2mg – 10mg/7 ‘ often failed to simplify further or made errors when attempting to do so. An answer in terms of m and g was required and so use of g = 9.8 m s-2 lost the final mark. The only acceptable answer for part (b) was that the accelerations are the same because the string is not just in part (a). Attempts to eliminate T and solve for k often revealed some poor algebraic processing. There were several possible acceptable answers to part (d). The assumptions of the model were listed

Common mistakes

• equation of motion for P: 2mg − T = 2m × 5g/7, so T = 4mg/7 in terms of m and g (substituting g = 9.8 loses the A mark)
• sign errors and missing m are common

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The only acceptable answer for part (b) was that the accelerations are the same because the string is inextensible. Extra irrelevant or incorrect comments relating to, for example, light string, smooth pulley, constant tension etc. were penalised. In part (c) there were some reasonable attempts at an

Common mistakes

• because the string is inextensible
• any extra or irrelevant reason (light string, smooth pulley, constant tension) scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

pulley, constant tension etc. were penalised. In part (c) there were some reasonable attempts at an equation of motion for Q and, to gain credit, it had to be seen or used in this part of the question and not just in part (a). Attempts to eliminate T and solve for k often revealed some poor algebraic

Common mistakes

• equation for Q: T − kmg = km × 5g/7 (or whole system 2mg − kmg = (2m + km) × 5g/7)
• substitute T = 4mg/7 so m cancels: k = 1/3
• the equation must be seen in this part
• poor algebra

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

There were several possible acceptable answers to part (d). The assumptions of the model were listed in the question and any one of these (such as smooth pulley, light string, modelling the balls as particles) could therefore be used to describe a limitation of the model. Reference to the accuracy of g = 9.8 m s-2 (a numerical value was not required for the question) was not credited.

Common mistakes

• any listed assumption: pulley not smooth or light, string not light or not inextensible, balls not particles, air resistance
• a more accurate value of g scores 0
• an extra wrong answer scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

sufficient and accurate but generally this was well done. In part (b), most used a horizontal resolution and the given expression for T to calculate a value of x. Errors in the straightforward algebraic manipulation were surprisingly common, sometimes leading to a dimensionally incorrect value. The most popular approach to part (c) was to resolve vertically to find the vertical component of the

Common mistakes

• resolve horizontally T cos α = 2Mg using the given T (or moments about C)
• x = 2a/3
• algebra slips often gave a dimensionally wrong answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

most popular approach to part (c) was to resolve vertically to find the vertical component of the reaction. Occasionally a weight was omitted or the tension was not resolved, but there were a fair number of valid attempts. Again, slips in simplifying the expression were fairly prevalent and sometimes an incorrect value of x was being carried forward. Those who reached an answer which was a multiple of Mg generally gained the final method mark for dividing the components of reaction to find tan beta . Although there were some entirely correct solutions seen, there was evidence of confusion and lack of a systematic approach in other attempts.

Common mistakes

• resolve vertically (or moments about B) to find Y = 5Mg/2
• tan β = Y/X with X = 2Mg gives 5/4
• errors: omitting a weight, not resolving T