M8.2 Newton’s second law

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

Examiner's comment

Part (a) was mostly correctly answered, but some candidates were confused as to whether to add or subtract the forces, with 9i + 14j being the most common incorrect answer. This was usually carried through to part (b) but then often replaced in part (c) with the correct answer of 3i – 4j. Many candidiates worked with column vectors but most gave the answer in the required format. Very occasionally the answer was seen in (xi, yj) format or the column vector form of this.

Common mistakes

• add the force vectors: (−3i + 4j) N
• 9i + 14j (subtracting) was the most common error
• answer must be in i, j form

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

Examiner's comment

In part (b), most candidates used Pythagoras correctly but failed to use Newton’s 2nd Law. A significant number, of those that did, multiplied the resultant force by 0.25 instead of dividing it by 0.25.

Common mistakes

• Newton's second law F = ma with m = 0.25 then Pythagoras: magnitude of a = 4 × 5 = 20 m s⁻²
• most used Pythagoras but forgot F = ma, or multiplied by 0.25 instead of dividing
• 20 then "√20" loses the A mark

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

Examiner's comment

Part (c) not always correct even if the rest of the question was and often left blank. There was the occasional correct follow through from part (a).

Common mistakes

• constant velocity means the resultant force is zero (Newton's first law), so F3 = −(their (a)) = (3i − 4j) N
• often left blank

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

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

In part (b), most responses scored both marks by correctly finding the value of the driving force of the engine. Many started by quoting F = ma and working with a complete equation. These responses were generally successful, although some made an arithmetic slip when rearranging their formula to make D the subject. Others did not start by formally stating the equation but approached it in a fragmented manner. These responses were far more likely to demonstrate the common error of subtracting the 400 N force from their ‘ma’ term instead of adding it, leading to an incorrect answer of 1200 N. A few responses included g with the mass of 800 in their calculations possibly due to confusion of mass and weight.

Common mistakes

• form the full equation D − 400 = 800 × 2 first
• common error 1200 (subtracting 400 instead of adding)
• mass not weight, no g in the ma term

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

Examiner's comment

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.

Common mistakes

• F = 0.2 × 19.6 (use the new reaction, not 16.6)
• −F = 2a gives a = −0.2g
• wrong a = μ or 0.4g
• then v² = u² + 2as
• answer 4.1 or 4.08 m to 2 or 3 s.f., not 40/g

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

Examiner's comment

In part (a) most responses scored the first mark by using F = ma with the given mass and acceleration vector. A few responses omitted the mass. The question stipulated that the answer should be given in terms of i’s and j’s and most learners did so correctly.

Common mistakes

• F = ma = 0.5(2i − 2.4j) = i − 1.2j
• omitting the mass
• final answer must be simplified in i and j

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

Few scored the mark in part (c). Although often a correct equation was used to find a = 5/6 or a = -5/6, it was not always made explicitly clear that 5/6 was the deceleration. Those who had achieved few marks in the previous parts of the question were sometimes able to use a correct suvat method in part (d) to find the distance travelled by the trailer once the

Common mistakes

• after the break only resistance acts on the trailer: 500 = 600a so a = 5/6
• must state explicitly that the deceleration is 5/6 (a = −5/6 alone is not enough)

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

Examiner's comment

to use a correct suvat method in part (d) to find the distance travelled by the trailer once the tow bar had broken. Occasionally the acceleration from before the break was used but generally this was well done with the correct numerical value obtained. The question required the value of ‘d’ so some indication was expected that the calculated distance was actually d.

Common mistakes

• v² = u² + 2as: 0 = 12.5² − 2(5/6)d gives d = 93.75 (94)
• using the acceleration 0.6 from before the break
• state that the distance is d

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

Examiner's comment

In part (e), it was fairly rare to see two acceptable correct reasons for why the distance was likely to be different from that calculated. Many scored one mark for ‘resistance unlikely to be constant’ or ‘deceleration not constant’; however, these are not independent factors, so they were not credited as different reasons. Other common correct responses included that the mass of the broken tow bar was not considered, and the trailer would be unbalanced. The most common incorrect responses involved claiming that the model did not consider friction or air resistance. Also, some explanations included the car whereas this part was specifically about the trailer. It should be remembered that two reasons were asked for and if a candidate listed more than two then incorrect extra answers were penalised.

Common mistakes

• two different reasons: resistance not constant (or more air resistance out of the car's slipstream), mass of the towbar ignored, trailer may not travel in a straight line, uneven road, trailer unstable/tips
• "resistance not constant" and "deceleration not constant" count as one reason
• answers about the car or "model ignores friction/air resistance" are wrong
• extra wrong answers are penalised

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. (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 .

Common mistakes

• mg sin α − F = ma with F = μ × (12mg/13)
• setting mg sin α = F (no acceleration)
• using μ = 5/12
• missing m or g
• both trig ratios substituted as fractions before the given answer

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

candidates who made no attempt at this part. The most common error in the final part was to make no reference to the modelling of the resistance. Comments about friction and air resistance abounded, along with ones about the rope not being light/inextensible, tyres, road surfaces (which at least had an inkling of the right idea) and the weather. Relatively few scored this final mark.

Common mistakes

• the resistance is unlikely to be constant (e.g. varies with speed)
• must refer to the resistance model
• friction/air resistance, rope, tyres or weather comments score 0

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

Examiner's comment

were very many correct answers for the second part also, but some ignored the 28N force = completely and gave 5 1.4 , as their answer or made a sign error and ended up with 7 = 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 didn’t give it to 2 sf and lost the mark anyway.

Common mistakes

• 28 − F = 5 × 1.4 gives F = 21
• ignoring the 28 N force (F = 7) or a sign error (F = 35)

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 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 =

Common mistakes

• X removed so R must be re-resolved: R = 5g cos α, not 68.6
• do not reuse F = 9.8 from (a)
• F = 0.5R
• 5g sin α − F = 5a gives a = 1.96 m s⁻²

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

Examiner's comment

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 mu when substituting λ and there was occasional use of only one of the forces.

Common mistakes

• μ = 3 so resultant 6i + 2j and a = (6i + 2j)/4
• u = 0 as P starts at rest (u = 3i + j is wrong)
• using the force as a displacement
• using only one force
• AB = √160 = 4√10 ≈ 12.6 m

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

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 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

Common mistakes

• straight line from the origin
• no vertical line at the end (loses the mark)
• the explanation (acceleration is constant) was often omitted

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

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

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

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

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.

Common mistakes

• compare Fmax = 16mg/13 with weight component down the slope 10mg/13 (or μ with tan α) numerically
• no tension once B has landed
• consider A at rest, not just after B lands

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

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

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

Common mistakes

• resolve vertically R + 40 sin α = 20g (vertical component of tension often omitted, losing 3 marks)
• 40 cos α − F = 20a with F = 0.14R
• a = 0.40 m s⁻²

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

Examiner's comment

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.

Common mistakes

• pushing increases R so the available friction increases, so acceleration decreases
• must mention R and conclude about a
• confusion of reaction with resultant