Vectors Pure

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

Examiner's comment

In part (a), a good proportion of candidates arrived at the correct value for OB as 40 or 2 10 , either in part (a) or (b). Some candidates mistakenly found OB as OA AB - rather than OA AB + whilst others found OB but not its magnitude, although they were allowed recovery if they subsequently found the magnitude in part (b). A few just found AB in this part and others slipped up on one of the components of OB .

Common mistakes

• OB = OA + AB = −2i − 6j (not OA − AB)
• exact magnitude √40 = 2√10
• stopping at the vector OB or finding the magnitude of AB

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

Examiner's comment

In part (b), many drew a diagram to help and there were many correct answers. Those who did not draw a diagram found this more challenging. The vast majority achieved at least one mark for finding the length of OA or OB . Most understood the need to use the cosine rule, and this was often the rearranged form with cos(OAB) as the subject. A surprising number of candidates found the wrong angle of the triangle by substituting their vector lengths into the wrong places in the cosine rule formula, especially those without a diagram. Others made slips in the rearranging. Some candidates incorrectly assumed that the triangle was right-angled and attempted to find angle OAB using trigonometry in a right-angled triangle. For those with a good diagram, the 1 tan- approach worked well, and many successful strategies were seen. A few students used the scalar product approach, but again those without a diagram often used incorrect vectors and found the wrong angle.

Common mistakes

• OA = √73, AB = √29, OB = √40
• cosine rule with √40 opposite angle A gives 47.6°
• lengths in the wrong places find the wrong angle
• do not assume a right angle (sin⁻¹(√40/√73) = 47.75° is wrong)
• draw a diagram

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

Examiner's comment

This question tested learners’ knowledge of vectors and proof of the properties of a triangle. In part (a), many gained both marks by successfully adding the vectors, correctly identifying and combining the i, j, and k components. Common errors included misreading the vector QR as 6 6 + i j, or subtracting the two given vectors. It should be noted that subtracting vectors was seen in a significant minority of responses. This was likely due to confusion arising from a misinterpretation of the vectors as position vectors, where PR OR OP = - . Learners should be encouraged to sketch a diagram as a guide.

Common mistakes

• PR = PQ + QR = 8i + 8j + 4k (add, do not subtract: they are not position vectors)
• misreading QR as 6i + 6j

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

Examiner's comment

This question tested learners’ knowledge of vectors and proof of the properties of a triangle. It should be noted that in part (b) that, despite it being possible to gain full credit by finding the length of vectors and Pythagoras’ Theorem in a few lines, many learners chose to use more involved methods. This often led to them not gaining full marks due to errors in their working. Part (b) was well answered by most learners. Many responses lacked clarity when attempting to prove that the triangle was both isosceles and right-angled. In demonstrating that the triangle is isosceles, some calculated the magnitudes of the sides correctly but did not gain the accuracy mark as they just stated it was isosceles without reference two sides being equal. It seemed that learners found it harder to prove the triangle was right-angled. Most successful responses used Pythagoras’ Theorem to show that the triangle was right-angled and most of these responses gained full marks. When learners used the scalar dot product method, many of them did not show the calculation and therefore did not gain credit, whilst others were not very successful in showing the triangle was right-angled, with some learners incorrectly using it to show a perpendicular gradient of 1 - instead of 0. The least successful responses tended to involve SOHCAHTOA, mainly because this approach was more involved and thus required a fuller explanation, which was not always in evidence. As this was a ‘show’ question, a conclusion (either overall or separate for each property) and an assumption that the triangle was-right angled was required to score full marks. There were lots of incorrect attempts at the cosine rule, which learners either could not remember correctly, or did not rearrange correctly. A significant minority of responses attempted to show the right-angled property by splitting the triangle in half to create two right angled triangles and then used trigonometry. 15

Common mistakes

• magnitudes PQ = QR = √72 = 6√2 (state that two sides are equal) and PR = 12 with 72 + 72 = 144 (Pythagoras) or scalar product = 0 shown
• conclusion for both properties needed
• longer cosine-rule or trigonometry methods led to errors

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

Examiner's comment

This question about vectors proved to be a good differentiator between learners. Many were only able to make progress in part (a), whilst learners who had a good knowledge of the techniques required progressed through part (b) as well. Most learners made some progress, but also found this question to be challenging. Throughout, responses saw varied methods, and many made effective use of the given diagram to help answer the question. In general, those that wrote the route down using good vector notation fared better than those who did not in both parts of the question. In part (a), most successful responses attempted to use AD or DA , often finding an equation such as ( ) 4 15 5 2 3 k λ - + = - + + a b a b a b or ( ) 6 3 AD k = + - a b gaining the first mark. Some responses worked with the components of a and b from the start, rather than writing down expressions with vectors, but often produced correct work. A common mistake, however, was to incorrectly use AD AB DB = + and responses doing this made no progress in part (a). Those who had made a successful start were then often able to equate the a and b components of their expression to produce two simultaneous equations and used the a component to establish a value of their parameter λ, and the b component to find k. Others simply compared ( ) 6 3 k + - a b and 15a – 5b by inspection and were able to find the correct scale factor between them using it to find k with sufficient working. Regardless of the method used, some learners with incorrect work often wrote down the given answer, k = 5, without realising that their work did not lead to this value.

Common mistakes

• AD = AB − DB = 6a + (3 − k)b (not AB + DB)
• AD parallel to BC: compare 6a + (3 − k)b with 15a − 5b (scale factor 2.5)
• writing k = 5 from wrong work

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

Examiner's comment

This question about vectors proved to be a good differentiator between learners. Many were only able to make progress in part (a), whilst learners who had a good knowledge of the techniques required progressed through part (b) as well. Most learners made some progress, but also found this question to be challenging. Throughout, responses saw varied methods, and many made effective use of the given diagram to help answer the question. In general, those that wrote the route down using good vector notation fared better than those who did not in both parts of the question. Learners found part (b) of the question significantly harder. Many responses earned the first mark for identifying that 1 , 5 BN BC = and this was often the only mark scored in this question. A common error, however, was to misinterpret the ratio BN : NC, and to incorrectly use 1 4 BN BC = . Most learners who made progress found two expressions for one of the vectors BX ± , AX ± , NX ± or DX ± , one using a multiple of AN and one using a multiple of DB . Where responses found a second vector with a different parameter, they were typically successful in solving simultaneous equations and identifying the correct ratio. Common errors included: only finding one expression or using the same vector ( AN or DB ) in both of their expressions or using the same parameter in both expressions. There were other strategies employed, but invariably they had the same issues as the strategy covered in the main scheme. A neat approach was to establish that triangles AXD and NXB were similar and since AD = 2BN it followed immediately that BX : XD = 1 : 2. There were some responses that attempted to use a similar strategy, but often the work was not detailed enough, simply stating that AD and BN were parallel without further justification that they understood the triangles were similar.

Common mistakes

• BN = ⅕BC = 3a − b (not ¼BC)
• two expressions for BX with different parameters (λ along BD, μ along AN) then compare coefficients
• BX : XD = 1 : 2
• similar triangles AXD and NXB (AD = 2BN) need justifying

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) 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

In part (b) most responses scored both marks by quoting and correctly using t = + v u a with their vector quantities. A few used their F for a. Arithmetical slips were occasionally seen. It was relatively rare to see the method of integration deployed. This was more mixed in terms of levels of success with some not considering the constant of integration. Others made arithmetical slips, the longer method providing more opportunity to make an error. Many responses demonstrated a preference for column vectors and used them in this part.

Common mistakes

• v = u + at with vectors
• use a, not F
• if integrating, include the constant from v at t = 0

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

Examiner's comment

Finding the bearing of the direction of motion in part (c) proved more challenging for many learners. A good number were able to find an acute angle relating to their velocity vector. Use of basic trigonometry or the properties of a right-angled isosceles triangle were seen in roughly equal measure, but gaining the final mark for a correct bearing was often not achieved. Unrepresentative diagrams and a misunderstanding of bearings appeared to be the cause of this. A significant number did not need to use trigonometry, realising that the components indicated a north west direction and writing down 315o directly. A few learners first worked out the displacement vector and attempted to find the bearing of the position of the particle. This was not what the question had asked for and gained no credit.

Common mistakes

• v = −3i + 3j points north-west so bearing 315°
• unrepresentative diagrams and misread bearings lose the final mark
• using the displacement instead of the velocity scores nothing

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

Examiner's comment

Part (d) was well answered with a suvat equation being the preferred and most successful method for the majority. A few attempted to use integration, and with the constant of integration being the zero vector, this was generally successful albeit a longer method. Almost all responses gave the answer in terms of i’s and j’s, as required by the question.

Common mistakes

• r = ut + ½at² with t = 5 and u = −7i + 7.8j (not u = 0)
• answer −10i + 9j in i and j

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

Examiner's comment

In part (b) Pythagoras’ Theorem was generally applied correctly to find the magnitude. An exact (surd) answer was required so those who only stated a decimal approximation did not sqrt 322 -162 thereby not gaining both marks. gain the accuracy mark. A few did not account properly for the negative sign and evaluated In the third part the majority of learners recognised the requirement to differentiate to find the

Common mistakes

• exact answer 16√5 needed (decimal loses A1)
• square the −16 correctly, not √(32² − 16²)

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

Examiner's comment

Overall, this question proved fairly tricky with lots of candidates losing marks throughout. In part (a), most candidates knew they needed to subtract the vectors, there were a few signs slips when doing this and quite a few candidates subtracted the wrong way round so found QP instead. There were a small but significant number who added the vectors.

Common mistakes

• PQ = q − p = −6i + 3j
• subtracting the wrong way round (QP) or adding the vectors

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

Examiner's comment

Overall, this question proved fairly tricky with lots of candidates losing marks throughout. In part (b), quite a few candidates seemed unclear as to how to show 90 PQR = ; some made no attempt whilst others used the given fact within their proof often finding two angles in a triangle and showing that they were complementary, but they had assumed a right angle in order to utilise trigonometry which achieved no marks. The candidates who tended to be most successful at this were the ones who attempted to show the required property using gradients. There were a small number of signs slips and incorrect attempts at gradients but overall, this approach was usually successful. Quite a lot attempted to use Pythagoras’ theorem but there were a significant number who would only find PQ and QR and either did not attempt to find PR or had a poor attempt at finding it. It is of note that for those who successfully used Pythagoras’ theorem to complete the proof, diagrams proved useful. There were a small number who attempted the dot product or cosine rule and usually these attempts were good enough to achieve the marks. It was disappointing that the most common reasons for the M1A0 score were due to incorrect/incomplete solutions or simply a lack of conclusion.

Common mistakes

• gradients −1/2 and 2 multiply to −1
• or PQ² + QR² = PR² with PR found
• must conclude angle PQR = 90°
• assuming the right angle to use trigonometry scores 0

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

Examiner's comment

Overall, this question proved fairly tricky with lots of candidates losing marks throughout. Part (c) proved quite tricky for a lot of candidates many of whom did not seem to have fully understood the shape with which they were working. Those who were successful usually had a good diagram identifying it as a trapezium and were able to find the area. There were a number of attempts at splitting the shape up into a rectangle and triangle or 2 triangles, but often there was a misunderstanding about which lengths were which and how they should be combined to find the area. The most common error in the rectangle and triangle approach was that the candidates failed to spot that the triangle was a right-angled triangle so could use QP as the ‘height.’ The two triangles approach would also often have the same issue. There were quite a few attempts that treated the shape either as just a rectangle or just a triangle. It was pleasing however to see that whilst a large number of candidates were unsure of how to proceed to find the area many did at least acknowledge the need to attempt the lengths of various vectors, which generally resulted in two marks being awarded.

Common mistakes

• PQRS is a trapezium (PS parallel to QR) with height PQ = 3√5
• QR = 9√5, PS = 27√5
• area = ½(9√5 + 27√5)(3√5) = 270
• a diagram helps
• treating the shape as a rectangle or parallelogram
• not spotting the right angle

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

Examiner's comment

This question required problem solving skills and focused on coordinate geometry and circles. Part (b) required candidates to find the equation of the circle. Most candidates found both values of k but ignored the information in the question that k was a positive constant. Those who did appreciate this and used 5 k = went on with good success to find the coordinates of P, giving them the centre of the circle and Q which allowed them to calculate the radius and complete the question. This part was more often successfully answered than the “show” part (a), which showed good accessibility, as well as good exam technique by many candidates to still attempt a later part, even if they were unable to successfully answer the earlier part.

Common mistakes

• k = 5 only (k positive, reject −2)
• centre (−1, 13)
• r² = PQ² = 4² + 2² = 20
• (x + 1)² + (y − 13)² = 20

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

Examiner's comment

This question proved to be a differentiator between candidates. Some were unable to make any meaningful progress while others demonstrated a very intuitive grasp of the topic and were able to produce answers with very little working. Most candidates were successful in part (a) although there were a small number of incorrect responses which had arithmetic errors, used addition of vectors rather than subtraction or had poor notation.

Common mistakes

• AB = b − a = 3i + 9j + 3k (subtract, not add)
• a vector, not coordinates (3, 9, 3)

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

Examiner's comment

This question proved to be a differentiator between candidates. Some were unable to make any meaningful progress while others demonstrated a very intuitive grasp of the topic and were able to produce answers with very little working. In part (b), many candidates did not appreciate the scaffolding in the question and did not use their answer to part (a) to help them with a strategy for part (b). Relatively few candidates used the most straightforward method to find both two possible positions of P, e.g., 2 OP OA AB = + and 2 3 OP OA AB = + . Some candidates were able to find one of the positions (usually 8i+15j+11k) but not the other, with a common misconception being that the second point was on the opposite side of A, attempting 2 OP OA AB = - . There was also some confusion with the ratio with some candidates using 1 3 OP OA AB = + . Many candidates failed to appreciate that there were two points satisfying 2 AP BP = . Relatively few candidates drew a diagram or used the given diagram on the question paper to give them some idea of where P could be. It was much more common for candidates to take an unnecessarily complicated algebraic approach. Many responses calculated |𝐴𝐵⃗⃗⃗⃗⃗ |(= 3 sqrt 11) and progressed no further. Many responses attempted to express 2 AP BP = in terms of variables representing components of an unknown vector, showing a good understanding of how to calculate the magnitude, but struggled to formulate suitable equations. Some candidates used a fully correct method of considering the point P as the general point on the line OP OA AB = + and then setting up the equation 2 AP BP = . These candidates tended to arrive at two final answers for P, having found two log( ) log( ) log( ) ab a b = values for λ, but many had made processing errors and could only gain the method marks as a result. Other candidates set up a general point P in the form xi + yj + zk, then used 2 AP BP = , extracted and solved ( ) ( ) 2 2 2 4 5 , x x - = - ( ) ( ) 2 2 3 4 6 y y + = - and ( ) ( ) 2 2 5 4 8 z z - = - to generate 3 pairs of possible values for the x, y and z components. These candidates then tended to struggle to determine the possible positions of P from their 6 values. Vector notation was very variable with a wide variety of non-standard notations being encountered. It would be helpful for candidates to bear in mind that a four-mark question should not involve pages of work and that there are more efficient strategies. Whatever method was adopted, work was sometimes difficult to follow as there was little explanation and unclear notation used. In part (b), because of the volume of work they were producing, candidates often needed more space to complete their solution. Candidates are strongly advised that if they run out of space, they should use supplementary paper rather than complete their answer within the space of another question.

Common mistakes

• two positions: OA + 2AB = 8i + 15j + 11k and OA + (2/3)AB = 4i + 3j + 7k
• the second point is between A and B, not OA − 2AB
• a diagram helps
• finding the modulus of AB (3√11) alone gets nowhere
• long algebraic methods error-prone

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

(a) This part of the question was worth three marks and required candidates to produce a rigorous proof to show that c = 3, either by using trigonometry or an isosceles triangle or both. It seemed that many candidates were telling themselves that the answer was obvious but were uncertain as to how to go about justifying the fact. Most candidates struggled to produce a rigorous proof, instead just stating, without justification, that for a bearing of 135o, the distance travelled in the x-direction would have the same magnitude as the distance travelled in the negative y-direction or equivalent. They were able to score two of the three marks available if they then went on the show that c = 3. Those candidates who drew a diagram and labelled it with all the information from the question were generally more successful in scoring all three marks than those who either did not draw a diagram or whose diagram did not include some of the information given. Some - 6 = o tan135 candidates correctly used or the equivalent, without a diagram. Common 2c errors included not being able to correctly sketch a bearing of 135o, or omissions of angles or sides in the diagram. The negative sign caused issues in several instances and was sometimes glossed over. There was also some poor algebra. A few candidates incorrectly wrote 2c + 6 = 0 to obtain c = 3.

Common mistakes

• at t = 4, r = 2ci − 6j
• bearing 135° means a 45° angle below east so 2c = 6, justified by tan 45° or an isosceles right-angled triangle on a correct diagram
• stating equal components without justification loses a mark
• 2c + 6 = 0 is wrong

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

Examiner's comment

(b) The majority of candidates realised the need to differentiate the given expression for r, with c = 3 substituted, to obtain the velocity, and most of these differentiated correctly. They then substituted t = 4 and used Pythagoras to find the speed, giving their answer either as an exact value or an appropriate decimal equivalent. A minority correctly found the velocity but then stopped and lost two marks. A few candidates delayed substituting c = 3 until the end and a few delayed putting t = 4 which made the working a lot more demanding. About half worked with column vectors instead of using i - j form. Common errors included calculation slips when evaluating the speed or when substituting t = 4, misreading the expression such as losing the negative sign in the j-component and losing the i and j and not recovering to use components. A few candidates tried to use a suvat approach not realising that this was not a constant acceleration situation. (c) Many of those who differentiated in (b), successfully differentiated their velocity vector to obtain the acceleration vector. However, a significant number of candidates did not know how to proceed from there and the second stage of the process proved to be a discriminator. It was only the stronger candidates who realised that their acceleration vector had to be a multiple of the given vector. Those who did make this link generally were able to complete the question successfully. A number of unsuccessful candidates just set the i-component of the acceleration equal to -1 (or more rarely to -27). Most of k -- those who compared a to ( 27 ) i j were able to reach an equation in t only which they then tried to solve. Fewer candidates used a ratio approach but a significant minority of these used the reciprocal of the correct ratio but were still able to score the method mark. Having reached a correct equation in t or T only, some then struggled to obtain the correct answer. There were a few candidates who tried to use a suvat approach for this part not realising that this was not a constant acceleration situation and scored nothing.

Common mistakes

• differentiate r (powers of t decrease by 1): v = (3/2)t^(−1/2)i − (3/4)tj
• put t = 4 in both components then Pythagoras: speed = 3√17/4 ≈ 3.09 m s⁻¹
• stopping at the velocity vector
• suvat invalid (acceleration not constant)

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

Examiner's comment

(c) Many of those who differentiated in (b), successfully differentiated their velocity vector to obtain the acceleration vector. However, a significant number of candidates did not know how to proceed from there and the second stage of the process proved to be a discriminator. It was only the stronger candidates who realised that their acceleration vector had to be a multiple of the given vector. Those who did make this link generally were able to complete the question successfully. A number of unsuccessful candidates just set the i-component of the acceleration equal to -1 (or more rarely to -27). Most of k -- those who compared a to ( 27 ) i j were able to reach an equation in t only which they then tried to solve. Fewer candidates used a ratio approach but a significant minority of these used the reciprocal of the correct ratio but were still able to score the method mark. Having reached a correct equation in t or T only, some then struggled to obtain the correct answer. There were a few candidates who tried to use a suvat approach for this part not realising that this was not a constant acceleration situation and scored nothing.

Common mistakes

• a = −(3/4)t^(−3/2)i − (3/4)j must be a multiple of (−i − 27j): use the ratio of components (not the reciprocal)
• setting the i component equal to −1 is wrong
• T = 9

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

Examiner's comment

This was another short question on two-dimensional vectors with many candidates able to score 2 out of the 3 marks. Part (a) was generally answered accurately by all candidates. Sometimes the mark was lost because the candidate omitted the square root and wrote 25 9 4 OA = + + . Sometimes the notation was not made clear or i, j and k appeared.

Common mistakes

• must show how 38 is formed from the three components

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

Examiner's comment

This was another short question on two-dimensional vectors with many candidates able to score 2 out of the 3 marks. In part (b), the most common approach was to proceed to 2 18 a or 2 18 a = , and the majority of candidates were able to achieve this. However, a significant number of candidates did not conclude that 5 a = , with some claiming that e.g. 18 is an integer or missing the requirement that “a” had to be an integer. Some other incorrect answers included 19 a = following 18 a and the answer 5 a was not condoned, which lots the final mark for a number of candidates. A common processing error was the simplification of 2 20 a + to 20 a + . Very few candidates used the substitution method and some of those did not reach any conclusion in the end.

Common mistakes

• a must be a positive integer: a² > 18 gives a = 5, not a ≥ 5 or √19
• √(20 + a²) wrongly split into √20 + a

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

Examiner's comment

Given that candidates often find vector problems quite challenging, this question proved to be quite accessible, and most candidates made good progress through parts (a) and (b) of this question, with part (c) more mixed. In part (a), most candidates successfully reached 18 12 - + i j although a few subtracted the vectors the wrong way round or added the position vectors of A and B instead.

Common mistakes

• AB = OB − OA = −18i + 12j
• subtracting the wrong way round or adding

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

Examiner's comment

Given that candidates often find vector problems quite challenging, this question proved to be quite accessible, and most candidates made good progress through parts (a) and (b) of this question, with part (c) more mixed. Most were then able to calculate |AB| in (b) using their answer from (a), and many correctly reached 6 13 , which was allowed even if candidates had achieved an incorrect vector in (a), e.g., 18 12 + i j Attempted methods in (c)(i) were many and varied but most did not use vectors. The most common method was to establish the equation of the straight line through A and B, and this was then used with 2 x = - to find 5 p = . Other methods simply looked logically at the information given and found the appropriate scaling factor between, for example, the vectors AB and BC . Methods like this were often poorly written down but were clearly correct and gained full marks.

Common mistakes

• √(18² + 12²) = √468 = 6√13, fully simplified surd

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

Examiner's comment

Given that candidates often find vector problems quite challenging, this question proved to be quite accessible, and most candidates made good progress through parts (a) and (b) of this question, with part (c) more mixed. Many candidates did not successfully reach p = 5. Typical errors in (c)(i) were assuming AB = BC , setting the sum of the position vectors of A, B and C equal to zero or assuming that ( ) 18 12 2 k p - + = - + i j i j . A significant number of candidates omitted (c)(ii) after an unsuccessful attempt at (c)(i), and when responses were seen they were often lengthy and involved unnecessary attempts at calculating the areas of the triangles AOC and AOB. A common incorrect answer was 4:9, reached after assuming incorrectly that the two triangles were similar. Some candidates did, however, successfully reach a ratio of 2:3 either after area calculations, or by, for example, examining the relative lengths of AB and AC. Part (ii) did not rely on an answer to part (i) and so a reasonable number of candidates scored the final mark having been unable to find p, with some doing so very concisely. It was notable that many of the concise solutions to both parts came from a simple, well-constructed and clear diagram.

Common mistakes

• (i) AB parallel to BC (or gradient/line AB) gives p = 5
• wrong assumptions: AB = BC, position vectors summing to zero
• (ii) triangles share height from O so ratio = AC : AB = 2 : 3 (4 : 9 from assuming similarity is wrong)
• a clear diagram helps

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

Examiner's comment

This question was relatively well attempted. Most candidates attempted both parts and scored highly overall. In part (a) the majority of candidates correctly identified 10i + 24j, and 50i + 120j. Those who didn’t, often mistakenly added the position vectors instead of subtracting them. Most then went on to show that one was a multiple of the other (or that they were both different multiples of the same vector). However, some wrote the vectors the wrong way round for their multiple, which lost the second mark. Some candidates used other successful methods involving ratios or “gradients”. A common error here was candidates attempting to conclude the vectors were parallel by calculating and using their magnitudes. The most common reason for the loss of a mark in part (a) was with candidates correctly showing that the two vectors were multiples of each other but then not drawing a conclusion afterwards. It is important that candidates state what is shown by their working and provide a conclusion.

Common mistakes

• AD = d − a = 10i + 24j, BC = 50i + 120j (subtract, not add, position vectors)
• BC = 5AD so parallel, with a conclusion
• equal magnitudes do not show parallel

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

Examiner's comment

This question was relatively well attempted. Most candidates attempted both parts and scored highly overall. In part (b) most candidates correctly identified the lengths of AB, BC, CD and AD, but many left out their calculations by not showing the application of Pythagoras’ Theorem. This is an important step that should be included to fully demonstrate where their answers came from. There was a misunderstanding by some students who thought they could add the 4 vectors of the sides together to calculate a total vector and work out the length of the resulting vector. A few instead found the lengths of the position vectors and added them. Some also assumed that there were two pairs of equal length lines and/or two pairs of parallel lines and just used the two lengths from part (a) then doubled them. Despite this, candidates often achieved the first two marks. The calculation of the average speed was also done well by many candidates. Although several candidates based their calculation on a single lap and arrived at half the correct answer, the majority completed it correctly and clearly showed division by a fraction equivalent to 5 60 (most commonly 1 12 ) or multiplication by its reciprocal. Candidates should be aware of the general instruction in the rubric to give non-exact answers to 3 significant figures; it was common to see 7 or 7.0 instead of the correct final answer.

Common mistakes

• lengths 20, 130, 28√17 ≈ 115.5, 26 by Pythagoras (show it)
• 2 laps = 583 m in 5 min
• speed 6.99 km/h
• one lap only gives half
• adding the vectors or using lengths of position vectors
• assuming opposite sides equal

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

Examiner's comment

Part (a) was generally a well answered question. Some found the velocity vector at t = 0 and went no further, either forgetting, or not being aware that they needed to use Pythagoras’ Theorem to find the speed. A small number of candidates scored the M mark only, if they had an incorrect velocity vector but used Pythagoras’ Theorem correctly. The majority of candidates who attempted the second part used a correct method, setting

Common mistakes

• substitute t = 0 to get 7i − 3j, then speed is the magnitude √58 = 7.6
• stopping at the velocity vector

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

Examiner's comment

The majority of candidates who attempted the second part used a correct method, setting the i and j components of the velocity equal to each other or both equal to a constant which was then eliminated. A few also successfully used trial and error. However, a significant number set both i and j coefficients equal to 1, solved one or two separate equations and scored nothing.

Common mistakes

• equate the i and j components (t² − 3t + 7 = 2t² − 3), t = 2 only
• setting each component equal to 1 scores nothing

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

Examiner's comment

Part (a) was generally a well answered question. Some found the velocity vector at t = 0 and went no further, either forgetting, or not being aware that they needed to use Pythagoras’ Theorem to find the speed. A small number of candidates scored the M mark only, if they had an incorrect velocity vector but used Pythagoras’ Theorem correctly. The majority of candidates who attempted the second part used a correct method, setting the i and j components of the velocity equal to each other or both equal to a constant which was then eliminated. A few also successfully used trial and error. However, a significant number set both i and j coefficients equal to 1, solved one or two separate equations and scored nothing.

Common mistakes

• perpendicular to i means the i-component of acceleration is zero: 2t − 3 = 0, t = 1.5
• setting both components to zero scores M0

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

Examiner's comment

The first part was usually well attempted with most candidates scoring all four marks. The vast majority used v = u + at to find the velocity but a few lost the second two marks by not going on to find the magnitude of their velocity vector. Parts (b) and (c), however,

Common mistakes

• v = u + at with t = 5 gives −4i + 2j
• then the magnitude √20 = 4.47
• stopping at the velocity vector loses the last two marks

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

Examiner's comment

Many candidates tackled this question well, scoring full marks. Only a few made either no attempt or scored 0 for their efforts. The use of diagrams was widespread and seemed to aid those candidates in parts (a) and (c) to clarify their work. In part (a), errors were either adding the given vectors, subtracting the incorrect way round, or arithmetic errors involving negative numbers. Many candidates chose to use PQ PR - + rather than PR PQ - and seemed to have more success with the j component.

Common mistakes

• QR = PR − PQ (not PQ + PR)
• sign slips with the j component
• 10i − 20j

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

Examiner's comment

Many candidates tackled this question well, scoring full marks. Only a few made either no attempt or scored 0 for their efforts. The use of diagrams was widespread and seemed to aid those candidates in parts (a) and (c) to clarify their work. In part (b), the concept and calculation of the magnitude of a vector was clearly understood by the vast majority and this was evaluated with very few errors seen; the A mark was usually only being lost by candidates with a component error in (a). A minority of candidates attempted to construct a scale drawing: most of these were unsuccessful at providing sufficient evidence to score the method mark.

Common mistakes

• square and add before square rooting
• simplified surd 10√5
• scale drawing is not a method

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

Examiner's comment

Many candidates tackled this question well, scoring full marks. Only a few made either no attempt or scored 0 for their efforts. The use of diagrams was widespread and seemed to aid those candidates in parts (a) and (c) to clarify their work. Part (c) did provide problems for a number of candidates, although a pleasing number provided full solutions which were often concise. Some continued down the line of attempting to calculate magnitudes using the answer to (b) and magnitudes of the given vectors, whilst others failed to find the appropriate proportion of QR (usually using 2 3 or 1 3 ). The other main error seen was candidates failing to take proper account of the direction of RS . Others just found the appropriate proportion of QR but did not provide a full method and scored no marks as a result.

Common mistakes

• PS = PQ + (3/5)QR (not 2/3 or 1/3 of QR)
• take care with the direction of RQ if going via R
• working with magnitudes is wrong
• 9i − 7j

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

Examiner's comment

This vector question was set within the context of a rhombus. Surprisingly, many candidates were unable to use (or remember) the basic geometric of a rhombus and over 30% did not manage to make any progress. Part (a) required candidates to show that parallelogram PQRS was a rhombus. Although the requirement to 'show' is becoming more familiar to candidates, a significant number in this case didn't recognise the level of detail expected. So a lack of conclusion, or stating the sides were of length Ö29 without evidence of calculation, led to loss of credit. It was also necessary to conclude that the sides were equal, and not simply find the lengths and leave it to the reader to decide.

Common mistakes

• modulus of PQ = modulus of QR = √29 with calculations shown and conclusion "so rhombus"
• adjacent sides equal

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

Examiner's comment

This vector question was set within the context of a rhombus. Surprisingly, many candidates were unable to use (or remember) the basic geometric of a rhombus and over 30% did not manage to make any progress. A vast array of approaches were in evidence in part (b) to find the area of the rhombus. The easiest and most direct approach was to find 1 2 PR QS x . A more common method used involved finding the length of one diagonal, followed by the cosine rule to find an internal angle and then use of the formula 'ab sin C '. Unfortunately most attempts using this method only scored 3 marks as it was usual for answers to be given as 22.7 rather than the one required, Ö517. Other methods seen involved use of Heron’s formula, the “shoelace” determinant method, scalar product and vector product….. with many successfully used. Students who added a “narrative” – so explaining what they attempted to do – were in a better position to gain credit if they suffered a processing error early in their solution. This has to be considered good practice. The most common incorrect solution, seen many times over was in stating area = 29 29 29 PQ PS x = x =

Common mistakes

• diagonals PR = PQ + QR = 7i + 3j − 6k, QS = −PQ + QR = 3i − 3j + 2k
• area = ½ × PR × QS = √517 (exact)
• the cosine-rule/½ab sin C route often gave 22.7 not √517
• over 30% made no progress

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

Examiner's comment

This question was a good differentiator, with the majority of candidates gaining some credit, but only the strongest achieving full marks. Overall, candidates often demonstrated good reasoning skills and were rewarded with good marks on this question but a significant number of candidates struggled with this topic. Throughout, methods of solution were many and varied, and often involved non vector approaches that were sometimes difficult to follow. In part (a) candidates using a vector method often attempted to subtract the given position vectors to form at least two of the vectors ±𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗, ±𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ or ± 𝐵𝐵𝐵𝐵 ሬሬሬሬሬ⃗ although errors in their components were common. Candidates unfortunately often misread 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗ as ൭ 4 0 6 ൱ or ൭ 4 6 0 ൱. This was as far as many candidates reached, but some were able to successfully complete this approach by establishing 𝐵𝐵𝐵𝐵 ሬሬሬሬሬ⃗= 4𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗, 𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗= 5𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ or 5𝐵𝐵𝐶𝐶 ሬሬሬሬሬሬሬሬ⃗= 4𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ and reached p = 32. Some candidates incorrectly used methods like 𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗= 𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ or assumed that the original position vectors were parallel. Candidates attempting to work out the magnitudes of vectors usually gained no marks. Others who made progress with vector methods, established the vector equation of the straight line through A, B and C and used this to find p. Candidates often used non vector methods as well. A common approach was to ignore one component and compare ‘gradients’ using the other two e.g. 4-(-3) 0-4 = 𝑝𝑝-4 -16-0 . This was judged to be an acceptable approach, although errors with the substitution of values were common. Some even established an ‘equation’ such as 𝑦𝑦= - 7 4 𝑥𝑥+ 4 and then substituted in 𝑥𝑥= -16 to find p, which, again, was acceptable, but it was a shame that so many candidates resorted to such methods and seemed to lack confidence in their ability to construct a solution using vectors. In some cases where the first method mark was awarded, candidates went on to an incorrect equation such as 𝑝𝑝= 4 x 7 and were much less likely to be awarded the second method mark, as it was often unclear which vector they were working with for this step. Many fully correct answers for (a) were seen, sometimes with minimal working. It should be noted here that candidates should try to make their intention clear, as there were many occasions where examiners struggled to follow the structure of candidates’ work and were unable to award marks as a result.

Common mistakes

• form two of AB, AC, BC by subtraction (OB = 4j + 6k, no i component)
• parallel vectors are multiples: AC = 5AB gives p + 3 = 35, p = 32
• magnitudes are no help

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

Examiner's comment

This question was a good differentiator, with the majority of candidates gaining some credit, but only the strongest achieving full marks. Overall, candidates often demonstrated good reasoning skills and were rewarded with good marks on this question but a significant number of candidates struggled with this topic. Throughout, methods of solution were many and varied, and often involved non vector approaches that were sometimes difficult to follow. Part (b) was found to be challenging by most candidates, with few gaining any marks and with many not attempting it. Many candidates again worked with magnitudes of vectors and made no progress. Those candidates who did gain credit usually used the fact that 𝑘𝑘𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗-𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗= ൭ 16 4𝑘𝑘-𝑝𝑝 6𝑘𝑘-10 ൱ was parallel to 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗ to find 𝑘𝑘= 5 giving 𝑂𝑂𝑂𝑂 ሬሬሬሬሬሬ⃗= ൭ 0 20 30 ൱. Many candidates failed to realise that 𝑂𝑂𝑂𝑂 ሬሬሬሬሬሬ⃗= 𝑘𝑘൭ 0 4 6 ൱, and made no progress, whilst others incorrectly assumed that ൭ 16 4𝑘𝑘-𝑝𝑝 6𝑘𝑘-10 ൱= 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗. Candidates often drew diagrams to visualise the problem but full marks in this part were rarely awarded. A significant number of candidates who had found the correct vector, failed to find its magnitude.

Common mistakes

• OD = λOB and CD = OD − OC parallel to OA gives λ = 5, OD = 20j + 30k
• modulus 10√13 (fully simplified)
• CD is not equal to OA
• many stopped at the vector

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

Examiner's comment

In part (a) most candidates correctly substituted t = 2 into the expression for the velocity and went on to use Pythagoras correctly to find an exact answer or an answer correct to the i and j and then finding 12-6 sqrt 2 rather than apply Pythagoras. A small number 2sf or better. Common mistakes were to stop once the velocity had been found, losing subtracted the squared values rather than adding in their Pythagoras. In the second part, most candidates identified that this was variable acceleration and differentiated correctly

Common mistakes

• substitute t = 2 then use Pythagoras for speed (6√6, about 14.7)
• stopping at the velocity vector
• finding 12 − 6√2 instead of the magnitude
• subtracting the squares

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

Examiner's comment

In part (a), the majority of candidates were able to achieve the first mark for adding the two forces and collecting the i and j components. Many just equated this resultant force to the vector (3i + j) rather than using (3i + j) to define the direction. Some set up equations λ + 4 = 3 and mu - 1 = 1 and proceeded to solve them simultaneously gaining no further credit. Those who did use ratio usually went on to derive the required equation correctly; however, it should be remembered that, if required to derive a given answer, this must be written as a final conclusion in the form exactly as stated in the question. Another successful approach was to equate the i and j coefficients to 3k and k and solve as simultaneous equations. A small minority of candidates applied F = 4a or used integration to find v at this stage and then used ratio appropriately to obtain the required answer. In the second part, many candidates were able to find the resultant force and

Common mistakes

• resultant (4 + λ)i + (−1 + μ)j is parallel to 3i + j so use ratios
• equating the resultant to 3i + j (λ + 4 = 3, μ − 1 = 1) scores no more
• final line must be λ − 3μ + 7 = 0 exactly

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

This question using vectors was generally well answered with most candidates scoring more than half of the marks available. In part (a) candidates seemed unsure how to go about proving that the stone passed through the origin. The most successful attempts typically involved finding the equation of the line for the stone which was 5 12 y x = and showed that the origin was on this line or stated that this linear equation would pass through the origin. Other attempts which candidates proceeded with involved either comparing the position vectors and showing that one was a multiple of the other or showing that the gradients of the position vectors were the same. Plotting the points on a graph was not an acceptable method to prove.

Common mistakes

• show AO and OB are parallel, e.g. −24i − 10j = −2(12i + 5j), and conclude that as the stone moves in a straight line it passes through O
• or show the line y = (5/12)x passes through the origin
• plotting points is not a proof

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

Examiner's comment

This question using vectors was generally well answered with most candidates scoring more than half of the marks available. In part (b) most candidates were able to find the vector AB and proceeded to find the magnitude of their vector. Some forgot to then find the speed by dividing by 9, whilst others omitted or gave the wrong unit, or just gave the generic units per second as their unit for speed.

Common mistakes

• AB = √(36² + 15²) = 39 m
• divide by the 4 seconds: speed 9.75 m s⁻¹
• units required (not "units per second")
• forgetting to divide by the time

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

Examiner's comment

This was another question where the modal mark was full marks, this time achieved by over 40% of the candidates Part(a) A very significant majority of candidates obtained vector AC correctly. Only a few subtracted rather than adding the given vectors AB and BC.

Common mistakes

• AC = AB + BC = −2i − 3j − k
• add, not subtract

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

Examiner's comment

This was another question where the modal mark was full marks, this time achieved by over 40% of the candidates Part (b) The vast majority of candidates made some progress in this part, realising that the lengths of the vectors would be required to solve this problem. A small minority then simply stopped, having done no more than finding the lengths of the sides of the triangle ABC, with some assuming that the triangle was right angled. However most then went on to use the cosine rule to try to show the required result. In almost all cases a correct cosine rule was stated and used, so that following on from a previous accuracy error, the method marks could be awarded. Fully correct calculations were necessary in order to gain the final accuracy mark and very many candidates managed to do this and so gained full marks for this question. A few candidates, presumably those who had studied further maths, used the scalar correct way around. Many using this method showed that cos ABC = - 9 product. This method would be perfectly acceptable, as long as the vectors are used the 10 but failed to explain why the answer should be positive.

Common mistakes

• find magnitudes √50, √18, √14 and use the cosine rule with the sides in the right positions
• scalar product (BA.BC, not AB.BC) gives −9/10 if the direction is wrong

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

Examiner's comment

9 This question provided a good range of marks between the candidates, although nearly all candidates struggled to proceed to the final answer in part (c). There were also several blank responses on this question which may have been down to candidates either running out of time or just getting overwhelmed towards the end of a paper. Part (c) was either not attempted at all or candidates made an attempted but abandoned due to the expressions formed appearing so complicated that it seemed as though they were convinced they had gone wrong. Usually, candidates opted to substitute their part (a) answer into their part (b) equation of the circle. To score the first mark candidates had to multiply out the brackets and collect terms together to form a 3TQ where k was in both the b and c coefficients of the quadratic. Some candidates stopped before getting this far, although it was pleasing that some continued to persevere and pick up the first mark. Those who continued, realised the discriminant needed to be used, but disappointingly once they had their expression also decided to stop even if they had a quadratic in k which they could solve. Therefore only 1 or no marks was scored in nearly all responses. The easiest way was the vector approach, which was seen, but far too frequently, and even then, it was only a few candidates who were able to score full marks on this question.

Common mistakes

• the other tangent touches at the opposite end of the diameter through P: (13/2, 11/2), then k = 10/3
• via simultaneous equations the quadratic in x has b and c depending on k and b² − 4ac = 0 (k = 0 is rejected)
• candidates often abandoned the messy algebra

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

Examiner's comment

Part (a) was well answered with most students making use of v = u + at to arrive at the correct answer 8 6 = - v i j. It is important to note that this uses the initial velocity of 4i. The alternative method was to use integration, so the equivalent of the first stage was to integrate and use the 4i when calculating the constant of integration. A few made careless arithmetic errors and others forgot to use the 4i in the integration. A small number either differentiated or made no progress.

Common mistakes

• v = u + at with u = 4i (if integrating, the 4i is the constant of integration and is often forgotten)
• keep i and j (M0 if lost)
• answer 8i − 6j

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

Examiner's comment

Part (b) proved to be more challenging. Although there were many good solutions employing 2 1 2 t t = + r u a to find the change in displacement from t = 0 to t = 3, earning the method mark, a sizeable number of candidates forgot to add on i + j. The equivalent to the first stage using integration was 2 1 (2 3 ) 4 2 t t = - + r i j i . It was essential to include use of the 4i to earn the method mark, but the use of i + j was only required for the A mark. Some students chose to answer the question in stages, working out position vectors at t = 1, 2 and then t = 3. Other errors involved not keeping the question in i and j or column vector form, which lost the M mark, or incorrectly trying to turn the problem into a scalar problem.

Common mistakes

• s = ut + ½at² gives the displacement only: add the initial position i + j (commonly forgotten)
• stay in vector form, not a scalar problem
• answer 22i − 12.5j

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

Examiner's comment

Candidates were instructed to use g = 10 in this question but were only penalised once if candidates who used 𝑣𝑣2 = 𝑢𝑢2 + 2𝑎𝑎𝑎𝑎 to find the vertical component of velocity of the in 2/4 marks being scored for this part. A few candidates did not resolve the speed of 65 m s-. This lost the M mark and consequently they scored zero for part (a). In part (b), 1 particle at A were generally more successful than those who used the time found in part (a). Many solutions finished here and did not consider the need to look at the horizontal component as well. Those who had both components usually went on to apply Pythagoras theorem and calculate the actual speed of 75 m s-. For part (c), there were several 1

Common mistakes

• need both components: horizontal 65 cos α = 60 (often omitted) and vertical 45 (v² = u² + 2as is more reliable than using t)
• Pythagoras gives 75 m/s

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

Examiner's comment

Part (a) was well answered by most candidates who realised that they needed to differentiate although a few candidates did omit the vector notation and therefore achieved zero marks. Part (b) proved to be the least popular part of this question.

Common mistakes

• a = dv/dt: differentiate both components (powers decrease by 1)
• keep the vector notation (M0 if i, j disappear)
• answer (3/2)t^(−½) i − 2j

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

Examiner's comment

achieved zero marks. Part (b) proved to be the least popular part of this question. 1 = 2t and then 3t Successful candidates employed a ratio method to obtain the equation 2 solved it. However, a significant minority of candidates incorrectly equated the i and j parts and then solved the two equations to obtain the correct answer. Candidates should note that this approach scored no marks. Other errors included using the acceleration vector or the displacement vector rather than the velocity vector when dealing with direction of travel. Part (c) tended to be well answered with a good number of candidates

Common mistakes

• moving in direction i − j means the i and j components of v are equal and opposite: 3t^½ = 2t
• equating v to i − j scores nothing
• use v, not a or r
• t = 9/4

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

Examiner's comment

direction of travel. Part (c) tended to be well answered with a good number of candidates integrating and gaining 2 or 3 marks here. It was disappointing to see a few using constant acceleration formulae. Part (d) allowed those who had made little progress on the first 3

Common mistakes

• r = ∫v dt, both powers increase by 1
• use r = −j at t = 1 to get C = −2i
• suvat is invalid (acceleration not constant)

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

Examiner's comment

acceleration formulae. Part (d) allowed those who had made little progress on the first 3 parts, to score some marks as it only required the use of the given velocity vector to get started. For those who attempted it, the use of Pythagoras’ theorem on the velocity components was very accessible and a number of candidates solved the quadratic to obtain t = 4. Sign errors here were common, however, with many ending up with a wrong t value. Some candidates then scored a further two M marks for substituting their value of t into their position vector and then applying Pythagoras to find the distance. The question required an exact distance and so only surd answers earned the final mark.

Common mistakes

• speed 10 means 9t + 4t² = 100, t = 4 (sign errors common)
• substitute into r: 14i − 16j
• distance is the magnitude of r
• exact answer 2√113 (√452), decimals lose the final mark

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

Examiner's comment

Candidates found the first part of this question challenging, but they were more successful with the second part, with many fully correct responses. There were a number of common mistakes in candidates’ approaches to part (a). Firstly, many candidates tried to find the resultant vector by adding the two given vectors, rather than subtracting them. Secondly, candidates seemed unclear on exactly what angle the question was looking for, with lots of time being spent on finding various different (but unfortunately not relevant) angles in their diagrams. Finally, some candidates took a rather long-winded 4 approach to find the correct angles (through use of the cosine rule to find one part of the required answer, then additional work using tan) as they had not spotted the more straightforward use of 7 arctan 3 . Even those who found a relevant angle were often unable to proceed to find a bearing.

Common mistakes

• displacement is (−3i − 5j) − (4i − 2j) = −7i − 3j (subtract, not add)
• find the right angle (arctan 7/3) and convert to a bearing 180° + 66.8° = 246.8°

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

Examiner's comment

Candidates found the first part of this question challenging, but they were more successful with the second part, with many fully correct responses. Part (b) was generally answered successfully, with a few candidates mistakenly stating that 2 hours and 45 minutes was equal to 2.45 hours, rather than 2.75 hours.

Common mistakes

• distance √58 km
• time 2 h 45 min = 2.75 h, not 2.45 h
• speed ≈ 2.77 km/h

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

Examiner's comment

In part (a) most candidates were able to successfully subtract vectors the correct way round gaining both marks. Very few candidates added the two vectors together.

Common mistakes

• AB = b − a (subtract the right way round) = i − 8j + 2k
• coordinates (1, −8, 2) not accepted

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

Examiner's comment

In (b) those who gained both marks were able to compare the two vectors AB and OC to show that they were parallel and conclude that OABC is a trapezium. Usually candidates made a sketch of a trapezium so they could visualize which vectors or sides they would compare. This is a very good idea and should be encouraged by centres. Marks were lost when • candidates did not give a minimal conclusion after finding OC and AB were parallel. • candidates incorrectly stating that 2 AB OC = x • candidates finding the length of each side OABC rather than their directions

Common mistakes

• OC = 2AB so OC is parallel to AB, hence a trapezium
• a reason and minimal conclusion needed
• finding side lengths instead of directions scores nothing

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

Examiner's comment

This proved to be a demanding question for many candidates, with false starts and seemingly random statements. Those who were able to make some progress generally gained the initial method mark for expressing one of the line segments in terms of the difference of two position vectors, but many candidates were unable to make further progress. Unfortunately, the addition of two position vectors was a common error. Diagrams helped with some responses, but they frequently depicted OP, OQ and OR as collinear, which tended to cause more confusion. The most common approach amongst those who were able to make further progress were either 𝑃𝑃𝑃𝑃 ሬሬሬሬሬ⃗= 1 3 𝑃𝑃𝑃𝑃 ሬሬሬሬሬ⃗ or 𝑂𝑂𝑂𝑂 ሬሬሬሬሬሬ⃗= 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗+ 1 3 𝑃𝑃𝑃𝑃 ሬሬሬሬሬ⃗. Vector notation was often poor, for example Q being used to denote a vector rather than a point, leading to such confused statements as Q = 1/3 PR.

Common mistakes

• PQ = q − p (subtract, do not add, position vectors)
• use PR = 3PQ or q = p + ⅓(r − p)
• careful vector notation (Q is a point, not a vector)

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

Examiner's comment

In the first part, the majority of candidates used v = u + at with t = 2 to obtain a correct constant, in this case, the initial velocity. In part (b), most used r = ut + 1 velocity. Some chose to integrate, often successfully, although some failed to include a 2 at2 to find a displacement vector. Again, some chose to integrate, mostly correctly. Having equated their displacement expression to λi – 4.5j, some did not know how to proceed. A few tried to use magnitudes and reduce the problem to scalars and some had mixed up velocity and displacement. Those who realised that equating j components gave an equation in T generally solved the resulting quadratic and found the correct answer. Part (c) required

Common mistakes

• v = u + at with vectors: (−2i + 2j) + 2(4i − 5j) = 6i − 8j
• if integrating, the initial velocity is the constant
• velocity is a vector, the speed is not needed

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

Examiner's comment

In the first part, the majority of candidates used v = u + at with t = 2 to obtain a correct constant, in this case, the initial velocity. In part (b), most used r = ut + 1 velocity. Some chose to integrate, often successfully, although some failed to include a 2 at2 to find a displacement vector. Again, some chose to integrate, mostly correctly. Having equated their displacement expression to λi – 4.5j, some did not know how to proceed. A few tried to use magnitudes and reduce the problem to scalars and some had mixed up velocity and displacement. Those who realised that equating j components gave an equation in T generally solved the resulting quadratic and found the correct answer. Part (c) required the marks in part (b) went on to gain the two marks here. A method mark was available for those carrying through a wrong value for T provided that it had been found from a valid method in part (b).

Common mistakes

• r = ut + ½at² (not with u = 0)
• equate the j components only: −4.5 = 2T − (5/2)T² gives T = 1.8
• do not take magnitudes or mix up velocity and displacement

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

Examiner's comment

generally solved the resulting quadratic and found the correct answer. Part (c) required the value for T to be substituted into the i component and virtually all those who achieved the marks in part (b) went on to gain the two marks here. A method mark was available

Common mistakes

• substitute T = 1.8 into the i component: λ = −2T + 2T² = 2.88 (72/25)
• reject a negative T

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

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• variable acceleration so suvat scores nothing
• integrate each component and include the constant 36i
• v = 8i − (28/3)j

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

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• perpendicular to i means the i component of v is zero: t − 2t² + 36 = 0 gives t = 4.5
• equating the j component to zero (or i to j) scores nothing
• use velocity, not displacement

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

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• differentiate r to get v = (2t − 1)i + 3j
• speed is the magnitude: (2t − 1)² + 3² = 5² gives t = 2.5
• equating the vector to 5 or dropping i and j is wrong
• not √(x² − y²)

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

Examiner's comment

This was a poorly answered question at all levels. In part (a) it was expected that candidates would recognise the fact that the vectors would need to lie in the same direction but many attempted to write out the given expression in words. There was slightly more success in part (b) with most attempting a diagram of sorts to get a feel for the problem. Unfortunately a lack of understanding of vectors 6 m-n 3 n 3 m meant that and was usually followed by . There were however, some excellent and well thought out solutions showing a full understanding of the question,

Common mistakes

• a and b are parallel and in the same direction
• do not just restate the equation in words

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

Examiner's comment

This was a poorly answered question at all levels. In part (a) it was expected that candidates would recognise the fact that the vectors would need to lie in the same direction but many attempted to write out the given expression in words. There was slightly more success in part (b) with most attempting a diagram of sorts to get a feel for the problem. Unfortunately a lack of understanding of vectors 6 m-n 3 n 3 m meant that and was usually followed by . There were however, some excellent and well thought out solutions showing a full understanding of the question,

Common mistakes

• draw the triangle with 6 opposite the 30° and 3 adjacent
• sine rule sin 30°/6 = sin θ/3 gives θ = 14.5°
• the angle between m and m − n is 180 − 30 − 14.5 = 135.5°
• no negative lengths

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

Examiner's comment

Q10 discriminated well between the medium and higher ability candidates, with lower ability candidates struggling to gain any creditable access. Q10 increased in difficulty as the candidates progressed through the question, and only a small proportion of candidates scored full marks. In Q10(a), most candidates followed a correct method to achieve the correct answer 3 1 CM Of the candidates who identified a suitable route from C to M, the . a b 2 2 1 . CM CA AM CA AB most common approach was Less efficient routes such 2 as CM CO OB BM or even CM CO OA AM were also observed. Some candidates struggled to apply the vectors a and b to the routes they had chosen. Incorrect methods included applying CA a (instead of CA a ) or applying AB AB ). b a a b (instead of There were a few candidates who incorrectly deduced the position of C to be in the opposite direction to OA and they were mostly 1 , CM CO OA AM CO OA AB able to apply the approach, to gain some 2 credit. Q10(b) proved to be demanding, with some candidates struggling to understand what the question required them to do, and many seemed to be confused by the introduction of the scalar multiple . Only a minority deduced that CN CM and applied this . ON OC CN deduction to the route Most of these candidates progressed well to the given answer unless they had found an incorrect CM in Q10(a). Many candidates used ON OC CM MB BN or ON OA AB BN and made no creditable progress. These candidates usually produced solutions which led to the elimination of from the a component of the resulting vector for . ON A few candidates used ON OM MN and proceeded to obtain a correct expression 1 1 3 1 ON a b a b (or ), in terms but only a few deduced that this 2 2 2 2 (or ). 1 could lead to the stated solution by setting 1 Q10(c) proved to be challenging with many candidates failing to make any creditable is parallel to OB progress. The crucial step was to recognise that ON and hence is a multiple of b with no a component. Only a few candidates set the a component ON 3 4 , 2 2 0. to zero to give the equation Almost all of these candidates found 3 2 ON b with many finding and giving a correct explanation of the given ratio. 3

Common mistakes

• CM = CA + ½AB with CA = −a (not a) and AB = b − a: −3/2a + ½b, simplified

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

Examiner's comment

Q10 discriminated well between the medium and higher ability candidates, with lower ability candidates struggling to gain any creditable access. Q10 increased in difficulty as the candidates progressed through the question, and only a small proportion of candidates scored full marks. In Q10(a), most candidates followed a correct method to achieve the correct answer 3 1 CM Of the candidates who identified a suitable route from C to M, the . a b 2 2 1 . CM CA AM CA AB most common approach was Less efficient routes such 2 as CM CO OB BM or even CM CO OA AM were also observed. Some candidates struggled to apply the vectors a and b to the routes they had chosen. Incorrect methods included applying CA a (instead of CA a ) or applying AB AB ). b a a b (instead of There were a few candidates who incorrectly deduced the position of C to be in the opposite direction to OA and they were mostly 1 , CM CO OA AM CO OA AB able to apply the approach, to gain some 2 credit. Q10(b) proved to be demanding, with some candidates struggling to understand what the question required them to do, and many seemed to be confused by the introduction of the scalar multiple . Only a minority deduced that CN CM and applied this . ON OC CN deduction to the route Most of these candidates progressed well to the given answer unless they had found an incorrect CM in Q10(a). Many candidates used ON OC CM MB BN or ON OA AB BN and made no creditable progress. These candidates usually produced solutions which led to the elimination of from the a component of the resulting vector for . ON A few candidates used ON OM MN and proceeded to obtain a correct expression 1 1 3 1 ON a b a b (or ), in terms but only a few deduced that this 2 2 2 2 (or ). 1 could lead to the stated solution by setting 1 Q10(c) proved to be challenging with many candidates failing to make any creditable is parallel to OB progress. The crucial step was to recognise that ON and hence is a multiple of b with no a component. Only a few candidates set the a component ON 3 4 , 2 2 0. to zero to give the equation Almost all of these candidates found 3 2 ON b with many finding and giving a correct explanation of the given ratio. 3

Common mistakes

• ON = OC + λCM (CN is a multiple of CM)
• routes via M, B or A eliminate λ and get nowhere

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

Examiner's comment

Q10 discriminated well between the medium and higher ability candidates, with lower ability candidates struggling to gain any creditable access. Q10 increased in difficulty as the candidates progressed through the question, and only a small proportion of candidates scored full marks. In Q10(a), most candidates followed a correct method to achieve the correct answer 3 1 CM Of the candidates who identified a suitable route from C to M, the . a b 2 2 1 . CM CA AM CA AB most common approach was Less efficient routes such 2 as CM CO OB BM or even CM CO OA AM were also observed. Some candidates struggled to apply the vectors a and b to the routes they had chosen. Incorrect methods included applying CA a (instead of CA a ) or applying AB AB ). b a a b (instead of There were a few candidates who incorrectly deduced the position of C to be in the opposite direction to OA and they were mostly 1 , CM CO OA AM CO OA AB able to apply the approach, to gain some 2 credit. Q10(b) proved to be demanding, with some candidates struggling to understand what the question required them to do, and many seemed to be confused by the introduction of the scalar multiple . Only a minority deduced that CN CM and applied this . ON OC CN deduction to the route Most of these candidates progressed well to the given answer unless they had found an incorrect CM in Q10(a). Many candidates used ON OC CM MB BN or ON OA AB BN and made no creditable progress. These candidates usually produced solutions which led to the elimination of from the a component of the resulting vector for . ON A few candidates used ON OM MN and proceeded to obtain a correct expression 1 1 3 1 ON a b a b (or ), in terms but only a few deduced that this 2 2 2 2 (or ). 1 could lead to the stated solution by setting 1 Q10(c) proved to be challenging with many candidates failing to make any creditable is parallel to OB progress. The crucial step was to recognise that ON and hence is a multiple of b with no a component. Only a few candidates set the a component ON 3 4 , 2 2 0. to zero to give the equation Almost all of these candidates found 3 2 ON b with many finding and giving a correct explanation of the given ratio. 3

Common mistakes

• ON is parallel to OB so the a-component is 0: λ = 4/3, ON = ⅔b, NB = ⅓b

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

Examiner's comment

In part (a), although many candidates realised that finding an expression for the velocity was relevant to determining the direction of motion, only a minority could use the fact that it was specified as being in the direction of 3i – 4j. The most common error was in equating the actual vectors rather than either equating the ratios of components or equating to a multiple of 3i – 4j. A fair number attempted to use a displacement vector showing a lack of understanding of the situation. The second part was generally

Common mistakes

• v = (−1 + 2T)i + (4 − 3T)j
• parallel to 3i − 4j means ratio of components (4 − 3T)/(−1 + 2T) = −4/3, not equating the vectors themselves
• using displacement is wrong
• T = 8

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

Examiner's comment

displacement vector showing a lack of understanding of the situation. The second part was generally handled more successfully with many correct answers seen for the displacement vector (using s = ut 1 2 at2) although not all proceeded to find the distance by calculating its magnitude. Although most used + suvat formulae throughout the question, some successfully integrated the given constant acceleration to find the velocity and displacement. Use of magnitudes of vectors throughout and some invalid multiplication/division of vectors were seen on occasion.

Common mistakes

• s = ut + ½at² with u = −i + 4j and t = 4 gives 12i − 8j
• distance is its magnitude 4√13 ≈ 14.4 m
• 4(2i − 3j) is not the displacement

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

Examiner's comment

In part (a), the majority of candidates obtained the first mark for the horizontal component and most candidates also made a good attempt to find the vertical component, although sometimes it was seen as part of their working for part (b). Of those who found the horizontal and vertical components of the

Common mistakes

• horizontal 20cos 30° and vertical 20sin 30° − 2g = −9.6
• need both speed (19.8 m/s) and direction (29° below horizontal)
• do not give the answer in i and j (not defined)

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

Examiner's comment

This question was very accessible to the vast majority of students with many scoring full marks. Errors seen in part (a) included finding the sum of the two vectors instead of the difference, or, finding the difference ‘the wrong way around’. Correct vector notation was seen in all but a few cases with incorrect notation such as (-9i,3j) being penalised.

Common mistakes

• AB = OB − OA = −9i + 3j
• common errors: adding the vectors or OA − OB
• incorrect notation such as (−9i, 3j) penalised

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

Examiner's comment

This question was very accessible to the vast majority of students with many scoring full marks. Part (b) was more successful with the vast majority of students knowing how to calculate the length of a vector and simplify the surd. A follow through on their vector from part (a) enabled most students access to both marks.

Common mistakes

• modulus √((−9)² + 3²) = √90 = 3√10 as a simplified surd
• ±3√10 loses the A mark

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

Examiner's comment

students scoring the 3 marks. Part (b) required the students to find the equation of the circle C using the information given. Again most students attempted this with ease, finding firstly the coordinates of point P, and then the radius PA, before writing down the equation of the circle. Errors witnessed tended to be arithmetic, although some students complicated the question by attempting to solve 2 2 simultaneously in an attempt to 2 and 7 5 x y r 2 1 y x find r. There were numerous ways to find the value of k in part (c). One of the best involved

Common mistakes

• solve y = 2x + 1 with 2y + x = 17 to get P(3, 7)
• r² = PA² = 20
• equation (x − 7)² + (y − 5)² = 20, not (√20)²

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

Examiner's comment

find r. There were numerous ways to find the value of k in part (c). One of the best involved using vector geometry to find the coordinates of the point N on C, where meets C. 2 y x k Using the fact that PA enabled students to find N = (11,3) and then k by substituting AN this point into Again simultaneous equation methods were common, some more 2 . y x k successful than other. The most common correct attempt involved solving with 2 y x k 2 2 and then finding where the resulting equation had one root. Care was 7 5 20 x y required with the algebra, especially the squaring of terms, as well as the application of the 2 discriminant This proved to be too great a challenge for many. 4 0. b ac

Common mistakes

• best route: other tangent point N = A + PA vector = (11, 3), then k = −19
• discriminant route 5x² + (4k − 34)x + k² − 10k + 54 = 0 with b² − 4ac = 0 has algebra/squaring errors
• reject k = 1

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

Examiner's comment

This was a well-answered question with most students scoring at least 4 of the 5 marks available. Part (a) was often completed successfully. Most students found AB and used the given result AB BD to deduce and apply . OD OB AB A few students applied 2 . OD OA AB There were some sign, manipulation and bracketing errors seen in this part and students who used column vectors were generally more successful than those who used the , , i j k notation.

Common mistakes

• OD = OB + BD = OB + AB = 2OB − OA = 6i − 7j + 10k
• sign, bracketing errors with i j k notation (column vectors safer)

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

Examiner's comment

This was a well-answered question with most students scoring at least 4 of the 5 marks available. In part (b), most students found the difference between OC and , OA and applied Pythagoras’ Theorem to 4 AC to give the correct 2 2 2 2 ( 2) (2) (2) (4) . a The resulting quadratic equation was generally solved correctly but not always efficiently. Too often, students expanded to form a 3-term quadratic rather than spotting that their quadratic was already in a ‘completed square’ form. Despite the question stating that 0, a some students gave both solutions 2 2 2 a as their final answer, while others mistakenly rejected 2 2 2 a to leave 2 2 2. a

Common mistakes

• AC = OC − OA
• (a − 2)² + 2² + 2² = 16 is already completed-square form, (a − 2)² = 8
• reject 2 + 2√2 because a < 0, answer a = 2 − 2√2

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

Examiner's comment

The vast majority of students recognised the requirement to integrate to find the position vector and subsequently to substitute the given values of t. Occasionally, these were substituted straight into the velocity vector but such instances were rare and much correct working was seen. Most proceeded to find the magnitude of the vector joining the two points to give the exact distance as a surd as required; only a very small minority went straight to a rounded decimal value.

Common mistakes

• integrate v to get r = 2t^½ i − 2t² j (+ C)
• positions at t = 1 and t = 4, not substituting into v
• exact distance √904 = 2√226, not a rounded decimal

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

Examiner's comment

The first part was generally well done with most students using a constant acceleration formula in 1 2 at2) to find a, using the given velocity and position vectors. The few who chose to vectors (r = ut + integrate a constant acceleration vector, and then use the given information to identify the constants of integration, mostly did so successfully. The magnitude of a was almost invariably calculated correctly but, since the answer was given, it was important that there was some working shown. Those who had no valid method for finding a tended to achieve the method mark for calculating the magnitude of their vector. A small minority of students attempted to apply suvat equations to the magnitudes of the given vectors; by chance this led to an answer which rounded to the given 2.5 m s-2. However, such attempts achieved no credit.

Common mistakes

• r = ut + ½at² in vector form gives a = 1.5i − 2j, magnitude 2.5
• no credit for suvat on the magnitudes or separate components

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

Examiner's comment

Part (b) provided a greater challenge. Here it was necessary to find v after 2 seconds and then use this as the initial velocity for the second part of the motion. The given direction of motion meant that the coefficients of the i and j components of velocity were equal which in turn led to an equation in t. The main errors were in either trying to equate coefficients for a position rather than a velocity vector or in using 7i -10j (a position vector) as the initial velocity. Some students made no valid progress at all with this part of the question. There were, nevertheless, a fair number of entirely correct solutions seen

Common mistakes

• velocity at A = (2i − 3j) + 2(1.5i − 2j) = 5i − 7j
• north-east means i and j components of velocity equal: 5 + 4t = 8.8t − 7, t = 2.5 s
• common errors: equating components of a position vector, using 7i − 10j as the initial velocity