Edexcel A-level Maths · Topic M7
Practice questions covering kinematics from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
6 questions on Kinematics, each with its mark scheme and worked method. Reveal the answer when you have had a go.
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(3)
(Total for Question 1 is 3 marks)
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(3)
(Total for Question 2 is 3 marks)
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(4)
(Total for Question 3 is 4 marks)
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(4)
(Total for Question 4 is 4 marks)
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(5)
(Total for Question 5 is 5 marks)
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(4)
(Total for Question 6 is 4 marks)
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Printable question pack — M7 KinematicsAll 50 questions across 5 specification points — 44 more than the 6 above — with the full mark scheme. Guest accessible · browser print.17 questions shown as printed, 12 reference-only · newest series first
Question and mark-scheme extracts © Pearson Education Ltd. Reproduced for educational use; the original paper is linked on every question. Contact hello@oscarsongtutoring.co.uk for removal.
Edexcel 9MA0 · June 2024 · Paper 32 · Q2 · 8 marks
8 marks
Examiner report
This question was accessible to all candidates, particularly parts (a) and (b). (a) This was generally answered very well either using a suvat equation or by finding the gradient of the first section of the graph. Some lost the second mark for writing incorrect units or, occasionally, omitting the units altogether ( 1 m s− being the most common error with the units). A few candidates confused acceleration with distance and found the area under the graph during the first six seconds. (b) The vast majority scored all three marks here. The most popular approach was to sum the separate areas of the triangle and the rectangle or alternatively to find the area of the trapezium. Occasionally the ‘ 1 2 ’ was missing from either the area of the triangle or the trapezium. This error scored no marks for use of an incorrect formula. Those who used suvat methods tended to be less successful than those using an area approach. (c) This part was more challenging. Most candidates recognised that they needed to subtract their answer to (b) from 200 to get the distance travelled from t = 18 to t = 24. Whilst some used 1 2 ( ) s u v t = + , the most common successful approaches were to find an expression for the sum of the area of the triangle and rectangle or the area of the trapezium and equate that to 40. However, a significant number of candidates used the area of a triangle rather than a trapezium and wrote 1 2 (10 ) 6 40 U − = leading to 10 3 U = − then ignored the sign and wrote it as a positive answer which coincidentally was the correct answer but they scored no marks for an incorrect method. A small number of students attempted to find U by using suvat methods for the whole race, usually without success.
An athlete runs 200 m in 24 s. A speed-time graph shows constant acceleration from rest to 10 m s^-1 over 0 to 4 s, constant speed 10 m s^-1 from 4 to 18 s, then constant deceleration to speed U m s^-1 at 24 s.
Edexcel 9MA0 · June 2024 · Paper 32 · Q4 · 11 marks
11 marks
Examiner report
(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 candidates correctly used o 6 tan135 2c − = or the equivalent, without a diagram. Common 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. (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 those who compared a to ( 27 ) k −− 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.
In this question i is due east and j is due north. At time t seconds, t >= 1, a particle has position vector r = ct^(1/2)i - (3/8)t^2 j metres, where c is constant. At t = 4, its bearing from the origin is 135 degrees.
Edexcel 9MA0 · June 2024 · Paper 32 · Q5 · 12 marks
12 marks
Examiner report
(a)This part required the derivation of an equation for the flight path of a stone moving freely under gravity given the initial speed and angle of projection. There were many excellent responses seen where candidates used horizontal and vertical components of the initial velocity in 2 1 2 s ut at = + to find expressions for x and y in terms of t, with the majority realising that the acceleration was zero in the horizontal direction. Occasionally the velocities were not resolved or the acceleration in both directions was taken as g, but such instances were rare. Most proceeded to eliminate t correctly and obtain the printed equation. Since it was a given answer any slips in working tended to be rectified including any previous cos/sin confusion. The final mark did require 9.8 for the value of g to be seen being used explicitly in the working. Although full marks were often achieved in this part of the question, there were candidates who failed to make any valid progress, often because they equated their vertical distance to zero, possibly proceeding to calculate t and then realising they had no strategy for finding a general equation. Those who quoted the general equation for a trajectory and substituted in the given angle achieved no credit, but this was only very rarely seen. Some omitted this part of the question entirely, but it did not preclude them from accessing all marks in the subsequent parts. An alternative but significantly less common approach was to quote the general equation of a parabola as y = ax2 + bx + c and use the coordinates of three specific points to find the constants. Most realised that c = 0 and achieved the first two marks. Generally, the points corresponding to the range and the greatest height were found using suvat equations, although other points were possible. Some attempted to use completing the square with the maximum point at (60, 22.5) but failed to deal properly with the coefficients. Occasionally the derivative at x = 0 was used successfully to find the value of b. There were a fair number of entirely correct solutions seen using this method but also some which lacked organisation and a clear strategy. (b) Most candidates scored the two available marks in this part for finding the range OA. The most common method was to set y = 0 in the given equation and solve to find x. Sometimes a two-step suvat approach was adopted, generally successfully. Those who made no progress in part (a) often achieved the marks here. (c) This part was generally done well with many correct answers seen. Often a suvat approach was used to find H, the greatest height. Alternatively, many used the symmetry of the parabola and substituted x = 60 (half the range) into the given equation to find y. Some differentiated to find the maximum (or used a calculator) but occasionally gave 60 (the x value) rather than 22.5 (the y value) as their final answer. Completing the square was also employed successfully on occasion. (d) In this part the majority of candidates recognised that H would be greater than K i.e the particle would reach a greater height in a model without air resistance than it would in a model including air resistance. Not all, however, gave a satisfactory reason with some just claiming ‘because of air resistance’ or ‘air resistance limits the maximum height reached’ rather than explaining that air resistance opposes the motion of the particle or causes it to reduce its velocity more quickly. A few referred to the initial velocity being less which was incorrect since the speed of projection was given and was not part of the model. (e) There were many correct answers for possible limitations of the refined model in this part. The most popular included not accounting for spin, wind effects, dimensions of the stone and the inaccuracy of using g = 9.8 m s-2. Continuing to model the stone as a particle was an acceptable limitation, but those who thought that this meant mass was ignored did not gain the mark. Also, the ground not being entirely flat and horizontal was not credited since it was not part of the model. Only one limitation was required, and extra incorrect answers were penalised to avoid rewarding those who made a list of possible answers hoping that at least one might be correct. However, the instances of this happening appeared to be fewer than in previous years.
A small stone is projected from horizontal ground at 35 m s^-1 at angle alpha, where tan(alpha) = 3/4. Initially it is modelled as a particle moving freely under gravity with g = 9.8 m s^-2.
Edexcel 9MA0 · June 2023 · Paper 32 · Q1 · 3 marks
3 marks
Examiner report
This proved to be a very friendly starter. In part (a), almost all of the candidates were able to score the mark. In some rare instances it was lost for incorrect subsequent working such as 16 4 = . In the second part, the vast majority used a correct method with most using 2 1 2 s ut at = + but other suvat formulae were seen. Area under a v-t graph and integration were employed very occasionally but usually with success. The most common mistake was to assume constant velocity and a few made an error with the arithmetic and lost the final mark. A surprising number wrote 3.2 for acceleration but used 3.5 in their calculation. Candidates would be well advised to state the formula being used before doing their calculation, so that method marks can still be awarded in the event of an arithmetical slip.
A car is initially at rest on a straight horizontal road and then accelerates along the road with constant acceleration 3.2 m s^-2.
Edexcel 9MA0 · June 2023 · Paper 32 · Q3 · 9 marks
9 marks
Examiner report
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.
At time t seconds, where t >= 0, a particle P has velocity v = (t^2 - 3t + 7)i + (2t^2 - 3)j m s^-1.
Edexcel 9MA0 · June 2023 · Paper 32 · Q4 · 10 marks
10 marks
Examiner report
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, proved to be significantly more challenging. A few candidates wrote down a correct vector equation but were unable to make any further progress without focussing on separate components. Those who used integration to find r were often more successful, because the initial position of A was the integration constant. Of those who used r = ut + 0.5at2, many forgot to include the initial position of A and/or made a sign error and lost marks in both (b) and (c). In part (b), occasionally the quadratic equation was solved correctly but the wrong value of T was selected and in part (c) there was often a sign error on the 30 term. It was very rare to see candidates working with B as their initial position.
In this question i and j are horizontal unit vectors and position vectors are relative to a fixed origin O. A particle P moves on a smooth horizontal plane with constant acceleration (2.4i + j) m s^-2. At t = 0 it passes through A with velocity (-16i - 3j) m s^-1. At t = 5 seconds it passes through B. The position vector of A is (44i - 10j) m. At t = T seconds, where T > 5, P passes through C, whose position vector is (4i + cj) m.
Edexcel 9MA0 · June 2023 · Paper 32 · Q5 · 11 marks
11 marks
Examiner report
In part (a), the vast majority of candidates recognised that the horizontal component of velocity of the particle remained constant throughout the motion; this was then used, together with the horizontal distance, to form an equation in T and α and to derive the given result. The second part, however, provided a greater challenge and, although many realised that they needed to consider the vertical motion, a few did not know how to proceed, with some solving the given quadratic in tanα rather deriving it. However, many did make progress, using 2 1 2 s ut at = + with a vertical distance of 20 although there were occasional sign errors. They mostly then substituted the expression for T from part (a) to obtain an equation in α although slips were not uncommon at this stage. Those starting with different suvat equations tended to make little significant progress. To achieve the given result it was necessary to use the trig identity sec2 α = 1 + tan2 α (or equivalent work) to obtain an unsimplified quadratic equation in tan α before reaching the given equation. This was the main stumbling block for many and it was not unusual to see candidates make more than one attempt. Since the final equation was given and the demand of the question was “show that”, it was essential that all steps in the working were correct and the final answer was stated exactly as printed in order to achieve all the marks. In part (c) most candidates realised they should solve the given quadratic in tanα to give two possible angles of projection. Consideration of the situation should have indicated that the larger of the two angles would lead to the greatest possible height. Although many recognised that the greatest height is reached when the vertical component of velocity is zero, some chose to use the smaller 45o angle. Some found the greatest heights for both angles but then failed to select the relevant one. Other errors included using the time to greatest height as half the answer in part (a) or choosing sin α = 1 to give the maximum height in an otherwise correct suvat equation. Rounding errors were not uncommon leaving the final answer incorrect to 3 significant figures. In the final part, most candidates were able to suggest an acceptable limitation of the model such as it not including spin, dimensions of the ball, effect of wind or a more accurate value for g. The most common incorrect answers were that the model did not include weight/mass of the ball and the ground was not perfectly flat which was not specified as a modelling assumption in the question.
A small ball is projected from point O on horizontal ground with speed 28 m s^-1 at angle alpha above the ground. After T seconds it passes through A, which is 40 m horizontally and 20 m vertically from O. The ball is modelled as a particle moving freely under gravity.
Edexcel 9MA0 · June 2022 · Paper 32 · Q1 · 8 marks
8 marks
Examiner report
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 2sf or better. Common mistakes were to stop once the velocity had been found, losing the i and j and then finding 12-6√2 rather than apply Pythagoras. A small number 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 to find the acceleration vector. Common mistakes were losing either or both i and j and not recovering, integrating instead of differentiating and, despite the question specifying the answer was to be given in i j form, final answers given in column vector form were occasionally seen. Many candidates integrated to achieve the first two marks in part (c), but not all used the displacement at t = 4 to find the constant of integration. Those that found a vector equation containing a constant of integration C, went on to find C and then substituted t = 1 to reach the correct answer although some responses contained numerical errors involving the negative signs. A common error was to use suvat formulae or r = r0 + vt. rather than integrate.
Relative to a fixed origin, at time t seconds where t > 0 a particle P has velocity v = 3t^2 i - 6t^(1/2) j m s^-1. At t = 4 seconds its position vector is (i - 4j) m.
Edexcel 9MA0 · June 2022 · Paper 32 · Q3 · 9 marks
9 marks
Examiner report
In part (a), the majority of candidates were able to achieve the first mark for adding the two forces and collecting the i and j components. Many just equated this resultant force to the vector (3i + j) rather than using (3i + j) to define the direction. Some set up equations λ + 4 = 3 and µ - 1 = 1 and proceeded to solve them simultaneously gaining no further credit. Those who did use ratio usually went on to derive the required equation correctly; however, it should be remembered that, if required to derive a given answer, this must be written as a final conclusion in the form exactly as stated in the question. Another successful approach was to equate the i and j coefficients to 3k and k and solve as simultaneous equations. A small minority of candidates applied F = 4a or used integration to find v at this stage and then used ratio appropriately to obtain the required answer. In the second part, many candidates were able to find the resultant force and acceleration vectors. Some candidates found the magnitude of the force or acceleration at this stage and proceeded correctly with suvat formulae from there, while others continued with vectors. A number of candidates did not identify the initial velocity as zero, reading back to the information given at the start of the question being important here, so failed to use suvat formulae correctly. Use of u = 3i + j was not uncommon. Although most used suvat formulae, some successfully integrated the constant acceleration to find the velocity and displacement vectors. The majority of candidates knew how to find the magnitude of a vector using Pythagoras. The most significant errors seen included finding the resultant force but then using it as a displacement or velocity, showing a lack of understanding of the situation. A small number of candidates found an incorrect value for µ when substituting λ and there was occasional use of only one of the forces.
In this question i and j are horizontal unit vectors. A particle P of mass 4 kg is initially at rest at point A on a smooth horizontal plane. At t = 0, forces F1 = (4i - j) N and F2 = (lambda i + mu j) N are applied. P moves in the direction 3i + j. At t = 4 seconds it passes through B.
Edexcel 9MA0 · June 2022 · Paper 32 · Q5 · 12 marks
12 marks
Examiner report
In part (a), the majority of candidates found an equation for the horizontal motion in terms of t. Most then used this to substitute t =120/Ucosα into 0 = Usinαt – ½ gt2, the vertical motion equation for the whole trajectory. This was generally rearranged to give the required answer although cancelling the common factor of t from the quadratic before substitution led to a neater solution. A minority chose to consider just half the motion with v = u + at and v = 0 at the greatest height which also led to a neat solution. Some candidates failed to identify that the vertical displacement would be zero over the whole motion, and some were inconsistent in considering the time for the whole or half trajectory. A small but surprising number of answers were seen where 120 m was used as the vertical distance. It should again be emphasised that candidates, when required to derive a given result, should write the final answer in exactly the same form as that stated in the question. In the second part, many candidates correctly arrived at an equation in Usinα using v2 = u2 + 2as although a failure to resolve the vertical velocity giving U2 = 196 was not uncommon. Most were unable to progress further and so only gained 2 out of a possible 4 marks. For example, they just left it there or found the “correct” answer without any justification for the value of α. Some did not realise they could use the answer from part (a) and, of those who did, some struggled with the required algebra. Many who correctly identified the value of α as tan-1 (1/3) used the decimal value for the angle in degrees rather than calculating an exact value for sinα with which to substitute, and lost the final mark. A few candidates set up an equation by using s = ut + ½at2 with s = 10 and substituting in the time to travel 60 m horizontally, but again had trouble finishing off for the last 2 marks having more difficulty with the slightly more complicated algebra required. Part (c) required a comparison of the initial velocity in the model without air resistance (U) and the initial velocity in a model which took account of air resistance (V). Since the horizontal range and greatest height were fixed, the expected answer was that V > U as air resistance would have to be overcome. However, it was very common to see ‘U is greater because air resistance will decrease the velocity’ or something similar. A few perhaps confused V with the final speed and so said that U would be greater as air resistance would slow the final speed down. A correct possible refinement was often given in part (d). The most common correct suggestions were to take spin into account, consider the effect of wind or to model the particle as having dimensions. A fair number of candidates incorrectly believed that considering mass or weight was an acceptable answer and that a particle has no mass. Others made suggestions regarding the ground not being level but this, however, was not part of the model as defined in the question.
A golf ball starts at rest at A on horizontal ground. It is struck with initial speed U m s^-1 at angle alpha above the ground and first returns to the ground at B, where AB = 120 m. It is modelled as a particle moving freely under gravity.
Edexcel 9MA0 · October 2021 · Paper 32 · Q1 · 4 marks
4 marks
Examiner report
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. 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.
A particle P moves with constant acceleration (2i - 3j) m s^-2. At time t = 0, P has velocity 4i m s^-1 and position vector (i + j) m relative to a fixed origin O.
Edexcel 9MA0 · October 2021 · Paper 32 · Q2 · 12 marks
12 marks
Examiner report
Candidates need to make sure that they read and then answer questions carefully. In part (a) an equation of motion for A was required. Most realized that this required the use of Newton’s second law on the 3m mass. However, some ignored this requirement and went straight to a whole system approach to the problem. Errors in part (a) were mainly due to the omission of a relevant force or a misunderstanding of direction. It was essential that the frictional force on A and the tension were in the same direction i.e. had the same sign in the equation. In part (b) either an equation of motion for B was required or an equation of motion for the whole system. This was generally well answered provided that the reaction force from the plane on A had been found and used with F R µ = to give 1 6 F R = . The question was a “Show that” question so full marks were only awarded if there was a fully correct solution, with all the steps included. Errors involved sin/cos confusion, the omission of g, an incorrect direction for the reaction force, omission of a force in an equation of motion or an incorrect reaction such as R = 3mg. Some candidates did make algebraic errors in solving their equations. A few candidates assumed that the tension in the string was mg. Those candidates who attempted a whole system approach were mostly successful. In part (c), most of the graphs given were correct but there are still candidates who add a solid vertical line at the end of the graph and this lost the mark. However, the second mark here was frequently lost. All that was required was a comment that the acceleration was constant. For the final part, candidates needed to answer the question which asked for “ how this would affect the working in part (b)”. Those who referred to how it would affect the answer received no credit.
A small stone A of mass 3m rests on a rough plane inclined at angle alpha, where tan(alpha) = 3/4. It is connected by a light inextensible string, parallel to a line of greatest slope and passing over a small smooth pulley P at the top of the plane, to a freely hanging stone B of mass m. The coefficient of friction between A and the plane is 1/6. A is released from rest and moves down the plane, so B moves upwards, until B reaches the pulley. Both stones are modelled as particles.
Edexcel 9MA0 · October 2021 · Paper 32 · Q4 · 10 marks
10 marks
Examiner report
Candidates were instructed to use g = 10 in this question but were only penalised once if another value was used. In part (a), the majority of successful candidates employed the most efficient method using a single application of 2 1 2 s ut at = + to find the time to the ground. Other successful methods found the time to the top of the flight and then the time to go from the top to the ground and added. However, sign errors were common resulting in 2/4 marks being scored for this part. A few candidates did not resolve the speed of 65 1 m s−. This lost the M mark and consequently they scored zero for part (a). In part (b), candidates who used 𝑣𝑣2 = 𝑢𝑢2 + 2𝑎𝑎𝑎𝑎 to find the vertical component of velocity of the 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 1 m s−. For part (c), there were several possible answers, and candidates should always focus their attention on refining the model that has been used, in this case g = 10 and the stone being modelled as a particle i.e. the dimensions of the particle ignored (not the mass of the particle, which was a common wrong answer). Candidates should note that they should concentrate on giving a single correct answer to this type of question as extra incorrect answer(s) will lose the mark. The most common correct answer was ‘to use a more accurate value for g’.
A small stone is projected at 65 m s^-1 from a point O at the top of a vertical cliff. Point O is 70 m vertically above point N on horizontal ground. The stone is projected at angle alpha above the horizontal, where tan(alpha) = 5/12, and hits the ground at A. It is modelled as a particle moving freely under gravity with g = 10 m s^-2.
Edexcel 9MA0 · October 2021 · Paper 32 · Q5 · 14 marks
14 marks
Examiner report
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. Successful candidates employed a ratio method to obtain the equation = 2t and then 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 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 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. 1 2 3t
At time t seconds, where t > 0, a particle P has velocity v = 3t^(1/2)i - 2tj m s^-1. Its position vector relative to a fixed origin O is r metres, and when t = 1, r = -j.
Edexcel 9MA0 · October 2020 · Paper 32 · Q2 · 8 marks
8 marks
Examiner report
In the first part, the majority of candidates used v = u + at with t = 2 to obtain a correct velocity. Some chose to integrate, often successfully, although some failed to include a constant, in this case, the initial velocity. In part (b), most used r = ut + 1 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 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 for those carrying through a wrong value for T provided that it had been found from a valid method in part (b).
A particle P has constant acceleration (4i - 5j) m s^-2. At t = 0 it has velocity (-2i + 2j) m s^-1 and passes through the origin O. At time T > 0 it passes through A, whose position vector is (lambda i - 4.5j) m.
Edexcel 9MA0 · October 2020 · Paper 32 · Q3 · 12 marks
12 marks
Examiner report
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.
(i) For t >= 0, particle P has acceleration a = (1 - 4t)i + (3 - t^2)j m s^-2. At t = 0 its velocity is 36i m s^-1. (ii) For t >= 0, particle Q has position vector r = (t^2 - t)i + 3tj metres relative to a fixed origin O.
Edexcel 9MA0 · October 2020 · Paper 32 · Q5 · 11 marks
11 marks
Examiner report
In part (a), many candidates made some valid progress in creating horizontal and vertical distance equations in terms of t and U. Most tried to eliminate or find t to solve for U but sometimes lacked systematic working and gave up before reaching an answer or simply stated the given answer following incorrect or incomplete work. Those who did not complete a solution for U could go on to gain full marks for part (b) which was generally well attempted. Nearly all recognised that the vertical velocity was zero at the highest point. A few failed to use the vertical component of U and so did not achieve the method mark. Others assumed the highest point occurred when the horizontal distance was 50 m, half the range, which was not the case here since the start and end points were not on the same level. The most common error was in giving the answer as a height above the level of projection rather than above ground level. For part (c) there were certain statements in the question that were not part of the model. These included that AN =100 m. Therefore, if air resistance were included in the model, the ball would have to be thrown with a speed greater than 28 if it were to still reach the same point, A. The vast majority of candidates failed to realise this with most stating that it would be less. Although a reason was not required for the mark, many stated that it would be less because air resistance slows the ball down; however, this should have led to the conclusion that U should be greater and not less than 28. In part (d) one refinement, apart from air resistance, was required for this final mark. The most popular acceptable answers were to include spin, wind effects, use a more accurate value for g, take account of the size of the ball. The most common wrong answer was to take account of the mass/weight of the ball. Also, comments about the ground not being exactly horizontal or the angle of projection not exactly 45o did not relate to this specific model. It should be remembered that, when only one refinement is required, any extra answers would all have to be correct for the mark to be awarded.
A small ball is projected at speed U m s^-1 from point O at the top of a vertical cliff. O is 25 m vertically above point N on horizontal ground. The ball is projected at 45 degrees above the horizontal and hits the ground at A, where AN = 100 m. It is initially modelled as a particle moving freely under gravity, using g = 9.8 m s^-2.
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