Questions — OCR (4619 questions)

1 In this question you must show detailed reasoning. The number \(N\) is written as 28 A 3 B in base-12 form. Express \(N\) in decimal (base-10) form.

2 The points \(A\) and \(B\) have position vectors \(\mathbf { a }\) and \(\mathbf { b }\) relative to the origin \(O\). It is given that \(\mathbf { a } = \left( \begin{array} { c } 2
4
3 \lambda \end{array} \right)\) and \(\mathbf { b } = \left( \begin{array} { r } \lambda
- 4
6 \end{array} \right)\), where \(\lambda\) is a real parameter.

1. In the case when \(\lambda = 3\), determine the area of triangle \(O A B\).
2. Determine the value of \(\lambda\) for which \(\mathbf { a } \times \mathbf { b } = \mathbf { 0 }\).

3 The surface \(S\) has equation \(z = f ( x , y )\), where \(f ( x , y ) = 4 x ^ { 2 } y - 6 x y ^ { 2 } - \frac { 1 } { 12 } x ^ { 4 }\) for all real values of \(x\) and \(y\). You are given that \(S\) has a stationary point at the origin, \(O\), and a second stationary point at the point \(P ( a , b , c )\), where \(\mathrm { c } = \mathrm { f } ( \mathrm { a } , \mathrm { b } )\).

1. Determine the values of \(a , b\) and \(c\).
2. Throughout this part, take the values of \(a\) and \(b\) to be those found in part (a).
   1. Evaluate \(\mathrm { f } _ { x }\) at the points \(\mathrm { U } _ { 1 } ( \mathrm { a } - 0.1 , \mathrm {~b} , \mathrm { f } ( \mathrm { a } - 0.1 , \mathrm {~b} ) )\) and \(\mathrm { U } _ { 2 } ( \mathrm { a } + 0.1 , \mathrm {~b} , \mathrm { f } ( \mathrm { a } + 0.1 , \mathrm {~b} ) )\).
   2. Evaluate \(\mathrm { f } _ { y }\) at the points \(\mathrm { V } _ { 1 } ( \mathrm { a } , \mathrm { b } - 0.1 , \mathrm { f } ( \mathrm { a } , \mathrm { b } - 0.1 ) )\) and \(\mathrm { V } _ { 2 } ( \mathrm { a } , \mathrm { b } + 0.1 , \mathrm { f } ( \mathrm { a } , \mathrm { b } + 0.1 ) )\).
   3. Use the answers to parts (b)(i) and (b)(ii) to sketch the portions of the sections of \(S\), given by
      • \(z = f ( x , b )\), for \(| x - a | \leqslant 0.1\),
3. \(z = f ( a , y )\), for \(| y - b | \leqslant 0.1\).

4 The first five terms of the Fibonacci sequence, \(\left\{ \mathrm { F } _ { \mathrm { n } } \right\}\), where \(n \geqslant 1\), are \(F _ { 1 } = 1 , F _ { 2 } = 1 , F _ { 3 } = 2 , F _ { 4 } = 3\) and \(F _ { 5 } = 5\).

1. Use the recurrence definition of the Fibonacci sequence, \(\mathrm { F } _ { \mathrm { n } + 1 } = \mathrm { F } _ { \mathrm { n } } + \mathrm { F } _ { \mathrm { n } - 1 }\), to express \(\mathrm { F } _ { \mathrm { n } + 4 }\) in terms of \(\mathrm { F } _ { \mathrm { n } }\) and \(\mathrm { F } _ { \mathrm { n } - 1 }\).
2. Hence prove by induction that \(\mathrm { F } _ { \mathrm { n } }\) is a multiple of 3 when \(n\) is a multiple of 4 .

5 The set \(S\) consists of all \(2 \times 2\) matrices having determinant 1 or - 1 . For instance, the matrices \(\mathbf { P } = \left( \begin{array} { c c } \frac { 1 } { 2 } & \frac { \sqrt { 3 } } { 2 }
- \frac { \sqrt { 3 } } { 2 } & \frac { 1 } { 2 } \end{array} \right) , \mathbf { Q } = - \left( \begin{array} { c c } \frac { 1 } { 2 } & \frac { \sqrt { 3 } } { 2 }
- \frac { \sqrt { 3 } } { 2 } & \frac { 1 } { 2 } \end{array} \right)\) and \(\mathbf { R } = \left( \begin{array} { r r } 1 & 0
0 & - 1 \end{array} \right)\) are elements of \(S\). It is given that \(\times _ { \mathbf { M } }\) is the operation of matrix multiplication.

1. State the identity element of \(S\) under \(\times _ { \mathbf { M } }\). The group \(G\) is generated by \(\mathbf { P }\), under \(\times _ { \mathbf { M } }\).
2. Determine the order of \(G\). The group \(H\) is generated by \(\mathbf { Q }\) and \(\mathbf { R }\), also under \(\times _ { \mathbf { M } }\).
3. 1. By finding each element of \(H\), determine the order of \(H\).
   2. List all the proper subgroups of \(H\).
4. State whether each of the following statements is true or false. Give a reason for each of your answers.
   • \(G\) is abelian
   • \(G\) is cyclic
   • \(H\) is abelian
   • \(H\) is cyclic

6 For positive integers \(n\), let \(f ( n ) = 1 + 2 ^ { n } + 4 ^ { n }\).

1. 1. Given that \(n\) is a multiple of 3 , but not of 9 , use the division algorithm to write down the two possible forms that \(n\) can take.
   2. Show that when \(n\) is a multiple of 3 , but not of 9 , \(f ( n )\) is a multiple of 73 .
2. Determine the value of \(\mathrm { f } ( n )\), modulo 73 , in the case when \(n\) is a multiple of 9 .

7 In a long-running biochemical experiment, an initial amount of 1200 mg of an enzyme is placed into a mixture. The model for the amount of enzyme present in the mixture suggests that, at the end of each hour, one-eighth of the amount of enzyme that was present at the start of that hour is used up due to chemical reactions within the mixture. To compensate for this, at the end of each six-hour period of time, a further 500 mg of the enzyme is added to the mixture.

1. Let \(n\) be the number of six-hour periods that have elapsed since the experiment began. Explain how the amount of enzyme, \(\mathrm { E } _ { \mathrm { n } } \mathrm { mg }\), in the mixture is given by the recurrence system \(E _ { 0 } = 1200\) and \(E _ { n + 1 } = \left( \frac { 7 } { 8 } \right) ^ { 6 } E _ { n } + 500\) for \(n \geqslant 0\).
2. Solve the recurrence system given in part (a) to obtain an exact expression for \(\mathrm { E } _ { \mathrm { n } }\) in terms of \(n\).
3. Hence determine, in the long term, the amount of enzyme in the mixture. Give your answer correct to \(\mathbf { 3 }\) significant figures.
4. In this question you must show detailed reasoning. The long-running experiment is then repeated. This time a new requirement is added that the amount of enzyme present in the mixture must always be at least 500 mg . Show that the new requirement ceases to be satisfied before 12 hours have elapsed. \section*{END OF QUESTION PAPER} OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (\href{http://www.ocr.org.uk}{www.ocr.org.uk}) after the live examination series.
   If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.
   For queries or further information please contact The OCR Copyright Team, The Triangle Building, Shaftesbury Road, Cambridge CB2 8EA.
   OCR is part of Cambridge University Press \& Assessment, which is itself a department of the University of Cambridge.

1 The points \(A , B\) and \(C\) have position vectors \(\mathbf { a } = \left( \begin{array} { l } 3
0
0 \end{array} \right) , \mathbf { b } = \left( \begin{array} { l } 0
4
0 \end{array} \right)\) and \(\mathbf { c } = \left( \begin{array} { l } 0
0
1 \end{array} \right)\) respectively, relative to the origin \(O\).

1. 1. Calculate \(\mathbf { a } \times \mathbf { b }\), giving your answer as a multiple of \(\mathbf { c }\).
   2. Explain, geometrically, why \(\mathbf { a } \times \mathbf { b }\) must be a multiple of \(\mathbf { c }\).
2. Use a vector product method to calculate the area of triangle \(A B C\).

2 The surface \(S\) is given by \(z = x ^ { 2 } + 4 x y\) for \(- 6 \leqslant x \leqslant 6\) and \(- 2 \leqslant y \leqslant 2\).

1. 1. Write down the equation of any one section of \(S\) which is parallel to the \(x\)-z plane
   2. Sketch the section of (a)(i) on the axes provided in the Printed Answer Booklet.
2. Write down the equation of any one contour of \(S\) which does not include the origin.

3 For positive integers \(n\), the sequence of Fibonacci numbers, \(\left\{ \mathrm { F } _ { \mathrm { n } } \right\}\), starts with the terms \(F _ { 1 } = 1 , F _ { 2 } = 1 , F _ { 3 } = 2 , \ldots\) and is given by the recurrence relation \(\mathrm { F } _ { \mathrm { n } } = \mathrm { F } _ { \mathrm { n } - 1 } + \mathrm { F } _ { \mathrm { n } - 2 } ( \mathrm { n } \geqslant 3 )\).

1. Show that \(\mathrm { F } _ { 3 \mathrm { k } + 3 } = 2 \mathrm {~F} _ { 3 \mathrm { k } + 1 } + \mathrm { F } _ { 3 \mathrm { k } }\), where \(k\) is a positive integer.
2. Prove by induction that \(\mathrm { F } _ { 3 n }\) is even for all positive integers \(n\).

4

1. Let \(a = 1071\) and \(b = 67\).
   1. Find the unique integers \(q\) and \(r\) such that \(\mathrm { a } = \mathrm { bq } + \mathrm { r }\), where \(q > 0\) and \(0 \leqslant r < b\).
   2. Hence express the answer to (a)(i) in the form of a linear congruence modulo \(b\).
2. Use the fact that \(358 \times 715 - 239 \times 1071 = 1\) to prove that 715 and 1071 are co-prime.

5 A trading company deals in two goods. The formula used to estimate \(z\), the total weekly cost to the company of trading the two goods, in tens of thousands of pounds, is
\(z = 0.9 x + \frac { 0.096 y } { x } - x ^ { 2 } y ^ { 2 }\),
where \(x\) and \(y\) are the masses, in thousands of tonnes, of the two goods. You are given that \(x > 0\) and \(y > 0\).

1. In the first week of trading, it was found that the values of \(x\) and \(y\) corresponded to the stationary value of \(z\). Determine the total cost to the company for this week.
2. For the second week, the company intends to make a small change in either \(x\) or \(y\) in order to reduce the total weekly cost. Determine whether the company should change \(x\) or \(y\). (You are not expected to say by how much the company should reduce its costs.)

6 The set \(S\) consists of the following four complex numbers.
\(\begin{array} { l l l l } \sqrt { 3 } + \mathrm { i } & - \sqrt { 3 } - \mathrm { i } & 1 - \mathrm { i } \sqrt { 3 } & - 1 + \mathrm { i } \sqrt { 3 } \end{array}\)
For \(z _ { 1 } , z _ { 2 } \in S\), the binary operation \(\bigcirc\) is defined by \(z _ { 1 } \bigcirc z _ { 2 } = \frac { 1 } { 4 } ( 1 + i \sqrt { 3 } ) z _ { 1 } z _ { 2 }\).

1. 1. Complete the Cayley table for \(( S , \bigcirc )\) given in the Printed Answer Booklet.
   2. Verify that ( \(S , \bigcirc\) ) is a group.
   3. State the order of each element of \(( S , \bigcirc )\).
2. Write down the only proper subgroup of ( \(S , \bigcirc\) ).
3. 1. Explain why ( \(S , \bigcirc\) ) is a cyclic group.
   2. List all possible generators of \(( S , \bigcirc )\).

7

1. Let \(f ( n ) = 2 ^ { 4 n + 3 } + 3 ^ { 3 n + 1 }\). Use arithmetic modulo 11 to prove that \(\mathrm { f } ( n ) \equiv 0 ( \bmod 11 )\) for all integers \(n \geqslant 0\).
2. Use the standard test for divisibility by 11 to prove the following statements.
   1. \(10 ^ { 33 } + 1\) is divisible by 11
   2. \(10 ^ { 33 } + 1\) is divisible by 121

8 A sequence \(\left\{ \mathrm { u } _ { \mathrm { n } } \right\}\) is defined by the recurrence system
\(u _ { 1 } = 1\) and \(\mathrm { u } _ { \mathrm { n } + 1 } = \mathrm { a } - \frac { \mathrm { a } ^ { 2 } } { 2 \mathrm { u } _ { \mathrm { n } } }\) for \(n \geqslant 1\), where \(a\) is a positive constant.
Determine with justification the behaviour of the sequence for all possible values of \(a\). \section*{END OF QUESTION PAPER}

2
\includegraphics[max width=\textwidth, alt={}, center]{60f72141-4a99-4907-93b1-adb0cd66948e-2_211_1276_1427_365} Three particles \(A , B\) and \(C\) are free to move in the same straight line on a large smooth horizontal surface. Their masses are \(1.2 \mathrm {~kg} , 1.8 \mathrm {~kg}\) and \(m \mathrm {~kg}\) respectively (see diagram). The coefficient of restitution in collisions between any two of them is \(\frac { 3 } { 4 }\). Initially, \(B\) and \(C\) are at rest and \(A\) is moving with a velocity of \(4.0 \mathrm {~ms} ^ { - 1 }\) towards \(B\).
a) Show that immediately after the collision between \(A\) and \(B\) the speed of \(B\) is \(2.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
b) Find the velocity of \(A\) immediately after this collision.
\(B\) subsequently collides with \(C\).
c) Find, in terms of \(m\), the velocity of \(B\) after its collision with \(C\).
d) Given that the direction of motion of \(B\) is reversed by the collision with \(C\), find the range of possible values of \(m\). The car is attached to a trailer of mass 200 kg by a light rigid horizontal tow bar. The greatest steady speed of the car and trailer on the road is now \(30 \mathrm {~ms} ^ { - 1 }\). The resistance to motion of the trailer may also be assumed constant.
(b) Find the magnitude of the resistance force on the trailer. The car and trailer again travel along the road. At one instant their speed is \(15 \mathrm {~ms} ^ { - 1 }\) and their acceleration is \(0.57 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
(c) (i) Find the power of the engine of the car at this instant.
(ii) Find the magnitude of the tension in the tow bar at this instant. In a refined model of the motion of the car and trailer the resistance to the motion of each is assumed to be zero until they reach a speed of \(10 \mathrm {~ms} ^ { - 1 }\). When the speed is \(10 \mathrm {~ms} ^ { - 1 }\) or above the same constant resistance forces as in the first model are assumed to apply to each. The car and trailer start at rest on the road and accelerate, using maximum power.
(d) Without carrying out any further calculations,
(i) explain whether the time taken to attain a speed of \(20 \mathrm {~m} ^ { - 1 }\) would be predicted to be lower, the same or higher using the refined model compared with the original model,
(ii) explain whether the greatest steady speed of the system would be predicted to be lower, the same or higher using the refined model compared with the original model. \section*{Total Marks for Question Set 1: 31} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

• When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
• When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

• If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
• If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
• if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h] \end{table}

2 A particle moves in a straight line with constant acceleration. Its initial and final velocities are \(u\) and \(v\) respectively and at time \(t\) its displacement from its starting position is \(s\). An equation connecting these quantities is \(s = k \left( u ^ { \alpha } + v ^ { \beta } \right) t ^ { \gamma }\), where \(k\) is a dimensionless constant.

1. Use dimensional analysis to find the values of \(\alpha , \beta\) and \(\gamma\).
2. By considering the case where the acceleration is zero, determine the value of \(k\).

3
Two particles \(A\) and \(B\) are connected by a light inextensible string. Particle \(A\) has mass 1.2 kg and moves on a smooth horizontal table in a circular path of radius 0.6 m and centre \(O\). The string passes through a small smooth hole at \(O\). Particle \(B\) moves in a horizontal circle in such a way that it is always vertically below \(A\). The angle that the portion of the string below the table makes with the downwards vertical through \(O\) is \(\theta\), where \(\cos \theta = \frac { 4 } { 5 }\) (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{75f629e7-969d-43ae-8222-031875ae54ae-02_453_696_1571_552}

1. Find the time taken for the particles to perform a complete revolution.
2. Find the mass of \(B\). \section*{Total Marks for Question Set 2: 29} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
   b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
   A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
   c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
   d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
   e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
   • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
   • When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
   Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
   Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
   f Rules for replaced work and multiple attempts:
   • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
   • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
   • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
   For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
   If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Abbreviations}

   Abbreviations used in the mark scheme	Meaning
   dep*	Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
   cao	Correct answer only
   ое	Or equivalent
   rot	Rounded or truncated
   soi	Seen or implied
   www	Without wrong working
   AG	Answer given
   awrt	Anything which rounds to
   BC	By Calculator
   DR	This question included the instruction: In this question you must show detailed reasoning.

   \end{table}

   Question		Answer	Marks	AOs	Guidance
   \multirow[t]{8}{*}{2}	\multirow[t]{8}{*}{(a)}	\(\alpha = \beta\)	B1	2.2a	soi - does not need justification	\multirow{8}{*}{Allow \(\mathrm { L } = \mathrm { L } ^ { \alpha } \mathrm { T } ^ { - \alpha } \mathrm { T } ^ { \gamma } + \mathrm { L } ^ { \beta } \mathrm { T } ^ { - \beta } \mathrm { T } ^ { \gamma }\) with consistent indices, must be expanded, use BOD}
   		\([ \mathrm { u } ] = \mathrm { LT } ^ { - 1 }\) or \([ \mathrm { v } ] = \mathrm { LT } ^ { - 1 }\)	B1	3.3	Seen
   		\(\mathrm { L } = \mathrm { L } ^ { \alpha } \mathrm { T } ^ { - \alpha } \mathrm { T } ^ { \gamma }\) or \(\mathrm { L } ^ { \alpha } \mathrm { T } ^ { \gamma - \alpha }\)	M1	1.1a	No \(k\)
   					Could be \(\beta\)
   		\(\alpha = 1\)	A1	1.1	or \(\beta = 1\)
   		\(\gamma - \alpha = 0\)	M1	3.4
   		\(\gamma = 1\) and \(\beta = 1\)	A1	1.1	or \(\alpha = 1\) if \(\beta\) found
   			[6]
   	(b)	If \(a = 0\) then \(u = v\) and \(s = 2 k u t . .\). ...but "dist \(=\) speed × time" so \(k = 1 / 2\)	M1	2.1		\multirow{3}{*}{Do not accept use of prior knowledge of uvast}
   			A1	2.2a	Must include justification
   			[2]

   Question		Answer	Marks	AOs	Guidance
   3	(a)	\(\begin{aligned} T \cos \theta = m _ { B } g T \sin \theta = m _ { B } \times 0.6 \times \omega ^ { 2 } \tan \theta = \left( 0.6 \omega ^ { 2 } \right) / g \tan \theta = \frac { 3 } { 4 } \mathrm { oe } \omega = 3.5 t = \frac { 2 \pi } { 3.5 } \end{aligned}\) Time for one revolution is 1.8 seconds	\(\begin{aligned} \text { M1* } \text { M1* } \end{aligned}\) M1dep B1 A1 M1	3.1b 3.3 3.1b 1.1 1.1 1.1 3.2a	Balancing vertical forces on \(B\) NII for \(B\) with \(r = 0.6\) (could use \(v ^ { 2 }\) / 0.6) Combining equations and eliminating \(T\) May be implied. Accept \(\theta = 36.9\) Their 3.5	Or \(t = 2 \pi r / v \quad ( v = 2.1 \mathrm {~m} / \mathrm { s } )\)
   	(b)	\(T = 1.2 \times 0.6 \omega ^ { 2 } ( = 8.82 )\) \(8.82 \cos \theta = m _ { B } g\) or \(8.82 \sin \theta = m _ { B } \times 0.6 \omega ^ { 2 }\) \(m _ { B } = 0.72\)	M1 М1 A1 [3]	2.2a 3.1b 1.1	NII for \(A\) and for realising that \(\omega\) is the same for \(A\) and \(B\). Substituting their \(T\) into either of their equations of motion for \(B\).	Could be seen in (a)

1
\includegraphics[max width=\textwidth, alt={}, center]{d6a0d7a6-4166-4c26-a461-39b2414c0412-02_494_390_251_255} A smooth wire is shaped into a circle of radius 2.5 m which is fixed in a vertical plane with its centre at a point \(O\). A small bead \(B\) is threaded onto the wire. \(B\) is held with \(O B\) vertical and is then projected horizontally with an initial speed of \(8.4 \mathrm {~ms} ^ { - 1 }\) (see diagram).

1. Find the speed of \(B\) at the instant when \(O B\) makes an angle of 0.8 radians with the downward vertical through \(O\).
2. Determine whether \(B\) has sufficient energy to reach the point on the wire vertically above \(O\).

2 A student is studying the speed of sound, \(u\), in a gas under different conditions.
He assumes that \(u\) depends on the pressure, \(p\), of the gas, the density, \(\rho\), of the gas and the wavelength, \(\lambda\), of the sound in the relationship \(u = k p ^ { \alpha } \rho ^ { \beta } \lambda ^ { \gamma }\), where \(k\) is a dimensionless constant. (The wavelength of a sound is the distance between successive peaks in the sound wave.)

1. Use the fact that density is mass per unit volume to find \([ \rho ]\).
2. Given that the units of \(p\) are \(\mathrm { Nm } ^ { - 2 }\), determine the values of \(\alpha , \beta\) and \(\gamma\).
3. Comment on what the value of \(\gamma\) means about how fast sounds of different wavelengths travel through the gas. The student carries out two experiments, \(A\) and \(B\), to measure \(u\). Only the density of the gas varies between the experiments, all other conditions being unchanged. He finds that the value of \(u\) in experiment \(B\) is double the value in experiment \(A\).
4. By what factor has the density of the gas in experiment \(A\) been multiplied to give the density of the gas in experiment \(B\) ? Particles \(A\) of mass \(2 m\) and \(B\) of mass \(m\) are on a smooth horizontal floor. \(A\) is moving with speed \(u\) directly towards a vertical wall, and \(B\) is at rest between \(A\) and the wall (see diagram).
   \includegraphics[max width=\textwidth, alt={}, center]{d6a0d7a6-4166-4c26-a461-39b2414c0412-03_211_795_285_244}
   \(A\) collides directly with \(B\). The coefficient of restitution in this collision is \(\frac { 1 } { 2 }\).
   \(B\) then collides with the wall, rebounds, and collides with \(A\) for a second time.
5. Show that the speed of \(B\) after its second collision with \(A\) is \(\frac { 1 } { 2 } u\). The first collision between \(A\) and \(B\) occurs at a distance \(d\) from the wall. The second collision between \(A\) and \(B\) occurs at a distance \(\frac { 1 } { 5 } d\) from the wall.
6. Find the coefficient of restitution for the collision between \(B\) and the wall. \section*{Total Marks for Question Set 3: 28} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
   b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
   A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
   c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
   d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
   e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
   • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
   • When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
   Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
   Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
   f Rules for replaced work and multiple attempts:
   • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
   • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
   • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
   For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
   If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Abbreviations}

   Abbreviations used in the mark scheme	Meaning
   dep*	Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
   cao	Correct answer only
   ое	Or equivalent
   rot	Rounded or truncated
   soi	Seen or implied
   www	Without wrong working
   AG	Answer given
   awrt	Anything which rounds to
   BC	By Calculator
   DR	This question included the instruction: In this question you must show detailed reasoning.

   \end{table}

   Question			Answer	Marks	AOs	Guidance
   1	(a)		Initial (kinetic) energy \(= \frac { 1 } { 2 } \times m \times 8.4 ^ { 2 }\) Energy at \(0.8 \mathrm { rad } = \frac { 1 } { 2 } m v ^ { 2 } + m \times 9.8 \times 2.5 ( 1 - \cos 0.8 )\) \(=\) Initial energy \(v ^ { 2 } = 55.698 \ldots \Rightarrow\) speed is \(7.46 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)	B1 M1 A1 [3]	1.1a 1.1 1.1	\(35.28 m\) Attempt to find \(\mathrm { KE } + \mathrm { PE }\) at 0.8 rad (or \(45.8 ^ { \circ }\) ) and equate to initial kinetic energy (KE must use correct formula) \(\mathrm { NB } \Delta h = 0.7582 \ldots\) SC1 for use of constant acceleration without justification	\(m\) may be implied \(\frac { 1 } { 2 } m v ^ { 2 } + 7.4306 \ldots m\) Or subtract PE from initial KE (to give final KE) (Final KE is \(27.849 . . . m\) )
   1	(b)		Minimum energy to reach top \(= m \times 9.8 \times ( 2 \times 2.5 )\) \(= 49 \mathrm {~m}\) \(49 m > 35.28 m\) so insufficient energy to reach top	M1 A1 A1ft	1.1a 1.1 2.2a	Or attempt to find angle when \(v = 0 35.28 m = 24.5 m ( 1 - \cos \theta ) \left( + \frac { 1 } { 2 } m ( 0 ) ^ { 2 } \right)\) Condone missing \(m \theta = 2.03 \mathrm { rads }\) or \(116 ^ { \circ }\) Comparison between their numerical multiples of \(m\) ( \(m\) could be missing) Allow \(\neq\) and consistent ft conclusion	Or attempt to find \(h\) when \(v = 0 \left( h = \frac { 35.28 } { g } \right)\) \(h = 3.6\) or comparison of their angle with \(2 \pi\) or \(180 ^ { \circ }\) Or show that \(h = 3.6 < 5\) or show that \(v ^ { 2 } = - 27.44 < 0\) (is not valid)

   2	(a)		\([ \rho ] = \mathrm { ML } ^ { - 3 }\)	B1 [1]	3.3	If M, L and T not used B0, but do not penalise any further instances of non-standard notation as long as it is used consistently.
   2	(b)		\([ p ] = \mathrm { MLT } ^ { - 2 } \mathrm {~L} ^ { - 2 } = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 }\) \(\mathrm { LT } ^ { - 1 } = \mathrm { M } ^ { \alpha } \mathrm { L } ^ { - \alpha } \mathrm { T } ^ { - 2 \alpha } \mathrm { M } ^ { \beta } \mathrm { L } ^ { - 3 \beta } \mathrm {~L} ^ { \gamma }\) M: \(\alpha + \beta = 0\) \(\begin{aligned} \text { T: } - 2 \alpha = - 1 \alpha = \frac { 1 } { 2 } , \beta = - \frac { 1 } { 2 } \end{aligned}\) \(\mathrm { L } : \quad 1 = - \alpha - 3 \beta + \gamma\) \(\gamma = 0 \quad\) www	B1 B1ft M1 M1 A1 M1 A1	2.1 3.3 3.3 3.4 3.4 1.1 1.1 3.4 1.1	Allow this mark as long as the equations for M and T are correct.	Do not allow any marks for using addition instead of multiplication

   2	(c)	Sounds of any wavelength have the same speed through the gas	E1FT	2.2b	Follow from their \(\gamma\) : If \(\gamma > 0\) then speed increases as wavelength increases (or better - e.g. \(\gamma = 1 / 2 \rightarrow\) speed is proportional to \(\sqrt { } \lambda\) ); if \(\gamma < 0\) then speed decreases as wavelength increases (or better)
   			[1]
   2	(d)	\(u \propto \frac { 1 } { \sqrt { \rho } } \text { or } u = k \sqrt { \frac { p } { \rho } } \text { oe }\) \(\frac { 1 } { 4 }\)	M1 A1 [2]	3.4 1.1	\(2 = \left( \frac { \rho _ { B } } { \rho _ { A } } \right) ^ { - \frac { 1 } { 2 } }\) Award if no working seen provided \(\beta = - \frac { 1 } { 2 }\)	Using their value of \(\beta\) Allow missing k or use of \(=\) instead of proportion symbol

   3	(a)	\(1 ^ { \text {st } }\) collision for \(A \ B : 2 m u = 2 m v _ { A } + m v _ { B }\) \(\frac { 1 } { 2 } = \frac { v _ { B } - v _ { A } } { u }\) \(v _ { A } = \frac { 1 } { 2 } u\) \(2 ^ { \text {nd } }\) collision for \(A \ B : 2 m \times \frac { 1 } { 2 } u + m U _ { B } = 2 m V _ { A } + m V _ { B }\) \(\begin{aligned} \frac { 1 } { 2 } = \frac { V _ { B } - V _ { A } } { \frac { 1 } { 2 } u - U _ { B } } u + U _ { B } = 2 V _ { A } + V _ { B } \text { and } u - 2 U _ { B } = 4 V _ { B } - 4 V _ { A } \Rightarrow 3 u = 6 V _ { B } \Rightarrow V _ { B } = \frac { 1 } { 2 } u \end{aligned}\)	M1 М1 A1 М1 M1 A1 [6]	3.1b 1.1a 1.1 1.1 1.1 2.1	Conservation of momentum Restitution Conservation of momentum Restitution AG Intermediate work towards cancellation must be seen	May see \(- U _ { B }\) or \(\pm e u\) Do not allow assumed value of \(U _ { B }\) e.g. \(\frac { 1 } { 2 } u\) or \(u\). Do not allow assumed value of \(U _ { B }\) e.g. \(\frac { 1 } { 2 } u\) or \(u\). SC1 if assumed value for \(V _ { B }\) has been used (giving M0M0), provided \(\left| U _ { B } \right| \leq u\), direction of travel is towards A and equations are otherwise correct.
   3	(b)	\(v _ { B } = u\) Collision for \(B \ \) wall: \(e = \pm \frac { U _ { B } } { u }\) or \(U _ { B } = \pm e u\) \(\frac { \frac { 4 } { 5 } d } { \frac { 1 } { 2 } u } = \frac { d } { u } + \frac { \frac { 1 } { 5 } d } { e u }\) \(\frac { 3 } { 5 } = \frac { 1 } { 5 e }\) So coefficient of restitution between \(B\) and wall is \(\frac { 1 } { 3 }\)	B1 M1 M1 M1 A1 [5]	1.1 3.1b 3.1b 1.1 3.2a	Restitution May see \(V _ { 2 B }\) or similar instead of \(\pm e u\) with use of restitution at the end. Seeing that \(A\) travels \(\frac { 4 } { 5 } d\) at \(\frac { 1 } { 2 } u\) in the same time as \(B\) travels \(d\) at \(u\) and \(\frac { 1 } { 5 } d\) at \(e u\) Correctly cancelling \(d\) and \(u\) and simplifying their 3 term equation including e in the denominator	Award if seen in (a). Award if seen in (a) Do not allow assumed rebound speed

1 A particle \(A\) of mass 3.6 kg is attached by a light inextensible string to a particle \(B\) of mass 2.4 kg .
\(A\) and \(B\) are initially at rest, with the string slack, on a smooth horizontal surface. \(A\) is projected directly away from \(B\) with a speed of \(7.2 \mathrm {~ms} ^ { - 1 }\).

1. Calculate the speed of \(A\) after the string becomes taut.
2. Find the impulse exerted on \(A\) at the instant that the string becomes taut.
3. Find the loss in kinetic energy as a result of the string becoming taut.

2 A car of mass 1500 kg has an engine with maximum power 60 kW . When the car is travelling at \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along a straight horizontal road using maximum power, its acceleration is \(3.3 \mathrm {~ms} ^ { - 2 }\). In an initial model of the motion of the car it is assumed that the resistance to motion is constant.

1. Using this initial model, find the greatest possible steady speed of the car along the road. In a refined model the resistance to motion is assumed to be proportional to the speed of the car.
2. Using this refined model, find the greatest possible steady speed of the car along the road. The greatest possible steady speed of the car on the road is measured and found to be \(21.6 \mathrm {~ms} ^ { - 1 }\).
3. Explain what this value means about the models used in parts (a) and (b).
   \includegraphics[max width=\textwidth, alt={}, center]{aa25b8a6-9a5a-4de2-9534-18db8a175c34-03_583_378_169_255} As shown in the diagram, \(A B\) is a long thin rod which is fixed vertically with \(A\) above \(B\). One end of a light inextensible string of length 1 m is attached to \(A\) and the other end is attached to a particle \(P\) of mass \(m _ { 1 } \mathrm {~kg}\). One end of another light inextensible string of length 1 m is also attached to \(P\). Its other end is attached to a small smooth ring \(R\), of mass \(m _ { 2 } \mathrm {~kg}\), which is free to move on \(A B\). Initially, \(P\) moves in a horizontal circle of radius 0.6 m with constant angular velocity \(\omega \mathrm { rads } ^ { - 1 }\). The magnitude of the tension in string \(A P\) is denoted by \(T _ { 1 } \mathrm {~N}\) while that in string \(P R\) is denoted by \(T _ { 2 } \mathrm {~N}\).
4. By considering forces on \(R\), express \(T _ { 2 }\) in terms of \(m _ { 2 }\).
5. Show that
   1. \(T _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right)\),
   2. \(\omega ^ { 2 } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } }\).
6. Deduce that, in the case where \(m _ { 1 }\) is much bigger than \(m _ { 2 } , \omega \approx 3.5\). In a different case, where \(m _ { 1 } = 2.5\) and \(m _ { 2 } = 2.8 , P\) slows down. Eventually the system comes to rest with \(P\) and \(R\) hanging in equilibrium.
7. Find the total energy lost by \(P\) and \(R\) as the angular velocity of \(P\) changes from the initial value of \(\omega \mathrm { rads } ^ { - 1 }\) to zero. \section*{Total Marks for Question Set 4: 32} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
   b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
   A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
   c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
   d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
   e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
   • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
   • When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
   Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
   Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
   f Rules for replaced work and multiple attempts:
   • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
   • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
   • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
   For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
   If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Abbreviations}

   Abbreviations used in the mark scheme	Meaning
   dep*	Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
   cao	Correct answer only
   ое	Or equivalent
   rot	Rounded or truncated
   soi	Seen or implied
   www	Without wrong working
   AG	Answer given
   awrt	Anything which rounds to
   BC	By Calculator
   DR	This question included the instruction: In this question you must show detailed reasoning.

   \end{table}

   1	(a)	\(3.6 \times 7.2 = 3.6 v _ { A } + 2.4 v _ { B }\) \(v _ { A } = v _ { B }\) \(4.32 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)	M1 M1 A1 [3]	1.1a 1.1 1.1	Conservation of momentum soi May be -4.32 if the initial velocity is counted as negative.	(25.92)
   1	(b)	\(\pm 3.6 \times 4.32 \mp 3.6 \times 7.2\) \(- 10.4 \mathrm { Ns } \left( \right.\) or \(\left. \mathrm { kg } \mathrm { m } \mathrm { s } ^ { - 1 } \right)\)	M1 A1 [2]	1.1a 1.1	Using their 4.32 from 2(a) provided c.o.m. used Or 10.4 Ns s towards \(B\) Must be opposite sign to the initial velocity.	Or \(- ( 2.4 \times 4.32 )\) Deduct final mark if correct direction not soi
   1	(c)	\(\pm \left( \frac { 1 } { 2 } \times 3.6 \times 7.2 ^ { 2 } - \frac { 1 } { 2 } \times ( 3.6 + 2.4 ) \times 4.32 ^ { 2 } \right)\) 37.3 J	M1 A1 [2]	1.1a 1.1	Using their 4.32 from 2(a) provided c.o.m. used Allow one slip in substitution other than sign error; must have 3 terms Allow -37.3J	\(93.31 \ldots - ( 33.59 \ldots +\) 22.39 ...)

   \begin{center}

   2	(a)	\begin{tabular}{l} \(\frac { 60000 } { 10 } - R = 1500 \times 3.3\)
   \(R = 1050\) \(\frac { 60000 } { v } = 1050\)
   The greatest speed is \(57.1 \mathrm {~ms} ^ { - 1 }\)

   &

   M1

   A1 M1

   A1 [4]

   &

   3.3

   1.1 3.4

   1.1

   &

   = 4950

   May be -1050

   &
   \hline 2 & (b) &

   \(\frac { 60000 } { 10 } - k \times 10 = 1500 \times 3.3 k = 105\)

   \(\frac { 60000 } { v } = 105 v\) \(v ^ { 2 } = 571.4 \ldots\)

   \(v = 23.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)

   &

   M1

   A1

   M1

   A1

   A1

   [5]

   &

   3.3

   1.1

   3.4

   1.1

   1.1

   &

   Or \(1050 = 10 k\)

   Must be positive

   &
   \hline 2 & (c) &

   The constant resistance model does not seem to be very accurate

   The refined (linear) model (is not perfect but) gives a much more accurate answer than the constant resistance model

   &

   B1ft

   B1ft

   & \begin{tabular}{l} 3.5a

4
2.4
\end{tabular} &

B1 for each of two correct statements about the models.

If commenting on the accuracy of (a), must emphasise that (a) is very inaccurate or at least quite inaccurate

Do not allow e.g.

- model (a) is not very effective

- Neither model is accurate

- (a) and (b) are not very accurate

Clear comparison between the accuracy of the two models (must emphasise that (b) is fairly accurate or considerably more accurate than (a)), or other suitable distinct second comment

Do not allow e.g.

- model (b) is more accurate than model (a)

- (b) is not accurate

Do not allow statement claiming that resistance is proportional to speed, or to speed \({ } ^ { 2 }\)

&

Suitable comments for (a):

- is very inaccurate

- predicted speed is nearly three times the actual value

- constant resistance is not a suitable model

- both models underestimate the resistance (as top speed is lower than expected)

For the linear model (b)

- is fairly accurate (but probably underestimates the resistance at higher speeds)

- resistance is not proportional to speed but is a much better model than constant resistance

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