Questions (30808 questions)

2. Prove by induction that, for all positive integers \(n\), $$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { 1 } { 3 } n \left( 4 n ^ { 2 } - 1 \right)$$ [BLANK PAGE]

3. The figure below shows the curve with cartesian equation \(\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = x y\).
\includegraphics[max width=\textwidth, alt={}, center]{f42517a5-d7ed-40f3-bb04-faea97d4b19b-08_830_997_228_262}

1. Show that the polar equation of the curve is \(r ^ { 2 } = a \sin b \theta\), where \(a\) and \(b\) are positive constants to be determined.
2. Determine the exact maximum value of \(r\).
3. Determine the area enclosed by one of the loops.
   [0pt] [BLANK PAGE] \section*{4. In this question you must show detailed reasoning.}
   1. The curves with equations $$y = \frac { 3 } { 4 } \sinh x \text { and } y = \tanh x + \frac { 1 } { 5 }$$ intersect at just one point \(P\)
4. Use algebra to show that the \(x\) coordinate of \(P\) satisfies the equation $$15 \mathrm { e } ^ { 4 x } - 48 \mathrm { e } ^ { 3 x } + 32 \mathrm { e } ^ { x } - 15 = 0$$
5. Show that \(\mathrm { e } ^ { x } = 3\) is a solution of this equation.
6. Hence state the exact coordinates of \(P\).
   (ii) Show that $$\int _ { - 4 } ^ { 0 } \frac { \mathrm { e } ^ { \frac { 1 } { x } } } { x ^ { 2 } } \mathrm {~d} x = \mathrm { e } ^ { - \frac { 1 } { 4 } }$$ [BLANK PAGE]

5. Use the method of differences to prove that for \(n > 2\) $$\sum _ { r = 2 } ^ { n } \frac { 4 } { r ^ { 2 } - 1 } = \frac { ( p n + q ) ( n - 1 ) } { n ( n + 1 ) }$$ where \(p\) and \(q\) are constants to be determined.
[0pt] [BLANK PAGE]

6.

1. $$z _ { 1 } = a + b \mathrm { i } \text { and } z _ { 2 } = c + d \mathrm { i }$$ where \(a , b , c\) and \(d\) are real constants.
   Given that
   • \(b > d\)
   • \(z _ { 1 } + z _ { 2 }\) is real
   • \(\left| z _ { 1 } \right| = \sqrt { 13 }\)
   • \(\left| z _ { 2 } \right| = 5\)
   • \(\operatorname { Re } \left( z _ { 2 } - z _ { 1 } \right) = 2\)
     show that \(a = 2\) and determine the value of each of \(b , c\) and \(d\)
   • (a) On the same Argand diagram
   • sketch the locus of points \(z\) which satisfy \(| z - 12 | = 7\)
   • sketch the locus of points \(w\) which satisfy \(| w - 5 \mathrm { i } | = 4\) showing the coordinates of any points of intersection with the axes.
     (b) Determine the range of possible values of \(| z - w |\)
     [0pt] [BLANK PAGE]

1. In this question you must show detailed reasoning.

Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 2 } { x ^ { 2 } - x + 1 } \mathrm {~d} x\). Give your answer in exact form.
[0pt] [BLANK PAGE] \section*{8. In this question you must show detailed reasoning.} The diagram shows the curve with equation \(y = \frac { x + 3 } { \sqrt { x ^ { 2 } + 9 } }\).
\includegraphics[max width=\textwidth, alt={}, center]{f42517a5-d7ed-40f3-bb04-faea97d4b19b-18_890_1010_367_242} The region R , shown shaded in the diagram, is bounded by the curve, the \(x\)-axis, the \(y\)-axis, and the line \(x = 4\).

1. Determine the area of R . Give your answer in the form \(p + \ln q\) where \(p\) and \(q\) are integers to be determined. The region R is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Determine the volume of the solid of revolution formed. Give your answer in the form \(\pi \left( a + b \ln \left( \frac { c } { d } \right) \right)\) where \(a , b , c\) and \(d\) are integers to be determined.
   [0pt] [BLANK PAGE]

10 November 2025 Instructions

• Answer all the questions.
• Use black or blue ink. Pencil may be used for graphs and diagrams only.
• There are blank pages at the end of the paper for additional working. You must clearly indicate when you have moved onto additional pages on the question itself. Make sure to include the question number.
• You are permitted to use a scientific or graphical calculator in this paper.
• Where appropriate, your answer should be supported with working. Marks might be given for using a correct method, even if your answer is wrong.
• Give non-exact numerical answers correct to 3 significant figures unless a different degree of accuracy is specified in the question.
• The acceleration due to gravity is denoted by \(g \mathrm {~ms} ^ { - 2 }\). When a numerical value is needed use \(g = 9.8\) unless a different value is specified in the question.

Information

• The total mark for this paper is \(\mathbf { 7 5 }\) marks.
• The marks for each question are shown in brackets.
• You are reminded of the need for clear presentation in your answers.
• You have \(\mathbf { 7 5 }\) minutes for this paper.

\section*{Arithmetic series} \(S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}\) \section*{Geometric series} \(S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }\)
\(S _ { \infty } = \frac { a } { 1 - r }\) for \(| r | < 1\) \section*{Binomial series} \(( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )\),
where \({ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }\)
\(( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )\) \section*{Series} \(\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 ) , \sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }\) \section*{Maclaurin series} \(\mathrm { f } ( x ) = \mathrm { f } ( 0 ) + \mathrm { f } ^ { \prime } ( 0 ) x + \frac { \mathrm { f } ^ { \prime \prime } ( 0 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { \mathrm { f } ^ { ( r ) } ( 0 ) } { r ! } x ^ { r } + \ldots\)
\(\mathrm { e } ^ { x } = \exp ( x ) = 1 + x + \frac { x ^ { 2 } } { 2 ! } + \ldots + \frac { x ^ { r } } { r ! } + \ldots\) for all \(x\)
\(\ln ( 1 + x ) = x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 } - \ldots + ( - 1 ) ^ { r + 1 } \frac { x ^ { r } } { r } + \ldots ( - 1 < x \leq 1 )\)
\(\sin x = x - \frac { x ^ { 3 } } { 3 ! } + \frac { x ^ { 5 } } { 5 ! } - \ldots + ( - 1 ) ^ { r } \frac { x ^ { 2 r + 1 } } { ( 2 r + 1 ) ! } + \ldots\) for all \(x\)
\(\cos x = 1 - \frac { x ^ { 2 } } { 2 ! } + \frac { x ^ { 4 } } { 4 ! } - \ldots + ( - 1 ) ^ { r } \frac { x ^ { 2 r } } { ( 2 r ) ! } + \ldots\) for all \(x\)
\(( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )\) \section*{Differentiation} Quotient rule \(y = \frac { u } { v } , \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { v \frac { \mathrm {~d} u } { \mathrm {~d} x } - u \frac { \mathrm {~d} v } { \mathrm {~d} x } } { v ^ { 2 } }\) \section*{Differentiation from first principles} \(\mathrm { f } ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\) \section*{Integration} \(\int \frac { \mathrm { f } ^ { \prime } ( x ) } { \mathrm { f } ( x ) } \mathrm { d } x = \ln | \mathrm { f } ( x ) | + c\)
\(\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c\) Integration by parts \(\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x\)
Area of sector enclosed by polar curve is \(\frac { 1 } { 2 } \int r ^ { 2 } \mathrm {~d} \theta\) \section*{Numerical methods} Trapezium rule: \(\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}\), where \(h = \frac { b - a } { n }\)
The Newton-Raphson iteration for solving \(\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }\) \section*{Complex numbers} Circles: \(| z - a | = k\)
Half lines: \(\arg ( z - a ) = \alpha\)
Lines: \(| z - a | = | z - b |\) \section*{Small angle approximations} \(\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta\) where \(\theta\) is small and measured in radians \section*{Trigonometric identities} \(\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B\)
\(\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B\)
\(\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)\) \section*{Hyperbolic functions} $$\begin{aligned} & \cosh ^ { 2 } x - \sinh ^ { 2 } x = 1
& \sinh ^ { - 1 } x = \ln \left[ x + \sqrt { \left( x ^ { 2 } + 1 \right) } \right]
& \cosh ^ { - 1 } x = \ln \left[ x + \sqrt { \left( x ^ { 2 } - 1 \right) } \right] , x \geq 1
& \tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right) , - 1 < x < 1 \end{aligned}$$

1. The complex number \(z\) satisfies the equation \(z + 2 \mathrm { i } z ^ { * } + 1 - 4 \mathrm { i } = 0\).

You are given that \(z = x + \mathrm { i } y\), where \(x\) and \(y\) are real numbers.
Determine the values of \(x\) and \(y\).
2. Prove by induction that, for all positive integers \(n\), $$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { 1 } { 3 } n \left( 4 n ^ { 2 } - 1 \right)$$ [BLANK PAGE]
3. The figure below shows the curve with cartesian equation \(\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = x y\).
\includegraphics[max width=\textwidth, alt={}, center]{f42517a5-d7ed-40f3-bb04-faea97d4b19b-08_830_997_228_262}

1. Show that the polar equation of the curve is \(r ^ { 2 } = a \sin b \theta\), where \(a\) and \(b\) are positive constants to be determined.
2. Determine the exact maximum value of \(r\).
3. Determine the area enclosed by one of the loops.
   [0pt] [BLANK PAGE] \section*{4. In this question you must show detailed reasoning.}
   1. The curves with equations $$y = \frac { 3 } { 4 } \sinh x \text { and } y = \tanh x + \frac { 1 } { 5 }$$ intersect at just one point \(P\)
4. Use algebra to show that the \(x\) coordinate of \(P\) satisfies the equation $$15 \mathrm { e } ^ { 4 x } - 48 \mathrm { e } ^ { 3 x } + 32 \mathrm { e } ^ { x } - 15 = 0$$
5. Show that \(\mathrm { e } ^ { x } = 3\) is a solution of this equation.
6. Hence state the exact coordinates of \(P\).
   (ii) Show that $$\int _ { - 4 } ^ { 0 } \frac { \mathrm { e } ^ { \frac { 1 } { x } } } { x ^ { 2 } } \mathrm {~d} x = \mathrm { e } ^ { - \frac { 1 } { 4 } }$$ [BLANK PAGE]
   5. Use the method of differences to prove that for \(n > 2\) $$\sum _ { r = 2 } ^ { n } \frac { 4 } { r ^ { 2 } - 1 } = \frac { ( p n + q ) ( n - 1 ) } { n ( n + 1 ) }$$ where \(p\) and \(q\) are constants to be determined.
   [0pt] [BLANK PAGE]
   6.
   (i) $$z _ { 1 } = a + b \mathrm { i } \text { and } z _ { 2 } = c + d \mathrm { i }$$ where \(a , b , c\) and \(d\) are real constants.
   Given that
   • \(b > d\)
   • \(z _ { 1 } + z _ { 2 }\) is real
   • \(\left| z _ { 1 } \right| = \sqrt { 13 }\)
   • \(\left| z _ { 2 } \right| = 5\)
   • \(\operatorname { Re } \left( z _ { 2 } - z _ { 1 } \right) = 2\)
     show that \(a = 2\) and determine the value of each of \(b , c\) and \(d\)
     (ii) (a) On the same Argand diagram
   • sketch the locus of points \(z\) which satisfy \(| z - 12 | = 7\)
   • sketch the locus of points \(w\) which satisfy \(| w - 5 \mathrm { i } | = 4\) showing the coordinates of any points of intersection with the axes.
   • Determine the range of possible values of \(| z - w |\)
     [0pt] [BLANK PAGE]
   1. In this question you must show detailed reasoning.
   Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 2 } { x ^ { 2 } - x + 1 } \mathrm {~d} x\). Give your answer in exact form.
   [0pt] [BLANK PAGE] \section*{8. In this question you must show detailed reasoning.} The diagram shows the curve with equation \(y = \frac { x + 3 } { \sqrt { x ^ { 2 } + 9 } }\).
   \includegraphics[max width=\textwidth, alt={}, center]{f42517a5-d7ed-40f3-bb04-faea97d4b19b-18_890_1010_367_242} The region R , shown shaded in the diagram, is bounded by the curve, the \(x\)-axis, the \(y\)-axis, and the line \(x = 4\).
7. Determine the area of R . Give your answer in the form \(p + \ln q\) where \(p\) and \(q\) are integers to be determined. The region R is rotated through \(2 \pi\) radians about the \(x\)-axis.
8. Determine the volume of the solid of revolution formed. Give your answer in the form \(\pi \left( a + b \ln \left( \frac { c } { d } \right) \right)\) where \(a , b , c\) and \(d\) are integers to be determined.
   [0pt] [BLANK PAGE]
   9. Given that $$y = \cos x \sinh x \quad x \in \mathbb { R }$$
9. show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = k y$$ where \(k\) is a constant to be determined.
10. Hence determine the first three non-zero terms of the Maclaurin series for \(y\), giving each coefficient in simplest form.
    [0pt] [BLANK PAGE]
    10. The quartic equation $$2 x ^ { 4 } + A x ^ { 3 } - A x ^ { 2 } - 5 x + 6 = 0$$ where \(A\) is a real constant, has roots \(\alpha , \beta , \gamma\) and \(\delta\)
11. Determine the value of $$\frac { 3 } { \alpha } + \frac { 3 } { \beta } + \frac { 3 } { \gamma } + \frac { 3 } { \delta }$$ Given that \(\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } = - \frac { 3 } { 4 }\)
12. determine the possible values of \(A\)
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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

\hline \end{tabular} \end{center}

2 A particle \(P\) of mass 5.6 kg is attached to one end of a light rod of length 2.1 m . The other end of the rod is freely hinged to a fixed point \(O\). The particle is initially at rest directly below \(O\). It is then projected horizontally with speed \(5 \mathrm {~ms} ^ { - 1 }\). In the subsequent motion, the angle between the rod and the downward vertical at \(O\) is denoted by \(\theta\) radians, as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{2d0be306-be7a-419d-bf74-5a239e8eff65-02_483_304_936_242}

1. Find the speed of \(P\) when \(\theta = \frac { 1 } { 4 } \pi\).
2. Find the value of \(\theta\) when \(P\) first comes to instantaneous rest. A particle of mass \(m\) moves in a straight line with constant acceleration \(a\). Its initial and final velocities are \(u\) and \(v\) respectively and its final displacement from its starting position is \(s\). In order to model the motion of the particle it is suggested that the velocity is given by the equation
   \(v ^ { 2 } = p u ^ { \alpha } + q a ^ { \beta } s ^ { \gamma }\) where \(p\) and \(q\) are dimensionless constants.
3. Explain why \(\alpha\) must equal 2 for the equation to be dimensionally consistent.
4. By using dimensional analysis, determine the values of \(\beta\) and \(\gamma\).
5. By considering the case where \(s = 0\), determine the value of \(p\).
6. By multiplying both sides of the equation by \(\frac { 1 } { 2 } m\), and using the numerical values of \(\alpha , \beta\) and \(\gamma\), determine the value of \(q\).

4 Three particles \(A , B\) and \(C\) are free to move in the same straight line on a large smooth horizontal surface. Their masses are \(3.3 \mathrm {~kg} , 2.2 \mathrm {~kg}\) and 1 kg respectively. The coefficient of restitution in collisions between any two of them is \(e\). Initially, \(B\) and \(C\) are at rest and \(A\) is moving towards \(B\) with speed \(u \mathrm {~ms} ^ { - 1 }\) (see diagram). \(A\) collides directly with \(B\) and \(B\) then goes on to collide directly with \(C\).
\includegraphics[max width=\textwidth, alt={}, center]{2d0be306-be7a-419d-bf74-5a239e8eff65-03_216_1307_456_242}

1. The velocities of \(A\) and \(B\) immediately after the first collision are denoted by \(v _ { A } \mathrm {~ms} ^ { - 1 }\) and \(v _ { B } \mathrm {~ms} ^ { - 1 }\) respectively.
   • Show that \(v _ { A } = \frac { u ( 3 - 2 e ) } { 5 }\).
   • Find an expression for \(v _ { B }\) in terms of \(u\) and \(e\).
   • Find an expression in terms of \(u\) and \(e\) for the velocity of \(B\) immediately after its collision with \(C\).
   After the collision between \(B\) and \(C\) there is a further collision between \(A\) and \(B\).
2. Determine the range of possible values of \(e\). \section*{Total Marks for Question Set 6: 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}

   Question		Answer	Marks	AOs	Guidance
   1	(a)	\(\begin{aligned}	\text { At constant velocity, } F = 250 = 10000 / v
   	v = 40 \end{aligned}\)	\(\begin{gathered} \text { M1 }
   \text { A1 }
   { [ 2 ] } \end{gathered}\)	1.1 1.1	Tractive force \(= P / v =\) resistance
   1	(b)	\(\begin{aligned}	D = \frac { 10000 } { 30 }
   	10000 - 250 = 1200 a
   	30
   	\text { awrt } 0.069 \mathrm {~ms} ^ { - 2 } \end{aligned}\)	M1 M1 \(\begin{aligned} \text { A1 } { [ 3 ] } \end{aligned}\)	1.1 1.1 1.1	Use of \(P = D v\) where \(D\) is tractive force Attempt NII with 2 forces (one of which could be just " \(D\) ")
   2	(a)	Initial Energy \(= 1 / 2 \times 5.6 \times 5 ^ { 2 }\) \(\begin{aligned} \text { When } \theta = \frac { 1 } { 4 } \pi , P \text { 's PE } 5.6 g \times \left( 2.1 - 2.1 \cos \frac { 1 } { 4 } \pi \right) \end{aligned}\) So conservation of energy \({ } ^ { 1 } \times 5.6 v ^ { 2 } + 5.6 g \times \left( 2.1 - 2.1 \cos ^ { 1 } \pi \right) = 70\) \(v = \text { awrt } 3.6\)	B1 *M1 > M1 dep	1.1 1.1 1.1 1.1	Using \(u\) to find \(P\) 's initial (kinetic) energy (=70J) Finding \(P\) 's PE when \(\theta = \frac { 1 } { 4 } \pi\) (= 33.8J) Finding expression for \(P\) 's energy ( \(\mathrm { KE } + \mathrm { PE }\) ) and equating to initial energy	Allow 1 slip, but PE must not become negative Final speed must be < 3.6
   2	(b)	When \(v = 0 P\) 's energy \(= 5.6 g \times ( 2.1 - 2.1 \cos \theta ) = 70\) \(\theta =\) awrt 1.17 rads	M1 A1 [2]	3.1b 1.1	Finding expression for \(P\) 's final (potential) energy and equating to initial energy. cao	Allow in degrees awrt \(66.9 ^ { \circ }\)

   \begin{displayquote} M1 dep \end{displayquote}

   Question			Answer	Marks	AOs	Guidance
   3	(a)		\(\left[ v ^ { 2 } \right] = \left[ p u ^ { \alpha } \right]\) and \([ v ] = [ u ]\) and \(p\) is dimensionless \(\text { or } \mathrm { L } ^ { 2 } \mathrm {~T} ^ { - 2 } = \mathrm { L } ^ { \alpha } \mathrm { T } ^ { - \alpha } = > \alpha = 2\)	M1 A1 [2]	2.1 2.1	For the idea, that might be stated in words, that the dimensions of every term in the equation must be the same AG	Must relate LHS to RHS Must explicitly state that \(p\) is dimensionless to get this mark (this may be seen in part (b)).
   3	(b)		\(\mathrm { L } ^ { 2 } \mathrm {~T} ^ { - 2 } = \left( \mathrm { LT } ^ { - 2 } \right) ^ { \beta } \mathrm { L } ^ { \gamma }\) \(- 2 = - 2 \beta \Rightarrow \beta = 1\) \(\beta + \gamma = 2\) \(\gamma = 1\)	M1 A1 M1 A1 [4]	3.3 3.3 1.1 1.1	Equating dimensions of other term with [ \(v ^ { 2 }\) ] with \(q\) gone and \([ a ]\) and \([ s ]\) used Ignore term in [ \(u ^ { 2 }\) ] if present Equating powers of L	May be seen expanded ASM note - must be seen or strongly implied - see wording of question. SC2 for both correct
   3	(c)		If \(s = 0\) then \(v = u\) so \(u ^ { 2 } = p u ^ { 2 } + 0 = > p = 1\)	B1 [1]	3.4	Details must be shown	Do not allow use of prior knowledge of \(v ^ { 2 } = u ^ { 2 } + 2 a s\).
   3	(d)		\(\begin{aligned} 1 2 \end{aligned} m v ^ { 2 } - { } _ { 2 } ^ { 1 } m u ^ { 2 } = { } _ { 2 } ^ { 1 } q m a s = { } _ { 2 } ^ { 1 } q F s = { } _ { 2 } ^ { 1 } q W\) So we can see that this equation is a statement of the Work-Energy principle so \({ } _ { 2 } ^ { 1 } q = 1\) so \(q = 2\)	M1 A1 [2]	3.4 2.4	Where \(F\) must be the force acting and \(W\) the work done by this force	Do not allow use of prior knowledge of \(v ^ { 2 } = u ^ { 2 } + 2 a s\).

   Question		Answer	Marks	AOs	Guidance
   4	(a)	\(\begin{aligned}	1 ^ { \text {st } } \text { Collision: } 3.3 u = 3.3 v _ { A } + 2.2 v _ { B }
   	1 ^ { \text {st } } \text { Collision: } \pm e = v _ { B } - v _ { A }
   	u
   	3 u = 3 v _ { A } + 2 v _ { B } \text { and } 2 e u = 2 v _ { B } - 2 v _ { A }
   	\Rightarrow 5 v = 3 u - 2 e u \Rightarrow v = \frac { u ( 3 - 2 e ) } { A }
   	\begin{array} { c } 3 u = 3 v _ { A } + 2 v _ { B } \text { and } 3 e u = 3 v _ { B } - 3 v _ { A }
   \Rightarrow 5 v = 3 u + 3 e u \Rightarrow v _ { B } ^ { v } = \frac { 3 u ( 1 + e ) } { 5 } \end{array} \end{aligned}\)	M1 M1 A1 A1 [4]	3.1b 3.1b 1.1 1.1	Conservation of momentum NEL AG find \(v _ { B }\) by elimination or substitution	Must be seen Must be seen AEF - award if seen in (b)
   4	(b)	\(2 ^ { \text {nd } } \text { Collision: } 2.2 \times \frac { 3 u ( 1 + e ) } { 5 } = \underset { B } { 2.2 V } + \underset { C } { V }\) \(2 ^ { \text {nd } }\) Collision: \(\pm e = V _ { C } - V _ { B }\) \(( 3 u ( 1 + e ) )\) \(255 { } ^ { B } \quad { } ^ { C } \quad 5 \quad { } ^ { C } \quad { } ^ { B }\) \(\begin{aligned} \Rightarrow { } _ { 5 } ^ { 16 } V _ { B } = { } _ { 25 } ^ { 33 } u ( 1 + e ) - 3 e u ( 1 + e ) \Rightarrow V _ { B } = 3 u ( 1 + e ) ( 11 - 5 e ) 80 \end{aligned}\)	M1ft M1ft	3.3 3.3	Conservation of momentum ( ft their value of \(V _ { B }\) ) NEL ( ft their value of \(V _ { B }\) )	\(V _ { C } = \begin{gathered} 33 u ( 1 + e ) ^ { 2 } 80 \end{gathered}\) Must be in terms of \(e\) and \(u\) only.

   \begin{center} \begin{tabular}{|l|l|l|l|l|l|l|} \hline 4 & \multirow[t]{4}{*}{(c)} & \multirow[t]{2}{*}{\(\begin{gathered} u ( 3 - 2 e )

5 \end{gathered} > \begin{gathered} 3 u ( 1 + e ) ( 11 - 5 e )
80 \end{gathered}\)} &

M1ft

М1

& \multirow[t]{3}{*}{

3.4

1.1

2.2a

1.1

} & \multirow[t]{4}{*}{

Correct condition for further collision (ft their \(V _ { B }\) from (b)

Rearranging to 3 term quadratic inequality in \(e\) \(e < { } _ { 3 } ^ { 1 } \text { is not sufficient for } \mathrm { A } 1\)

} & \multirow{4}{*}{If B1 not awarded then award A1 for \(0 \leq e < \begin{aligned} & 1
& 3 \end{aligned}\)}
\hline & & & & & &
\hline & & \(A\) and \(B\) collide again \(= > e \neq 0\) \(\begin{aligned} & ( 3 e - 1 ) ( e - 3 ) > 0 \text { and } 0 \leq e \leq 1 \text { and } e \neq 0
& \Rightarrow 0 < e < 1 \end{aligned}\) & B1 & & &
\hline & & & [4] & & &
\hline \end{tabular} \end{center}

1 A particle \(P\) of mass 4.5 kg is moving in a straight line on a smooth horizontal surface at a speed of \(2.4 \mathrm {~ms} ^ { - 1 }\) when it strikes a vertical wall directly. It rebounds at a speed of \(1.6 \mathrm {~ms} ^ { - 1 }\).

1. Find the coefficient of restitution between \(P\) and the wall.
2. Determine the impulse applied to \(P\) by the wall, stating its direction.
3. Find the loss of kinetic energy of \(P\) as a result of the collision.
4. State, with a reason, whether the collision is perfectly elastic.

2 A particle \(P\) of mass 2.4 kg is moving in a straight line \(O A\) on a horizontal plane. \(P\) is acted on by a force of magnitude 30 N in the direction of motion. The distance \(O A\) is 10 m .

1. Find the work done by this force as \(P\) moves from \(O\) to \(A\). The motion of \(P\) is resisted by a constant force of magnitude \(R \mathrm {~N}\). The velocity of \(P\) increases from \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at \(O\) to \(18 \mathrm {~ms} ^ { - 1 }\) at \(A\).
2. Find the value of \(R\).
3. Find the average power used in overcoming the resistance force on \(P\) as it moves from \(O\) to \(A\). When \(P\) reaches \(A\) it collides directly with a particle \(Q\) of mass 1.6 kg which was at rest at \(A\) before the collision. The impulse exerted on \(Q\) by \(P\) as a result of the collision is 17.28 Ns .
4. 1. Find the speed of \(Q\) after the collision.
   2. Hence show that the collision is inelastic. It is required to model the motion of a car of mass \(m \mathrm {~kg}\) travelling at a constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) around a circular portion of banked track. The track is banked at \(30 ^ { \circ }\) (see diagram).
      \includegraphics[max width=\textwidth, alt={}, center]{b9741472-f230-4e2d-9c8b-47f7e168e938-03_355_565_269_274} In a model, the following modelling assumptions are made.
      • The track is smooth.
5. The car is a particle.
6. The car follows a horizontal circular path with radius \(r \mathrm {~m}\).
7. Show that, according to the model, \(\sqrt { 3 } v ^ { 2 } = g r\).
For a particular portion of banked track, \(r = 24\).
8. Find the value of \(v\) as predicted by the model. A car is being driven on this portion of the track at the constant speed calculated in part (b). The driver finds that in fact he can drive a little slower or a little faster than this while still moving in the same horizontal circle.
9. Explain
   • how this contrasts with what the model predicts,
   • how to improve the model to account for this.
   \section*{Total Marks for Question Set 6: 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}

   1	(a)		\(1.6 = e \times 2.4 = > e = \begin{aligned}	2
   	3 \end{aligned}\)	B1 [1]	1.1
   1	(b)		\(4.5 \times - 1.6 - 4.5 \times 2.4 = - 18\) so \(18 \mathrm { Ns } \left( \right.\) or \(\left. \mathrm { kg } \mathrm { ms } ^ { - 1 } \right) \ldots\) ...in the final direction of motion of \(P\)	M1 A1 B1 [3]	1.1 1.1 2.2a	Attempt at \(m v - m u\) Allow \(\pm 18\) could be shown on a diagram	Allow sign confusion e.g. 1.6-2.4 Ignore missing units
   1	(c)		\(\begin{aligned}	1 \times 4.5 \times 2.4 ^ { 2 } - { } _ { 2 } ^ { 1 } \times 4.5 \times 1.6 ^ { 2 }
   	2
   	7.2 \mathrm {~J} \end{aligned}\)	M1 A1 [2]	1.1 1.1	Attempt at \(\pm \begin{gathered} 1
   m v ^ { 2 } - { } ^ { 1 } m u ^ { 2 }
   2 \end{gathered} \quad 2\).
   1	(d)		Not perfectly elastic since KE is lost (due to the collision)	B1 [1]	1.2	or \(e < 1\) but valid reason must be given.	Must mention KE or collision, Not just e.g. "energy lost"

   Question			Answer	Marks	AOs	Guidance
   2	(a)		\(30 \times 10\) 300 J	M1 A1 [2]	1.1 1.1	Using Work done \(= F d\)
   2	(b)		\(\begin{aligned}	1 \times 2.4 \times 12 ^ { 2 } + 300 = { } ^ { 1 } \times 2.4 \times 18 ^ { 2 } + W
   	2
   	10 R = 84
   	R = 8.4 \end{aligned}\)	M1 M1 A1 [3]	1.1 1.1 1.1	where \(W\) is energy loss (could be eg \(10 R\) ) Use of energy loss \(= 10 R\)	Allow 1 slip e.g. \(R\) instead of \(W\) Must be e.g. \(10 R\), not \(R\)
   			Alternative method \(\begin{aligned}	a = 18 ^ { 2 } - 12 ^ { 2 }
   	\quad 2 \times 10
   	30 - R = 2.4 \times 9
   	R = 8.4 \end{aligned}\)	М1 M1 A1		Using \(v ^ { 2 } = u ^ { 2 } + 2 a s\) to find \(a\) Use of \(F = m a\) with their \(a\)
   2	(c)		\(t = \begin{aligned} 2 \times 10 18 + 12 \end{aligned}\) \(t = { } ^ { 2 }\) \({ } _ { 2 } \frac { 84 ^ { 3 } } { 3 } = 126 \mathrm {~W}\)	M1 A1 A1	1.1 1.1 1.1	Using \(s = \binom { v + u } { 2 } t\)	Or use Average power \(=\) Force × average speed, i.e. \(P = F _ { ( 2 ) } ^ { v + u } \left( { } _ { 2 } \right)\)
   			Alternative method \(t = \begin{gathered} 18 - 12 9 \end{gathered}\) \(t = { } ^ { 2 }\) \(2 _ { 3 } ^ { 3 } = 126 \mathrm {~W}\)	M1 A1 A1		Using \(v = u + a t\) to find \(t\)	NB \(a = 9\) from (b) \(\text { or N2L: } a = { } ^ { 30 - 8.4 } = 9\) 2.4
   				[3]

   Question			Answer	Marks	AOs	Guidance
   2	(d)	(i)	\(\begin{aligned}	17.28 = 1.6 \times v _ { Q }
   	v _ { Q } = 10.8 \mathrm {~m} \mathrm {~s} ^ { - 1 } \end{aligned}\)	M1 A1 [2]	1.1 1.1		Do not allow -10.8
   2	(d)	(ii)	\(2.4 \times 18 = 2.4 \times v _ { P } + 1.6 \times 10.8\) So \(v _ { P } = 10.8 = v _ { Q }\), so the particles coalesce and the collision is therefore inelastic.	M1 A1 [2]	3.1b 2.1	Attempt conservation of momentum Find equal velocities and conclude	Do not allow use of KE
   3	(a)		\(\text { } \uparrow C \cos 30 ^ { \circ } = m g\) \(- C \sin 30 ^ { \circ } = \begin{gathered} m v ^ { 2 } r \end{gathered}\) \(m v ^ { 2 }\) \(\begin{aligned} \Rightarrow \begin{array} { l } C \sin 30 ^ { \circ } = \quad r C \cos 30 ^ { \circ } m g \end{array} \Rightarrow \tan 30 ^ { \circ } = \frac { 1 } { 3 } ^ { v } { } ^ { v } \stackrel { 2 } { \Rightarrow g } 3 v ^ { 2 } = r g \end{aligned}\)	M1* M1* M1ft A1 [4]	3.3 3.3 1.1 1.1	where C is the (normal) contact force between the car and the track NII with centripetal acceleration Dividing so that \(C\) and \(m\) will cancel. May see \(\tan \theta\) or \(\tan 30\) instead of sin/cos AG	Allow sin/cos confusion Allow \(\theta\) instead of \(30 ^ { \circ }\). Allow sin/cos confusion Or rearrange one equation for \(C\) and substitute into the other one \(\theta\) must be clearly stated and correctly used to gain this mark
   3	(b)		\(3 v ^ { 2 } = 24 \times 9.8\) \(v =\) awrt 11.7	M1 A1 [2]	3.4 1.1	Using the formula from (a)
   3	(c)		The model implies that only a single value for the speed is possible for a given radius so any change in speed should cause the car to move in a different circle The track should be modelled as resisting sideways motion	B1 B1 B1 [3]	3.5b 2.2b 3.5c	Or equivalent e.g. to slide sideways Accept 'model track as rough' or 'include friction' etc without explicit reference to 'sideways'	Do not allow discussion of the assumptions here Or any equivalent comment about a possible consequence according to the model of a change in the speed or the radius Must be relevant to the question. Do not accept references that ignore friction

2
The position vector of point \(A\) is \(\mathbf { a } = - 9 \mathbf { i } + 2 \mathbf { j } + 6 \mathbf { k }\).
The line \(l\) passes through \(A\) and is perpendicular to \(\mathbf { a }\).

1. Determine the shortest distance between the origin, \(O\), and \(l\).
   \(l\) is also perpendicular to the vector \(\mathbf { b }\) where \(\mathbf { b } = - 2 \mathbf { i } + \mathbf { j } + \mathbf { k }\).
2. Find a vector which is perpendicular to both \(\mathbf { a }\) and \(\mathbf { b }\).
3. Write down an equation of \(l\) in vector form.
   \(P\) is a point on \(l\) such that \(P A = 2 O A\).
4. Find angle \(P O A\) giving your answer to 3 significant figures.
   \(C\) is a point whose position vector, \(\mathbf { c }\), is given by \(\mathbf { c } = p \mathbf { a }\) for some constant \(p\). The line \(m\) passes through \(C\) and has equation \(\mathbf { r } = \mathbf { c } + \mu \mathbf { b }\). The point with position vector \(9 \mathbf { i } + 8 \mathbf { j } - 12 \mathbf { k }\) lies on \(m\).
5. Find the value of \(p\). \section*{In this question you must show detailed reasoning.} You are given that \(\alpha , \beta\) and \(\gamma\) are the roots of the equation \(5 x ^ { 3 } - 2 x ^ { 2 } + 3 x + 1 = 0\).
6. Find the value of \(\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }\).
7. Find a cubic equation whose roots are \(\alpha ^ { 2 } , \beta ^ { 2 }\) and \(\gamma ^ { 2 }\) giving your answer in the form \(a x ^ { 3 } + b x ^ { 2 } + c x + d = 0\) where \(a , b , c\) and \(d\) are integers.

4 In this question you must show detailed reasoning.
\(\mathbf { M }\) is the matrix \(\left( \begin{array} { l l } 1 & 6
0 & 2 \end{array} \right)\).
Prove that \(\mathbf { M } ^ { n } = \left( \begin{array} { c c } 1 & 3 \left( 2 ^ { n + 1 } - 2 \right)
0 & 2 ^ { n } \end{array} \right)\), for any positive integer \(n\). \section*{Total Marks for Question Set 1: 30} \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} \section*{Appendix: Special cases for question 3 using the substitution method} \section*{Working is shown in part (a). If working shown in part (b) mark as mark scheme above. If little or no working then mark part (b) as:}

1. Working in (a) is fully correct. Then award:

4/4 If correct cubic equation is re-written and there is a reference to the working in part (a) such as "See working above" or arrows etc.
3/4 If correct cubic equation re-written and there is no reference to the working in part (a)
0/4 If an incorrect cubic is written down, or " \(= 0\) " missing
2. Working in (a) is not fully correct. Then award: 0/4 if B1 M1 M1 is not awarded in part (a) (i.e. at least three marks given for (a))
3/4 if at least three marks awarded in part (a) and reference is made to the working in part (a) and the same cubic as in part (a) is re-written in part
(b) with and " \(= 0\) " 2/4 if at least three marks awarded in part (a) and the same cubic as in part (a) is re-written in part (b) with and " \(= 0\) " but no reference is made to the working in part (a) \section*{Working is shown in part (b) and no working shown in part (a).}

1. Working in part (b) is fully correct. Then award:
   \(5 / 5\) if a reference to the working in part (b) is made, the correct cubic equation is written down, there is a comment stating that the roots of the new cubic are \(\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }\) and the correct answer is given.
   \(4 / 5\) as above but no comment about the roots of the new cubic equation being \(\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }\) or there is no reference to the working in part (b) (i.e. one element of explanation is missing).
   \(3 / 5\) if the correct cubic equation is written down and correct answer of \(\frac { 13 } { 25 }\)
   \(3 / 5\) if a reference is made to working in part (b) but the cubic equation is not re-written and no comment about the roots being \(\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }\) (and correct answer of \(\frac { 13 } { 25 }\) is given)
2. Working in part (b) is not fully correct.
   \(3 / 5\) if a reference to the working in part (b) is made, the same cubic equation from part (b) is written down, there is a comment stating that the roots of the new cubic are \(\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }\) and the correct follow through coefficient ratio is given.
   \(2 / 5\) as above but no comment about the roots of the new cubic equation being \(\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }\) or there is no reference to the working in part (b) (i.e. one element of explanation is missing).
   \(1 / 5\) if same cubic equation as in part (b) is written down and the correct follow through coefficient ratio is given.
3. If \(\frac { 13 } { 25 }\) appears as an answer in part (a) with no supporting working or comments then \(0 / 5\)

1 Matrices \(\mathbf { P }\) and \(\mathbf { Q }\) are given by \(\mathbf { P } = \left( \begin{array} { c c c } 1 & k & 0
- 2 & 1 & 3 \end{array} \right)\) and \(\mathbf { Q } = ( ( 1 + k ) - 1 )\) where \(k\) is a constant.
Exactly one of statements A and B is true.
Statement A: \(\quad \mathbf { P }\) and \(\mathbf { Q }\) (in that order) are conformable for multiplication.
Statement B: \(\quad \mathbf { Q }\) and \(\mathbf { P }\) (in that order) are conformable for multiplication.

1. State, with a reason, which one of A and B is true.
2. Find either \(\mathbf { P Q }\) or \(\mathbf { Q P }\) in terms of \(k\).

2 In this question you must show detailed reasoning. You are given that \(\mathrm { f } ( z ) = 4 z ^ { 4 } - 12 z ^ { 3 } + 41 z ^ { 2 } - 128 z + 185\) and that \(2 + \mathrm { i }\) is a root of the equation \(\mathrm { f } ( z ) = 0\).

1. Express \(\mathrm { f } ( z )\) as the product of two quadratic factors with integer coefficients.
2. Solve \(\mathrm { f } ( z ) = 0\). Two loci on an Argand diagram are defined by \(C _ { 1 } = \left\{ z : | z | = r _ { 1 } \right\}\) and \(C _ { 2 } = \left\{ z : | z | = r _ { 2 } \right\}\) where \(r _ { 1 } > r _ { 2 }\). You are given that two of the points representing the roots of \(\mathrm { f } ( z ) = 0\) are on \(C _ { 1 }\) and two are on \(C _ { 2 } \cdot R\) is the region on the Argand diagram between \(C _ { 1 }\) and \(C _ { 2 }\).
3. Find the exact area of \(R\).
4. \(\omega\) is the sum of all the roots of \(\mathrm { f } ( z ) = 0\). Determine whether or not the point on the Argand diagram which represents \(\omega\) lies in \(R\).

3 A transformation T is represented by the matrix \(\mathbf { T }\) where \(\mathbf { T } = \left( \begin{array} { c c } x ^ { 2 } + 1 & - 4
3 - 2 x ^ { 2 } & x ^ { 2 } + 5 \end{array} \right)\). A quadrilateral \(Q\), whose area is 12 units, is transformed by T to \(Q ^ { \prime }\). Find the smallest possible value of the area of \(Q ^ { \prime }\).

4 A transformation A is represented by the matrix \(\mathbf { A }\) where \(\mathbf { A } = \left( \begin{array} { c c c } - 1 & x & 2
7 - x & - 6 & 1
5 & - 5 x & 2 x \end{array} \right)\).
The tetrahedron \(H\) has vertices at \(O , P , Q\) and \(R\). The volume of \(H\) is 6 units. \(P ^ { \prime } , Q ^ { \prime } , R ^ { \prime }\) and \(H ^ { \prime }\) are the images of \(P , Q , R\) and \(H\) under A .

1. In the case where \(x = 5\)
   • find the volume of \(H ^ { \prime }\),
   • determine whether A preserves the orientation of \(H\).
   • Find the values of \(x\) for which \(O , P ^ { \prime } , Q ^ { \prime }\) and \(R ^ { \prime }\) are coplanar (i.e. the four points lie in the same plane).
   \section*{Total Marks for Question Set 2: 30} \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}