Questions — OCR FM1 AS (28 questions)

1 A climber of mass 65 kg climbs from the bottom to the top of a vertical cliff which is 78 m in height. The climb takes 90 minutes so the velocity of the climber can be neglected.

1. Calculate the work done by the climber in climbing the cliff.
2. Calculate the average power generated by the climber in climbing the cliff.

2 The universal law of gravitation states that \(F = \frac { G m _ { 1 } m _ { 2 } } { r ^ { 2 } }\) where \(F\) is the magnitude of the force between two objects of masses \(m _ { 1 }\) and \(m _ { 2 }\) which are a distance \(r\) apart and \(G\) is a constant. Find the dimensions of \(G\).

3
\includegraphics[max width=\textwidth, alt={}, center]{a1a43547-0a68-4346-884a-0c6d9302cf24-2_473_298_1037_884} A particle \(P\) of mass 1.5 kg is attached to one end of a light inextensible string of length 2.4 m . The other end of the string is attached to a fixed point \(O\). The particle is initially at rest directly below \(O\). A horizontal impulse of magnitude 9.3 Ns is applied to \(P\). In the subsequent motion the string remains taut and makes an angle of \(\theta\) radians with the downwards vertical at \(O\), as shown in the diagram.

1. Find the speed of \(P\) when \(\theta = \frac { 1 } { 6 } \pi\).
2. Determine whether \(P\) will reach the same horizontal level as \(O\).

4
\includegraphics[max width=\textwidth, alt={}, center]{a1a43547-0a68-4346-884a-0c6d9302cf24-3_216_1219_255_415}
\(A\) and \(B\) are two long straight parallel horizontal sections of railway track. An engine on track \(A\) is attached to a carriage of mass 6000 kg on track \(B\) by a light inextensible chain which remains horizontal and taut in the ensuing motion. The chain is 13 m in length and the points of attachment on the engine and carriage are a perpendicular distance of 5 m apart. The engine and carriage start at rest and then the engine accelerates uniformly to a speed of \(5.6 \mathrm {~ms} ^ { - 1 }\) while travelling 250 m . It is assumed that any resistance to motion can be ignored.

1. Find the work done on the carriage by the tension in the chain.
2. Find the magnitude of the tension in the chain. The mass of the engine is 10000 kg .
3. At a point further along the track the engine and the carriage are moving at a speed of \(8.4 \mathrm {~ms} ^ { - 1 }\) and the power of the engine is 68 kW . Find the acceleration of the engine at this instant.

5 Two discs, \(A\) and \(B\), have masses 1.4 kg and 2.1 kg respectively. They are sliding towards each other in the same straight line across a large sheet of horizontal ice. Immediately before the collision \(A\) has speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(B\) has speed \(3 \mathrm {~ms} ^ { - 1 }\). Immediately after the collision \(A\) 's speed is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Explain why it is impossible for \(A\) to be travelling in the same direction after the collision as it was before the collision.
2. Find the velocity of \(B\) immediately after the collision.
3. Calculate the coefficient of restitution between \(A\) and \(B\).
4. State what your answer to part (iii) means about the kinetic energy of the system. The discs are made from the same material. The discs will be damaged if subjected to an impulse of magnitude greater than 6.5 Ns .
5. Determine whether \(B\) will be damaged as a result of the collision.
6. Explain why \(A\) will be damaged if, and only if, \(B\) is damaged.

6
\includegraphics[max width=\textwidth, alt={}, center]{a1a43547-0a68-4346-884a-0c6d9302cf24-4_547_597_251_735} A particle of mass 0.2 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(O\) which is 1.8 m above a smooth horizontal table. The particle moves on the table in a circular path at constant speed with the string taut (see diagram). The particle has a speed of \(0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its angular velocity is \(0.625 \mathrm { rad } \mathrm { s } ^ { - 1 }\).

1. Show that the radius of the circular path is 0.8 m .
2. Find the magnitude of the normal contact force between the particle and the table. The speed is changed to \(v \mathrm {~ms} ^ { - 1 }\). At this speed the particle is just about to lose contact with the table.
3. Find the value of \(v\).

7 The masses of two particles \(A\) and \(B\) are \(m\) and \(2 m\) respectively. They are moving towards each other on a smooth horizontal table. Just before they collide their speeds are \(u\) and \(2 u\) respectively. After the collision the kinetic energy of \(A\) is 8 times the kinetic energy of \(B\). Find the coefficient of restitution between \(A\) and \(B\). \section*{END OF QUESTION PAPER}

1 A particle \(P\) of mass 2.4 kg is attached to one end of a light inextensible string of length 1.4 m . The other end of the string is attached to a fixed point \(O\) on a smooth horizontal table. \(P\) moves on the table at constant speed along a circular path with \(O\) at its centre. The magnitude of the tension in the string is 21 N .

1. (a) Find the magnitude of the acceleration of \(P\).
   (b) State the direction of the acceleration of \(P\).
2. Find the speed of \(P\).
3. Find the time taken for \(P\) to complete a single revolution.

2 A pump is pumping still water from the base of a well at a constant rate of 300 kg per minute. The well is 4.5 m deep and water is released from the pump at ground level in a horizontal jet with a speed of \(6.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Ignoring any energy losses due to resistance, calculate the power generated by the pump.

3 A student is investigating fluid flowing through a pipe.
In her first model she assumes a relationship of the form \(P = S \rho ^ { \alpha } g ^ { \beta } h ^ { \gamma }\) where \(\rho\) is the density of the fluid, \(h\) is the length of the pipe, \(P\) is the pressure difference between the ends of the pipe, \(g\) is the acceleration due to gravity and \(S\) is a dimensionless constant. You are given that \(\rho\) is measured in \(\mathrm { kg } \mathrm { m } ^ { - 3 }\).

1. Use the fact that pressure is force per unit area to show that \([ P ] = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 }\).
2. Find the values of \(\alpha , \beta\) and \(\gamma\). The density of the fluid the student is using is \(540 \mathrm {~kg} \mathrm {~m} ^ { - 3 }\). In her experiment she finds that when the length of the pipe is 1.40 m the pressure difference between the ends of the pipe is \(3.25 \mathrm { Nm } ^ { - 2 }\).
3. Find the length of the pipe for which her first model would predict a pressure difference between the ends of the pipe of \(4.65 \mathrm { Nm } ^ { - 2 }\). In an alternative model the student suggests a modified relationship of the form \(P = S \rho ^ { \alpha } g ^ { \beta } h ^ { \gamma } + \frac { 1 } { 2 } h v ^ { 2 }\), where \(v\) is the average velocity of the fluid in the pipe.
4. Use dimensional analysis to assess the validity of her alternative model.

4 A car has a mass of 850 kg and its engine can generate a maximum power of 35 kW . The total resistance to motion of the car is modelled as \(k v \mathrm {~N}\) where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the car and \(k\) is a constant. When the car is moving in a straight line on a straight horizontal road, the greatest constant speed that it can attain is \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(k = 56\).
2. Find the greatest possible acceleration of the car on the road at an instant when it is moving with a speed of \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A trailer of mass 240 kg is attached to the car by means of a light inextensible tow bar which is parallel to the surface of the road. The resistance to motion of the trailer is modelled as a constant force of magnitude 350 N . The car and trailer move on the horizontal road. At a certain instant the car's engine is working at a rate of 30 kW and the acceleration of the car is \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
3. (a) Find the speed of the car at this instant.
   (b) Find the magnitude of the tension in the tow bar at this instant. The car and trailer now move in a straight line on a straight road inclined at \(8 ^ { \circ }\) to the horizontal.
4. Find the difference between their greatest possible constant speed travelling up the slope and their greatest possible constant speed travelling down the slope.

5 Two particles \(A\) and \(B\) are on a smooth horizontal floor with \(B\) between \(A\) and a vertical wall. The masses of \(A\) and \(B\) are 4 kg and 11 kg respectively. Initially, \(B\) is at rest and \(A\) is moving towards \(B\) with a speed of \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). \(A\) collides directly with \(B\). The coefficient of restitution between \(A\) and \(B\) is \(e\).
\includegraphics[max width=\textwidth, alt={}, center]{bf86ac88-0fd1-4d49-a705-9b8d06fbac2a-3_209_803_1658_630}

1. Show that immediately after the collision the speed of \(B\) is \(\frac { 4 } { 15 } u ( 1 + e )\). After the collision between \(A\) and \(B\) the direction of motion of \(A\) is reversed. \(B\) subsequently collides directly with the vertical wall. The coefficient of restitution between \(B\) and the wall is \(\frac { 1 } { 2 } e\).
2. Given that there is a second collision between \(A\) and \(B\), find the range of possible values of \(e\).

6 A fairground game involves a player kicking a ball, \(B\), from rest so as to project it with a horizontal velocity of magnitude \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The ball is attached to one end of a light rod of length \(l \mathrm {~m}\). The other end of the rod is smoothly hinged at a fixed point \(O\) so that \(B\) can only move in the vertical plane which contains \(O\), a fixed barrier and a bell which is fixed \(l \mathrm {~m}\) vertically above \(O\). Initially \(B\) is vertically below \(O\). The barrier is positioned so that when \(B\) collides directly with the barrier, \(O B\) makes an angle \(\theta\) with the downwards vertical through \(O\) (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{bf86ac88-0fd1-4d49-a705-9b8d06fbac2a-4_643_659_584_724} The coefficient of restitution between \(B\) and the barrier is \(e . B\) rebounds from the barrier, passes through its original position and continues on a circular path towards the bell. The bell will only ring if the ball strikes it with a speed of at least \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The player wins the game if the player causes the bell to ring having kicked \(B\) so that it first collides with the barrier. You may assume that \(B\) and the bell are small and that the barrier has negligible thickness. Show that, whatever the position of the barrier, the player cannot win the game if \(u ^ { 2 } < 4 g l + \frac { V ^ { 2 } } { e ^ { 2 } }\). \section*{END OF QUESTION PAPER}

2
\includegraphics[max width=\textwidth, alt={}, center]{c397fca5-e7e8-4f3d-b519-cd92a983ebcc-02_810_743_831_644} A smooth wire is shaped into a circle of centre \(O\) and radius 0.8 m . The wire is fixed in a vertical plane. A small bead \(P\) of mass 0.03 kg is threaded on the wire and is projected along the wire from the highest point with a speed of \(4.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When \(O P\) makes an angle \(\theta\) with the upward vertical the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram).

1. Show that \(v ^ { 2 } = 33.32 - 15.68 \cos \theta\).
2. Prove that the bead is never at rest.
3. Find the maximum value of \(v\).
4. Write down the dimension of density. The workings of an oil pump consist of a right, solid cylinder which is partially submerged in oil. The cylinder is free to oscillate along its central axis which is vertical. If the base area of the pump is \(0.4 \mathrm {~m} ^ { 2 }\) and the density of the oil is \(920 \mathrm {~kg} \mathrm {~m} ^ { - 3 }\) then the period of oscillation of the pump is 0.7 s .
   A student assumes that the period of oscillation of the pump is dependent only on the density of the oil, \(\rho\), the acceleration due to gravity, \(g\), and the surface area, \(A\), of the circular base of the pump. The student attempts to test this assumption by stating that the period of oscillation, \(T\), is given by \(T = C \rho ^ { \alpha } g ^ { \beta } A ^ { \gamma }\) where \(C\) is a dimensionless constant.
5. Use dimensional analysis to find the values of \(\alpha , \beta\) and \(\gamma\).
6. Hence give the value of \(C\) to 3 significant figures.
7. Comment, with justification, on the assumption made by the student that the formula for the period of oscillation of the pump was dependent on only \(\rho , g\) and \(A\). A car of mass 1250 kg experiences a resistance to its motion of magnitude \(k v ^ { 2 } \mathrm {~N}\), where \(k\) is a constant and \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the car's speed. The car travels in a straight line along a horizontal road with its engine working at a constant rate of \(P \mathrm {~W}\). At a point \(A\) on the road the car's speed is \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and it has an acceleration of magnitude \(0.54 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). At a point \(B\) on the road the car's speed is \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and it has an acceleration of magnitude \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
8. Find the values of \(k\) and \(P\). The power is increased to 15 kW .
9. Calculate the maximum steady speed of the car on a straight horizontal road.

5
\includegraphics[max width=\textwidth, alt={}, center]{c397fca5-e7e8-4f3d-b519-cd92a983ebcc-04_221_1233_367_328} The masses of two spheres \(A\) and \(B\) are \(3 m \mathrm {~kg}\) and \(m \mathrm {~kg}\) respectively. The spheres are moving towards each other with constant speeds \(2 u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) respectively along the same straight line towards each other on a smooth horizontal surface (see diagram). The two spheres collide and the coefficient of restitution between the spheres is \(e\). After colliding, \(A\) and \(B\) both move in the same direction with speeds \(v \mathrm {~ms} ^ { - 1 }\) and \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), respectively.

1. Find an expression for \(v\) in terms of \(e\) and \(u\).
2. Write down unsimplified expressions in terms of \(e\) and \(u\) for
   (a) the total kinetic energy of the spheres before the collision,
   (b) the total kinetic energy of the spheres after the collision.
3. Given that the total kinetic energy of the spheres after the collision is \(\lambda\) times the total kinetic energy before the collision, show that $$\lambda = \frac { 27 e ^ { 2 } + 25 } { 52 }$$
4. Comment on the cases when
   (a) \(\lambda = 1\),
   (b) \(\lambda = \frac { 25 } { 52 }\).
   \includegraphics[max width=\textwidth, alt={}, center]{c397fca5-e7e8-4f3d-b519-cd92a983ebcc-05_789_981_324_543} The fixed points \(A\), \(B\) and \(C\) are in a vertical line with \(A\) above \(B\) and \(B\) above \(C\). A particle \(P\) of mass 2.5 kg is joined to \(A\), to \(B\) and to a particle \(Q\) of mass 2 kg , by three light rods where the length of rod \(A P\) is 1.5 m and the length of rod \(P Q\) is 0.75 m . Particle \(P\) moves in a horizontal circle with centre \(B\). Particle \(Q\) moves in a horizontal circle with centre \(C\) at the same constant angular speed \(\omega\) as \(P\), in such a way that \(A , B , P\) and \(Q\) are coplanar. The rod \(A P\) makes an angle of \(60 ^ { \circ }\) with the downward vertical, rod \(P Q\) makes an angle of \(30 ^ { \circ }\) with the downward vertical and rod \(B P\) is horizontal (see diagram).
5. Find the tension in the \(\operatorname { rod } P Q\).
6. Find \(\omega\).
7. Find the speed of \(P\).
8. Find the tension in the \(\operatorname { rod } A P\).
9. Hence find the magnitude of the force in rod \(B P\). Decide whether this rod is under tension or compression.

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 )