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]
\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) | | \(\mathrm { KE } = 1 / 2 \times m \times 1.2 ^ { 2 } ( = 0.72 m )\) | | PE difference \(= m g \times 3.2 \left( 1 - \cos 15 ^ { \circ } \right) ( = 1.0685 \ldots m ) 1 / 2 \times m \times v ^ { 2 } = m g \times 3.2 \left( 1 - \cos 15 ^ { \circ } \right) + 0.72 m\) | | 1.89 |
| | | Conservation of energy (in 3 terms) (condone if \(m\) cancelled) | |
| (b) | | \(m g \times 3.2 ( 1 - \cos \theta ) = 1.7885 \ldots m\) | | \(\theta = 19.4\) |
| | | Conservation of energy with \(v = 0\) (condone if \(m\) cancelled) Allow 19.5 from correct working | Their non-zero \(\frac { 1 } { 2 } m u ^ { 2 }\) |
| 2 | (a) | | \(\begin{aligned} | 1.2 \times 4 = 1.2 v _ { A } + 1.8 v _ { B } | | \frac { v _ { B } - v _ { A } } { 4 } = \frac { 3 } { 4 } \end{aligned}\) | | Attempt to solve for \(v _ { A }\) and \(v _ { B }\) \(v _ { B } = 2.8\) |
| | | | Conservation of momentum | | Restitution Allow sign error |
| Allow one minor slip, e.g. transpose masses |
| (b) | \(v _ { A } = - 0.2\) | B1 [1] | 1.1 | 0.2 in opposite direction | Allow "away from B" |
| (c) | | \(\begin{aligned} | 1.8 \times 2.8 = 1.8 V _ { B } + m V _ { C } | | \frac { V _ { C } - V _ { B } } { 2.8 } = \frac { 3 } { 4 } \end{aligned}\) | | Attempt to solve for \(V _ { B }\) in terms of \(m\) \(V _ { B } = \frac { 5.04 - 2.1 m } { 1.8 + m } \mathrm { oe }\) |
| | | | Conservation of momentum Restitution Allow sign error | | \(V _ { C }\) must be eliminated \(\frac { 8.82 } { 1.8 + m } - 2.1\) |
| Allow 1 minor slip NB \(\mathrm { v } _ { \mathrm { C } } > \mathrm { v } _ { \mathrm { B } }\) \(\frac { 25.2 - 10.5 m } { 5 m + 9 }\) |
| (d) | | Direction reversed ⇒ \(V _ { B } < 0\) | | \(m > 2.4\) |
| | | | Seen or implied by eg \(\frac { 5.04 - 2.1 m } { 1.8 + m } < 0\) | | Must be from an inequality |
| If \(\mathrm { V } _ { \mathrm { c } }\) found in error, \(\mathrm { V } _ { \mathrm { c } } <\) 2.1 or \(\frac { 8.82 } { 1.8 + m } < 2.1\) |
| Question | Answer | Marks | AOs | Guidance |
| 3 | (a) | | | \(R _ { \mathrm { C } } = 40000 / 42\) | | 952 N |
| | | | |
| (b) | | | \(R _ { \mathrm { T } } = 40000 / 30 - R _ { \mathrm { C } }\) | | 381 N |
| | | | |
| (c) | (i) | | \(D - R _ { \mathrm { C } } - R _ { \mathrm { T } } = 1400 \times 0.57\) \(P = D \times 15\) | | 32000 or 32 kW |
| | | | Attempt at " \(F = m a\) " for whole system (4 term equation) | | Allow \(1333.3 \ldots\) instead of \(\mathrm { R } _ { \mathrm { C } } + \mathrm { R } _ { \mathrm { T } }\) Correct equation (unsimplified) | | NB 31970W |
| or \(D - R _ { \mathrm { C } } - T = 1200 \times 0.57\) (" \(F = m a\) " for car) |
| (c) | (ii) | | \(T - R _ { \mathrm { T } } = 200 \times 0.57\) | | 495 |
| | | " \(F = m a\) " for trailer | Solution could use " \(F = m a\) " for car. Could be seen in (iii)(a). |
| (d) | (i) | | new model will predict a lower time to achieve a speed of \(20 \mathrm {~ms} ^ { - 1 }\). | | Because at low speeds new model has no resistance and so acceleration will be greater |
| | | Resistance and acceleration must be mentioned or implied | Allow e.g. "no resistance means reaching \(10 \mathrm {~m} / \mathrm { s }\) would occur faster" |
| (d) | (ii) | | New model predicts the same | | Greatest speed depends only on (final) resistance (and power) |
| | | | |