OCR FM1 AS (Further Mechanics 1 AS) 2021 June

Question 1
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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.
Question 2 16 marks
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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 schemeMeaning
    dep*Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
    caoCorrect answer only
    оеOr equivalent
    rotRounded or truncated
    soiSeen or implied
    wwwWithout wrong working
    AGAnswer given
    awrtAnything which rounds to
    BCBy Calculator
    DRThis 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
Question 4 14 marks
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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}
3(a)\(T _ { 2 } \cos \theta = m _ { 2 } g\) \(T _ { 2 } = \frac { m _ { 2 } \times 9.8 } { 0.8 } = 12.25 m _ { 2 }\)
M1
A1
[2]
1.1a
1.1
Resolving \(T _ { 2 }\) vertically and balancing forces on \(R\)
Do not allow extra forces present
Allow use of g, e.g. \(\frac { 5 } { 4 } g m _ { 2 }\)
In this solution \(\theta\) is the angle between \(R P\) and \(R A\) Sin may be seen instead if \(\theta\) is measured horizontally.
Do not allow incomplete expressions e.g. \(\frac { m _ { 2 } g } { \sin 53.13 }\)
3(b)(i)\(\begin{aligned}T _ { 2 } \cos \theta + m _ { 1 } g = T _ { 1 } \cos \theta
T _ { 1 } = T _ { 2 } + \frac { 9.8 m _ { 1 } } { 0.8 } =
\qquad 12.25 m _ { 2 } + 12.25 m _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right) \end{aligned}\)
M1
A1
[2]
3.1b
2.1
Vertical forces on \(P\); 3 terms including resolving of \(T _ { 1 }\); allow sign error
AG Dividing by \(\cos \theta ( = 0.8 )\), substituting their \(T _ { 2 }\) and rearranging
Allow 12.25 instead of \(\frac { 49 } { 4 }\)
Or \(T _ { 1 } \cos \theta = m _ { 1 } g + m _ { 2 } g\) (equation for the system as a whole)
At least one intermediate step must be seen
3(b)(ii)\(\begin{aligned}T _ { 1 } \sin \theta + T _ { 2 } \sin \theta = m _ { 1 } a
12.25 \left( m _ { 1 } + m _ { 2 } \right) \times 0.6 + 12.25 m _ { 2 } \times 0.6 = m _ { 1 } \times 0.6 \omega ^ { 2 }
\omega ^ { 2 } = \frac { 7.35 m _ { 1 } + 14.7 m _ { 2 } } { 0.6 m _ { 1 } } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } } \end{aligned}\)
M1
М1
A1
[3]
3.1b
1.1
2.1
NII horizontally for \(P ; 3\) terms including resolving of tensions; allow sign error
Substituting for \(T _ { 1 }\), their \(T _ { 2 } , \sin \theta\) and \(\alpha\)
AG Must see an intermediate step
Could see \(a\) or \(0.6 \omega ^ { 2 }\) or \(\frac { v ^ { 2 } } { 0.6 }\) or \(\omega ^ { 2 } r\) or \(\frac { v ^ { 2 } } { r } \sin \theta = 0.6\)
must be \(a = 0.6 \omega ^ { 2 }\)
3(c)\(\begin{aligned}\text { E.g } m _ { 1 } \gg m _ { 2 } \Rightarrow \frac { 2 m _ { 2 } } { m _ { 1 } } \approx 0 \text { or } \frac { 49 m _ { 2 } } { 4 m _ { 1 } } \approx 0
\omega \approx \sqrt { \frac { 49 m } { 4 m } } = 3.5 \end{aligned}\)
M1 A1
[2]
1.1
1.1
Allow argument such as if \(m _ { 1 } \gg m _ { 2 }\) then \(m _ { 1 } + 2 m _ { 2 } \approx m _ { 1 }\)
AG \(m\) may be missing
SC1 for result following argument that \(m _ { 2 }\) is negligible (by comparison with \(m _ { 1 }\) ) without justification, or using trial values of \(m _ { 1 }\) and \(m _ { 2 }\) with \(m _ { 1 } \gg m _ { 2 }\).
Do not allow the assumption that \(m _ { 2 } = 0\)
If using trial values, \(m _ { 1 }\) must be at least \(70 \times m _ { 2 }\) to give \(\omega = 3.5\) to 1 dp .
3\multirow{3}{*}{(d)}
\(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\)
Final energy \(= 2.5 \times g \times 1\) \(\text { Initial } \mathrm { KE } = \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }\)
Initial PE \(= 2.5 \times g \times 1.2 + 2.8 \times g \times 0.4\)
Energy loss \(= 17.8605 + 40.376 - 24.5 = 33.7365\)
M1
B1
М1
M1
A1
1.2
1.1
1.1
1.1
3.2a
Use of \(v = r \omega\) with values for \(m _ { 1 }\) and \(m _ { 2 }\)
(Assuming zero PE level at 2 m below \(A\); other values possible)
Do not allow use of \(\omega = 3.5\)
oe with different zero PE level awrt 33.7
( \(v = 3.78 , v ^ { 2 } = 14.2884\) )
NB \(\omega = 6.3\) (24.5)
(17.8605)
(40.376)
Alternate method \(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\)
Initial KE \(= \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }\)
\(\triangle P E\) for \(m _ { 1 } = \pm 2.5 \times 9.8 \times ( 0.8 - 1 )\)
\(\triangle P E\) for \(m _ { 2 } = \pm 2.8 \times 9.8 ( 1.6 - 2 )\)
Energy loss \(= 17.8605 + 4.9 + 10.976\)
М1
М1
M1
M1
A1
Use of \(v = r \omega\) with values for \(m _ { 1 }\) and \(m _ { 2 }\)
Or \(- \triangle P E\) \(= 2.5 \times 9.8 \times 0.2 + 2.8 \times 9.8 \times 0.4\)
awrt 33.7
( \(v = 3.78 , v ^ { 2 } = 14.2884\) ) \(\mathrm { NB } \omega = 6.3\)
(17.8605)
\(( \pm 4.9 )\)
\(( \pm 10.976 )\)
\(( \pm 15.876 )\)
Or 15.876 + 17.8605
[5]