OCR FM1 AS (Further Mechanics 1 AS) 2021 June

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

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

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

1. The velocities of \(A\) and \(B\) immediately after the first collision are denoted by \(v _ { A } \mathrm {~ms} ^ { - 1 }\) and \(v _ { B } \mathrm {~ms} ^ { - 1 }\) respectively.
   • Show that \(v _ { A } = \frac { u ( 3 - 2 e ) } { 5 }\).
   • Find an expression for \(v _ { B }\) in terms of \(u\) and \(e\).
   • Find an expression in terms of \(u\) and \(e\) for the velocity of \(B\) immediately after its collision with \(C\).
   After the collision between \(B\) and \(C\) there is a further collision between \(A\) and \(B\).
2. Determine the range of possible values of \(e\). \section*{Total Marks for Question Set 6: 32} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
   b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
   A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
   c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
   d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
   e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
   • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
   • When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
   Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
   Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
   f Rules for replaced work and multiple attempts:
   • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
   • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
   • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
   For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
   If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Abbreviations}

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

   \end{table}

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

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

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

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

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

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

M1ft

М1

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

3.4

1.1

2.2a

1.1

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

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

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

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