Questions - OCR Further Mechanics

2 A particle \(P\) of mass 4.5 kg is free to move along the \(x\)-axis. In a model of the motion it is assumed that \(P\) is acted on by two forces:

a constant force of magnitude \(f \mathrm {~N}\) in the positive \(x\) direction;
a resistance to motion, \(R \mathrm {~N}\), whose magnitude is proportional to the speed of \(P\).

At time \(t\) seconds the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). When \(t = 0 , P\) is at the origin \(O\) and is moving in the positive direction with speed \(u \mathrm {~ms} ^ { - 1 }\), and when \(v = 5 , R = 2\).

Show that, according to the model, \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 10 f - 4 v } { 45 }\).
1. By solving the differential equation in part (a), show that \(v = \frac { 1 } { 2 } \left( 5 f - ( 5 f - 2 u ) \mathrm { e } ^ { - \frac { 4 } { 45 } t } \right)\).
2. Describe briefly how, according to the model, the speed of \(P\) varies over time in each of the following cases.
  - \(u < 2.5 f\)
\(u = 2.5 f\)
\(u > 2.5 f\)
In the case where \(u = 2 f\), find in terms of \(f\) the exact displacement of \(P\) from \(O\) when \(t = 9\).
\includegraphics[max width=\textwidth, alt={}, center]{439ec1f0-60b8-4272-98cc-0c30d116c4cf-03_314_652_137_660}

Find an exact expression for the distance \(O F\). The acceleration due to gravity on and near the surface of the planet Earth is roughly \(6 \gamma\).

Explain whether \(O F\) would increase, decrease or remain unchanged if the action were repeated on the planet Earth. \section*{Total Marks for Question Set 2: 39} \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

(b)

(ii)

\(u < 2.5 f \Rightarrow v\) increases (from \(u\) ) and approaches 2.5f as \(t \rightarrow \infty\)

\(u = 2.5 f \Rightarrow v = 2.5 f\), constant

\(u > 2.5 f \Rightarrow v\) decreases (from \(u\) ) and approaches \(2.5 f\) as \(t \rightarrow \infty\)

3.4

Allow the idea that \(v = 2.5 f\) for large \(t\), and allow technically inaccurate statements (eg "v speeds up") provided that intent is clear

SC: If B0B1B0 or B0B0B0 awarded. If mentions \(v\) approaches \(2.5 f\) for cases 1 and 3 award B1 or if mentions \(v\) increases in case 1 and \(v\) decreases in case 3 award B1

3.4

See above

[3]

(c)

\multirow[b]{5}{*}{

*M1

Dep* M1

[4]

}

1.1

3.4

Could be in a definite integral

\(\begin{aligned}

\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 5 - \mathrm { e } ^ { - \frac { 4 } { 45 } t } \right) f \text { oe }

x = 2.5 f t + \frac { 45 } { 8 } f \mathrm { e } ^ { - \frac { 4 } { 45 } t } + c ^ { \prime }

\Rightarrow c ^ { \prime } = - \frac { 45 } { 8 } f \text { and use of } t = 9 \text { to find } x

x = \frac { 45 } { 8 } \left( 3 + \mathrm { e } ^ { - 0.8 } \right) f \end{aligned}\)

\includegraphics[max width=\textwidth, alt={}]{439ec1f0-60b8-4272-98cc-0c30d116c4cf-11_491_63_799_1606}

, or \(x = \left[ 2.5 f t + \frac { 45 } { 8 } f \mathrm { e } ^ { - \frac { 4 } { 45 } t } \right] _ { 0 } ^ { 9 }\) with

1.1

Question		Answer	Marks	AOs	Guidance
\multirow{10}{*}{3}	\multirow{10}{*}{(a)}		M1 A1	3.3 1.1	Conservation of energy	\(\theta\) is the angle between \(O P\) and the upward vertical
			B1	1.1	NII for \(P\) at point where it is about to lose contact with surface.	Could see contact force, \(C\), later set to 0
		\(\frac { 1 } { 2 } m v ^ { 2 } + m \gamma r \cos \theta = m \gamma r \cos \alpha\) oe \(v ^ { 2 } + 2 \gamma r \cos \theta = \frac { 3 } { 2 } \gamma r\) \(m \gamma \cos \theta \quad ( - C ) = m a\) \(a = \frac { v ^ { 2 } } { r }\) \(\cos \theta = \frac { a } { \gamma } = \frac { v ^ { 2 } } { \gamma r }\)	М1	2.2a	Use previous two results to relate \(v\) and \(\cos \theta\)
		\(\nu ^ { 2 } = \frac { 1 } { 2 } \gamma r\)	A1	2.2a	for \(v\) or \(v ^ { 2 }\) or \(\cos \theta = \frac { 1 } { 2 }\)
		\(r \cos \theta = \left( \sqrt { \frac { 1 } { 2 } \gamma r } \sin \theta \right) t + \frac { 1 } { 2 } \gamma t ^ { 2 }\)	M1	3.4	Use of \(s = u t + \frac { 1 } { 2 } a t ^ { 2 }\) using their vertical component of v as u where v has come from consideration of theta	Or use of trajectory eqn: \(y = x \tan \theta + \frac { \gamma x ^ { 2 } } { 2 v ^ { 2 } \cos ^ { 2 } \theta } \text { with }\)
		\(t ^ { 2 } + \sqrt { \frac { 3 r } { 2 \gamma } } t - \frac { r } { \gamma } = 0\) \(t = \frac { 1 } { 4 } \sqrt { \frac { r } { \gamma } } ( \sqrt { 22 } - \sqrt { 6 } )\)	*M1	1.1	Reduction to 3 term quadratic with numerical values for trig ratios	\(8 x ^ { 2 } + 2 \sqrt { 3 } r x - r ^ { 2 } = 0\)
			Dep* M1	1.1
		\(O F = r \sin \theta + \sqrt { \frac { 1 } { 2 } \gamma r } \cos \theta \times \frac { 1 } { 4 } \sqrt { \frac { r } { \gamma } } ( \sqrt { 22 } - \sqrt { 6 } )\)	Dep* M1	3.4	Find \(O F\)	\(x = \frac { \sqrt { 11 } - \sqrt { 3 } } { 8 } r\)
		\(O F = \frac { r } { 8 } ( \sqrt { 11 } + 3 \sqrt { 3 } ) \quad\) oe	A1	1.1
			[11]

Question

Answer

Marks

AOs

Guidance

(b)

Unchanged since \(O F\) does not depend on \(\gamma\)

[1]

3.5a

4 One end of a light elastic string of natural length \(l \mathrm {~m}\) and modulus of elasticity \(\lambda \mathrm { N }\) is attached to a particle \(A\) of mass \(m \mathrm {~kg}\). The other end of the string is attached to a fixed point \(O\) which is on a horizontal surface. The surface is modelled as being smooth and \(A\) moves in a circular path around \(O\) with constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The extension of the string is denoted by \(x \mathrm {~m}\).

Show that \(x\) satisfies \(\lambda x ^ { 2 } + \lambda l x - l m v ^ { 2 } = 0\).
By solving the equation in part (a) and using a binomial series, show that if \(\lambda\) is very large then \(\lambda x \approx m v ^ { 2 }\).
By considering the tension in the string, explain how the result obtained when \(\lambda\) is very large relates to the situation when the string is inextensible. The nature of the horizontal surface is such that the modelling assumption that it is smooth is justifiable provided that the speed of the particle does not exceed \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). In the case where \(m = 0.16\) and \(\lambda = 260\), the extension of the string is measured as being 3.0 cm .
Estimate the value of \(v\).

Explain whether the value of \(v\) means that the modelling assumption is necessarily justifiable in this situation. \section*{Total Marks for Question Set 4: 35} \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.

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.

\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

(a)

\(\binom { 2 } { 10 } \cdot \binom { 50 } { 140 }\)

\(2 \times 50 + 10 \times 140 = 1500 \mathrm {~J}\)

[2]

1.1

Using WD = F.x

(b)

\(1500 / 5 = 300 \mathrm {~W}\)

B1ft

[1]

1.1a

Their 1500

Must be a scalar value

\multirow[t]{3}{*}{1}

\multirow[t]{3}{*}{(c)}

\(\frac { 1 } { 2 } \times 1.25 v ^ { 2 } = \frac { 1 } { 2 } \times 1.25 \times 10 ^ { 2 } + 1500\)

\(50 \mathrm {~ms} ^ { - 1 }\)

1.1

Correct use of work-energy principle

Not ±

Can use their WD for M mark

Alternative method \(\binom { 50 } { 140 } = 5 \mathbf { u } + \frac { 1 } { 2 } \cdot \binom { \frac { 8 } { 5 } } { 8 } \cdot 5 ^ { 2 } \Rightarrow \mathbf { u } = \binom { 6 } { 8 }\)

Then \(\mathbf { v } = \binom { 6 } { 8 } + \binom { \frac { 8 } { 5 } } { 8 } .5 = \binom { 14 } { 48 }\)

\(| \mathbf { v } | = 50 \mathrm {~ms} ^ { - 1 }\)

М1

complete method involving constant acceleration formula(e)

[2]

Question

Answer

Marks

AOs

Guidance

(a)

\(\begin{aligned}

\bar { x } = \frac { \int _ { 0 } ^ { 5 } x ( 6 + \sin x ) \mathrm { d } x } { \int _ { 0 } ^ { 5 } ( 6 + \sin x ) \mathrm { d } x } = \frac { 72.622 \ldots } { 30.716 \ldots }

= \frac { 72.622 \ldots } { 30.716 \ldots } = 2.36 ( 3 \mathrm { sf } ) \end{aligned}\)

[2]

1.1

BC Attempt to use formula and either top or bottom correct soi

AG. Both must be seen, or correct \(2.364 \ldots\) seen

(b)

\(\bar { y } = \frac { \int _ { 0 } ^ { 5 } \frac { 1 } { 2 } ( 6 + \sin x ) ^ { 2 } \mathrm {~d} x } { \int _ { 0 } ^ { 5 } ( 6 + \sin x ) \mathrm { d } x } = \frac { 95.616 \ldots } { 30.716 \ldots }\)

[2]

1.1

BC Attempt to use formula and either top or bottom correct soi

(c)

The (part of the) binding (attached to the cover) is light oe

The CoM of the badge is at \(A\) oe

[2]

3.5b

eg The binding has no mass or the binding is very small so that the mass is concentrated at the hinge or the binding is smooth eg The badge is modelled as a particle or the badge is uniform

(d)

\(6 \times 2.36 \cos \frac { \pi } { 3 } + 2 \times 3 \cos \frac { \pi } { 3 }\)

\(= P \times 5\)

\(P = 2.02\)

[3]

3.4

1.1

Total 'clockwise' moment about binding axis (allow inclusion of \(g\) if consistent)... ...equals 'anticlockwise' moment

May use new \(\bar { x } = 2.523 \ldots\) \(8 \cos \frac { \pi } { 3 } \times \bar { x }\)

\(= 5 \mathrm { P }\)

Question

Answer

Marks

AOs

Guidance

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

\multirow[t]{3}{*}{(a)}

\(\begin{aligned}

u _ { A y } = \sqrt { 3 } \text { or awrt } 1.73

v _ { A y } = u _ { A y } ( = \sqrt { 3 } )

\binom { 1 } { \sqrt { 3 } } \cdot \binom { v _ { A x } } { \sqrt { 3 } } = 0 \Rightarrow v _ { A x } = - 3

3 \times 2 \cos 60 ^ { \circ } + 4 \times - 5 = 3 \times - 3 + 4 v _ { B x }

v _ { B x } = - 2

e = \frac { - 2 - - 3 } { 2 \cos 60 ^ { \circ } - - 5 }

e = \frac { 1 } { 6 } \text { or awrt } 0.17 \end{aligned}\)

М1

3.1b

2.2a

3.1b

1.1

3.1b

1.1

Correct perpendicular component of velocity of \(A\) before collision

This component of \(A\) 's velocity is unchanged by collision

Or \(\tan 30 ^ { \circ } = \frac { \sqrt { 3 } } { v _ { A x } }\). Using perpendicularity of \(A\) 's velocities to derive a value for \(v _ { A x }\)

Conservation of momentum with 4 terms

Restitution

or \(v _ { A } = 2 \sqrt { 3 }\) seen

Could be positive if shown in diagram

Allow one sign slip

Could be positive

Allow one sign slip

Alternative method for last 5 marks \(\begin{aligned}	3 \times 2 \cos 60 ^ { \circ } + 4 \times - 5 = 3 v _ { A x } + 4 v _ { B x }
	e = \frac { v _ { B x } - v _ { A x } } { 2 \cos 60 ^ { \circ } - - 5 }
	v _ { A x } = \frac { - 17 - 24 e } { 7 } \end{aligned}\) \(\binom { 1 } { \sqrt { 3 } } \cdot \binom { \frac { - 17 - 24 e } { 7 } } { \sqrt { 3 } } = 0\)
\(e = \frac { 1 } { 6 }\) or awrt 0.17

М1

Conservation of momentum Restitution \(v _ { B x } = \frac { - 17 + 18 e } { 7 }\)

Or \(\tan 30 ^ { \circ } = \frac { \sqrt { 3 } } { \left( \frac { 17 + 24 e } { 7 } \right) }\).

Using perpendicularity of \(A\) 's velocities.

\(\begin{aligned}

3 v _ { A x } + 4 v _ { B x } = - 17

v _ { B x } - v _ { A x } = 6 e \end{aligned}\)

[7]

Question

Answer

Marks

AOs

Guidance

\multirow[t]{12}{*}{3}

\multirow[t]{12}{*}{(b)}

3.1b

Attempt to find final speed (squared) of \(A\) vectorially

\multirow[t]{4}{*}{\(v _ { A } { } ^ { 2 } = 12\)}

\(\begin{aligned}

v _ { A } ^ { 2 } = ( \sqrt { 3 } ) ^ { 2 } + ( - 3 ) ^ { 2 }

\frac { 1 } { 2 } \times 4 \times 5 ^ { 2 } + \frac { 1 } { 2 } \times 3 \times 2 ^ { 2 }

\frac { 1 } { 2 } \times 4 \times 2 ^ { 2 } + \frac { 1 } { 2 } \times 3 \times 12 \end{aligned}\)

М1

1.1

Attempt to find total initial KE

1.1

Attempt to find total final KE

56 or 26

1.1

Either correctly calculated

\multirow{8}{*}{}

\(56 - 26 = 30 \mathrm {~J}\)

1.1

Not -30

\multirow{7}{*}}{

Alternative method for last 4 marks

\multirow[b]{3}{*}{

}

\multirow{3}{*}{}

\multirow[t]{2}{*}{

М1

}

\(\begin{aligned}

\frac { 1 } { 2 } \times 4 \times 5 ^ { 2 } - \frac { 1 } { 2 } \times 4 \times 2 ^ { 2 }

\frac { 1 } { 2 } \times 3 \times 2 ^ { 2 } - \frac { 1 } { 2 } \times 3 \times 12 \end{aligned}\)

\(42 \text { or } 12 \text { so loss } = 42 + ( - 12 )\)

Can be implied by correct

\(= 30 \mathrm {~J}\)

Not -30

[5]

Question

Answer

Marks

AOs

Guidance

(a)

\(T = \frac { \lambda x } { l }\) and \(r = l + x\) both used in solution \(T = \frac { m v ^ { 2 } } { l + x }\)

М1

[3]

3.3

Use of \(F = m a\) with centripetal acceleration

(b)

\(\begin{aligned}

x = \frac { - \lambda l + \sqrt { ( \lambda l ) ^ { 2 } - 4 \lambda \left( - l m v ^ { 2 } \right) } } { 2 \lambda }

x = \frac { l } { 2 } \left( \left( 1 + \frac { 4 m v ^ { 2 } } { \lambda l } \right) ^ { \frac { 1 } { 2 } } - 1 \right)

x = \frac { l } { 2 } \left( 1 + \frac { 1 } { 2 } \frac { 4 m v ^ { 2 } } { \lambda l } + \ldots - 1 \right)

x \approx \frac { l } { 2 } \left( \frac { 1 } { 2 } \frac { 4 m v ^ { 2 } } { \lambda l } \right) = \frac { m v ^ { 2 } } { \lambda } \Rightarrow \lambda x \approx m v ^ { 2 } \end{aligned}\)

М1

[3]

2.1

3.1b

1.1

Use of the quadratic equation formula

Reject negative route and rearranging to form with \(\sqrt { 1 + \ldots }\)

Use of binomial series

No need to mention \(\left| \frac { 4 m v ^ { 2 } } { \lambda l } \right| < 1\)

(c)

\(\lambda x \approx m v ^ { 2 } \Rightarrow \frac { \lambda x } { l } = T \approx \frac { m v ^ { 2 } } { l }\) and if the string were inextensible, corresponding to an infinite value of \(\lambda\) and \(x\) being 0 , then \(l\) would be the radius of the motion and so the RHS would be the centripetal force

[1]

3.5a

(d)

\(v \approx \sqrt { \frac { 260 \times 0.03 } { 0.16 } } =\) awrt 7.0

[1]

3.4

(6.982...)

Question			Answer	Marks	AOs	Guidance
4	(e)		While \(v\) in this situation is slightly below 7 nevertheless it is an estimate so we cannot be certain that the modelled value does not exceed 7 in which case the assumption would not be justified	B1	2.2b

OCR Further Mechanics 2021 June Q1

1 A car of mass 800 kg is driven with its engine generating a power of 15 kW .

The car is first driven along a straight horizontal road and accelerates from rest. Assuming that there is no resistance to motion, find the speed of the car after 6 seconds.
The car is next driven at constant speed up a straight road inclined at an angle \(\theta\) to the horizontal. The resistance to motion is now modelled as being constant with magnitude 150 N . Given that \(\sin \theta = \frac { 1 } { 20 }\), find the speed of the car.
The car is now driven at a constant speed of \(30 \mathrm {~ms} ^ { - 1 }\) along the horizontal road pulling a trailer of mass 150 kg which is attached by means of a light rigid horizontal towbar. Assuming that the resistance to motion of the car is three times the resistance to motion of the trailer, find
- the resistance to motion of the car,
- the magnitude of the tension in the towbar.

OCR Further Mechanics 2021 June Q2

2 One end of a light inextensible string of length 0.8 m is attached to a fixed point, \(O\). The other end is attached to a particle \(P\) of mass \(1.2 \mathrm {~kg} . P\) hangs in equilibrium at a distance of 1.5 m above a horizontal plane. The point on the plane directly below \(O\) is \(F\).
\(P\) is projected horizontally with speed \(3.5 \mathrm {~ms} ^ { - 1 }\). The string breaks when \(O P\) makes an angle of \(\frac { 1 } { 3 } \pi\) radians with the downwards vertical through \(O\) (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{0428f2f2-12c4-4e89-93ab-8cfe2c5aca4a-02_757_889_1482_251}

Find the magnitude of the tension in the string at the instant before the string breaks.
Find the distance between \(F\) and the point where \(P\) first hits the plane.

OCR Further Mechanics 2021 June Q3

37 marks

3 This question is about modelling the relation between the pressure, \(P\), volume, \(V\), and temperature, \(\theta\), of a fixed amount of gas in a container whose volume can be varied. The amount of gas is measured in moles; 1 mole is a dimensionless constant representing a fixed number of molecules of gas. Gas temperatures are measured on the Kelvin scale; the unit for temperature is denoted by K . You may assume that temperature is a dimensionless quantity. A gas in a container will always exert an outwards force on the walls of the container. The pressure of the gas is defined to be the magnitude of this force per unit area of the walls, with \(P\) always positive. An initial model of the relation is given by \(P ^ { \alpha } V ^ { \beta } = n R \theta\), where \(n\) is the number of moles of gas present and \(R\) is a quantity called the Universal Gas Constant. The value of \(R\), correct to 3 significant figures, is \(8.31 \mathrm { JK } ^ { - 1 }\).

Show that \([ P ] = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 }\) and \([ R ] = \mathrm { ML } ^ { 2 } \mathrm {~T} ^ { - 2 }\).
Hence show that \(\alpha = 1\) and \(\beta = 1\). 5 moles of gas are present in the container which initially has volume \(0.03 \mathrm {~m} ^ { 3 }\) and which is maintained at a temperature of 300 K .
Find the pressure of the gas, as predicted by the model. An improved model of the relation is given by \(\left( P + \frac { a n ^ { 2 } } { V ^ { 2 } } \right) ( V - n b ) = n R \theta\), where \(a\) and \(b\) are constants.
Determine the dimensions of \(b\) and \(a\). The values of \(a\) and \(b\) (in appropriate units) are measured as being 0.14 and \(3.2 \times 10 ^ { - 5 }\) respectively.
Find the pressure of the gas as predicted by the improved model. Suppose that the volume of the container is now reduced to \(1.5 \times 10 ^ { - 4 } \mathrm {~m} ^ { 3 }\) while keeping the temperature at 300 K .

By considering the value of the pressure of the gas as predicted by the improved model, comment on the validity of this model in this situation. \section*{Total Marks for Question Set 5: 37} \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.

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.

\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.

15000 \times 6 = \frac { 1 } { 2 } \times 800 v ^ { 2 }

v = \sqrt { 225 } = 15 \mathrm {~m} \mathrm {~s} ^ { - 1 } \end{aligned}\)

[2]

1.1a

1.1

Finding energy input using \(W = P t\) and equating to KE gained

\(\begin{aligned}

\text { Or } \frac { 15000 } { v } = 800 \frac { \mathrm {~d} v } { \mathrm {~d} t }

\Rightarrow \frac { 75 } { 2 } \int _ { 0 } ^ { 6 } \mathrm {~d} t = \int _ { 0 } ^ { v } 2 v \mathrm {~d} v \text { oe } \end{aligned}\)

(b)

Considering forces along the road: \(800 g \sin \theta + 150 = D\) \(15000 = D v\) \(v = \frac { 15000 } { 800 g \times \frac { 1 } { 20 } + 150 } = \text { awrt } 27.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)

[3]

1.1a

1.1

Where \(D\) is the driving force

Use of \(P = F v\)

27.675276....

Or \(W = 800 g d \sin \theta + 150 d\), where \(d\) is distance travelled …

\(\ldots\) in time \(t\) and \(W = 15000 t\) and \(v = \frac { d } { t }\)

(c)

\(\begin{aligned}	D = \frac { 15000 } { 30 } \quad ( = 500 )
	\frac { 15000 } { 30 } = R _ { \text {total } }
	R _ { \mathrm { C } } = \frac { 3 } { 4 } \times 500 = 375 \mathrm {~N} \end{aligned}\)
Car: \(500 = T + 375 \Rightarrow T = 125 \mathrm {~N}\)

... equals total resistance

Or Trailer: \(T = R _ { \mathrm { T } } = \frac { 1 } { 4 } \times 500 = 125 \mathrm {~N}\)

Question		Answer	Marks	AOs	Guidance
\multirow[t]{5}{*}{2}	\multirow[t]{5}{*}{(a)}	\(\frac { 1 } { 2 } m \times 3.5 ^ { 2 } = \frac { 1 } { 2 } m v ^ { 2 } + m g \times 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right)\)	M1	3.1b	Conservation of energy (could be general angle for RHS)
		\(v ^ { 2 } = 3.5 ^ { 2 } - 0.8 g = 4.41\)	A1	1.1	Or \(v = 2.1\)
		\(T - 1.2 g \cos \frac { 1 } { 3 } \pi = \frac { 1.2 v ^ { 2 } } { 0.8 }\)	M1	3.1b	Resolving weight and use of NII with correct centripetal acceleration
		So tension when string breaks is awrt 12.5 N	A1	3.2a
			[4]

\multirow[t]{3}{*}{2}

\multirow[t]{3}{*}{(b)}

\(v _ { y } = 2.1 \sin \frac { 1 } { 3 } \pi\) upwards

\(P\) starts freefall \(1.5 + 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right)\) above plane \(- \left( 1.5 + 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right) \right) = \left( 2.1 \sin \frac { 1 } { 3 } \pi \right) t - \frac { 1 } { 2 } g t ^ { 2 }\)

\(- 1.9 = ( 1.05 \sqrt { 3 } ) t - 4.9 t ^ { 2 }\) or \(4.9 t ^ { 2 } - 1.8186 \ldots t - 1.9 = 0\)

\(t\) = awrt 0.835

Horizontal distance in freefall is \(2.1 \cos \frac { 1 } { 3 } \pi \times 0.835 \ldots\)

\(0.8 \sin \frac { 1 } { 3 } \pi + 2.1 \cos \frac { 1 } { 3 } \pi \times 0.835 \ldots =\) awrt 1.57 m

Direction can be implied by eg use of -9.8

(=1.9)

Setting up 3 term quadratic equation in \(t\) with adjustment to \(\pm 1.5\) as displacement

Correct equation soi

0.877...

horizontal distance while attached plus horizontal distance in freefall.

\(\begin{aligned}

\frac { \sqrt { 331 } + 3 \sqrt { 3 } } { 28 }

\frac { 3 \sqrt { 331 } + 9 \sqrt { 3 } } { 80 } \end{aligned}\)

Alternative solution

Projection velocity is \(2.1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at \(\frac { 1 } { 3 } \pi\) above horizontal

\(P\) starts freefall \(1.5 + 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right)\) above plane

Equation of trajectory: \(y = x \tan \frac { 1 } { 3 } \pi - \frac { g x ^ { 2 } } { 2 \times 2.1 ^ { 2 } \cos ^ { 2 } \left( \frac { 1 } { 3 } \pi \right) }\) \(- 1.9 = x \sqrt { 3 } - \frac { 40 } { 9 } x ^ { 2 } \text { or } 4.444 \ldots x ^ { 2 } - 1.732 \ldots x - 1.9 = 0\)

\(x =\) awrt 0.877

\(0.8 \sin \frac { 1 } { 3 } \pi + 0.877 \ldots =\) awrt 1.57 m

Or suitably adjusted if not using \(P\) 's position when string breaks as the origin

Correct equation (with their 2.1) soi oe: formulation of explicit quadratic equation in \(x\)

\(\frac { 3 \sqrt { 331 } + 9 \sqrt { 3 } } { 80 }\)

{ [ F ] = [ m a ] = \mathrm { MLT } ^ { - 2 } }

{ [ P ] = \frac { [ F ] } { [ A ] } = \mathrm { MLT } ^ { - 2 } \mathrm {~L} ^ { - 2 } = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 } }

{ [ R ] = [ F ] [ d ] }

{ [ R ] = \mathrm { MLT } ^ { - 2 } \mathrm {~L} = \mathrm { ML } ^ { 2 } \mathrm {~T} ^ { - 2 } } \end{aligned}\)

Determining dimensions of energy

(b)

\(\begin{aligned}

\left( \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 } \right) ^ { \alpha } \mathrm { L } ^ { 3 \beta } = \mathrm { ML } ^ { 2 } \mathrm {~T} ^ { - 2 }

\mathrm { M } ^ { \alpha } = \mathrm { M } \Rightarrow \alpha = 1

\mathrm {~L} : - \alpha + 3 \beta = 2 \Rightarrow \beta = 1 \text { (AGG) } \end{aligned}\)

Using their dimensions or similarly with T

\(n\) and \(\theta\) dimensionless

(c)

\(\begin{aligned}

P = \frac { 5 \times 8.31 \times 300 } { 0.03 }

415500 \mathrm { Nm } ^ { - 2 } \end{aligned}\)

[2]

3.4

1.1

Correctly substituting values

With their \(\alpha\) and \(\beta\)

(d)

\(\begin{aligned}

{ [ b ] = \mathrm { L } ^ { 3 } }

{ [ a ] = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 } \left( \mathrm {~L} ^ { 3 } \right) ^ { 2 } }

{ [ a ] = \mathrm { ML } ^ { 5 } \mathrm {~T} ^ { - 2 } } \end{aligned}\)

[3]

2.2a

1.1

Using equality of \(\left[ a / V ^ { 2 } \right]\) and \([ P ]\)

With \(n\) dimensionless

(e)

\(P + \frac { 0.14 \times 5 ^ { 2 } } { 0.03 ^ { 2 } } = \frac { 5 \times 8.31 \times 300 } { 0.03 - 5 \times 3.2 \times 10 ^ { - 5 } }\)

awrt \(414000 \mathrm { Nm } ^ { - 2 }\)

[2]

3.4

1.1

Correctly substituting values

With their \(\alpha\) and \(\beta\)

(f)

\(P = \frac { 5 \times 8.31 \times 300 } { 1.5 \times 10 ^ { - 4 } - 5 \times 3.2 \times 10 ^ { - 5 } } - \frac { 0.14 \times 5 ^ { 2 } } { \left( 1.5 \times 10 ^ { - 4 } \right) ^ { 2 } } \approx - 1.4 \times 10 ^ { 9 }\)

A negative value is not a valid value for pressure...

...and so the model is not valid for these values

OR from \(V - n b < 0\), so since \(R , \theta , n\) and \(V\) are all positive \(P < 0\)

OCR Further Mechanics 2021 June Q2

2 Three particles, \(A , B\) and \(C\), of masses \(2 \mathrm {~kg} , 3 \mathrm {~kg}\) and 5 kg respectively, are at rest in a straight line on a smooth horizontal plane with \(B\) between \(A\) and \(C\). Collisions between \(A\) and \(B\) are perfectly elastic. The coefficient of restitution for collisions between \(B\) and \(C\) is \(e\).
\(A\) is projected towards \(B\) with a speed of \(5 u \mathrm {~ms} ^ { - 1 }\) (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{709f3a7a-d857-4813-98ab-de6b41a3a8dc-02_190_885_1151_260} Show that only two collisions occur.

OCR Further Mechanics 2021 June Q3

3 A particle \(P\) of mass 8 kg moves in a straight line on a smooth horizontal plane. At time \(t \mathrm {~s}\) the displacement of \(P\) from a fixed point \(O\) on the line is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). Initially, \(P\) is at rest at \(O\).
\(P\) is acted on by a horizontal force, directed along the line away from \(O\), with magnitude proportional to \(\sqrt { 9 + v ^ { 2 } }\). When \(v = 1.25\), the magnitude of this force is 13 N .

Show that \(\frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 }\).
Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
Find an expression for \(x\) in terms of \(t\) for \(t \geqslant 0\).

OCR Further Mechanics 2021 June Q4

36 marks

4 Particles \(A , B\) and \(C\) of masses \(2 \mathrm {~kg} , 3 \mathrm {~kg}\) and 5 kg respectively are joined by light rigid rods to form a triangular frame. The frame is placed at rest on a horizontal plane with \(A\) at the point \(( 0,0 )\), \(B\) at the point ( \(0.6,0\) ) and \(C\) at the point ( \(0.4,0.2\) ), where distances in the coordinate system are measured in metres (see Fig. 1). \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{709f3a7a-d857-4813-98ab-de6b41a3a8dc-03_311_661_338_258} \captionsetup{labelformat=empty} \caption{Fig. 1}

\end{figure} \(G\), which is the centre of mass of the frame, is at the point \(( \bar { x } , \bar { y } )\).

- Show that \(\bar { x } = 0.38\).
- Find \(\bar { y }\).
- Explain why it would be impossible for the frame to be in equilibrium in a horizontal plane supported at only one point.
A rough plane, \(\Pi\), is inclined at an angle \(\theta\) to the horizontal where \(\sin \theta = \frac { 3 } { 5 }\). The frame is placed on \(\Pi\) with \(A B\) vertical and \(B\) in contact with \(\Pi . C\) is in the same vertical plane as \(A B\) and a line of greatest slope of \(\Pi . C\) is on the down-slope side of \(A B\). The frame is kept in equilibrium by a horizontal light elastic string whose natural length is \(l \mathrm {~m}\) and whose modulus of elasticity is \(g \mathrm {~N}\). The string is attached to \(A\) at one end and to a fixed point on \(\Pi\) at the other end (see Fig. 2). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{709f3a7a-d857-4813-98ab-de6b41a3a8dc-03_605_828_1525_248} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} The coefficient of friction between \(B\) and \(\Pi\) is \(\mu\).
Show that \(l = 0.3\).

Show that \(\mu \geqslant \frac { 14 } { 27 }\). \section*{Total Marks for Question Set 6: 38} \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.

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.

\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.

\mathrm { KE } = \frac { 1 } { 2 } \times 2 \times \binom { - 19.5 } { - 60 } \cdot \binom { - 19.5 } { - 60 } = 19.5 ^ { 2 } + 60 ^ { 2 }

\text { awrt } 3980 \mathrm {~J} \end{aligned}\)

Using KE \(= \frac { 1 } { 2 } m \mathbf { v } . \mathbf { v }\) or \(\frac { 1 } { 2 } m v ^ { 2 }\) and attempting to find dot product or magnitude

(b)

\(\begin{aligned}	\binom { 4 t } { - 2 } = 2 \mathbf { a } = 2 \frac { \mathrm {~d} \mathbf { v } } { \mathrm {~d} t }
	\Rightarrow \mathbf { v } = \binom { t ^ { 2 } } { - t } + \mathbf { u } \end{aligned}\)
\(t = 0 \Rightarrow \mathbf { u } = \binom { - 19.5 } { - 60 }\) so \(\mathbf { v } = \binom { t ^ { 2 } - 19.5 } { - t - 60 }\) oe \(P = \mathbf { F } \cdot \mathbf { v } = \binom { 4 t } { - 2 } \cdot \binom { t ^ { 2 } - 19.5 } { - t - 60 } = 4 t ^ { 3 } - 76 t + 120\)
\(4 t ^ { 3 } - 76 t + 120 = 0\)
\(t = 2,3\)

Using \(\mathbf { F } = m \mathbf { a }\) with \(\mathbf { a } = \frac { \mathrm { d } \mathbf { v } } { \mathrm { d } t }\) soi

Integrating to find \(\mathbf { v }\) (u may be missing or may look like a scalar at this stage)

Using \(P =\) F.v and attempt at dot product

\(\mathbf { B C }\) If \(t = - 5\) included then A0

Must be reasonable attempt at integration

Question

Answer

Marks