Questions - OCR - A-Level Maths

OCR FS1 AS 2021 June Q2

23 marks

2 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by $X$.

State a further assumption needed for $X$ to be well modelled by a Poisson distribution. Assume now that $X$ can be well modelled by the distribution $\operatorname { Po } ( 0.7 )$.
Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws. The number of flaws in 100 m of FOC of a larger diameter has the distribution $\mathrm { Po } ( 1.6 )$.
Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4 . Judith believes that mathematical ability and chess-playing ability are related. She asks 20 randomly chosen chess players, with known British Chess Federation (BCF) ratings $X$, to take a mathematics aptitude test, with scores $Y$. The results are summarised as follows. $$n = 20 , \Sigma x = 3600 , \Sigma x ^ { 2 } = 660500 , \Sigma y = 1440 , \Sigma y ^ { 2 } = 105280 , \Sigma x y = 260990$$
Calculate the value of Pearson's product-moment correlation coefficient $r$.
State an assumption needed to be able to carry out a significance test on the value of $r$.
Assume now that the assumption in part (ii) is valid. Test at the $5 \%$ significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\text { ELO rating } = 8 \times \text { BCF rating } + 650$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.
Calculate the value of Pearson's product-moment correlation coefficient $r$.
State an assumption needed to be able to carry out a significance test on the value of $r$.
Assume now that the assumption in part (b) is valid. Test at the $5 \%$ significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.

There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\mathrm { ELO } \text { rating } = 8 \# \mathrm { BCF } \text { rating } + 650 .$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion. An environmentalist measures the mean concentration, $c$ milligrams per litre, of a particular chemical in a group of rivers, and the mean mass, $m$ pounds, of fish of a certain species found in those rivers. The results are given in the table. \end{table}

Question

Answer

Marks

AO

Guidance

1

(a)

\(\begin{aligned}

0.25 + 0.36 + x + x ^ { 2 } = 1

x ^ { 2 } + x - 0.39 = 0

x = 0.3 \text { (or } - 1.3 \text { ) }

x \text { cannot be negative }

\mathrm { E } ( W ) = 2.23

\mathrm { E } \left( W ^ { 2 } \right) = \Sigma w ^ { 2 } \mathrm { p } ( w ) \quad [ = 5.83 ]

\text { Subtract } [ \mathrm { E } ( W ) ] ^ { 2 } \text { to get } \mathbf { 0 . 8 5 7 1 } \end{aligned}\)

\(\begin{gathered} \text { M1 }

\text { A1 }

\text { B1ft }

\text { B1 }

\text { M1 }

\text { A1 }

{ [ 7 ] } \end{gathered}\)

3.1a

1.1b

2.3

1.1b

1.1

2.1

Equation using $\Sigma p = 1$

Correct simplified quadratic Correctly obtain $x = 0.3$

Explicitly reject other solution

2.23 or exact equivalent only Use $\Sigma w ^ { 2 } \mathrm { p } ( w )$

Correctly obtain given answer, www

Can be implied

Method needed ft on their quadratic Allow for $\mathrm { E } ( W ) ^ { 2 } = 4.9729$

Need 2.23 or 4.9729 and 5.83 or full numerical $\Sigma w ^ { 2 } \mathrm { p } ( w )$

1

(b)

$9 \times 0.8571 = 7.7139$

B1

[1]

1.1b

Allow 7.71 or 7.714

2

(a)

Flaws must occur at constant average rate (uniform rate)

B1

[1]

1.2

Context (e.g. "flaws") needed

Extra answers, e.g. "singly": B0

Not "constant rate" or "average constant rate".

2

(b)

$\operatorname { Po(2.1)~or~ } e ^ { - \lambda } \frac { \lambda ^ { 3 } } { 3 ! }$

M1

A1

[2]

1.1

1.1b

Po(2.1) stated or implied, or formula with $\lambda = 2.1$ stated Awrt 0.189

2

(c)

Po(3)

$1 - \mathrm { P } ( \leq 3 )$

M1

A1

[3]

1.1

1.1b

$\operatorname { Po } ( 2 \times 0.7 + 1.6 )$ stated or implied

Allow $1 - \mathrm { P } ( \leq 4 ) = 0.1847$, or from wrong $\lambda$

Awrt 0.353

Or all combinations $\leq 3$

$1 -$ above, not just $= 3$

Question

Answer

Marks

AO

Guidance

3

(a)

0.4(00)

B2

[2]

1.1

1.1b

SC: if B0, give SC B1 for two of $S _ { x x } = 12500 , S _ { y y } = 1600 , S _ { x y } = 1790$ and $S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)$

Also allow SC B1 for equivalent methods using Covariance \

SDs

3

(b)

Data needs to have a bivariate normal distribution

B1

[1]

1.2

Needs "bivariate normal" or clear equivalent. Not just "both normally distributed"

Allow "scatter diagram forms ellipse"

3

(c)

$\mathrm { H } _ { 0 }$ : higher maths scores are not associated with higher BCF grading; $\mathrm { H } _ { 1 }$ : positively associated

CV 0.3783

$0.400 > 0.3783$ so reject $\mathrm { H } _ { 0 }$

Significant evidence that higher maths scores are associated with higher BCF grading

B1

M1ft

A1ft

[4]

2.5

1.1b

2.2b

3.5a

Needs context and clearly onetailed $O R \rho$ used and defined Not "evidence that ..."

Allow 0.378

Reject/do not reject $\mathrm { H } _ { 0 }$

Contextualised, not too definite Needn't say "positive" if $\mathrm { H } _ { 1 } \mathrm { OK }$

SC 2-tail: B0; 0.4438, or 0.3783 B1; then M1A0

$\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho > 0$ where $\rho$ is population pmcc (not $r$ )

FT on their $r$, but not CV

Not "scores are associated

...". FT on their $r$ only

3

(d)

It makes no difference as this is a linear transformation

B1

[1]

2.2a

Need both "unchanged" oe and reason, need "linear" or exact equivalent

"oe" includes "their 0.4"

4

(a)

Neither

B1

[1]

2.5

OE

Not "neither is independent of the other"

4

(b)

$c = 2.848 - 0.1567 m \quad \mathbf { B C }$

B1

[3]

1.1

Correct $a$, awrt 2.85

Correct $b$, awrt 0.157

Letters correct from correct method

(If both wrongly rounded, e.g. $c = 2.84 - 0.156 m$, give B2)

$\mathrm { SC } : m$ on $c$ :

$m = 15.65 - 4.832 c$ : B2

$y = 15.65 - 4.832 x$ : B1

$c = 15.65 - 4.832 m : \mathrm { B } 1$

If B0B0, give B1 for correct letters from valid working

Question

Answer

Marks

AO

Guidance

4

(c)

$a$ unchanged, $b$ multiplied by 2.2 (allow " $a$ unchanged, $b$ increases", etc)

B1 [1]

2.2a

oe, e.g. $c = 2.848 - 0.345 m$; $m = 7.114 - 2.196 c$

SC: $m$ on $c$ in (b): Both divided by 2.2 B1

4

(d)

Draw approximate line of best fit

Draw at least one vertical from line to point

Say that "Best fit" line minimises the sum of squares of these distances

M1

A1

[3]

1.1

2.4

Needs M2 and "minimises" and "sums of squares" oe

SC: Horizontal(s):

full marks (indept of (b))

OCR FS1 AS 2021 June Q1

1 On any day, the number of orders received in one randomly chosen hour by an online supplier can be modelled by the distribution $\mathrm { Po } ( 120 )$.

Find the probability that at least 28 orders are received in a randomly chosen 10 -minute period.
Find the probability that in a randomly chosen 10-minute period on one day and a randomly chosen 10-minute period on the next day a total of at least 56 orders are received.
State a necessary assumption for the validity of your calculation in part (b).

OCR FS1 AS 2021 June Q2

2 The members of a team stand in a random order in a straight line for a photograph. There are four men and six women.

Find the probability that all the men are next to each other.
Find the probability that no two men are next to one another.

OCR FS1 AS 2021 June Q3

28 marks

3 Sixteen candidates took an examination paper in mechanics and an examination paper in statistics.

For all sixteen candidates, the value of the product moment correlation coefficient $r$ for the marks on the two papers was 0.701 correct to 3 significant figures. Test whether there is evidence, at the $5 \%$ significance level, of association between the marks on the two papers.
A teacher decided to omit the marks of the candidates who were in the top three places in mechanics and the candidates who were in the bottom three places in mechanics. The marks for the remaining 10 candidates can be summarised by
$n = 10 , \Sigma x = 750 , \Sigma y = 690 , \Sigma x ^ { 2 } = 57690 , \Sigma y ^ { 2 } = 49676 , \Sigma x y = 50829$.
1. Calculate the value of $r$ for these 10 candidates.
2. What do the two values of $r$, in parts (a) and (b)(i), tell you about the scores of the sixteen candidates? A bag contains a mixture of blue and green beads, in unknown proportions. The proportion of green beads in the bag is denoted by $p$.
Sasha selects 10 beads at random, with replacement. Write down an expression, in terms of $p$, for the variance of the number of green beads Sasha selects. Freda selects one bead at random from the bag, notes its colour, and replaces it in the bag. She continues to select beads in this way until a green bead is selected. The first green bead is the $X$ th bead that Freda selects.
Assume that $p = 0.3$. Find
1. $\mathrm { P } ( X \geqslant 5 )$,
2. $\operatorname { Var } ( X )$.

In fact, on the basis of a large number of observations of $X$, it is found that $\mathrm { P } ( X = 3 ) = \frac { 4 } { 25 } \times \mathrm { P } ( X = 1 )$. Estimate the value of $p$. \section*{Total Marks for Question Set 4: 29} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. 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. 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'. 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 $\mathbf { 3 ~ s . f . }$ 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. 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
oe	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

AO

Guidance

1

(a)

$1 - \mathrm { P } ( \leq 27 )$

$= 0.0525 \quad \mathbf { B C }$

M1

A1

[2]

3.4

1.1

Allow M1 for $1 - 0.9657 = 0.0343$

In range [0.0524, 0.0525] BC

1

(b)

Po(40)

$1 - \mathrm { P } ( \leq 55 )$

$= 0.00968$

M1*

depM1

A1

[3]

3.1b

1.1a

1.1

$\operatorname { Po( } 2 \times$ their 20) stated or implied

Allow M1 for $1 - \mathrm { P } ( \leq 56 ) = 0.00658$ or $1 - 0.990 = 0.01$ or 0.0097

Awrt 0.00968

1

(c)

Orders on one day are independent of orders on the other

B1

[1]

3.2b

Use "orders independent", clearly referred to the two different days, needs context [not "events"], and nothing else

Not anything affecting given separate Poissons, such as "orders must be independent" or "constant average rate".

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

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

\multirow{3}{*}{}

$7 ! \times 4$ !

$\div 10$ ! $= \frac { 120960 } { 3628800 } = \frac { 1 } { 30 }$

M1

A1

2.1

1.1a

1.1

Allow for $6 ! \times 4$ ! or $6 ! \times 4 ! \times 2$

Divide by 10!, needs at least one factorial in numerator

Answer, exact or awrt 0.0333

3/3 for $\frac { 1 } { 30 }$ www

Alternative: $7 \times \frac { 6 } { 10 } \times \frac { 4 } { 9 } \times \frac { 5 } { 8 } \times \frac { 3 } { 7 } \times \frac { 4 } { 6 } \times \frac { 2 } { 5 } \times \frac { 3 } { 4 } \times \frac { 1 } { 3 } \times \frac { 2 } { 2 } \times \frac { 1 } { 1 }$

M1

A1

no 7, one other error

only one error

correct answer

[3]

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

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

\multirow{3}{*}{}

Women placed in 6 ! ways, men in 4 ! $[ = 720 \times 24 ]$

4 slots $m$ in $m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m = { } ^ { 7 } C _ { 4 }$ ${ } ^ { 7 } C _ { 4 } \times \frac { 6 ! \times 4 ! } { 10 ! }$

$= \frac { 1 } { 6 }$

B1

М1

A1

2.1

3.1b

1.1a

1.1

$6 ! \times 4 !$ anywhere, or $6 ! \times$ attempt at ${ } ^ { 7 } P _ { 4 }$

Or ${ } ^ { 7 } P _ { 4 }$. Allow for $m$ and $W$ reversed

Needs attempt at both terms

Or 0.167 or 0.1667 etc

Or $6 ! \times 7 \times 6 \times 5 \times 4 \quad$ B2

${ } ^ { 7 } P _ { 4 } \times 6$ !: B1M1

$4 \times ( 6 ! \times 4 ! ) / 10 ! = 2 / 105 :$ В 1 М 1

Alternative: PIE

$( 10 ! - 12 \times 9 ! + ( 3 \times 4 \times 8 ! + 12 \times 2 \times 8 ! ) - 24 \times 7 ! ) / 10 !$

$[ = ( 3628800 - 4354560 + 1451520 - 120960 ) / 10 ! ]$

M2

A1

Signs alternating, at least one term $\sqrt { }$ Allow one term omitted or wrong Correct answer

[3]

Three together: $7 \times 6 \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 5 }$

Two pairs: $\frac { 7 \times 6 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! }$

$\frac { 1 } { 10 }$

One pair: $7 \times \frac { 6 \times 5 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 2 }$

Question

Answer

Marks

AO

Guidance

3

(a)

$\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho \neq 0$, where $\rho$ is population pmcc

0.701 > 0.4973

Reject $\mathrm { H } _ { 0 }$. There is significant evidence of association between the marks on the two papers

B1

M1ft

A1

[4]

2.5

1.1a

1.1

2.2b

Must use symbols. Allow no definition of letter if $\rho$ used

Correct CV stated, allow 0.497

FT on wrong CV

Not FT. Needs context, and not too definite.

Not " $\mathrm { H } _ { 0 }$ : there is no assoc' n , $\mathrm { H } _ { 1 }$ : there is association"

Not There is association ..."

3

(b)

(i)

-0.534

B2

[2]

1.1a

1.1

SC: if B0, give B1 for two of 1440, 2066, -921 and $S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)$

-0.53: B1

3

(b)

(ii)

6 candidates did very well or very badly on both papers; middle 10 tended to do badly on one paper and well on the other

B1

[1]

2.4

Correct inference about scores oe, not "correlation/association/value of $r$ ". Not "outliers" or "anomalies".

Allow inference for one group only, provided it is clearly for only one group \

any ref to other group is not wrong

4

(a)

$10 p ( 1 - p )$

B1

[1]

1.2

Allow $10 p q$ oe, e.g. $10 p - 10 p ^ { 2 }$

Not just $n p ( 1 - p )$

4

(b)

(i)

$0.7 ^ { 4 }$

= 0.240(1)

M1

A1

[2]

1.1a

1.1

$0.7 ^ { 5 } = 0.168$ or $0.7 ^ { 6 } = 0.118$ : M1

Allow 0.24

Or $1 - 0.3 \left( 1 + 0.7 + 0.7 ^ { 2 } + 0.7 ^ { 3 } \right)$ Allow M1 if also $0.3 \times 0.7 ^ { 4 }$ [0.15 is from binomial]

4

(b)

(ii)

$q / p ^ { 2 } = \frac { 70 } { 9 }$ or $7.777 \ldots$

B1

[1]

1.1

Allow 7.78, 7.778, etc

Allow 8 only if evidence, e.g. ( $1 - 0.3$ )/ $0.3 ^ { 2 }$

4

(c)

\(\begin{aligned}

( 1 - p ) ^ { 2 } p = \frac { 4 } { 25 } p

p = 0 \text { or } ( 1 - p ) ^ { 2 } = \frac { 4 } { 25 } \quad ( p \neq 0 )

( 1 - p ) = \pm \frac { 2 } { 5 }

p \neq \frac { 7 } { 5 }

p = \frac { 3 } { 5 } \end{aligned}\)

B1

M1

B1ft

A1

[5]

1.1

1.1a

1.1

2.3

2.1

Correct equation

Reduce to quadratic/cubic and solve

Obtain two non-zero solutions

Explicitly discard one solution, either here or in line 2 (not enough to give 2 answers and then only 1 )

Exact final answer exact (0.6) no others left, allow from ± omitted

e.g. $p \left( p ^ { 2 } - 2 p + \frac { 21 } { 25 } \right) = 0$ ± omitted: M0B0A1
Allow " $p = 0 , \frac { 3 } { 5 } , \frac { 7 } { 5 }$ but $p \leq 1$ "
SC binomial: B0 then \(75 p ^ { 2 } = ( 1 - p ) ^ { 2 } \	\) solve M1 $0.104 [ 0.1035 ] \quad$ A1 Explicitly reject 0 or - 0.13 B1 SC Poisson: 0

OCR FS1 AS 2021 June Q1

1 Five observations of bivariate data $( x , y )$ are given in the table.

$x$	7	8	12	6	4
$y$	20	16	7	17	23

Find the value of Pearson's product-moment correlation coefficient.
State what your answer to part (a) tells you about a scatter diagram representing the data.
A new variable $a$ is defined by $a = 3 x + 4$. Dee says "The value of Pearson's product-moment correlation coefficient between $a$ and $y$ will not be the same as the answer to part (a)." State with a reason whether you agree with Dee. An investor obtains data about the profits of 8 randomly chosen investment accounts over two one-year periods. The profit in the first year for each account is $p \%$ and the profit in the second year for each account is $q \%$. The results are shown in the table and in the scatter diagram.
Account A B C D E F G H
$p$ 1.6 2.1 2.4 2.7 2.8 3.3 5.2 8.4
$q$ 1.6 2.3 2.2 2.2 3.1 2.9 7.6 4.8
$n = 8 \quad \Sigma p = 28.5 \quad \Sigma q = 26.7 \quad \Sigma p ^ { 2 } = 136.35 \quad \Sigma q ^ { 2 } = 116.35 \quad \Sigma p q = 116.70$
\includegraphics[max width=\textwidth, alt={}, center]{4c7546b9-03ee-47a1-915f-41e2b4ca19c0-03_762_1248_906_260}
State which, if either, of the variables $p$ and $q$ is independent.
Calculate the equation of the regression line of $q$ on $p$.
1. Use the regression line to estimate the value of $q$ for an investment account for which $p = 2.5$.
2. Give two reasons why this estimate could be considered reliable.
Comment on the reliability of using the regression line to predict the value of $q$ when $p = 7.0$.

OCR FS1 AS 2021 June Q3

38 marks

3 At a cinema there are three film sessions each Saturday, "early", "middle" and "late". The numbers of the audience, in different age groups, at the three showings on a randomly chosen Saturday are given in Table 1. \begin{table}[h] \end{table}

Question

Solution

Marks

AOs

Guidance

1

(a)

-0.954 BC

B2 [2]

1.1 1.1

SC: If B0, give B1 if two of 7.04, 29.0[4], -13.6[4] (or 35.2, 145[.2], -68.2) seen

1

(b)

Points lie close to a straight line Line has negative gradient

B1 B1 [2]

2.2b 1.1

Must refer to line, not just "negative correlation"

1

(c)

No, it will be the same as $x \rightarrow a$ is a linear transformation

B1 [1]

2.2a

OE. Either "same" with correct reason, or "disagree" with correct reason. Allow any clear valid technical term

2

(a)

Neither

B1 [1]

1.2

2

(b)

$q = 1.13 + 0.620 p$

B1B1 B1 [3]

1.1,1.1 1.1

0.62(0) correct; both numbers correct Fully correct answer including letters

2

(c)

(i)

2.68

B1ft [1]

1.1

awrt 2.68, ft on their (b) if letters correct

2

(c)

(ii)

2.5 is within data range, and points (here) are close to line/well correlated

B1 B1 [2]

2.2b 2.2b

At least one reason, allow "no because points not close to line" Full argument, two reasons needed

2

(d)

Not much data here/points scattered/ possible outliers

So not very reliable

M1 A1 [2]

2.3 1.1

Reason for not very reliable (not "extrapolation") Full argument and conclusion, not too assertive (not wholly unreliable!)

3

(a)

Expected frequency for Middle/25 to 60 is 4.4 which is < 5 so must combine cells

B1*ft depB1 [2]

2.4 3.5b

Correctly obtain this $F _ { E }$, ft on addition errors " < 5" explicit and correct deduction

3

(b)

Early	Middle	Late
29.4	23.1	31.5
26.6	20.9	28.5

Early	Middle	Late
0.9918	0.4160	2.2937
1.0962	0.4598	2.5351

B1

1.1

Both, allow 28.4 for 28.5

awrt 2.29, but allow 2.3 In range [2.53, 2.54]

Question

Solution

Marks

AOs

Guidance

3

(c)

$\mathrm { H } _ { 0 }$ : no association between session and age group. $\mathrm { H } _ { 1 }$ : some association

$\Sigma X ^ { 2 } = 7.793$

$v = 2 , \chi ^ { 2 } ( 2 ) _ { \text {crit } } = 5.991$

Reject $\mathrm { H } _ { 0 }$.

Significant evidence of association between session attended and age group.

B1

M1ft

A1ft [5]

1.1

2.2b

Both. Allow "independent" etc

Correct value of $X ^ { 2 }$, awrt 7.79 (allow even if wrong in (b))

Correct CV and comparison

Correct first conclusion, FT on their TS only

Contextualised, not too assertive

3

(d)

The two biggest contributions to $\chi ^ { 2 }$ are both for the late session ... ... when the proportion of younger people is higher, and of older people is lower, than the null hypothesis would suggest.

M1ft

A1ft

[2]

1.1

2.4

Refer to biggest contribution(s), FT on their answers to (b), needs "reject $\mathrm { H } _ { 0 }$ "

Full answer, referring to at least one cell (ignore comments on next highest cells)

\multirow[t]{2}{*}{4}

\multirow{2}{*}{}

\multirow{2}{*}{OR:}

$\frac { { } ^ { 2 m } C _ { 2 } \times m } { { } ^ { 3 m } C _ { 3 } }$

$= \frac { 2 m ( 2 m - 1 ) } { 2 } \times m \div \frac { 3 m ( 3 m - 1 ) ( 3 m - 2 ) } { 6 }$

$= \frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) }$ $\frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) } = \frac { 28 } { 55 }$

$\Rightarrow 16 m ^ { 2 } - 71 m + 28 = 0$

$m = 4$ BC

Reject $m = \frac { 7 } { 16 }$ as $m$ is an integer

M1

A1

M1

A1

M1

A1

[7]

3.1b

2.1

3.1a

2.1

1.1

3.2a

Use ${ } ^ { 2 m } C _ { 2 }$ and $m$
Divide by ${ } ^ { 3 m } C _ { 3 }$
Correct expression in terms of $m$ (allow with $m$ not cancelled yet)
Equate to $\frac { 28 } { 55 }$ \	simplify to three-term quadratic
Correct simplified quadratic, or (quadratic) $\times m , = 0$, aef Solve to get both 4 and $\frac { 7 } { 16 }$
Explicitly reject $m = \frac { 7 } { 16 }$

$\frac { 2 m ( 2 m - 1 ) \times m \times 3 ! } { 3 m ( 3 m - 1 ) ( 3 m - 2 ) \times 2 }$ then as above

Multiplication method can get full marks, but if no 3 or 3 !, max

M1M0A0 M1A0M0A0

OCR FS1 AS 2021 June Q1

1 Every time a spinner is spun, the probability that it shows the number 4 is 0.2 , independently of all other spins.

A pupil spins the spinner repeatedly until it shows the number 4 . Find the mean of the number of spins required.
Calculate the probability that the number of spins required is between 3 and 10 inclusive.
Each pupil in a class of 30 spins the spinner until it shows the number 4 . Out of the 30 pupils, the number of pupils who require at least 10 spins is denoted by $X$. Determine the variance of $X$.

OCR FS1 AS 2021 June Q2

27 marks

2 After a holiday organised for a group, the company organising the holiday obtained scores out of 10 for six different aspects of the holiday. The company obtained responses from 100 couples and 100 single travellers. The total scores for each of the aspects are given in the following table. \end{table}

Question

Answer

Mark

AO

Guidance

1

(a)

$\frac { 1 } { 0.2 } = 5$

M1 A1 [2]

3.3 1.1

Geometric distribution soi 5 (or $5.00 \ldots$ ) only

1

(b)

$0.8 ^ { 2 } - 0.8 ^ { 10 }$ $= \mathbf { 0 . 5 3 3 } \quad ( 0.5326258 \ldots )$

М1 A1 [2]

1.1 3.4

Allow for powers 2, 3, 4 and 9, 10, 11 .

Awrt 0.533, www. [5201424/976562]

Or $0.2 \left( 0.8 ^ { 2 } + \ldots . + 0.8 ^ { 9 } \right) , \pm 1$ term at either end [0.506, 0.378, 0.275, 0.405, 0.302, 0.554, 0.426, 0.324]

1

(c)

$\mathrm { P } ( \geq 10 ) = 0.8 ^ { 9 }$

$= 0.1342 \ldots$

B(30, 0.1342...)

Variance $= n p q$ = 3.486…

M1

A1

М1

A1ft [4]

3.1b

1.1

3.1b

1.1

Or $0.8 ^ { 10 }$. Can be implied by correct $p$

[0.10737… is M1A0 here]

Stated or implied, their $0.8 ^ { 9 }$ or $0.8 ^ { 10 }$

In range [3.48, 3.49]

SC: 0.134(2) oe not properly shown: B2 for correct final answer.

SC: 2.875 from $0.8 ^ { 10 }$ : M1A0M1A1ft

Question

Answer

Mark

AO

Guidance

2

(a)

Test is for rankings/rankings arbitrary/not bivariate normal etc

B1 [1]

2.4

OE

2

(b)

$\mathrm { H } _ { 0 } : \rho _ { s } = 0 , \mathrm { H } _ { 1 } : \rho _ { s } > 0$, where $\rho _ { s }$ is the population rank correlation coefficient

Ranks 543612

512643

$\Sigma d ^ { 2 } = 20$

$r _ { s } = 1 - \frac { 6 \times 20 } { 6 \times 35 }$

$= 3 / 7$ or $0.42857 \ldots$

<0.9429

B1

M1

A1

B1

1.1

Allow $\rho _ { s }$ not defined; allow $\rho$.

Allow: $\mathrm { H } _ { 0 }$ : no association between rankings.

$\mathrm { H } _ { 1 }$ : positive association (but not $\mathrm { H } _ { 1 }$ : association)

Do not reject $\mathrm { H } _ { 0 }$

Insufficient evidence of association between ranking given by the two categories

M1ft

A1ft

[7]

1.1

2.2b

FT on their $\Sigma d ^ { 2 }$ only

2

(c)

Not dependent on any distributional assumptions

B1

[1]

1.2

Oe (cf. Specification, 5.08f)

Question

Answer

Mark

AO

Guidance

3

(a)

Failures occur to no fixed pattern/are not predictable

B1 [1]

1.1

OE. NOT "independent"

3

(b)

Failures occur independently of one another and at constant average rate

B1

[2]

1.1

Not recoverable from (a) if independence not restated here; must be contextualised

Ignore "singly". Allow "uniform" rate, not "constant rate" or "constant probability"; must be contextualised

3

(c)

Variance (1.6384) $\approx$ mean

So suggests that it is likely to be well modelled

M1

A1

[2]

1.1

3.5a

Compare variance (or SD). Allow square/square-root confusion

Correct comparison and conclusion, 1.64 or better seen

3

(d)

$\mathrm { e } ^ { - 1.61 }$

B1

[1]

3.4

Exact needed, allow even if $0 !$ or $1.61 ^ { 0 }$ or both left in

3

(e)

1	$\geq 2$
0.322	0.478

B1

[2]

3.4

1.1

One correct e.g. 0.3218

Other correct e.g. 0.4783

3

(f)

$\mathrm { P } ( F = 1 )$ will be smaller as single failures are less likely

B1*

depB1

[2]

3.5c

3.3

OE. Partial answer: B1

OCR Further Mechanics 2021 June Q2

2 A solenoid is a device formed by winding a wire tightly around a hollow cylinder so that the wire forms (approximately) circular loops along the cylinder (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{91098ecb-fb4a-44aa-9e59-6c6fe3704966-02_164_697_1484_233} When the wire carries an electrical current a magnetic field is created inside the solenoid which can cause a particle which is moving inside the solenoid to accelerate. A student is carrying out experiments on particles moving inside solenoids. His professor suggests that, for a particle of mass $m$ moving with speed $v$ inside a solenoid of length $h$, the acceleration $a$ of the particle can be modelled by a relationship of the form $a = k m ^ { \alpha } v ^ { \beta } h ^ { \gamma }$, where $k$ is a constant. The professor tells the student that $[ k ] = \mathrm { MLT } ^ { - 1 }$.

Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
The mass of an electron is $9.11 \times 10 ^ { - 31 } \mathrm {~kg}$ and the mass of a proton is $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$. For an electron and a proton moving inside the same solenoid with the same speed, use the model to find the ratio of the acceleration of the electron to the acceleration of the proton.
The professor tells the student that $a$ also depends on the number of turns or loops of wire, $N$, that the solenoid has. Explain why dimensional analysis cannot be used to determine the dependence of $a$ on $N$.

OCR Further Mechanics 2021 June Q3

3 A right circular cone $C$ of height 4 m and base radius 3 m has its base fixed to a horizontal plane. One end of a light elastic string of natural length 2 m and modulus of elasticity 32 N is fixed to the vertex of $C$. The other end of the string is attached to a particle $P$ of mass 2.5 kg .
$P$ moves in a horizontal circle with constant speed and in contact with the smooth curved surface of $C$. The extension of the string is 1.5 m .

Find the tension in the string.
Find the speed of $P$.

OCR Further Mechanics 2021 June Q4

33 marks

4 Two particles $A$ and $B$, of masses $m \mathrm {~kg}$ and 1 kg respectively, are connected by a light inextensible string of length $d \mathrm {~m}$ and placed at rest on a smooth horizontal plane a distance of $\frac { 1 } { 2 } d \mathrm {~m}$ apart. $B$ is then projected horizontally with speed $v \mathrm {~ms} ^ { - 1 }$ in a direction perpendicular to $A B$.

Show that, at the instant that the string becomes taut, the magnitude of the instantaneous impulse in the string, $I \mathrm { Ns }$, is given by $I = \frac { \sqrt { 3 } m v } { 2 ( 1 + m ) }$.
Find, in terms of $m$ and $v$, the kinetic energy of $B$ at the instant after the string becomes taut. Give your answer as a single algebraic fraction.

In the case where $m$ is very large, describe, with justification, the approximate motion of $B$ after the string becomes taut. \section*{Total Marks for Question Set 1: 36} \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

4

(a)

$\cos \theta = \frac { 1 } { 2 }$ or $\sin \theta = \frac { \sqrt { 3 } } { 2 }$

M1

3.1b

Must be clear where it comes from ie. Shown in diagram or clear use of distances. May be other way round

$\theta$ is the angle between the string and the direction of $A B$ when the string initially becomes taut

A1

3.1b

Working must be seen

\(\begin{aligned}

v \sin \theta = ( 1 + m ) V

V = \frac { \sqrt { 3 } v } { 2 ( 1 + m ) }

I = m V \text { or } I = \pm ( V - v \sin \theta )

V = \frac { v \times \frac { 1 } { 2 } \sqrt { 3 } } { 1 + m } \Rightarrow I = \frac { \sqrt { 3 } m v } { 2 ( 1 + m ) } \end{aligned}\)

A1

2.2a

AG Justification of using Impulse = change in momentum

[4]

4

(b)

\(\begin{aligned}

\mathrm { KE } = \frac { 1 } { 2 } \left( V ^ { 2 } + ( v \cos \theta ) ^ { 2 } \right) \text { soi }

\mathrm { KE } = \frac { 1 } { 2 } \left( \left( \frac { \sqrt { 3 } v } { 2 ( 1 + m ) } \right) ^ { 2 } + \left( v \times \frac { 1 } { 2 } \right) ^ { 2 } \right)

\mathrm { KE } = \frac { v ^ { 2 } \left( 4 + 2 m + m ^ { 2 } \right) } { 8 ( 1 + m ) ^ { 2 } } \end{aligned}\)

M1

3.1b

Transverse component unchanged and using their longitudinal component

Condone consideration of speed squared or inclusion of m in KE

М1

1.1

Substituting in for $V$ and $\cos \theta$ (could just be in speed equation)

A1

1.1

Or any equivalent single algebraic fraction

[3]

4

(c)

$V = \frac { \sqrt { 3 } v } { 2 ( 1 + m ) } \rightarrow 0 \text { as } m \rightarrow \infty$

$B$ moves (approximately) in a circle around $A$

B1

3.2a

If m is very large then $A$ is approximately stationary or $B$ has only its transverse velocity of $\frac { 1 } { 2 } v$ after the string becomes taut

B1

2.2b

[2]

OCR Further Mechanics 2021 June Q1

1 A particle $Q$ of mass $m \mathrm {~kg}$ is acted on by a single force so that it moves with constant acceleration $\mathbf { a } = \binom { 1 } { 2 } \mathrm {~ms} ^ { - 2 }$. Initially $Q$ is at the point $O$ and is moving with velocity $\mathbf { u } = \binom { 2 } { - 5 } \mathrm {~ms} ^ { - 1 }$. After $Q$ has been moving for 5 seconds it reaches the point $A$.

Use the equation $\mathbf { v } . \mathbf { v } = \mathbf { u } . \mathbf { u } + 2 \mathbf { a } . \mathbf { x }$ to show that at $A$ the kinetic energy of $Q$ is 37 m J .
1. Show that the power initially generated by the force is - 8 mW W.
2. The power in part (b)(i) is negative. Explain what this means about the initial motion of $Q$.
1. Find the time at which the power generated by the force is instantaneously zero.
2. Find the minimum kinetic energy of $Q$ in terms of $m$.

OCR Further Mechanics 2021 June Q2

44 marks

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

2

(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 $t \rightarrow \infty$

B1

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

B1

3.4

B1

3.4

See above

[3]

2

(c)

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

B1

*M1

Dep* M1

A1

[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

3

(b)

Unchanged since $O F$ does not depend on $\gamma$

E1

[1]

3.5a

OCR Further Mechanics 2021 June Q1

1 A bungee jumper of mass 80 kg steps off a high bridge with an elastic rope attached to her ankles. She is assumed to fall vertically from rest and the air resistance she experiences is modelled as a constant force of 32 N . The rope has natural length 4 m and modulus of elasticity 470 N . By considering energy, determine the total distance she falls before first coming to instantaneous rest.

OCR Further Mechanics 2021 June Q2

2 One end of a light inextensible string of length 0.75 m is attached to a particle $A$ of mass 2.8 kg . The other end of the string is attached to a fixed point $O . A$ is projected horizontally with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 0.75 m vertically above $O$ (see Fig. 2). When $O A$ makes an angle $\theta$ with the upward vertical the speed of $A$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
$\xrightarrow [ A \text { a } ] { 6 \mathrm {~m} \mathrm {~s} ^ { - 1 } }$ Fig. 2

Show that $v ^ { 2 } = 50.7 - 14.7 \cos \theta$.
Given that the string breaks when the tension in it reaches 200 N , find the angle that $O A$ turns through between the instant that $A$ is projected and the instant that the string breaks.

OCR Further Mechanics 2021 June Q3

3 The resistive force, $F$, on a sphere falling through a viscous fluid is thought to depend on the radius of the sphere, $r$, the velocity of the sphere, $v$, and the viscosity of the fluid, $\eta$. You are given that $\eta$ is measured in $\mathrm { Nm } ^ { - 2 } \mathrm {~s}$.

By considering its units, find the dimensions of viscosity. A model of the resistive force suggests the following relationship: $F = 6 \pi \eta ^ { \alpha } r ^ { \beta } v ^ { \gamma }$.
Explain whether or not it is possible to use dimensional analysis to verify that the constant $6 \pi$ is correct.
Use dimensional analysis to find the values of $\alpha , \beta$ and $\gamma$. A sphere of radius $r$ and mass $m$ falls vertically from rest through the fluid. After a time $t$ its velocity is $v$.
By setting up and solving a differential equation, show that $\mathrm { e } ^ { - k t } = \frac { g - k v } { g }$ where $k = \frac { 6 \pi \eta r } { m }$. As the time increases, the velocity of the sphere tends towards a limit called the terminal velocity.
Find, in terms of $g$ and $k$, the terminal velocity of the sphere. In a sequence of experiments the sphere is allowed to fall through fluids of different viscosity, ranging from small to very large, with all other conditions being constant. The terminal velocity of the sphere through each fluid is measured.
Describe how, according to the model, the terminal velocity of the sphere changes as the viscosity of the fluid through which it falls increases.

OCR Further Mechanics 2021 June Q4

40 marks

4 Fig. 4.1 shows a uniform lamina in the shape of a sector of a circle of radius $r$ and angle $2 \theta$ where $\theta$ is in radians. The sector consists of a triangle $O A B$ and a segment bounded by the chord $A B$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{8859baf3-f8e8-4fbf-b54f-34f550b02c26-03_358_545_543_255} \captionsetup{labelformat=empty} \caption{Fig. 4.1}

\end{figure}

Explain why the centre of mass of the segment lies on the radius through the midpoint of $A B$.
Show that the distance of the centre of mass of the segment from $O$ is $\frac { 2 r \sin ^ { 3 } \theta } { 3 ( \theta - \sin \theta \cos \theta ) }$. A uniform circular lamina of radius 5 units is placed with its centre at the origin, $O$, of an $x - y$ coordinate system. A component for a machine is made by removing and discarding a segment from the lamina. The radius of the circle from which the segment is formed is 3 units and the centre of this circle is $O$. The centre of the straight edge of the segment has coordinates $( 0,2 )$ and this edge is perpendicular to the $y$-axis (see Fig. 4.2). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8859baf3-f8e8-4fbf-b54f-34f550b02c26-03_748_743_1594_251} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
\end{figure}
Find the $y$-coordinate of the centre of mass of the component, giving your answer correct to 3 significant figures.
$C$ is the point on the component with coordinates $( 0,5 )$. The component is now placed horizontally and supported only at $O$. A particle of mass $m \mathrm {~kg}$ is placed on the component at $C$ and the component and particle are in equilibrium.

Find the mass of the component in terms of $m$. Total Marks for Question Set 3: 40 \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

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

\end{table}

Question

Answer

Marks

AOs

Guidance

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

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

$\mathrm { e } ^ { - k t } = \frac { g - k v } { g } \quad$ so $t \rightarrow \infty , \Rightarrow v _ { T } = \frac { g } { k }$

B1

3.4

Alternative method $\frac { \mathrm { d } v } { \mathrm {~d} t } = 0 \Rightarrow m g - 6 \pi \eta r v _ { T } = 0 \Rightarrow v _ { T } = \frac { m g } { 6 \pi \eta r } = \frac { g } { k }$

B1

[1]

3

(f)

As the viscosity increases the terminal velocity decreases...

...and as the viscosity tends to infinity the terminal velocity tends to 0

B1

[2]

2.2a

Question

Answer

Marks

AOs

Guidance

4

(a)

The central radius is a line of symmetry of the shape.

B1 [1]

2.4

Allow equal area each side

4

(b)

Area/mass of sector $\frac { 1 } { 2 } r ^ { 2 } \times 2 \theta$ and CoM of sector at $\frac { 2 r \sin \theta } { 3 \theta }$ from centre used

Area/mass of triangle $( - ) \frac { 1 } { 2 } r ^ { 2 } \sin ( 2 \theta )$ and CoM of triangle at $\frac { 2 } { 3 } r \cos \theta$ from centre used $\frac { \frac { 1 } { 2 } r ^ { 2 } \times 2 \theta \times \frac { 2 r \sin \theta } { 3 \theta } - \frac { 1 } { 2 } r ^ { 2 } \sin 2 \theta \times \frac { 2 } { 3 } r \cos \theta } { \frac { 1 } { 2 } r ^ { 2 } \times 2 \theta - \frac { 1 } { 2 } r ^ { 2 } \sin 2 \theta }$

B1

M1

A1

[4]

1.2

3.1b

1.1

Must be used

Using $\bar { x } = \frac { \Sigma m _ { i } x _ { i } } { \Sigma m _ { i } }$ with sector and triangle of-ve mass, oe AG

At least one intermediate step must be seen.

Must see change from double angle

4

(c)

Sector angle is $2 \cos ^ { - 1 } \frac { 2 } { 3 }$ or $\theta = \cos ^ { - 1 } \frac { 2 } { 3 }$

Area/Mass of component: $\pi \times 5 ^ { 2 } - \frac { 1 } { 2 } \times 3 ^ { 2 } ( 1.682 - \sin 1.682 )$

$= 75.44$

$\bar { y } = \frac { ( - 3.097 \ldots \times 2.406 \ldots ) } { 75.44 \ldots }$

$= - 0.0988 ( 3 \mathrm { sf } )$

B1

М1

A1

M1

A1

[5]

2.2a

1.1

3.1b

1.1

1.682...

Attempting to find mass of component using 'negative' mass

Their 3.097, 2.406 and 75.44

$0.841 \ldots$

Allow use of 0.841

If $\theta = 2 \cos ^ { - 1 } \frac { 2 } { 3 }$ used this gives 1.095... instead of 2.406...

4

(d)

$- 0.0988 M + 5 m = 0$

So the mass of the component is 50.6 m kg

M1

A1

[2]

1.1

May see $M g$ and $m g$

OCR Further Mechanics 2021 June Q2

2 The cover of a children's book is modelled as being a uniform lamina $L . L$ occupies the region bounded by the $x$-axis, the curve $y = 6 + \sin x$ and the lines $x = 0$ and $x = 5$ (see Fig. 2.1). The centre of mass of $L$ is at the point $( \bar { x } , \bar { y } )$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{d6bf2fa5-2f29-4632-b27d-ed8c5a0379cf-02_650_534_1030_255} \captionsetup{labelformat=empty} \caption{Fig. 2.1}

\end{figure}

Show that $\bar { x } = 2.36$, correct to 3 significant figures.
Find $\bar { y }$, giving your answer correct to 3 significant figures. The side of $L$ along the $y$-axis is attached to the rest of the book and the book is placed on a rough horizontal plane. The attachment of the cover to the book is modelled as a hinge. The cover is held in equilibrium at an angle of $\frac { 1 } { 3 } \pi$ radians to the horizontal by a force of magnitude $P \mathrm {~N}$ acting at $B$ perpendicular to the cover (see Fig. 2.2). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d6bf2fa5-2f29-4632-b27d-ed8c5a0379cf-03_412_213_402_525} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
\end{figure}
State two additional modelling assumptions, one about the attachment of the cover and one about the badge, which are necessary to allow the value of $P$ to be determined.
Using the modelling assumptions, determine the value of $P$ giving your answer correct to 3 significant figures.

OCR Further Mechanics 2021 June Q3

3 Two smooth circular discs $A$ and $B$ are moving on a horizontal plane. The masses of $A$ and $B$ are 3 kg and 4 kg respectively. At the instant before they collide

the velocity of $A$ is $2 \mathrm {~ms} ^ { - 1 }$ at an angle of $60 ^ { \circ }$ to the line joining their centres,
the velocity of $B$ is $5 \mathrm {~ms} ^ { - 1 }$ towards $A$ along the line joining their centres (see Fig. 3).

\begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{d6bf2fa5-2f29-4632-b27d-ed8c5a0379cf-03_479_1025_1466_248} \captionsetup{labelformat=empty} \caption{Fig. 3}

\end{figure} Given that the velocity of $A$ after the collision is perpendicular to the velocity of $A$ before the collision, find

the coefficient of restitution between $A$ and $B$,
the total loss of kinetic energy as a result of the collision.

OCR Further Mechanics 2021 June Q4

34 marks

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.

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)

$\binom { 2 } { 10 } \cdot \binom { 50 } { 140 }$

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

M1

A1

[2]

1.1

Using WD = F.x

1

(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)}

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

$50 \mathrm {~ms} ^ { - 1 }$

M1

A1

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

A1

complete method involving constant acceleration formula(e)

[2]

Question

Answer

Marks

AOs

Guidance

2

(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}\)

M1

A1

[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

2

(b)

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

M1

A1

[2]

1.1

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

2

(c)

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

The CoM of the badge is at $A$ oe

B1

[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

2

(d)

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

$= P \times 5$

$P = 2.02$

M1

A1

[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}\)

B1

М1

A1

M1

A1

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

A1

М1

A1

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)}

M1

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

M1

1.1

Attempt to find total final KE

56 or 26

A1

1.1

Either correctly calculated

\multirow{8}{*}{}

$56 - 26 = 30 \mathrm {~J}$

A1

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

$= 30 \mathrm {~J}$

A1

Not -30

[5]

Question

Answer

Marks

AOs

Guidance

4

(a)

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

M1

М1

A1

[3]

3.3

Use of $F = m a$ with centripetal acceleration

AG

4

(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}\)

M1

М1

A1

[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

AG

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

4

(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

B1

[1]

3.5a

4

(d)

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

B1

[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.

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.

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

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

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}\)

1

(b)

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

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 $v = \frac { d } { t }$

1

(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

BC

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 $x$

BC

$\frac { 3 \sqrt { 331 } + 9 \sqrt { 3 } } { 80 }$

{ [ 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

3

(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

3

(c)

\(\begin{aligned}

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

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

Correctly substituting values

With their $\alpha$ and $\beta$

3

(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}\)

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

With $n$ dimensionless

3

(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 $414000 \mathrm { Nm } ^ { - 2 }$

Correctly substituting values

With their $\alpha$ and $\beta$

3

(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 $t \geqslant 0$.

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Account	A	B	C	D	E	F	G	H
\(p\)	1.6	2.1	2.4	2.7	2.8	3.3	5.2	8.4
\(q\)	1.6	2.3	2.2	2.2	3.1	2.9	7.6	4.8