Questions - OCR FM1 AS - A-Level Maths

2 A particle \(P\) of mass 2.4 kg is moving in a straight line \(O A\) on a horizontal plane. \(P\) is acted on by a force of magnitude 30 N in the direction of motion. The distance \(O A\) is 10 m .

Find the work done by this force as \(P\) moves from \(O\) to \(A\). The motion of \(P\) is resisted by a constant force of magnitude \(R \mathrm {~N}\). The velocity of \(P\) increases from \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at \(O\) to \(18 \mathrm {~ms} ^ { - 1 }\) at \(A\).
Find the value of \(R\).
Find the average power used in overcoming the resistance force on \(P\) as it moves from \(O\) to \(A\). When \(P\) reaches \(A\) it collides directly with a particle \(Q\) of mass 1.6 kg which was at rest at \(A\) before the collision. The impulse exerted on \(Q\) by \(P\) as a result of the collision is 17.28 Ns .
1. Find the speed of \(Q\) after the collision.
2. Hence show that the collision is inelastic. It is required to model the motion of a car of mass \(m \mathrm {~kg}\) travelling at a constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) around a circular portion of banked track. The track is banked at \(30 ^ { \circ }\) (see diagram).
  \includegraphics[max width=\textwidth, alt={}, center]{b9741472-f230-4e2d-9c8b-47f7e168e938-03_355_565_269_274} In a model, the following modelling assumptions are made.
  - The track is smooth.
The car is a particle.
The car follows a horizontal circular path with radius \(r \mathrm {~m}\).
Show that, according to the model, \(\sqrt { 3 } v ^ { 2 } = g r\).
Find the value of \(v\) as predicted by the model. A car is being driven on this portion of the track at the constant speed calculated in part (b). The driver finds that in fact he can drive a little slower or a little faster than this while still moving in the same horizontal circle.

Explain

how this contrasts with what the model predicts,
how to improve the model to account for this.

\section*{Total Marks for Question Set 6: 28} \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}

(a)

\(1.6 = e \times 2.4 = > e = \begin{aligned}

3 \end{aligned}\)

B1 [1]

1.1

(b)

\(4.5 \times - 1.6 - 4.5 \times 2.4 = - 18\) so \(18 \mathrm { Ns } \left( \right.\) or \(\left. \mathrm { kg } \mathrm { ms } ^ { - 1 } \right) \ldots\) ...in the final direction of motion of \(P\)

A1 B1 [3]

1.1 1.1 2.2a

Attempt at \(m v - m u\) Allow \(\pm 18\) could be shown on a diagram

Allow sign confusion e.g. 1.6-2.4 Ignore missing units

(c)

\(\begin{aligned}

1 \times 4.5 \times 2.4 ^ { 2 } - { } _ { 2 } ^ { 1 } \times 4.5 \times 1.6 ^ { 2 }

7.2 \mathrm {~J} \end{aligned}\)

A1 [2]

1.1

Attempt at \(\pm \begin{gathered} 1

m v ^ { 2 } - { } ^ { 1 } m u ^ { 2 }

2 \end{gathered} \quad 2\).

(d)

Not perfectly elastic since KE is lost (due to the collision)

[1]

1.2

or \(e < 1\) but valid reason must be given.

Must mention KE or collision, Not just e.g. "energy lost"

Question

Answer

Marks

AOs

Guidance

(a)

\(30 \times 10\) 300 J

M1 A1 [2]

1.1 1.1

Using Work done \(= F d\)

(b)

\(\begin{aligned}

1 \times 2.4 \times 12 ^ { 2 } + 300 = { } ^ { 1 } \times 2.4 \times 18 ^ { 2 } + W

10 R = 84

R = 8.4 \end{aligned}\)

[3]

1.1

where \(W\) is energy loss (could be eg \(10 R\) )

Use of energy loss \(= 10 R\)

Allow 1 slip e.g. \(R\) instead of \(W\)

Must be e.g. \(10 R\), not \(R\)

Alternative method \(\begin{aligned}

a = 18 ^ { 2 } - 12 ^ { 2 }

\quad 2 \times 10

30 - R = 2.4 \times 9

R = 8.4 \end{aligned}\)

М1

Using \(v ^ { 2 } = u ^ { 2 } + 2 a s\) to find \(a\)

Use of \(F = m a\) with their \(a\)

(c)

\(t = \begin{aligned}	2 \times 10
	18 + 12 \end{aligned}\)
\(t = { } ^ { 2 }\)
\({ } _ { 2 } \frac { 84 ^ { 3 } } { 3 } = 126 \mathrm {~W}\)

1.1

Using \(s = \binom { v + u } { 2 } t\)

Or use Average power \(=\) Force × average speed, i.e. \(P = F _ { ( 2 ) } ^ { v + u } \left( { } _ { 2 } \right)\)

Alternative method \(t = \begin{gathered} 18 - 12

9 \end{gathered}\)

\(t = { } ^ { 2 }\) \(2 _ { 3 } ^ { 3 } = 126 \mathrm {~W}\)

Using \(v = u + a t\) to find \(t\)

NB \(a = 9\) from (b) \(\text { or N2L: } a = { } ^ { 30 - 8.4 } = 9\)

2.4

[3]

Question

Answer

Marks

AOs

Guidance

(d)

(i)

\(\begin{aligned}

17.28 = 1.6 \times v _ { Q }

v _ { Q } = 10.8 \mathrm {~m} \mathrm {~s} ^ { - 1 } \end{aligned}\)

M1 A1 [2]

1.1 1.1

Do not allow -10.8

(d)

(ii)

\(2.4 \times 18 = 2.4 \times v _ { P } + 1.6 \times 10.8\)

So \(v _ { P } = 10.8 = v _ { Q }\), so the particles coalesce and the collision is therefore inelastic.

[2]

3.1b

2.1

Attempt conservation of momentum

Find equal velocities and conclude

Do not allow use of KE

(a)

\(\text { } \uparrow C \cos 30 ^ { \circ } = m g\) \(- C \sin 30 ^ { \circ } = \begin{gathered} m v ^ { 2 }
r \end{gathered}\)
\(m v ^ { 2 }\) \(\begin{aligned}	\Rightarrow \begin{array} { l } C \sin 30 ^ { \circ } = \quad r
C \cos 30 ^ { \circ }
m g \end{array}
	\Rightarrow \tan 30 ^ { \circ } = \frac { 1 } { 3 } ^ { v } { } ^ { v } \stackrel { 2 } { \Rightarrow g } 3 v ^ { 2 } = r g \end{aligned}\)

M1*

M1ft

[4]

3.3

1.1

where C is the (normal) contact force between the car and the track

NII with centripetal acceleration

Dividing so that \(C\) and \(m\) will cancel. May see \(\tan \theta\) or \(\tan 30\) instead of sin/cos

Allow sin/cos confusion Allow \(\theta\) instead of \(30 ^ { \circ }\).

Allow sin/cos confusion

Or rearrange one equation for \(C\) and substitute into the other one

\(\theta\) must be clearly stated and correctly used to gain this mark

(b)

\(3 v ^ { 2 } = 24 \times 9.8\)

\(v =\) awrt 11.7

[2]

3.4

1.1

Using the formula from (a)

(c)

The model implies that only a single value for the speed is possible for a given radius so any change in speed should cause the car to move in a different circle

The track should be modelled as resisting sideways motion

[3]

3.5b

2.2b

3.5c

Or equivalent

e.g. to slide sideways

Accept 'model track as rough' or 'include friction' etc without explicit reference to 'sideways'

Do not allow discussion of the assumptions here

Or any equivalent comment about a possible consequence according to the model of a change in the speed or the radius

Must be relevant to the question. Do not accept references that ignore friction

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