Notes

Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.

B1: Must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\)
Do not accept \(0.5 \times (2\mathbf{i} - 2.4\mathbf{j})\)
Do not need \(\mathbf{F} =\)

Notes

Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.

M1: Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) or any other complete method, e.g. integration, to find an unsimplified expression for \(\mathbf{v}\) at \(t = 2\), condone sign errors.
N.B. If using integration, must have used \(t = 0\), \(\mathbf{v} = (-7\mathbf{i} + 7.8\mathbf{j})\) to find the constant of integration and included it in their expression for \(\mathbf{v}\) at \(t = 2\).

A1: cao.   N.B. Ignore if they go on and find the speed.

Notes

Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.

M1: Use trig to find an equation in a relevant angle for their \(\mathbf{v}\)

A1: Correct equation.

A1: Cao (with or without the degree sign), must come from a correct \(\mathbf{v}\)
N.B. \(\pm 45^\circ, \pm 135^\circ, \pm 225^\circ, -315^\circ\) with no working scores M1A1
\(315^\circ\) with no working scores all 3 marks.

Notes

Accept column vectors throughout apart from in the final answer for (a) and (d) which must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\), but only penalise ONCE for the whole question.
N.B. For final answers to (a), (b) and (d), penalise incorrect vector notation.

M1: Use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) with \(t = 5\) or any other complete method, e.g. integration of their \(\mathbf{v} = f(t)\) and use of \(t = 5\), to find an unsimplified expression for \(\overrightarrow{OA}\) (No need to show that \(\mathbf{C} = \mathbf{0}\))
M0 if they use \(\mathbf{u} = \mathbf{0}\)

A1: Must be in terms of \(\mathbf{i}\) and \(\mathbf{j}\)
N.B. Ignore if they go on and find the length \(OA\)

Notes

M1: Use of \(v = u + at\) or any other complete method to form an equation in \(v\) only. Condone sign errors.
e.g. May find \(s\) first: \(s = (15 \times 4) + \dfrac{1}{2} \times 2 \times 4^2 = 76\) then
either: \(76 = 4v - \dfrac{1}{2} \times 2 \times 4^2\)
or: \(76 = \dfrac{(15 + v)}{2} \times 4\)
or: \(v^2 = 15^2 + 2 \times 2 \times 76\)
N.B. If they use \(v = u + at\) and have \(15 = u + (2 \times 4)\), M1A0
M0 if they clearly use an incorrect suvat formula or \(u = 0\)

A1: cao. Must be positive.
A0 if they get 23 but go on and do something with it. i.e. not ISW
N.B. 23 with no working can score both marks.

Notes

M1: Use of \(F = ma\) to form an equation in \(D\) or a complete expression for \(D\), with correct number of terms, condone sign errors and \(g\) in the \(ma\) term.

A1: cao
N.B. 2000 with no working can score both marks.

Notes

M1: Correct no. of terms, condone sin/cos confusion and sign errors, dimensionally correct.
Allow use of a different symbol for the angle.

A1: Accept \(0.92mg\) or better. Must be positive.

Notes

M1: Correct no. of terms, condone sin/cos confusion and sign errors
(M0 if they use \(g\) for \(a\))
N.B. Must be using \(mg\) or \(m\) not \(W\) for the weight.

A1: Any correct equation e.g. \(mg\sin\alpha = ma + F\)
N.B. \(F\) does not need to be substituted.

M1: \(\mu \times\) their \(R\) seen (possibly on a diagram), any trig does not need to be replaced.
M0 if they use \(\mu = \dfrac{5}{12}\)
M0 for just \(\mu R\) with \(R\) not replaced.

A1*: Given answer correctly obtained, with at least one further line of working with both trig ratios substituted as fractions.
Allow \(\dfrac{g}{13}(5 - 12\mu)\) or \(\dfrac{g(5 - 12\mu)}{13}\)

Notes

B1: e.g. \(P\) (or the particle or the object) would stay at rest, \(P\) would not slide down, would not roll down the plane, static equilibrium, equilibrium at rest
Allow ‘it’ for \(P\) or just ‘stays at rest oe’
B0 for wouldn’t accelerate, would be in equilibrium (only), would stop
N.B. Ignore reasons but not contradictions.

Notes

Ignore units in this question.

N.B. Use of \(g = 9.81\) should only be penalised once for the whole question as should two fractional answers (a) 49/10 (b) 7/5.

Penalise the use of \(g = 9.81\) the first time you see it.

N.B.
If \(g = 9.81\) is used in (a), B0. If it is then used again in part (b), and they give the answer as 1.4 or 1.40, they can score M1A1 in part (b).
If \(g = 9.81\) is only used in (b), they can score max M1A0 for (b).

B1: cao. (must be positive)
B0 for a fraction \(\left(\dfrac{49}{10}\right)\)
B0 if they have \(0.5g\) and then clearly use \(g = 9.81\) i.e. NOT isw
This answer must appear in (a) to earn this mark. If no labelling, give BOD and award as many marks as possible.

Notes

Ignore units in this question.

N.B. Use of \(g = 9.81\) should only be penalised once for the whole question as should two fractional answers (a) 49/10 (b) 7/5.

Penalise the use of \(g = 9.81\) the first time you see it.

N.B.
If \(g = 9.81\) is used in (a), B0. If it is then used again in part (b), and they give the answer as 1.4 or 1.40, they can score M1A1 in part (b).
If \(g = 9.81\) is only used in (b), they can score max M1A0 for (b).

M1: \(\dfrac{2}{7} \times\) their (a)   (must be a numerical value)
If no answer for (a) or if (a) is incorrect and they don’t use it, allow a correct restart i.e.
\(\dfrac{2}{7} \times 4.9\) or \(\dfrac{2}{7} \times 0.5g\) or \(\dfrac{2}{7} \times 0.5\) (missing \(g\) is not an M error)

A1: A0 for a fraction
N.B. \(X =\) is not needed BUT
A0 for \(F\) (or \(P\) or horizontal force) = a correct answer if they don’t go on to state \(X = \ldots\)
If they obtain the correct answer for \(F\) and they have said \(F = X\) they can score A1.

N.B. 1.4, 1.40 or \(\frac{1}{7}g\) with no working, scores M1A1

Notes

B1: Allow either \(5g\) or 49. No penalty for using \(g = 9.81\) or 10.
Ignore any working. Must be a positive number.
B0 if \(m\) is involved.
N.B. Could be seen on a diagram, provided it’s clearly the reaction.

Notes

M1: Equation with correct terms, dimensionally correct, condone sign errors.

A1: cao but allow \(\dfrac{15g}{7}\). Ignore units.

Notes

B1ft: \(\mu = \dfrac{\text{their (b)}}{\text{their (a)}}\). Answer must be a positive number given to 2sf.
N.B.
B0 if they use \(g = 9.81\) or 10 in this part of the question.
Do not allow restarts.
Allow \(\mu > 1\).

Notes

Accept column vectors throughout

M1: Adding the two forces, \(\mathbf{i}\)’s and \(\mathbf{j}\)’s must be collected (or must be a single column vector) seen or implied

M1: Must be using ratios; Ignore an equation e.g. \((4 + \lambda)\mathbf{i} + (-1 + \mu)\mathbf{j} = 3\mathbf{i} + \mathbf{j}\) if they go on to use ratios.
However, if they write \(4 + \lambda = 3\) and \(-1 + \mu = 1\) then \(3(-1 + \mu) = 3\) so \(4 + \lambda = 3(-1 + \mu)\) with no use of a constant, it’s M0
They may use the acceleration, with a factor of \(\dfrac{1}{4}\) top and bottom, see alternative
Allow one side of the equation to be inverted

A1: Correct equation

A1*: Given answer correctly obtained. Must see at least one line of working, with the LH fraction ‘removed’.

Alternative 1 for last two M marks

Alternative 2 for last two M marks

Notes

Accept column vectors throughout

M1: Adding \(\mathbf{F}_1\) and \(\mathbf{F}_2\) to find the resultant force, \(\lambda\) and \(\mu\) must be substituted
N.B. M0 if they use \(\mu = 2\) coming from \(-1 + \mu = 1\) in part (a).

M1: Use of \(\mathbf{F} = 4\mathbf{a}\) Or \(|\mathbf{F}| = 4a\), where \(\mathbf{F}\) is their resultant. (including \(3\mathbf{i} + \mathbf{j}\))
This is an independent mark, so could be earned, for example, if they have subtracted the forces to find the ‘resultant’
N.B. M0 if only using \(\mathbf{F}_1\) or \(\mathbf{F}_2\)

DM1: Dependent on previous M mark for
Either: use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) with \(\mathbf{u} = \mathbf{0}\), their \(\mathbf{a}\) and \(t = 4\) to produce a displacement vector
Or : integrate twice, with \(\mathbf{u} = \mathbf{0}\), their \(\mathbf{a}\) and \(t = 4\) to produce a displacement Vector
Or: use of \(s = ut + \dfrac{1}{2}at^2\) with \(u = 0\), their \(a\) and \(t = 4\) to produce a length

M1: Use of Pythagoras, with square root, to find the magnitude of their displacement vector, \(\mathbf{a}\) or \(\mathbf{F}\) (M0 if only using \(\mathbf{F}_1\) or \(\mathbf{F}_2\)) depending on which method they have used.

A1: cao

Notes

(a)(i)

M1: Complete method to obtain an equation in \(F\) only.
For each equation used, correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved.

A1: Correct equation in \(F\) only, trig does not need to be substituted

A1: cao (must be positive)

(a)(ii)

A1: cao. Note that this mark is dependent on an answer of 9.8 or -9.8 for (a)(i) from a fully correct solution unless they have used \(g = 9.81\), in which case the answer will be 9.7 or – 9.7 (2sf) see SC2 below.
N.B. Allow this mark, if their answer to (a)(i) is fully correct apart from a small error due to use of inaccurate trig i.e using an angle \(36.9^\circ\)

SC 1: If they use \(\mu R\) at any point (with an unknown \(\mu\)) for \(F\) in part (a), can score
(a)(i) max M1A1A0
(a) (ii) A1, where they must have obtained \(\mu R = 9.8\) or \(-9.8\), from correct working.

SC 2:
If \(g = 9.81\) is used consistently throughout 2(a), (leading to \(X = 48.9\ldots\) and \(F = 9.7\) (2sf)) can score max (a)(i) M1A1A0 (a)(ii) A1

Notes

M1: Correct no.of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved.

A1: Correct equation for their F.

M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved.
(N.B. M0 if \(R = 68.6\) (N) is used in this equation)

A1: Correct equation

M1: Could be seen on a diagram (N.B. M0 if \(R = 68.6\) (N) is used)

A1: Cao. Must be positive.

Typical solution

Typical solution

Typical solution

The sledge is modelled as a particle.

Typical solution

\(a\) = 8 and \(b\) = 1

Typical solution

3 m s⁻²

Typical solution

Typical solution

Typical solution

0.2\(g\) N

Typical solution

1.63 m s⁻²

Notes

M1*: Correct number of dimensionally correct terms – must be using \(m = 10\) for M1

Allow sign errors and sin/cos mix only

A1: oe e.g. \(R = 49\sqrt{3} - 0.5T\) or \(R = 84.8(7048\ldots) - 0.5T\) etc.

\(R\) is the normal contact force

M1*: Correct number of dimensionally correct terms – must be using \(m = 10\) for M1

Allow sign errors and sin/cos mix only

A1: oe \(F = T \times \frac{\sqrt{3}}{2} + 49\) or \(F = 0.866(0254\ldots)T + 49\) etc.
May have replaced \(F\) with \(0.8R\) and be using their (possibly incorrect) \(R\)

\(F\) is the frictional contact force
For reference only: \(R = 77.4076\ldots\) and \(F = 61.92608\ldots\)

M1dep*: Apply \(F = 0.8R\) to obtain an equation in \(T\) only. A correct value of \(T\) to at least 4 significant figures rot can imply this and the next mark www

A correct equation in \(T\) implies the first 5 marks

A1: AG – awrt 14.9 www - sufficient working must be shown as answer given – for this mark we must see either the correct answer to at least 4 significant figures rot or a correct explicit expression for \(T\) e.g. \(T = \dfrac{8g\cos 30 - 10g\sin 30}{\cos 30 + 0.8\sin 30}\) before the given answer of 14.9

14.925760…

Alternative for part (a) – resolving vertically and horizontally

M1*: Correct number of dimensionally correct terms – must be using \(m = 10\) for M1

Allow sign errors and sin/cos mix only

A1: oe e.g. \(F + R\sqrt{3} = 20g\) or \(0.8R + R\sqrt{3} = 20g\) etc.

M1*: Correct number of dimensionally correct terms

Allow sign errors and sin/cos mix only

A1: oe e.g. \(2T + R = F\sqrt{3}\) or \(2T + R = 0.8R\sqrt{3}\) etc.

M1dep*: Apply \(F = 0.8R\) to obtain an equation in \(T\) only. A correct value of \(T\) to at least 4 significant figures rot can imply this and the next mark www

For reference: \(R = 77.4076\ldots\) and \(F = 61.92608\ldots\)

A1: AG – awrt 14.9 www - sufficient working must be shown as answer given – for this mark we must see either the correct answer to at least 4 significant figures or a correct explicit expression for \(T\) e.g. \(T = \dfrac{8g\cos 30 - 10g\sin 30}{\cos 30 + 0.8\sin 30}\) before the given answer of 14.9

14.925760…

Notes

M1*: Applying N2L for the upper block – allow sign errors and sin/cos mix only - must be using \(m = 4\) for M1 (if correct then \(F_1 = 5.6\))

where \(F_1\) is the frictional contact force between the two blocks

B1: Correctly resolve forces perpendicular to the line of contact between the two blocks

where \(R_1\) is the normal contact force between the two blocks

M1dep*: Apply \(F = \mu R\) with their \(F_1\) to obtain an un-simplified equation/expression in \(\mu\) only where \(R\) must be replaced by either \(4g\cos 30\) or \(4g\cos 60\) or \(4g\sin 30\) or \(4g\sin 60\)

A1: awrt 0.165 or allow exact \(\frac{2\sqrt{3}}{21}\)

0.16495721…

Notes

B1*: Correct expression for the maximum frictional force between the lower block and the plane – note that \(77.4204\ldots \times 0.8\) is B1

If expressions not seen then values must be correct to at least 3 sf
\(77.4076\ldots \times 0.8\) is B1

B1*: Correct expression for the resultant force acting down the plane for the lower block

Do not penalise if a more accurate correct value of \(T\) is used (14.925760…)

B1dep*: Correctly show that the lower block does not move or correctly showing that \(a\), or even \(6a\) is negative

Correct values to at least 3 sf rot must be seen for this mark

Notes

M1: Applying \(\mathbf{F} = m\mathbf{a}\) for both components – no misreads of vectors but allow sign errors only

Must be \(m = 3\) for both components. M0 if including the weight as a force in \(\mathbf{F}\)

B1: One correct explicit linear equation not in vector form

This mark can be implied by one correct value for \(a\) or \(b\)

A1: ISW if correct \(a\) and \(b\) stated in working and then miscopied onto the answer lines at the bottom of the PAB

Correct answers with no working scores full marks. One correct answer with no working scores M0 B1 A0

Notes

B1: Applying \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) correctly - no misreads of vectors. Or consider two correct linear equations i.e. \(7 = u_1 + 5 \times 2\) and \(-6 = u_2 - 5 \times 3\) (condone \(u\) for both \(u_1, u_2\))

oe correct expression/equation for \(\mathbf{u}\) if using calculus

B1: Condone given as a column vector

ISW if magnitude of correct \(\mathbf{u}\) is then considered
Correct answer with no working scores full marks

Notes

B1: oe e.g. \(\begin{pmatrix} -4 \\ 2 \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix}\)

M1: Sets their \(\mathbf{R}\) equal to \(k\) times \(-\mathbf{i} + 3\mathbf{j}\) (or \(k\) times \(\mathbf{R}\)) where \(k\) is non-numerical/unknown
Implied by a correct equation in \(a\) and \(b\) e.g. \(\frac{a - 4}{2 + b} = -\frac{1}{3}\) but not from stating \(-4 + a = -1\) and \(2 + b = 3\) (so must have come from a ‘gradient’ approach if \(k\) not seen)

A1: AG Eliminate \(k\) and derive given result
Using an assumed value of \(k\) is A0

Notes

M1*: Calculate \(|\mathbf{R}|\) or \(|\mathbf{R}|^2\) for their \(\mathbf{R}\)

M1dep*: N2L applied to the magnitude of their \(\mathbf{R}\) with \(5\sqrt{10}\)

A1: www 0.4 (oe) but square roots cancelled

Notes

M1: Allow for suvat equation(s) used leading to a value for \(v\) or \(v^2\)
Allow sign errors

A1: Allow even if the sign of \(u\) does not match the sign of \(s\) and \(a\)

Notes

M1: Allow for suvat equation(s) used leading to a value for \(a\)
Allow for \(s = 3\) used.
FT their (a) Allow sign errors

A1: Need not be evaluated

M1: Use of N2L allow one error or omission

A1: Fully correct equation
FT their acceleration. Weight must be included.

A1: Must be rounded to 3 sf. Accept 34.8 or 34.9 only

Further Maths students may attempt an energy method for (b)

M1: Attempt to calculate change in KE or GPE

M1: Allow if GPE is not included

Notes

B1: Driving force and common tension correct and labelled. Allow \(T_1\) and \(T_2\) if later seen equated. Ignore any vertical forces shown. Also allow for common thrust.

B1: Both resistances correct and labelled and no extra horizontal forces

Notes

M1: Attempt to find resultant horizontal force. All forces included and no extras Allow sign errors.

M1: Attempts N2L equation with their total mass and their resultant. Do not allow for weight used instead of mass.

A1: cao

Alternative solution

M1: Attempt at N2L for at least one part of the system Do not allow for weight used instead of mass. Allow sign errors. All forces included and no extras
(Use the mass to determine which part of the system considered)

M1: Attempt to eliminate \(T\) from their two equations

A1: cao

Notes

B1: Allow equivalent column vectors in all part questions

Notes

B1: May be implied by correct resultant force

M1: Must be vector equation. allow one missing or incorrect force.

A1: ISW if the magnitude is also given

Alternative method

B1: 3 N force used in the horizontal equation and not used in the vertical equation

M1: Considers motion in two directions. Allow one missing or incorrect force

A1: ISW if the magnitude is also given

Notes

M1: Using suvat equation(s) leading to a vector \(\mathbf{v}\). Do not award if scalar added to vector

A1: Mark final answer. Must be vector \(\mathbf{v}\) and not speed.
FT their vector acceleration

Notes

B1: Cao. ISW if the magnitude is also given

Notes

(i) M1: Newton’s second law; condone missing or incorrect resistance or incorrect total mass
Allow for \(T - R = ma\) if the correct substitution of values seen in part 9aii

A1: Fully correct equation need not be simplified

(ii) B1: cao

Notes

B1: Correct vertical forces
Allow ‘weight’ or \(120g\) N oe for \(0.12g\) N

B1: 12 N and 2.5 N correctly drawn and labelled with no extra forces
Allow ‘resistance’ for 2.5 N and ‘tension’ for 12 N providing it is clear that the tension in the string is distinct from the tension in the ribbon

B1: Tension in the ribbon correctly drawn and labelled

Notes

M1: N2L with correct mass and their acceleration. Allow missing or incorrect resistance
Allow for mass in grams if same error seen in part (a)

A1: cao
Allow A1 for correct answer from consistent use of grams

Notes

M1: Attempt to find the mass

M1: Use of ‘\(mg\)’ with Mars gravity

A1: Cao