Resolving Forces

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June 2025 Paper 3 Mechanics Q6

EdexcelCurrent spec11 marksMoments Resolving Forces

Figure 4: rod AB with end A on horizontal ground and end B against a vertical wall, at angle alpha to the wall, with a particle C of mass 2M on the rod — Figure 4

A uniform rod \(AB\) has mass \(M\) and length \(2a\).

A particle of mass \(2M\) is attached to the rod at the point \(C\), where \(AC = 0.5a\)

The rod rests with end \(A\) on rough horizontal ground and end \(B\) against a vertical wall.

The rod lies in a vertical plane which is perpendicular to the wall.

The rod is in equilibrium at an angle \(\alpha\) to the wall, as shown in Figure 4.

In an initial model

the vertical wall is modelled as being smooth
the magnitude of the normal reaction of the ground on the rod at \(A\) is \(R\)
the magnitude of the force exerted on the rod by the wall at \(B\) is \(S\)

Using the model,

(a) find \(R\) in terms of \(M\) and \(g\) (1)

(b) show that \(S = Mg\tan\alpha\) (3)

In a refined model

the vertical wall is modelled as being rough
the magnitude of the normal reaction of the ground on the rod at \(A\) is \(R_1\)

Figure 5: the rod AB, now with a particle of mass 3M at B as well as the particle C of mass 2M, with end A on the ground and end B against the wall, at angle beta to the wall — Figure 5

A second particle of mass \(3M\) is now attached to the rod at \(B\).

The rod again rests with end \(A\) on rough horizontal ground and end \(B\) against the vertical wall.

The rod lies in a vertical plane which is perpendicular to the wall.

The rod is now in limiting equilibrium at an angle \(\beta\) to the wall, as shown in Figure 5.

The vertical wall is again modelled as being smooth.

The coefficient of friction between the rod and the ground is \(\mu\)

Given that \(\tan\beta = \dfrac{1}{2}\)

(d) use the model to show that \(\mu = \dfrac{1}{3}\) (6)

force diagram: S horizontal at B, R up and F horizontal at A, 2Mg down at C, Mg down at G, angle alpha between rod and wall — Mark scheme (b)

Scheme	Marks	AO

N.B. Allow consistent use of a different letter for the angle in the working but must be \(\alpha\) in the final answer. BOD if their \(a\)'s and \(\alpha\)'s look the same.
EITHER:
Obtain an equation in \(S\), \(M\), \(a\) and \(\alpha\) only, by taking moments about \(A\)	M1	3.1a
M(\(A\)): \(S \times 2a\cos\alpha = Mga\sin\alpha + 2Mg \times 0.5a\sin\alpha\)	A1	1.1b
OR
Obtain an equation in \(S\), \(M\), \(a\) and \(\alpha\) only, by taking moments about \(B\), \(C\) or \(G\) and resolving once and eliminating \(F\) and \(R\) (N.B. \(R = 3Mg\) from (a))	M1	3.1a
Correct equation in \(S\), \(M\), \(a\) and \(\alpha\) only	A1	1.1b
Possible Moments equations: M(\(B\)): \(R \times 2a\sin\alpha = Mga\sin\alpha + 2Mg \times 1.5a\sin\alpha + F \times 2a\cos\alpha\) or M(\(C\)): \(F \times 0.5a\cos\alpha + S \times 1.5a\cos\alpha = R \times 0.5a\sin\alpha + Mg \times 0.5a\sin\alpha\) or M(\(G\)): \(Sa\cos\alpha + Fa\cos\alpha + 2Mg \times 0.5a\sin\alpha = Ra\sin\alpha\) Possible resolution equations: \((\rightarrow)\): \(F = S\) or \((\nearrow)\): \(R\cos\alpha + F\sin\alpha = S\sin\alpha + (2Mg + Mg)\cos\alpha\) or \((\nwarrow)\): \(R\sin\alpha + S\cos\alpha = (2Mg + Mg)\sin\alpha + F\cos\alpha\)
\(S = Mg\tan\alpha\) *	A1*	2.2a
	(3)

Mark scheme (c)

Scheme	Marks	AO
The friction force at \(B\) will provide an extra upward (not just vertical) force so \(R_1\) will be less i.e. \(R\) will be greater. \(R\) (is greater than \(R_1\)) as the friction act against the weight of the rod. \(R\) (is greater) as the friction acts in the same direction as \(R_1\) \(R\) (is greater) as the friction holds the rod up as well. \(R = 3Mg = R_1 + F\) so \(R\) will be greater N.B. Allow \(R_1\) will be smaller with a correct reason. B0 \(R\) is greater as there is no friction from the wall.	B1	3.5a
	(1)

Notes

B1: Correct explanation
B0 if any incorrect extras

force diagram: P horizontal and 3Mg down at B, N up and F1 horizontal at A, 2Mg down at C, Mg down at G, angle beta between rod and wall — Mark scheme (d)

Scheme	Marks	AO
N.B. Allow consistent use of a different letter for the angle and could score FULL marks.

EITHER: Moments about \(B\) or \(D\), the intersection of the normals	M1	3.4
M(\(B\)): \(N \times 2a\sin\beta = Mga\sin\beta + 2Mg \times 1.5a\sin\beta + F_1 \times 2a\cos\beta\)	A1	1.1b
or M(\(D\)): \(F_1 \times 2a\cos\beta = Mga\sin\beta + 2Mg \times 0.5a\sin\beta + 3Mg \times 2a\sin\beta\)	A1	1.1b
and resolve vertically: \((\uparrow)\ N = Mg + 2Mg + 3Mg\)	B1	3.3
OR: Moments about \(A\) or \(C\) or \(G\)	M1	3.4
A correct moments equation.	A1	1.1b
A correct resolution (N.B. Treat as a B mark so M0A0A1 is a possible score)	A1	1.1b
A correct second resolution or a correct second moments equation	B1	3.3
Possible Moments equations: M(\(A\)): \(2Mg \times 0.5a\sin\beta + Mga\sin\beta + 3Mg \times 2a\sin\beta = P \times 2a\cos\beta\) N.B. (You may see the first 2 terms collected as a single term) or M(\(C\)): \(0.5aF_1\cos\beta + 1.5Pa\cos\beta = 0.5Na\sin\beta + 0.5Mga\sin\beta + 3Mg \times 1.5a\sin\beta\) or M(\(G\)): \(F_1a\cos\beta + Pa\cos\beta + 2Mg \times 0.5a\sin\beta = Na\sin\beta + 3Mga\sin\beta\) Possible resolution equations: \((\uparrow)\): \(N = Mg + 2Mg + 3Mg\) \((\rightarrow)\): \(F_1 = P\) (or \(P = \mu N\)) \((\nearrow)\): \(F_1\sin\beta + N\cos\beta = P\sin\beta + (3Mg + Mg + 2Mg)\cos\beta\) \((\nwarrow)\): \(N\sin\beta + P\cos\beta = (3Mg + Mg + 2Mg)\sin\beta + F_1\cos\beta\)
(N.B. \(F_1 = P = 2Mg\))
Use of \(F_1 = \mu N\), their equations (which must be correct) and \(\tan\beta = \dfrac{1}{2}\), to obtain an equation in \(\mu\) only. N.B. This mark is only available if the previous 4 marks have all been earned.	dM1	3.1a
\(\mu = \dfrac{1}{3}\) *	A1*	2.2a
	(6)
	(11 marks)

June 2025 Paper 3 Mechanics Q2

EdexcelCurrent spec10 marksResolving Forces SUVAT

Figure 2: box B of mass 2 kg on a horizontal plane pulled by a force of 5 N at angle alpha above the horizontal — Figure 2

A small box \(B\) of mass 2 kg is dragged in a straight line, along a rough horizontal plane, at a constant speed by a force of magnitude 5 N.

The line of action of the force makes an angle \(\alpha\) with the plane, where \(\sin\alpha = \dfrac{3}{5}\), as shown in Figure 2.

(a) Show that the magnitude of the normal reaction of the plane on the box is 16.6 N. (3)

At the instant when \(B\) is at the point \(O\) on the plane, the force of magnitude 5 N is removed.

(b) Describe the motion of the box after the force of magnitude 5 N is removed. (1)

Given that after the force of magnitude 5 N is removed

the box is modelled as a particle
air resistance is modelled as being negligible
the coefficient of friction between the box and the plane is modelled as 0.2
the speed of the box as it passes through \(O\) is \(4\ \text{m s}^{-1}\)
the box comes to rest at the point \(X\) on the plane

(d) use the model to find the length \(OX\). (4)

(e) State one limitation of the model, apart from ignoring air resistance, that could affect your answer to part (d). (1)

Mark scheme (a)

Scheme	Marks	AO
Resolve vertically	M1	3.4
\((\uparrow)\ 5\sin\alpha + R = 2g\) oe	A1	1.1b
\((R =)\ 19.6 - 3 = 16.6\) (N)*	A1*	1.1b
	(3)

Notes

N.B. Penalise the use of \(g = 9.81\) ONCE (the first time it is used) for the whole question.

M1: Form an equation in \(R\) and \(\alpha\) only, with correct number of terms, condone sign errors and sin/cos confusion, \(\sin\alpha\) may or may not be substituted.
N.B. \(3 + R = 2g\) oe, with no evidence of resolving the 5 N, e.g. not using \(\sin\alpha\), is M0.

A1: Correct equation, \(\sin\alpha\) may or may not be be substituted.

A1*: Given answer correctly obtained, with at least one line of working.

Mark scheme (b)

Scheme	Marks	AO
The box decelerates, slows down, loses speed or velocity, will come to a stop or rest, loses momentum	B1	2.4
	(1)

Notes

B1: Any equivalent statement.
B0 if any incorrect extras e.g. box is stationary or at rest, doesn’t move, will stop moving.

Mark scheme (c)

Scheme	Marks	AO
\((S =)\ 19.6\) or 20 (N)	B1	3.3
	(1)

Notes

N.B. Penalise the use of \(g = 9.81\) ONCE (the first time it is used) for the whole question.

B1: Accept \(2g\) if \(g\) is not substituted for.
B0 if they use \(g = 9.81\) unless they have already been penalised in part (a).

Mark scheme (d)

Scheme	Marks	AO
Equation of motion along the plane: \(-F = 2a\) or \(F = 2a\)	M1	3.4
\(F = 0.2 \times\) their \(S\)	B1	1.1b
\(0^2 = 4^2 - 2 \times 0.2g \times d\) or \(0^2 = 4^2 + 2 \times 0.2g \times d\) or \(4^2 = 0^2 + 2 \times 0.2g \times d\)	M1	3.1b
\((OX =)\ 4.1\) (m) or 4.08 (m)	A1	1.1b
OR
Work done against friction \(= Fd\)	M1	3.4
\(F = 0.2 \times\) their \(S\)	B1	1.1b
\(0.2 \times 2g \times d = \dfrac{1}{2} \times 2 \times 4^2\)	M1	3.1b
\((d = OX =)\ 4.1\) (m) or 4.08 (m)	A1	1.1b
	(4)

Notes

N.B. Penalise the use of \(g = 9.81\) ONCE (the first time it is used) for the whole question.

M1: Correct number of terms. Condone an extra \(g\) in \(ma\) term.
M0 if 2 is missing.
N.B. M0 if they use a vertical force for \(F\) e.g. \(2g\) or 16.6

B1: Seen (e.g. on a diagram) or implied.

M1: Complete method to form an equation in \(d\) (= \(OX\)) only, condone sign errors, using their calculated acceleration from an attempt at using \(F = ma\)
M0 if clearly using \(\mu\) (0.2) or \(-\mu\) for \(a\) without any calculation.
e.g. may find \(t\) first: \(0 = 4 - 0.2gt \Rightarrow t = \dfrac{20}{g}\ \left(= \dfrac{100}{49}\right)\)
then \(d = 4 \times \dfrac{20}{g} - \dfrac{1}{2} \times 0.2g \times \left(\dfrac{20}{g}\right)^2\)
or \(d = 0 - \dfrac{1}{2} \times (-0.2g) \times \left(\dfrac{20}{g}\right)^2\)
or \(0^2 = 4^2 + 2 \times (-0.2g) \times d\)

A1: Either answer. A0 for \(\dfrac{40}{g}\)

Mark scheme (e)

Scheme	Marks	AO
the box (it) has been modelled as a particle the box (it) will have size or shape or dimensions the coefficient of friction has been modelled as being constant the coefficient of friction may not be exactly 0.2 or may vary the friction may vary B0: the ground may not be horizontal, an inaccurate value of \(g\) has been used, has not considered wind, the angle may not be accurate, any reference to the particle having mass (or not having mass). Ignore any reference to air resistance.	B1	3.5b
	(1)
	(10 marks)

Notes

B1: Any equivalent statement which refers to the model. B0 if incorrect extras.

June 2024 Paper 3 Mechanics Q6

EdexcelCurrent spec9 marksMoments Resolving Forces

Figure 5: rod AB with end A on horizontal ground at angle theta, resting on a peg P at point C where AC = 1.5a — Figure 5

Figure 5 shows a uniform rod \(AB\) of mass \(M\) and length \(2a\).

the rod has its end \(A\) on rough horizontal ground
the rod rests in equilibrium against a small smooth fixed horizontal peg \(P\)
the point \(C\) on the rod, where \(AC = 1.5a\), is the point of contact between the rod and the peg
the rod is at an angle \(\theta\) to the ground, where \(\tan\theta = \dfrac{4}{3}\)

The rod lies in a vertical plane perpendicular to the peg.

The magnitude of the normal reaction of the peg on the rod at \(C\) is \(S\).

(a) Show that \(S = \dfrac{2}{5}Mg\) (3)

The coefficient of friction between the rod and the ground is \(\mu\).

Given that the rod is in limiting equilibrium,

(b) find the value of \(\mu\). (6)

force diagram: rod AB at angle theta to the ground, R up and F horizontal at A, Mg down at G with AG = a, S perpendicular to the rod at C with AC = 1.5a — Mark scheme (a)

Scheme	Marks	AO

Take moments about \(A\)	M1	3.1a
\(S \times 1.5a = Mga\cos\theta = \left(Mga \times \dfrac{3}{5}\right)\)	A1	1.1b
\(S = \dfrac{2}{5}Mg\) *	A1*	2.2a
	(3)

Scheme	Marks	AO

N.B. Marks for the equations should be awarded in the order in which they appear on the script.
Resolve horizontally:	M1	3.4
\(F = S\sin\theta\)	A1	1.1b
Resolve vertically:	M1	3.3
\(R = Mg - S\cos\theta\)	A1	1.1b
Other possible equations: (any of which is worth max M1A1) (parallel to the rod): \(F\cos\theta + R\sin\theta = Mg\sin\theta\) (perp to the rod): \(F\sin\theta + Mg\cos\theta = S + R\cos\theta\) M(\(B\)): \((S \times 0.5a) + (R \times 2a\cos\theta) = (Mg \times a\cos\theta) + (F \times 2a\sin\theta)\) M(\(C\)): \((R \times 1.5a\cos\theta) = (Mg \times 0.5a\cos\theta) + (F \times 1.5a\sin\theta)\) M(\(G\)): \((R \times a\cos\theta) = (S \times 0.5a) + (F \times a\sin\theta)\) N.B. If they have more than two equations, mark only those that they use to try to find \(\mu\)
\(F = \mu R\) and two of their equations used to solve for \(\mu\)	DM1	3.1a
\(\mu = \dfrac{8}{19} = 0.42105\ldots\)	A1	2.2a
	(6)
	(9 marks)

June 2024 Paper 3 Mechanics Q3

EdexcelCurrent spec7 marksForces & Newton's Laws Resolving Forces

Figure 3: particle P of mass m on a plane inclined at angle alpha to the horizontal — Figure 3

A particle \(P\) of mass \(m\) is held at rest at a point on a rough inclined plane, as shown in Figure 3.

It is given that

the plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \dfrac{5}{12}\)
the coefficient of friction between \(P\) and the plane is \(\mu\), where \(\mu \lt \dfrac{5}{12}\)

The particle \(P\) is released from rest and slides down the plane.
Air resistance is modelled as being negligible.

Using the model,

(a) find, in terms of \(m\) and \(g\), the magnitude of the normal reaction of the plane on \(P\), (2)

(b) show that, as \(P\) slides down the plane, the acceleration of \(P\) down the plane is\[\frac{1}{13}g(5 - 12\mu)\] (4)

Mark scheme (a)

Scheme	Marks	AO
\((R =)\, mg\cos\alpha\)	M1	3.4
\(= \dfrac{12}{13}mg\)	A1	1.1b
	(2)

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.

Mark scheme (b)

Scheme	Marks	AO
Equation of motion down the plane	M1	2.1
\(mg\sin\alpha - F = ma\) or \(mg\sin\alpha - F = -ma\)	A1	1.1b
\((F =)\ \mu \times \text{their } R\)	M1	3.4
\(\dfrac{1}{13}g(5 - 12\mu)\) *	A1*	2.2a
	(4)

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 \(\dfrac{g(5 - 12\mu)}{13}\)

Mark scheme (c)

Scheme	Marks	AO
\(P\) wouldn’t move	B1	2.4
	(1)
	(7 marks)

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.

June 2024 Paper 3 Mechanics Q1

EdexcelCurrent spec3 marksForces & Newton's Laws Resolving Forces

Figure 1: particle P of mass 0.5 kg resting on a horizontal line — Figure 1

Figure 1 shows a particle \(P\) of mass 0.5 kg at rest on a rough horizontal plane.

(a) Find the magnitude of the normal reaction of the plane on \(P\). (1)

The coefficient of friction between \(P\) and the plane is \(\dfrac{2}{7}\)

A horizontal force of magnitude \(X\) newtons is applied to \(P\).

Given that \(P\) is now in limiting equilibrium,

(b) find the value of \(X\). (2)

Mark scheme (a)

Scheme	Marks	AO
\(0.5g,\ \dfrac{1}{2}g\) or \(4.9\) (N) seen	B1	3.4
	(1)

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.

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{2}{7} \times 4.9\) oe seen	M1	3.1a
1.4, 1.40 or \(\frac{1}{7}g\)	A1	1.1b
	(2)
	(3 marks)

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.

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

June 2023 Paper 3 Mechanics Q6

EdexcelCurrent spec13 marksMoments Resolving Forces

Figure 3: rod AB with end A on horizontal ground and end B against a vertical wall, making angle theta with the ground — Figure 3

A rod \(AB\) has mass \(M\) and length \(2a\).

The rod has its end \(A\) on rough horizontal ground and its end \(B\) against a smooth vertical wall.

The rod makes an angle \(\theta\) with the ground, as shown in Figure 3.

The rod is at rest in limiting equilibrium.

(a) State the direction (left or right on Figure 3 above) of the frictional force acting on the rod at \(A\). Give a reason for your answer. (1)

The magnitude of the normal reaction of the wall on the rod at \(B\) is \(S\).

In an initial model, the rod is modelled as being uniform.

Use this initial model to answer parts (b), (c) and (d).

(b) By taking moments about \(A\), show that\[S = \frac{1}{2}Mg\cot\theta\] (3)

The coefficient of friction between the rod and the ground is \(\mu\)

Given that \(\tan\theta = \dfrac{3}{4}\)

(d) find, in terms of \(M\) and \(g\), the magnitude of the resultant force acting on the rod at \(A\). (3)

In a new model, the rod is modelled as being non-uniform, with its centre of mass closer to \(B\) than it is to \(A\).

A new value for \(S\) is calculated using this new model, with \(\tan\theta = \dfrac{3}{4}\)

(e) State whether this new value for \(S\) is larger, smaller or equal to the value that \(S\) would take using the initial model. Give a reason for your answer. (1)

Mark scheme (a)

Scheme	Marks	AO
The normal reaction at \(B\) is acting to the left so it must act to the right, right as it needs to balance (oppose, counter) the force at \(B\), right as it prevents the rod from sliding (slipping, falling), right as the weight (mass) of the rod will mean the rod tends to slip left, mass or weight will be pushing the rod to the left so friction will oppose that. N.B. You may see an arrow on the diagram at \(A\), instead of ‘right’. B0 if they say the rod is moving oe Accept towards the wall instead of to the right.	B1	2.4
	(1)

Notes

B1: Any equivalent appropriate statement.

Mark scheme (b)

Scheme	Marks	AO
Take moments about \(A\)	M1	3.4
\(S \times 2a\sin\theta = Mga\cos\theta\)	A1	1.1b
\(S = \dfrac{1}{2}Mg\cot\theta\) *	A1*	2.2a
	(3)

Notes

M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors.
N.B. If \(a\)’s never appear, M0

A1: Correct equation

A1*: Correct given answer correctly obtained, with no wrong working seen.
Allow \(\dfrac{1}{2}Mg\cot\theta = S\) or \(S = \dfrac{Mg\cot\theta}{2}\) or \(\dfrac{Mg\cot\theta}{2} = S\) or \(S = \dfrac{Mg}{2}\cot\theta\) or similar
but NOT \(S = \dfrac{1}{2}\cot\theta\,\text{Mg}\) or similar
N.B. Allow \(m\) instead of \(M\)
Must be \(\theta\) in final answer but allow a different angle in the working.

Mark scheme (c)

Scheme	Marks	AO
Resolve vertically, \(R = Mg\)	B1	3.3
Resolve horizontally, \(F = S\)	B1	3.3
Other possible equations: Resolve along the rod, \(F\cos\theta + R\sin\theta = S\cos\theta + Mg\sin\theta\) Resolve perp to the rod, \(R\cos\theta + S\sin\theta = F\sin\theta + Mg\cos\theta\) M(\(B\)), \(R \times 2a\cos\theta = F \times 2a\sin\theta + Mga\cos\theta\) M(\(G\)), \(Ra\cos\theta = Fa\sin\theta + Sa\sin\theta\) N.B. When entering these two B marks on ePEN, First B1 is for a vertical resolution, second B1 is for a horizontal resolution, and if either is replaced by a different equation, enter appropriately. If both are replaced by other equations, enter in the order in which they appear in their working.
\(F = \mu R\)	B1	1.2
\(\dfrac{1}{2}Mg \times \dfrac{4}{3} = \mu Mg\)	dM1	2.1
\(\mu = \dfrac{2}{3}\) oe Accept 0.67 or better	A1	2.2a
S.C. For \(F \leqslant \mu R\), B0 \(\dfrac{1}{2}Mg \times \dfrac{4}{3} \leqslant \mu Mg\) M1 \(\dfrac{2}{3} \leqslant \mu\) A0 N.B. If \(\mu = \dfrac{2}{3}\) follows this, they could score all the marks.
	(5)

Notes

B1: cao

B1: Seen anywhere, e.g. on the diagram

dM1: Using \(F = \mu R\), their two equations and substitute for trig (not necessarily correctly) to produce an equation in \(\mu\) only.
This mark is dependent on the 3 previous B marks.

A1: Accept 0.67 or better

triangle of forces with sides X, Mg (vertical, downwards) and S (horizontal, to the left), with angle alpha between X and S — Mark scheme (d)

Scheme	Marks	AO
\(\sqrt{F^2 + R^2}\)	M1	3.1a
\(\sqrt{\left(\dfrac{2}{3}Mg\right)^2 + (Mg)^2}\)	M1	1.1b
\(\dfrac{1}{3}Mg\sqrt{13}\) or \(1.2Mg\) or better	A1	2.2a
	(3)

Scheme	Marks	AO
New value of \(S\) would be larger as the moment of the weight about A would be larger	B1	3.5a
	(1)
	(13 marks)

June 2023 Paper 3 Mechanics Q2

EdexcelCurrent spec4 marksForces & Newton's Laws Resolving Forces

Figure 1: particle P of mass 5 kg on a horizontal line, with a force of 28 N to the right and a force F N to the left — Figure 1

A particle \(P\) has mass 5 kg.

The particle is pulled along a rough horizontal plane by a horizontal force of magnitude 28 N.

The only resistance to motion is a frictional force of magnitude \(F\) newtons, as shown in Figure 1.

(a) Find the magnitude of the normal reaction of the plane on \(P\) (1)

The particle is accelerating along the plane at \(1.4\ \text{m s}^{-2}\)

(b) Find the value of \(F\) (2)

The coefficient of friction between \(P\) and the plane is \(\mu\)

Mark scheme (a)

Scheme	Marks	AO
Resolve vertically, \(R = 5g = 49\) (N)	B1	1.1b
	(1)

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.

Mark scheme (b)

Scheme	Marks	AO
Equation of motion: \(28 - F = 5 \times 1.4\)	M1	3.1a
\(F = 21\)	A1	1.1b
	(2)

Notes

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

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

Mark scheme (c)

Scheme	Marks	AO
\(\mu = 0.43\) (2sf required)	B1 ft	3.4
	(1)
	(4 marks)

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 \(\mu > 1\).

June 2022 Paper 3 Mechanics Q4

EdexcelCurrent spec11 marksMoments Resolving Forces

Figure 2: rod AB with end A on horizontal ground at angle theta, particle at C on the rod, and a string attached at B perpendicular to the rod — Figure 2

A uniform rod \(AB\) has mass \(M\) and length \(2a\)

A particle of mass \(2M\) is attached to the rod at the point \(C\), where \(AC = 1.5a\)

The rod rests with its end \(A\) on rough horizontal ground.

The rod is held in equilibrium at an angle \(\theta\) to the ground by a light string that is attached to the end \(B\) of the rod.

The string is perpendicular to the rod, as shown in Figure 2.

(a) Explain why the frictional force acting on the rod at \(A\) acts horizontally to the right on the diagram. (1)

The tension in the string is \(T\)

(b) Show that \(T = 2Mg\cos\theta\) (3)

Given that \(\cos\theta = \dfrac{3}{5}\)

The coefficient of friction between the rod and the ground is \(\mu\)

Given that the rod is in limiting equilibrium,

(d) show that \(\mu = \dfrac{8}{19}\) (4)

Mark scheme (a)

Scheme	Marks	AO
The horizontal component of \(T\) acts to the left and since the only other horizontal force is friction, it must act to the right oe	B1	2.4
	(1)

Notes

B1: Any equivalent explanation

Mark scheme (b)

Scheme	Marks	AO
Take moments about \(A\) or any other complete method to obtain an equation in T, M and \(\theta\) only. (see possible equations below that they may use)	M1	3.1b
\(T.2a = Mga\cos\theta + 2Mg \times 1.5a\cos\theta\) (A0 if \(a\)’s missing)	A1	1.1b
Other possible equations but \(F\) and \(R\) would need to be eliminated. \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\)
\(T = 2Mg\cos\theta\) *	A1*	1.1b
	(3)

Notes

M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors

A1: Correct equation, trig does not need to be substituted
(Allow: \(T.2a = Mga\cos\theta + 3Mga\cos\theta\))

A1*: Given answer correctly obtained with no wrong working seen.
Allow \(2Mg\cos\theta = T\)
But not \(T = 2\cos\theta Mg\)

Mark scheme (c)

Scheme	Marks	AO
e.g. Resolve vertically	M1	3.4
\((\uparrow)\), \(R + T\cos\theta = Mg + 2Mg\)	A1	1.1b
\(R = \dfrac{57Mg}{25}\) *	A1*	1.1b
Other possible equations but \(F\) would need to be eliminated. \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\)
	(3)

Notes

M1: For an equation in \(R\), \(M\), \(T\) and \(\theta\) only
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, \(T\) and trig do not need to be substituted

A1*: Given answer correctly obtained with no wrong working seen

Mark scheme (d)

Scheme	Marks	AO
Find an equation containing \(F\) e.g. Resolve horizontally	M1	3.4
\((\rightarrow)\), \(F = T\sin\theta\)	A1	1.1b
Other possible equations \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\)
\(F = \mu R\) used i.e. both \(F\) and \(R\) are substituted.	M1	3.1b
\(\mu = \dfrac{8}{19}\) *	A1*	2.2a
	(4)
	(11 marks)

Notes

M1: For any equation with \(F\) in it
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, trig does not need to be substituted

M1: Must be used i.e M0 if merely quoting it.

A1*: Given answer correctly obtained with no wrong working seen

June 2022 Paper 3 Mechanics Q2

EdexcelCurrent spec10 marksForces & Newton's Laws Resolving Forces

Figure 1: block B on a plane inclined at angle alpha to the horizontal, with a horizontal force X N acting on B towards the plane — Figure 1

A rough plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \dfrac{3}{4}\)

A small block \(B\) of mass 5 kg is held in equilibrium on the plane by a horizontal force of magnitude \(X\) newtons, as shown in Figure 1.

The force acts in a vertical plane which contains a line of greatest slope of the inclined plane.

The block \(B\) is modelled as a particle.

The magnitude of the normal reaction of the plane on \(B\) is 68.6 N.

Using the model,

(a)

(i) find the magnitude of the frictional force acting on \(B\), (3)

(ii) state the direction of the frictional force acting on \(B\). (1)

The horizontal force of magnitude \(X\) newtons is now removed and \(B\) moves down the plane.

Given that the coefficient of friction between \(B\) and the plane is 0.5

(b) find the acceleration of \(B\) down the plane. (6)

Mark scheme (a)

Scheme	Marks	AO
(i) Resolve vertically	M1	3.1b
\(F\) acting UP the plane: \((\uparrow)\ F\sin\alpha + 68.6\cos\alpha = 5g\) OR \(F\) acting DOWN the plane: \(-F\sin\alpha + 68.6\cos\alpha = 5g\) Other possible equations from which \(X\) would need to be eliminated to give an equation in \(F\) only to earn the M mark are shown below. The equation in \(F\) only must then be correct to earn the A mark. Possible equations: \((\nwarrow)\ 68.6 = X\sin\alpha + 5g\cos\alpha\) (leads to \(X = 49\) with \(g = 9.8\)) \(F\) acting UP the plane: \((\nearrow)\ F + X\cos\alpha = 5g\sin\alpha\) \((\rightarrow)\ F\cos\alpha + X = 68.6\sin\alpha\) OR \(F\) acting DOWN the plane: \(-F + X\cos\alpha = 5g\sin\alpha\) \(-F\cos\alpha + X = 68.6\sin\alpha\)	A1	1.1b
9.8 (N) (49/5 is A0) N.B. If sin and cos are interchanged in all equations, this leads to an answer of 9.8 in the wrong direction and can only score (a) (i)M1A0A0 (ii) A0	A1	1.1b
	(3)
(ii) Down the plane (Allow down or downwards or an arrow \(\swarrow\), but must appear as the answer to (a) (ii) not just on the diagram.)	A1	2.2a
	(1)

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

Mark scheme (b)

Scheme	Marks	AO
N.B. If they use \(R = 68.6\) in this part, the maximum they can score is M1A1M0A0M0A0 If they use \(F = 9.8\) or their \(F\) from (a) in this part, the maximum they can score is M1A1M0A0M0A0
Equation of motion down the plane	M1	2.1
\(5g\sin\alpha - F = 5a\) Allow (\(-a\)) instead of \(a\)	A1	1.1b
Resolve perpendicular to the plane	M1	3.1b
\(R = 5g\cos\alpha\)	A1	1.1b
\(F = 0.5R\) seen	M1	3.4
\(a = 1.96\) or 2.0 or 2 \((\text{m s}^{-2})\) or \(\dfrac{1}{5}g\)	A1	1.1b
	(6)
	(10 marks)

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.

October 2021 Paper 3 Mechanics Q3

EdexcelCurrent spec10 marksMoments Resolving Forces

Figure 2: beam AB of length 2a with A on horizontal ground and B against a vertical wall, inclined at angle theta to the horizontal — Figure 2

A beam \(AB\) has mass \(m\) and length \(2a\).

The beam rests in equilibrium with \(A\) on rough horizontal ground and with \(B\) against a smooth vertical wall.

The beam is inclined to the horizontal at an angle \(\theta\), as shown in Figure 2.

The coefficient of friction between the beam and the ground is \(\mu\)

The beam is modelled as a uniform rod resting in a vertical plane that is perpendicular to the wall.

Using the model,

(a) show that \(\mu \geqslant \dfrac{1}{2}\cot\theta\) (5)

A horizontal force of magnitude \(kmg\), where \(k\) is a constant, is now applied to the beam at \(A\).

This force acts in a direction that is perpendicular to the wall and towards the wall.

Given that \(\tan\theta = \dfrac{5}{4}\), \(\mu = \dfrac{1}{2}\) and the beam is now in limiting equilibrium,

(b) use the model to find the value of \(k\). (5)

force diagram: beam from A on the ground to B on the wall, R up and F towards the wall at A, mg down at G, S horizontal away from the wall at B; C is level with B above A, D is the foot of the wall — Mark scheme (a)

Scheme	Marks	AO
Part (a) is a ‘Show that..’ so equations need to be given in full to earn A marks

Moments equation: (M1A0 for a moments inequality)	M1	3.3
M\((A)\), \(mga\cos\theta = 2Sa\sin\theta\) M\((B)\), \(mga\cos\theta + 2Fa\sin\theta = 2Ra\cos\theta\) M\((C)\), \(F \times 2a\sin\theta = mga\cos\theta\) M\((D)\), \(2Ra\cos\theta = mga\cos\theta + 2Sa\sin\theta\) M\((G)\), \(Ra\cos\theta = Fa\sin\theta + Sa\sin\theta\).	A1	1.1b
\((\updownarrow)\ R = mg\) OR \((\leftrightarrow)\ F = S\)	B1	3.4
Use their equations (they must have enough) and \(F \leqslant \mu R\) to give an inequality in \(\mu\) and \(\theta\) only (allow DM1 for use of \(F = \mu R\) to give an equation in \(\mu\) and \(\theta\) only)	DM1	2.1
\(\mu \geqslant \dfrac{1}{2}\cot\theta\) *	A1*	2.2a
	(5)

force diagram: beam from A on the ground to B on the wall, R up at A, kmg towards the wall and one half mg away from the wall at A, mg down at G, N horizontal away from the wall at B; C is level with B above A, D is the foot of the wall — Mark scheme (b)

Scheme	Marks	AO

Moments equation:	M1	3.4
M\((A)\), \(mga\cos\theta = 2Na\sin\theta\) M\((B)\), \(mga\cos\theta + 2kmga\sin\theta = 2Ra\cos\theta + \dfrac{1}{2}mg2a\sin\theta\) M\((D)\), \(2Ra\cos\theta = mga\cos\theta + N2a\sin\theta\) M\((G)\), \(kmga\sin\theta + Na\sin\theta = \dfrac{1}{2}mga\sin\theta + Ra\cos\theta\)	A1	1.1b
S.C. M\((C)\), \(mga\cos\theta + \dfrac{1}{2}mg2a\sin\theta = kmg2a\sin\theta\) M1A1B1 \(1 + \dfrac{5}{4} = \dfrac{5k}{2}\) M1 \(k = 0.9\) A1
\(N = kmg - F\) OR \(R = mg\)	B1	3.3
Use their equations (they must have enough) to solve for \(k\) (numerical)	DM1	3.1b
\(k = 0.9\) oe	A1	1.1b
	(5)
	(10 marks)

October 2021 Paper 3 Mechanics Q2

EdexcelCurrent spec12 marksConnected Particles Resolving Forces

Figure 1: stone A of mass 3m on a plane inclined at angle alpha, attached by a string over a pulley P at the top of the plane to stone B of mass m hanging freely — Figure 1

A small stone \(A\) of mass \(3m\) is attached to one end of a string.

A small stone \(B\) of mass \(m\) is attached to the other end of the string.

Initially \(A\) is held at rest on a fixed rough plane.

The plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \dfrac{3}{4}\)

The string passes over a pulley \(P\) that is fixed at the top of the plane.

The part of the string from \(A\) to \(P\) is parallel to a line of greatest slope of the plane.

Stone \(B\) hangs freely below \(P\), as shown in Figure 1.

The coefficient of friction between \(A\) and the plane is \(\dfrac{1}{6}\)

Stone \(A\) is released from rest and begins to move down the plane.

The stones are modelled as particles.

The pulley is modelled as being small and smooth.

The string is modelled as being light and inextensible.

Using the model for the motion of the system before \(B\) reaches the pulley,

(a) write down an equation of motion for \(A\) (2)

(b) show that the acceleration of \(A\) is \(\dfrac{1}{10}g\) (7)

(c) sketch a velocity-time graph for the motion of \(B\), from the instant when \(A\) is released from rest to the instant just before \(B\) reaches the pulley, explaining your answer. (2)

In reality, the string is not light.

(d) State how this would affect the working in part (b). (1)

Mark scheme (a)

Scheme	Marks	AO
Equation of motion for \(A\)	M1	3.3
\(3mg\sin\alpha - F - T = 3ma\)	A1	1.1b
	(2)

Notes

Mark parts (a) and (b) together

N.B. If m’s are consistently missing treat as a MR, so max
(a) M1A0 (b) M1A0B0M1A1M1A1 (c) B1B1 (d) B1

For (a) and (b), allow verification, but must see full equations of motion.

M1: Equation in \(T\) and \(a\) with correct no. of terms, condone sign errors and sin/cos confusion (If one of the 3’s is missing, allow M1)
N.B. Treat \(\sin(3/5)\) etc as an A error but allow recovery

A1: Correct equation (allow \((-a)\) instead of \(a\) in both equations)

Mark scheme (b)

Scheme	Marks	AO
Resolve perpendicular to the plane	M1	3.4
\(R = 3mg\cos\alpha\)	A1	1.1b
\(F = \dfrac{1}{6}R\)	B1	1.2
Equation of motion for \(B\) OR for whole system	M1	3.3
\(T - mg = ma\) OR \(3mg\sin\alpha - F - mg = 3ma + ma\)	A1	1.1b
Complete method to solve for \(a\)	DM1	3.1b
\(a = \dfrac{1}{10}g\) *	A1*	2.2a
	(7)

Notes

Mark parts (a) and (b) together

N.B. If m’s are consistently missing treat as a MR, so max
(a) M1A0 (b) M1A0B0M1A1M1A1 (c) B1B1 (d) B1

For (a) and (b), allow verification, but must see full equations of motion.

M1: Correct no. of terms, condone sign errors and sin/cos confusion
Allow if appears in (a)

A1: Correct equation

B1: Seen anywhere in (a) or (b), including on a diagram

M1: Equation (for \(B\)) in \(T\) and \(a\) with correct no. of terms, condone sign errors and sin/cos confusion
OR Whole system equation with correct no. of terms, condone sign errors and sin/cos confusion

A1: Correct equation

DM1: Complete method (trig may not be substituted), dependent on M1 in (a) and second M1 in (b) if they use two equations, or second M1 in (b) if they use one equation.

A1*: Correct answer correctly obtained.

velocity-time graph: a straight line from the origin sloping upwards, axes v and t — Mark scheme (c)

Scheme	Marks	AO
	B1	1.1b
e.g. acceleration (of \(B\)) is constant; dependent on first B1	DB1	2.4
	(2)

Mark scheme (d)

Scheme	Marks	AO
e.g. the tensions in the two equations of motion would be different. Tension on \(A\) would be different to tension on \(B\)	B1	3.5a
	(1)
	(12 marks)

Notes

N.B. If m’s are consistently missing treat as a MR, so max
(a) M1A0 (b) M1A0B0M1A1M1A1 (c) B1B1 (d) B1

For (a) and (b), allow verification, but must see full equations of motion.

B1: B0 if incorrect extras

October 2020 Paper 3 Mechanics Q4

EdexcelCurrent spec10 marksMoments Resolving Forces

Figure 1: ladder AB of length 6a with A on horizontal ground, resting on a rail at C which is at height 4a above the ground, ladder inclined at angle alpha to the horizontal — Figure 1

A ladder \(AB\) has mass \(M\) and length \(6a\).

The end \(A\) of the ladder is on rough horizontal ground.

The ladder rests against a fixed smooth horizontal rail at the point \(C\).

The point \(C\) is at a vertical height \(4a\) above the ground.

The vertical plane containing \(AB\) is perpendicular to the rail.

The ladder is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{4}{5}\), as shown in Figure 1.

The coefficient of friction between the ladder and the ground is \(\mu\).

The ladder rests in limiting equilibrium.

The ladder is modelled as a uniform rod.

Using the model,

(a) show that the magnitude of the force exerted on the ladder by the rail at \(C\) is \(\dfrac{9Mg}{25}\) (3)

(b) Hence, or otherwise, find the value of \(\mu\). (7)

Mark scheme (a)

Scheme	Marks	AO
Take moments about \(A\)	M1	3.3
\(N \times \dfrac{4a}{\sin\alpha} = Mg \times 3a\cos\alpha\)	A1	1.1b
\(\dfrac{9Mg}{25}\) *	A1*	1.1b
	(3)

Notes

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors for an equation in \(N\) and \(Mg\) only.
For perp distance allow any of: \(\dfrac{4a}{\sin\alpha}, \dfrac{4a}{\cos\alpha}, 5a\) but
use of any of: \(6a, 5a\sin\alpha, 4a\cos\alpha, \ldots\) or anything involving \(\tan\alpha\) is M0
Also M0 if no \(a\)’s in their first equation.

A1: Correct equation, trig does not need to be substituted

A1*: Given answer correctly obtained.

Mark scheme (b)

Scheme	Marks	AO
Resolve horizontally	M1	3.4
\((\rightarrow)\ F = \dfrac{9Mg}{25}\sin\alpha\)	A1	1.1b
Resolve vertically	M1	3.4
\((\uparrow)\ R + \dfrac{9Mg}{25}\cos\alpha = Mg\)	A1	1.1b
Other possible equations: \((\nwarrow),\ R\cos\alpha + \dfrac{9Mg}{25} = Mg\cos\alpha + F\sin\alpha\) \((\nearrow),\ Mg\sin\alpha = F\cos\alpha + R\sin\alpha\) M\((C)\), \(Mg.2a\cos\alpha + F.5a\sin\alpha = R.5a\cos\alpha\) M\((G)\), \(\dfrac{9Mg}{25}.2a + F.3a\sin\alpha = R.3a\cos\alpha\) M\((B)\), \(Mg.3a\cos\alpha + F.6a\sin\alpha = R.6a\cos\alpha + \dfrac{9Mg}{25}a\) \(\left(F = \dfrac{36Mg}{125},\ R = \dfrac{98Mg}{125}\right)\)
\(F = \mu R\) used	M1	3.4
Eliminate \(R\) and \(F\) and solve for \(\mu\)	M1	3.1b
Alternative equations if they have at \(A\): \(X\) horizontally and \(Y\) perpendicular to the rod. \((\nwarrow),\ Y + \dfrac{9Mg}{25} = Mg\cos\alpha + X\sin\alpha\) \((\nearrow),\ Mg\sin\alpha = X\cos\alpha\) \((\uparrow),\ \dfrac{9Mg}{25}\cos\alpha + Y\cos\alpha = Mg\) \((\rightarrow),\ Y\sin\alpha + \dfrac{9Mg}{25}\sin\alpha = X\) M\((C)\), \(Mg.2a\cos\alpha + X.5a\sin\alpha = Y.5a\) M\((G)\), \(\dfrac{9Mg}{25}.2a + X.3a\sin\alpha = Y.3a\) M1A1 M1A1 M\((B)\), \(Mg.3a\cos\alpha + X.6a\sin\alpha = Y.6a + \dfrac{9Mg}{25}a\) \(\left(X = \dfrac{4Mg}{3},\ Y = \dfrac{98Mg}{75}\right)\) Then \(F = \mu R\) becomes: \(X - Y\sin\alpha = \mu Y\cos\alpha\) M1 Eliminate \(X\) and \(Y\) and solve for \(\mu\) M1
\(\mu = \dfrac{18}{49}\) (0.3673…..accept 0.37 or better)	A1	2.2a
	(7)
	(10 marks)

Notes

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1: Correct equation, trig does not need to be substituted but \(N\) does.

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1: Correct equation, trig does not need to be substituted but \(N\) does.

N.B. The above 4 marks are for any two equations, either resolutions or moments or one of each. Mark best two equations.
Equations may appear in part (a) but must be used in (b) to earn marks.

M1: Must be used, e.g. seen on the diagram. i.e. M0 if merely quoting it.
(M0 if \(F = \mu \times \dfrac{9Mg}{25}\) used)

M1: Must have 3 equations (and all 3 previous M marks)

A1: Accept 0.37 or better

October 2020 Paper 3 Mechanics Q1

EdexcelCurrent spec9 marksResolving Forces

1. A rough plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \dfrac{3}{4}\)

A brick \(P\) of mass \(m\) is placed on the plane.

The coefficient of friction between \(P\) and the plane is \(\mu\)

Brick \(P\) is in equilibrium and on the point of sliding down the plane.

Brick \(P\) is modelled as a particle.

Using the model,

(a) find, in terms of \(m\) and \(g\), the magnitude of the normal reaction of the plane on brick \(P\) (2)

(b) show that \(\mu = \dfrac{3}{4}\) (4)

For parts (c) and (d), you are not required to do any further calculations.

Brick \(P\) is now removed from the plane and a much heavier brick \(Q\) is placed on the plane.

The coefficient of friction between \(Q\) and the plane is also \(\dfrac{3}{4}\)

Brick \(Q\) is now projected with speed \(0.5\ \text{m s}^{-1}\) down a line of greatest slope of the plane.

Brick \(Q\) is modelled as a particle.

Using the model,

(d) describe the motion of brick \(Q\), giving a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks	AO
Resolve perpendicular to the plane	M1	3.4
\(R = mg\cos\alpha = \dfrac{4}{5}mg\)	A1	1.1b
	(2)

Notes

M1: Correct no. of terms, condone sin/cos confusion

A1: cao with no wrong working seen. \(mg\cos 36.86\) is A0

Mark scheme (b)

Scheme	Marks	AO
Resolve parallel to the plane or horizontally or vertically	M1	3.4
\(F = mg\sin\alpha\) or \(R\sin\alpha = F\cos\alpha\)	A1	1.1b
Use \(F = \mu R\) and solve for \(\mu\)	M1	2.1
\(\mu = \dfrac{3}{4}\) *	A1*	2.2a
	(4)

Notes

M1: Correct no. of terms, condone sin/cos confusion

A1: Correct equation

M1: Must use \(F = \mu R\) (not merely state it) to obtain a numerical value for \(\mu\).
This is an independent M mark.

A1*: Given answer correctly obtained

Mark scheme (c)

Scheme	Marks	AO
The forces acting on \(Q\) will still balance as the \(m\)’s cancel oe Other possibilities: e.g. the friction will increase in the same proportion as the weight component or force down the plane. The force pulling the brick down the plane increases by the same amount as the friction oe This mark can be scored if they do the calculation.	B1	2.4
	(1)

Notes

B1: Must have the 3 underlined phrases/word oe

Mark scheme (d)

Scheme	Marks	AO
Brick \(Q\) slides down the plane with constant speed.	B1	2.4
No resultant force down the plane (so no acceleration) oe	B1	2.4
These marks can be scored if they do the calculation.
	(2)
	(9 marks)

Notes

B1: Must say constant speed.

B1: Any appropriate equivalent statement

June 2019 Paper 3 Mechanics Q4

EdexcelCurrent spec11 marksMoments Resolving Forces

Figure 2: ramp AB with end A on horizontal ground at angle theta, resting on a cylindrical drum partly under the ground at C, where AC = 5 m and CB = 3 m — Figure 2

A ramp, \(AB\), of length 8 m and mass 20 kg, rests in equilibrium with the end \(A\) on rough horizontal ground.

The ramp rests on a smooth solid cylindrical drum which is partly under the ground. The drum is fixed with its axis at the same horizontal level as \(A\).

The point of contact between the ramp and the drum is \(C\), where \(AC = 5\) m, as shown in Figure 2.

The ramp is resting in a vertical plane which is perpendicular to the axis of the drum, at an angle \(\theta\) to the horizontal, where \(\tan\theta = \dfrac{7}{24}\)

The ramp is modelled as a uniform rod.

(a) Explain why the reaction from the drum on the ramp at point \(C\) acts in a direction which is perpendicular to the ramp. (1)

(b) Find the magnitude of the resultant force acting on the ramp at \(A\). (9)

The ramp is still in equilibrium in the position shown in Figure 2 but the ramp is not now modelled as being uniform.

Given that the centre of mass of the ramp is assumed to be closer to \(A\) than to \(B\),

Scheme	Marks	AO
Drum smooth, or no friction, (therefore reaction is perpendicular to the ramp)	B1	2.4
	(1)

force diagram: ramp AB at angle theta, R vertically up and F horizontal (towards the drum) at A, 20g vertically down, N perpendicular to the ramp at C on the drum — Mark scheme (b)

Scheme	Marks	AO
N.B. In (b), for a moments equation, if there is an extra \(\sin\theta\) or \(\cos\theta\) on a length, give M0 for the equation e.g. M(\(A\)): \(20g \times 4\cos\theta = 5N\sin\theta\) would be given M0A0

Possible equns \((\nearrow)\): \(F\cos\theta + R\sin\theta = 20g\sin\theta\) \((\nwarrow)\): \(N + R\cos\theta = 20g\cos\theta + F\sin\theta\) \((\uparrow)\): \(R + N\cos\theta = 20g\) \((\rightarrow)\): \(F = N\sin\theta\) M(\(A\)): \(20g \times 4\cos\theta = 5N\) M(\(B\)): \(3N + R \times 8\cos\theta = F \times 8\sin\theta + 20g \times 4\cos\theta\) M(\(C\)): \(R \times 5\cos\theta = F \times 5\sin\theta + 20g \times \cos\theta\) M(\(G\)): \(R \times 4\cos\theta = F \times 4\sin\theta + N\)	M1 A1 M1 A1 M1 A1	3.3 1.1b 3.4 1.1b 3.4 1.1b
(The values of the 3 unknowns are: \(N = 150.528\); \(F = 42.14784\); \(R = 51.49312\))
Solve their 3 equations for \(F\) and \(R\) OR \(X\) and \(Y\) OR \(H\) and \(S\)	M1	1.1b
\(\|\text{Force}\| = \sqrt{R^2 + F^2}\) Main scheme OR \(= \sqrt{X^2 + Y^2}\) Alternative 1 OR \(= \sqrt{(H^2 + S^2 - 2HS\cos(90^\circ - \theta))}\) Alternative 2	M1	3.1b
Magnitude = 67 or 66.5 (N)	A1	2.2a
	(9)

Scheme	Marks	AO
Possible equations: \((\nearrow)\): \(X = 20g\sin\theta\) \((\nwarrow)\): \(Y + N = 20g\cos\theta\) \((\uparrow)\): \(X\sin\theta + Y\cos\theta + N\cos\theta = 20g\) \((\rightarrow)\): \(X\cos\theta = Y\sin\theta + N\sin\theta\) M(\(A\)): \(20g \times 4\cos\theta = 5N\) M(\(B\)): \(20g \times 4\cos\theta = 8Y + 3N\) M(\(C\)): \(20g \times \cos\theta = 5Y\) M(\(G\)): \(4Y = N \times 1\)	M1 A1 M1 A1 M1 A1	3.3 1.1b 3.4 1.1b 3.4 1.1b
(The values of the 3 unknowns are: \(N = 150.528\); \(X = 54.88\); \(Y = 37.632\))

Scheme	Marks	AO
\((\nearrow)\): \(H\cos\theta = 20g\sin\theta\) \((\nwarrow)\): \(S + N = H\sin\theta + 20g\cos\theta\) \((\uparrow)\): \(S\cos\theta + N\cos\theta = 20g\) \((\rightarrow)\): \(H = S\sin\theta + N\sin\theta\) M(\(A\)): \(20g \times 4\cos\theta = 5N\) M(\(B\)): \(20g \times 4\cos\theta + H \times 8\sin\theta = 8S + 3N\) M(\(C\)): \(20g \times \cos\theta + H \times 5\sin\theta = 5S\) M(\(G\)): \(4S = N \times 1 + H \times 4\sin\theta\)	M1 A1 M1 A1 M1 A1	3.3 1.1b 3.4 1.1b 3.4 1.1b
(The values of the 3 unknowns are: \(N = 150.528\); \(H = 57.1666…\); \(S = 53.638666…\))

Scheme	Marks	AO
Magnitude of the normal reaction (at \(C\)) will decrease.	B1	3.5a
	(1)
	(11 marks)

June 2019 Paper 3 Mechanics Q3

EdexcelCurrent spec12 marksConnected Particles Resolving Forces

Figure 1: block A of mass 2m on a plane inclined at angle alpha to the horizontal, attached by a string over a pulley P at the top of the plane to block B of mass 3m hanging freely — Figure 1

Two blocks, \(A\) and \(B\), of masses \(2m\) and \(3m\) respectively, are attached to the ends of a light string.

Initially \(A\) is held at rest on a fixed rough plane.

The plane is inclined at angle \(\alpha\) to the horizontal ground, where \(\tan\alpha = \dfrac{5}{12}\)

The string passes over a small smooth pulley, \(P\), fixed at the top of the plane.

The part of the string from \(A\) to \(P\) is parallel to a line of greatest slope of the plane. Block \(B\) hangs freely below \(P\), as shown in Figure 1.

The coefficient of friction between \(A\) and the plane is \(\dfrac{2}{3}\)

The blocks are released from rest with the string taut and \(A\) moves up the plane.

The tension in the string immediately after the blocks are released is \(T\).

The blocks are modelled as particles and the string is modelled as being inextensible.

(a) Show that \(T = \dfrac{12mg}{5}\) (8)

After \(B\) reaches the ground, \(A\) continues to move up the plane until it comes to rest before reaching \(P\).

(b) Determine whether \(A\) will remain at rest, carefully justifying your answer. (2)

force diagram: on A (2m) on the plane, R perpendicular to the plane, T up the plane, F down the plane, 2mg vertically down; on B (3m), T up and 3mg down; plane at angle alpha — Mark scheme (a)

Scheme	Marks	AO

\(R = 2mg\cos\alpha\)	B1	3.4
\(F = \dfrac{2}{3}R\)	B1	1.2
Equation of motion for \(A\):	M1	3.3
\(T - F - 2mg\sin\alpha = 2ma\)	A1	1.1b
Equation of motion for \(B\):	M1	3.3
\(3mg - T = 3ma\)	A1	1.1b
Complete strategy to find an equation in \(T\), \(m\) and \(g\) only.	M1	3.1b
\(T = \dfrac{12mg}{5}\) *	A1*	2.2a
	(8)

Mark scheme (b)

Scheme	Marks	AO
\((F_{\max} =)\ \dfrac{16mg}{13} > \dfrac{10mg}{13}\)	M1	2.1
…… so \(A\) will not move.	A1	2.2a
	(2)

Notes

M1: Comparison of their \(F_{\max}\) \(\left(\dfrac{2}{3}R\right)\) and their component of weight down the slope, must be comparing numerical values. oe e.g. if they consider the difference
N.B. Allow comparison of \(\mu\) and \(\tan\alpha\) with numerical values

A1: Correctly justified conclusion and no errors seen
N.B. If they equate their difference to an ‘\(ma\)’ term then A0

Mark scheme (c)

Scheme	Marks	AO
Extensible string Weight of string Friction at pulley e.g. rough pulley Allow for the dimensions of the blocks e.g. “Do not model blocks as particles”; “(include) air resistance”; “include rotational effects of forces on blocks i.e. spin”	B1 B1	3.5c 3.5c
	(2)
	(12 marks)

Notes

B1 B1: Deduct 1 mark for each extra (more than 2) incorrect answer up to a maximum of 2 incorrect answers. Ignore extra correct answers.
e.g. two correct, one incorrect B1 B0
one correct, one incorrect B1 B0
one correct, two incorrect B0 B0
Ignore incorrect reasons or consequences.
Ignore any mention of wind or a general reference to friction.

June 2025 Paper 2 Q16

AQACurrent spec12 marksForces & Newton's Laws Resolving Forces

16 In this question use \(g = 9.8\ \text{m s}^{-2}\)

A sledge is pulled in a straight line up a rough path, as shown in the diagram.

The path is inclined at an angle of 20° to the horizontal.

The sledge is pulled by a light, inextensible rope inclined at an angle of 30° to the path.

A sledge on a path inclined at 20° to the horizontal, pulled up the slope by a rope at 30° to the path

The mass of the sledge is 10 kilograms.

(a) In one model, the sledge moves at a constant speed and experiences a combined resistance force of 15 newtons.

Find the tension in the rope for this model.

[4 marks]

(b) In a different model, the sledge experiences no air resistance.

The tension in the rope for this model is 54 N

The coefficient of friction between the sledge and the path is 0.2

Find the acceleration of the sledge for this model.

[7 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains \(10g\sin 20\) OE	B1	3.1b
Resolves parallel to the path to form a three-term equation with at least two correct terms	M1	3.3
Obtains \(10g\sin 20 + 15 = \mathbf{T}\cos 30\) OE	A1	1.1b
Obtains AWRT 56 Condone missing units.	A1	1.1b
	(4)

Typical solution

\[10g\sin 20 + 15 = \mathbf{T}\cos 30\]\[\mathbf{T} = 56.02\]\[\mathbf{T} = 56\ \text{N}\]

Mark scheme (b)

Scheme	Marks	AO
Uses \(\mathbf{F} = \mu\mathbf{R}\)	B1	1.1b
Obtains \(10g\cos 20\) OE	B1	1.1b
Resolves perpendicular to the plane to form a three-term dimensionally correct equation with at least two terms correct.	M1	3.3
Obtains a correct expression or value for R AWRT 65 PI by friction = AWRT \(\pm 13.02\)	A1	1.1b
Uses F = \(m\)a to form a four-term dimensionally correct equation, excluding a 15N force, with at least three terms correct. FT their R value Condone inclusion of a 15N force from 16(a) as a 5^th term	M1	3.4
Forms fully correct equation in a for the particle with all values substituted. FT their R value Note this alone obtains B1B1M1A1M1A1	A1F	1.1b
Obtains AWRT 0.023 \(\text{m s}^{-2}\)	A1	1.1b
	(7)

Typical solution

\[\mathbf{R} + 54\sin 30 = 10g\cos 20\]\[\mathbf{R} = 65.0899\ \text{N}\]\[54\cos 30 - \mu\mathbf{R} - 10g\sin 20 = 10a\]\[a = \frac{54\cos 30 - 0.2 \times 65.0899 - 98\sin 20}{10}\]\[a = 0.023\ \text{m s}^{-2}\]

June 2025 Paper 2 Q15

AQACurrent spec8 marksResolving Forces Vectors

15 A particle moves under the actions of two forces, \(\mathbf{F}_1\) and \(\mathbf{F}_2\)

\(\mathbf{F}_1\) has magnitude 17 newtons and acts due East.
\(\mathbf{F}_2\) has magnitude 26 newtons and acts at a bearing of 310°

The resultant of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) is \(\mathbf{R}\)

(a) Show that the magnitude of \(\mathbf{R}\) is 17.0 newtons, correct to three significant figures. [4 marks]

(b) Find the angle that \(\mathbf{R}\) makes with \(\mathbf{F}_1\)

Give your answer to the nearest degree.

[2 marks]

(i) State the magnitude of \(\mathbf{F}_3\) [1 mark]

(ii) State the bearing on which \(\mathbf{F}_3\) acts. [1 mark]

Mark scheme (a)

Scheme	Marks	AO
Resolves horizontally or vertically to find a correct expression for one of the i or j components of the resultant force. Or Uses the cosine rule with b = 17 and c = 26	M1	3.3
Resolves horizontally and vertically to find a correct expression for both i and j components of the resultant force PI by H: AWRT \(\pm\) 2.9 V: AWRT \(\pm\)16.7 Or Uses the cosine rule to obtain a² = AWRT 287.8	A1	1.1b
Finds the magnitude of the resultant vector of their horizontal and vertical components Or Finds the square root of their expression for b² + c² − 2bcCosA PI by AWFW [16.96, 16.97]	M1	1.1a
Completes reasoned argument showing a full method to obtain AWFW [16.96, 16.97] and concludes that the magnitude of R is 17.0 Condone 17 AG	R1	2.1
	(4)

Typical solution

Horizontally: \(17 - 26\cos(40)\)

\[= -2.917\ldots\]

Vertically: \(26\sin(40)\)

\[= 16.712\ldots\]\[\sqrt{(-2.917)^2 + 16.712^2} = 16.97\ \text{N}\]\[\approx 17.0\ \text{N}\]

Mark scheme (b)

Scheme	Marks	AO
Uses an appropriate trigonometric equation using their components for R Or Deduces that the two 17N forces make this an isosceles triangle, eg: 180 − (2 x 40) PI AWRT 79 or 80	M1	3.1a
Obtains any of the following angles AWRT 100, 101, \(\pm 259\), \(\pm 260\)	A1	1.1b
	(2)

Typical solution

\[\tan^{-1}\left(\frac{16.712}{2.917}\right) = 80^\circ\]\[180 - 80 = 100^\circ\]

June 2025 Paper 2 Q11

AQACurrent spec1 markResolving Forces

11 An object is pulled along a horizontal surface. The normal reaction between the surface and the object is 125 newtons.

The object experiences a constant frictional force of 40 newtons.

Find the value of the coefficient of friction.

Circle your answer. [1 mark]

0.245
0.32
0.5
3.125

June 2024 Paper 2 Q21

AQACurrent spec9 marksConnected Particles Resolving Forces

21 Two heavy boxes, \(M\) and \(N\), are connected securely by a length of rope.

The mass of \(M\) is 50 kilograms.
The mass of \(N\) is 80 kilograms.

\(M\) is placed near the bottom of a rough slope.
The slope is inclined at 60° above the horizontal.

The rope is passed over a smooth pulley at the top end of the slope so that \(N\) hangs with the rope vertical.

The boxes are initially held in this position, with the rope taut and running parallel to the line of greatest slope, as shown in the diagram below.

Box M on a slope inclined at 60° to the horizontal, connected by a rope parallel to the slope over a pulley at the top to box N hanging vertically

When the boxes are released, \(M\) moves up the slope as \(N\) descends, with acceleration \(a\) m s⁻²

The tension in the rope is \(T\) newtons.

(a) Explain why the equation of motion for \(N\) is\[80g - T = 80a\] [1 mark]

(b) Show that the normal reaction force between \(M\) and the slope is \(25g\) newtons. [1 mark]

Show that

\[a \geqslant \frac{(11 - 5\sqrt{3})g}{26}\] [6 marks]

(d) State one modelling assumption you have made throughout this question. [1 mark]

Mark scheme (c)

Scheme	Marks	AO
Resolves parallel to the slope to obtain \(50g\sin 60\) or better PI OE	B1	1.1b
Obtains \(\mu \times 25g\) for friction OE Or States \(\mu = 1\) and obtains \(25g\) for friction	B1	3.3
Forms a three or four-term equation of motion for \(M\) using \(F = ma\) parallel to slope Condone cos60 OE and sign errors	M1	3.3
Obtains single correct equation or inequality with \(\mu\) Or \(T - 50g\sin 60 - 25g = 50a\) where \(\mu = 1\) is stated Or \(T - 50g\sin 60 - 25g \leqslant 50a\) where \(\mu = 1\) is stated	A1	1.1b
Eliminates \(T\) using the equation in 21(a) and their equation or inequality with \(\mu\) Condone use of \(T = 80a + 80g\)	M1	1.1a
Completes reasoned argument to show \(a \geqslant \dfrac{(11 - 5\sqrt{3})g}{26}\) Must include clear reason for using \(\mu = 1\) or uses \(0 \leqslant \mu \leqslant 1\) or uses \(\mu \leqslant 1\) AG	R1	2.1
	(6)

Typical solution

parallel to slope

\[T - 50g\sin 60 - F = 50a\]\[F = 25g\mu\]\[T - 50g\sin 60 - 25g\mu = 50a\]

Using given equation for \(N\):

\[80g - T = 80a\]\[80g - 50g\sin 60 - 25g\mu = 130a\]\[a = \frac{80g - 50g\sin 60 - 25g\mu}{130}\]\[a = \frac{\left(16 - 5\sqrt{3} - 5\mu\right)g}{26}\]

Acceleration will be at its minimum when \(\mu = 1\)

Therefore

\[a \geqslant \frac{(11 - 5\sqrt{3})g}{26}\]

June 2023 Paper 2 Q19

AQACurrent spec12 marksConnected Particles Resolving Forces

19 A wooden toy comprises a train engine and a trailer connected to each other by a light, inextensible rod.

The train engine has a mass of 1.5 kilograms.
The trailer has a mass 0.7 kilograms.

A string inclined at an angle of 40° above the horizontal is attached to the front of the train engine.

The tension in the string is 2 newtons.

As a result the toy moves forward, from rest, in a straight line along a horizontal surface with acceleration 0.06 m s⁻² as shown in the diagram below.

Toy train engine pulling a trailer along a horizontal surface, with acceleration 0.06 m s^−2 to the right and a string force of 2 N at 40° above the horizontal attached to the front of the engine

As it moves the train engine experiences a total resistance force of 0.8 N

(a) Show that the total resistance force experienced by the trailer is approximately 0.6 N [4 marks]

(b) At the instant that the toy reaches a speed of 0.5 m s⁻¹ the string breaks.

As a result of this the train engine and trailer decelerate at a constant rate until they come to rest, having travelled a distance of \(h\) metres.

It can be assumed that the resistance forces remain unchanged.

(i) Find the tension in the rod after the string has broken. [4 marks]

(ii) Find \(h\) [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Resolves the 2 N force to obtain either \(2\cos 40\) AWRT 1.53 or \(2\sin 40\) AWRT 1.29 May be seen on the diagram.	M1	1.1a
Uses Newton’s 2nd Law to form a four term equation for the whole system. This may be seen with total resistance equivalent to \(0.8 + R\) or Uses Newton’s 2nd Law to form a three term equation for the trailer or a four term equation for the engine. Condone one incorrect sign.	M1	3.3
Obtains a fully correct equation for the whole system. or Obtains two fully correct equations for the train engine and the trailer. For example: \(2\cos 40 - T - 0.8 = 0.09\) \(T - R = 0.042\)	A1	3.3
Completes a reasoned argument to show that the given value for \(R\) is approximately 0.6 AG	R1	2.1
	(4)

Typical solution

Use \(F = ma\) for system

\[2\cos 40 - (0.8 + R) = 2.2(0.06)\]\[1.53 - 0.8 - R = 0.132\]\[R \approx 0.6\text{ N}\]

Mark scheme (b)

Scheme	Marks	AO
(i) Forms an equation of motion without a driving force for the engine or the combined system. PI by \(a = \dfrac{7}{11}\) or \(-\dfrac{7}{11}\)	M1	3.3
Obtains one of the following: \(\pm(T - 0.6) = 0.7a\) or \(\mp(0.8 + T) = 1.5a\) or \(\pm(0.8 + 0.6) = 2.2a\) Allow use of \(a = -\dfrac{7}{11}\)	M1	3.4
Obtains a correct pair of equations of motion for the trailer and the engine which are both moving in the same direction. or Obtains \(a = -\dfrac{7}{11}\) and one fully correct equation of motion involving \(T\)	A1	1.1b
Finds \(T\) AWFW [0.15 , 0.16]	A1	1.1b
	(4)
(ii) Obtains \(a = -\dfrac{7}{11}\) AWRT \(-0.64\) Condone missing or incorrect units	B1	3.1b
Selects an appropriate equation of constant acceleration to find \(s\) and substitutes \(u\) = 0.5 , \(v\) = 0 and their \(a\) Do not accept \(a = -g\)	M1	1.1a
Obtains required distance. AWRT 0.2 ISW	A1	1.1b
	(3)

Typical solution

(i)

\[-0.8 - T = 1.5a\]\[T - 0.6 = 0.7a\]\[T = \frac{17}{110}\text{ N}\]

(ii)

\[a = -\frac{7}{11}\]\[0 = 0.5^2 + 2ah\]\[h = \frac{11}{56} \approx 0.20\]

June 2023 Paper 2 Q15

AQACurrent spec4 marksForces & Newton's Laws Resolving Forces

15 In this question use \(g = 9.8\) m s⁻²

A particle, \(Q\), moves in a straight line across a rough horizontal surface.

A horizontal driving force of magnitude \(D\) newtons acts on \(Q\)

\(Q\) moves with a constant acceleration of 0.91 m s⁻²

\(Q\) has a weight of 0.65 N

The only resistance force acting on \(Q\) is due to friction.

The coefficient of friction between \(Q\) and the surface is 0.4

Find \(D\) [4 marks]

Mark scheme

Scheme	Marks	AO
Uses \(m = \dfrac{W}{g}\) PI by sight of AWRT 0.07	B1	1.1b
States or uses \(F = \mu R\) PI by sight of \(0.26\)	M1	1.1a
Forms a three term equation using \(a = 0.91\), their \(F\) and their \(m\)	M1	3.3
Forms fully correct equation with all values substituted correctly to obtain \(D\) = 0.32 AWRT 0.32	A1	1.1b
	(4 marks)

Typical solution

\[F = \mu R = 0.4 \times 0.65 = 0.26\]\[m = \frac{0.65}{9.8} = 0.066\]\[D - F = ma\]\[D - 0.26 = 0.066 \times 0.91\]\[D = 0.32\]

June 2022 Paper 2 Q19

AQACurrent spec11 marksResolving Forces SUVAT

19 In this question use \(g\) = 9.8 m s⁻²

A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal.

The bottom of the ramp is at the point \(O\).

A crate of mass 20 kg is at rest at the point \(A\) on the ramp.

The crate is pulled up the ramp using a rope attached to the crate.

Once in motion, the rope remains taut and parallel to the line of greatest slope of the ramp.

A ramp rising from O at 25 degrees to the horizontal; a crate at A on the ramp with a rope pulling it up the slope

(a) The tension in the rope is 230 N

The crate accelerates up the ramp at 1.2 m s⁻²

Find the coefficient of friction between the crate and the ramp. [7 marks]

(b)

(i) The crate takes 3.8 seconds to reach the top of the ramp.

Find the distance \(OA\). [3 marks]

(ii) Other than air resistance, state one assumption you have made about the crate in answering part (b)(i). [1 mark]

Mark scheme (a)

Scheme	Marks	AO
States \(F = \mu R\) seen anywhere PI by use of \(\mu R\) in their 4-term equation of motion or on diagram	B1	3.3
Resolves the weight parallel to the slope to obtain \(mg\sin 25\) or better	B1	1.1b
Resolves perpendicular to the slope to obtain \(R = mg\cos 25\) or better	B1	1.1b
Uses F = ma to form a four-term equation with consistent signs. eg \(T - \text{weight} - \text{Friction} = ma\) Condone omission of \(g\) in weight and friction component	M1	3.3
Substitutes \(T\) = 230 and \(F = \mu mg\cos 25\) into their four term \(F = ma\) equation with consistent signs. Condone ‘\(mga\)’ in \(F = ma\) for this mark	M1	1.1a
Obtains single correct equation with all numerical values substituted. eg \(230 - 196\sin 25 - 196\cos 25\,\mu = 24\) Scores B1B1B1M1M1A1	A1	1.1b
Obtains \(\mu = 0.69\) CAO	A1	3.2a
	(7)

Typical solution

\[T - \text{weight} - \text{Friction} = ma\]\[T - 20g\sin 25 - F = m \times 1.2\]\[T - 20g\sin 25 - 20g\cos 25\,\mu = m \times 1.2\]\[230 - 196\sin 25 - 196\cos 25\,\mu = 24\]\[\mu = 0.69\]

Mark scheme (b)

Scheme	Marks	AO
(i) Substitutes \(u\) = 0, \(a\) = 1.2 and \(t\) = 3.8 into \(s = ut + \dfrac{1}{2}at^2\) Or uses appropriate constant acceleration equations that forms a complete method to obtain \(s\)	M1	1.1a
Obtains AWRT 8.7	A1	1.1b
Obtains 10 − their 8.7 AWRT 1.3 or FT their 8.7 provided it is less than 10 Condone missing units	A1F	1.1b
	(3)
(ii) States the crate has been modelled as a particle OE	E1	3.5b
	(1)
	(11 marks)

Typical solution

(i)

\[s = ut + \frac{1}{2}at^2\]\[s = \frac{1}{2} \times 1.2 \times 3.8^2 = 8.664\]\[OA = 10 - 8.664 = 1.3 \text{ m}\]

(ii)

The crate is a particle

June 2022 Paper 2 Q18

AQACurrent spec8 marksResolving Forces

18 An object, \(O\), of mass \(m\) kilograms is hanging from a ceiling by two light, inelastic strings of different lengths.

The shorter string, of length 0.8 metres, is fixed to the ceiling at \(A\).

The longer string, of length 1.2 metres, is fixed to the ceiling at \(B\).

This object hangs 0.6 metres directly below the ceiling as shown in the diagram.

A horizontal ceiling with points A and B; strings AO (shorter) and BO (longer) meet at the object O, which hangs 0.6 m below the ceiling

(a) Show that the tension in the shorter string is over 30% more than the tension in the longer string. [4 marks]

(b) The tension in the longer string is known to be \(2g\) newtons.

Find the value of \(m\). [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains both \(\sin^{-1}\left(\dfrac{0.6}{0.8}\right)\) and \(\sin^{-1}\left(\dfrac{0.6}{1.2}\right)\) OE Accept complementary angles or exact values \(\dfrac{\sqrt{3}}{2}\) and \(\dfrac{\sqrt{7}}{4}\)	B1	1.1b
Resolves forces horizontally to form equilibrium equation, one component correct Or Uses a triangle of forces and applies the sine rule	M1	3.3
Obtains correct equation with angles substituted	A1	1.1b
Rearranges the correct equation to show that \(T_{OA} = k\,T_{OB}\) OE where \(1.305 \leqslant k \leqslant 1.325\) Completes argument to conclude that the tension in the shorter string is over 30% more than the tension in the longer string	R1	2.1
	(4)

Typical solution

Angle for \(OA = \sin^{-1}\left(\dfrac{0.6}{0.8}\right) = 48.59^\circ\)

Angle for \(OB = \sin^{-1}\left(\dfrac{0.6}{1.2}\right) = 30^\circ\)

\[T_{OA}\cos A = T_{OB}\cos B\]\[T_{OA} = T_{OB}\frac{\cos B}{\cos A}\]\[T_{OA} = T_{OB}\frac{\cos 30}{\cos 48.59}\]\[\therefore T_{OA} = 1.309T_{OB}\]\[\therefore T_{OA} \gt 1.3T_{OB}\]

So, the tension in the shorter string is more than 30% greater than the tension in the longer string

Mark scheme (b)

Scheme	Marks	AO
Obtains \(T_{OA} = 2g \times\) their ratio	B1	1.1b
Resolves forces vertically to form a three-term equilibrium equation, with at least two terms correct Or Uses a triangle of forces and applies the sine rule	M1	3.3
Forms fully correct equation of forces in equilibrium This mark can be awarded for \(mg = T_{OA}\sin A + T_{OB}\sin B\)	A1	1.1b
Substitutes their \(T_{OA}\) and \(T_{OB}\) and correct values for angles into the correct equation and obtains AWRT \(m\) = 3 Might come from 2.96.. FT their ratio from part (a) provided their \(m\) = AWRT 3	A1F	3.4
	(4)
	(8 marks)

Typical solution

\[mg = T_{OA}\sin A + T_{OB}\sin B\]\[mg = 2.6g\sin 48.59 + 2g\sin 30\]\[m = 3.0\]

June 2025 Paper 3 Q12

OCR ACurrent spec13 marksForces & Newton's Laws Resolving Forces

Fig. 1: rectangular block B resting on a plane labelled Pi inclined at 30 degrees to the horizontal, with a horizontal force T N applied to the block, pointing away from the slope — Fig. 1

A rectangular block \(B\) of mass 10 kg lies at rest in limiting equilibrium on a rough plane \(\Pi\) inclined at 30° to the horizontal. A horizontal force of magnitude \(T\) N, acting above a line of greatest slope, is applied to \(B\) (see Fig. 1).

The coefficient of friction between \(B\) and the plane is 0.8.

(a) Show that the value of \(T\) is 14.9, correct to 3 significant figures. [6]

For the remainder of the question, you should you use this value of \(T\).

Fig. 2: the block on the same 30 degree plane, now cut into an upper and a lower block along a line at 30 degrees to the horizontal (shown by a dashed horizontal reference line); the horizontal force T N acts on the lower block — Fig. 2

Block \(B\) is now cut at an angle of 30° to the horizontal into two smaller blocks. The upper block has a mass of 4 kg, and the lower block has a mass of 6 kg. The two blocks are held at rest with the lower block on \(\Pi\). The horizontal force of magnitude \(T\) N is now applied to the lower block (see Fig. 2).

The two blocks are released from rest and in the subsequent motion the upper block starts to move with acceleration 3.5 m s⁻².

(b) Determine the coefficient of friction between the two blocks. [4]

Mark scheme (a)

Scheme	Marks	AO
Resolving perpendicular to the plane to form an equation	M1*	3.3
\(R + T\sin 30 = 10g\cos 30\)	A1	1.1
Resolving parallel to the plane to form an equation	M1*	3.3
\(T\cos 30 + 10g\sin 30 = F\) or \(T\cos 30 + 10g\sin 30 = 0.8 \times \text{‘}R\text{’}\)	A1	1.1
\(T\cos 30 + 10g\sin 30 = 0.8(10g\cos 30 - T\sin 30)\)	M1dep*	3.4
\(T = \dfrac{40\sqrt{3} - 50}{5\sqrt{3} + 4}g = 14.9\) (N)	A1	2.1
	[6]

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

Scheme	Marks	AO
Resolving vertically to form an equation	M1*
\(F\sin 30 + R\cos 30 = 10g\)	A1
Resolving horizontally to form an equation	M1*
\(T + R\sin 30 = F\cos 30\)	A1
\(R(0.8\sin 30 + \cos 30) = 10g\) \(\Rightarrow T + \dfrac{10g \times \sin 30}{0.8\sin 30 + \cos 30} = \dfrac{10g \times 0.8\cos 30}{0.8\sin 30 + \cos 30}\)	M1dep*
\(T = \dfrac{40\sqrt{3} - 50}{5\sqrt{3} + 4}g = 14.9\) (N)	A1

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…

Mark scheme (b)

Scheme	Marks	AO
\(4g\sin 30 - F_1 = 4 \times 3.5\)	M1*	3.1b
\(R_1 = 4g\cos 30\)	B1	2.1
\(4g\sin 30 - 4 \times 3.5 = \mu \times 4g\cos 30\) \((\Rightarrow 5.6 = \mu \times 4g\cos 30)\)	M1dep*	3.4
\((\mu =)\ 0.165\)	A1	1.1
	[4]

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…

Mark scheme (c)

Scheme	Marks	AO
\((F_{\max} =)\ 0.8(10g\cos 30 - 14.9\sin 30)\ (= 61.9\ldots)\) or \((F_{\max} =)\ 14.9\cos 30 + 10g\sin 30\ (= 61.9\ldots)\)	B1*	3.1b
(Resultant force down the plane is) \(4g\sin 30 - 4 \times 3.5 + 14.9\cos 30 + 6g\sin 30\ (= 47.9\ldots)\) or \(5.6 + 14.9\cos 30 + 6g\sin 30\ (= 47.9\ldots)\)	B1*	3.1b
\(F_{\max} \gt\) Resultant force down the plane and so therefore the lower block does not move or \(47.9\ldots - 61.9\ldots = 6a \Rightarrow a \lt 0 \therefore\) doesn’t move	B1dep*	3.2a
	[3]

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

June 2025 Paper 3 Q10

OCR ACurrent spec8 marksMoments Resolving Forces

Horizontal rod AB with end A against a vertical wall, right angle at A; G marked on AB; a string from B to a point on the wall above A makes angle theta degrees with the rod at B

The diagram shows a non-uniform rod \(AB\) of mass 4 kg and length 2 m. The end \(A\) of the rod rests against a rough vertical wall. The rod is held in a horizontal position, perpendicular to the wall, by a light inextensible string attached to the rod at \(B\). The other end of the string is attached to the wall at a point vertically above \(A\). The string is inclined at an angle of \(\theta^\circ\) to the horizontal, where \(\sin\theta = \frac{4}{5}\).

The rod rests in equilibrium in a vertical plane perpendicular to the wall. The rod’s weight acts at the point \(G\) on \(AB\).

You are given that the magnitude of the moment of the rod’s weight about \(A\) is 47.04 N m.

(a) Show that the distance \(AG\) is 1.2 m. [1]

(b) Find the tension in the string. [2]

Mark scheme (a)

Scheme	Marks	AO
\(AG \times 4g = 47.04\) or \((AG =)\ \dfrac{47.04}{4g}\) or \(4g \times 1.2 = 47.04\)	B1	3.1b
	[1]

Notes

B1: Taking moments about \(A\) for the rod’s weight – allow \(\dfrac{47.04}{4 \times 9.8}\) oe (so using 9.8 for \(g\)) but B0 for only \(\dfrac{47.04}{39.2}\) oe
B0 for \(AB \times 4g = 47.04\) or similar

AG – www but need not see 1.2 explicitly stated

Mark scheme (b)

Scheme	Marks	AO
\(2 \times \frac{4}{5}T = 47.04\) or \(2 \times \frac{4}{5}T = 1.2 \times 4g\)	M1	3.3
\(T = 29.4\) (N)	A1	1.1
	[2]

Notes

M1: Taking moments about \(A\) for the rod – M1 for \(2 \times \frac{4}{5}T = 47.04\) or \(2 \times T\sin 53 = 47.04\) or \(2 \times \frac{3}{5}T = 47.04\) or \(2 \times T\cos 53 = 47.04\) only (or \(1.2 \times 4g\) for 47.04)

Allow use of 53 or better (53.1301…) for \(\theta\) - note that e.g. \(2 \times T\sin 37 = 47.04\) or \(2 \times T\cos 37 = 47.04\) (therefore using the complementary angle) is M1

A1: cao exact e.g. \(\frac{147}{5}\) or awrt 29.4

Mark scheme (c)

Scheme	Marks	AO
\(X = \pm\text{‘}29.4\text{’} \times \frac{3}{5}\) or \(\pm\text{‘}29.4\text{’} \times \cos 53 \quad (= \pm 17.64)\)	B1FT	3.3
Resolve vertically or take moments about \(B\) or \(G\) for the forces acting on the rod to form an equation or an expression for \(Y\)	M1*	3.3
\(\mathrm{R}(\uparrow)\): \(\text{‘}29.4\text{’} \times \frac{4}{5} + Y = 4g\) or \(\text{‘}29.4\text{’} \times \frac{4}{5} = 4g + Y\) or \(\mathfrak{M}(B)\): \(0.8 \times 4g = 2Y\) or \(\mathfrak{M}(G)\): \(0.8 \times \text{‘}29.4\text{’} \times \frac{4}{5} = Y \times 1.2\)	A1FT	1.1
\(\sqrt{17.64^2 + 15.68^2}\)	M1dep*	3.4
Magnitude is 23.6 (N)	A1	1.1
	[5]

Notes

B1FT: Follow through their value of \(T\) only. Allow use of 53 or better (53.1301…) for \(\theta\)

\(X\) is the horizontal component of the contact force at \(A\)

M1*: Must be the correct number of dimensionally correct terms but allow sign errors and sin/cos confusion only – all values must be substituted. If taking moments about \(B\) or \(G\) allow sin/cos confusion only

\(Y\) is the vertical contact force at \(A\)

A1FT: Correct equation for \(Y\) follow through their value of \(T\) only – if correct then \(Y = \pm 15.68\)

Allow \(\sin 53\) or \(\cos 37\) or better for \(\frac{4}{5}\)

M1dep*: Correct use of Pythagoras with their values of \(X\) and \(Y\) which must come from equations with the correct number of dimensionally correct relevant terms

Relevant for \(X\) requires \(\text{‘}29.4\text{’} \times \frac{3}{5}\) or \(\text{‘}29.4\text{’} \times \frac{4}{5}\) oe only

A1: awrt 23.6 or exact e.g. \(\dfrac{49\sqrt{145}}{25}\) oe

23.60152…

June 2024 Paper 3 Q13

OCR ACurrent spec12 marksProjectiles Resolving Forces

A particle at A at the bottom of a slope inclined at angle theta to horizontal ground, projected up the slope at 6 m s to the minus 1; the slope AB has length 1.375 m; a dashed curved path leaves B and lands at C on the ground

The points \(A\) and \(B\) are the lower and upper ends, respectively, of a line of greatest slope on a plane inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = 0.6\) and \(AB = 1.375\) m (see diagram).

A particle \(P\) is projected up the plane with speed \(6\,\mathrm{m\,s^{-1}}\) from \(A\) towards \(B\).

The plane at \(A\) is fixed to the ground which is horizontal.

The surface of the plane is rough and the coefficient of friction between \(P\) and the plane is 0.25.

(a) Show that the speed of \(P\) at \(B\) is \(3.8\,\mathrm{m\,s^{-1}}\). [6]

The particle leaves the slope at \(B\) and moves freely under gravity.

The particle first lands at a point \(C\) on the horizontal ground. The time taken for \(P\) to travel from \(A\) to \(C\) is \(T\) seconds.

(b) Determine the value of \(T\). [6]

Mark scheme (a)

Scheme	Marks	AO
\(R(\text{perp. plane}): R = mg\cos 36.86\ldots\) or \(mg \times 0.8\)	B1	3.3
\(F = 0.25 \times mg\cos 36.86\ldots\) or \(0.25 \times mg \times 0.8\)	M1*	3.4
N2L (parallel to plane): \(-0.25 \times mg \times 0.8 - mg \times 0.6 = ma\) or \(-0.25 \times mg \times \cos 36.86\ldots - mg \times \sin 36.86\ldots = ma\)	M1dep*	3.1b
\(a = -7.84\) or \(7.84\)	A1	1.1
\(v^2 = 6^2 + 2 \times -7.84 \times 1.375\)	M1dep*	3.4
\(v^2 = 14.44 \Rightarrow v = 3.8\ (\mathrm{m\,s^{-1}})\)	A1	2.2a
	[6]

Notes

B1: Allow using an angle of 37 or better
soi (possibly in N2L)

M1*: \(F = 0.25R\) with their \(R\) (with \(R \ne mg\)) – using an angle of 37 or better – allow sin/cos mix
\(R\) must be a component of weight soi (possibly in N2L)

M1dep*: Applying N2L parallel to the plane with correct number of terms and weight term resolved, dimensionally consistent – if down the plane is the positive direction expect \(0.25 \times mg \times 0.8 + mg \times 0.6 = ma\)
Allow sign errors and sin/cos mix, using an angle of 37 or better

A1: \(\pm 0.8g\) - allow awrt \(\pm 7.84\) or awrt \(\pm 7.85\) (using an angle of 37) www
Condone \(\pm 7.8\) or \(\pm 7.9\) www

M1dep*: Use of \(v^2 = u^2 + 2as\) with correct \(u\) and \(s\) and their negative \(a\) (\(\ne -9.8\)) - dependent on both previous M marks – must see the values substituted into the formula (as value for \(v\) is AG) – therefore just seeing \(v^2 = \frac{361}{25}\) or 14.44 is M0
Condone finding a positive value of \(a\) and then using the negative version

A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\) (so seeing e.g. \(\cos 36.86\ldots\) in their working loses this final A mark)
Must see either \(v^2 = 14.44\) or \(v^2 = 36 - 21.56\) or \(v^2 = 6^2 - 21.56\) or a correct expression for \(v\) before given answer

SC – if \(m\) absent throughout (so implying \(m = 1\)) or \(m\) given a numerical value then at most B0 M1 M1 A1 M1 A0 can be awarded.

Alternative for final two marks

Scheme	Marks
\(1.375 = 6t + \frac{1}{2} \times -7.84 \times t^2\) \(t = \frac{55}{196}\) (or \(t = 1.25\)) \(v = 6 + (-7.84) \times \frac{55}{196}\)	M1dep*
\(v = 3.8\ (\mathrm{m\,s^{-1}})\)	A1

M1dep*: Use of \(s = ut + \frac{1}{2}at^2\) with correct \(u\) and \(s\) and their negative \(a\) (\(\ne -9.8\)), and selects the smaller of the two positive \(t\) values and uses this value correctly in \(v = u + at\) with correct \(u\) and their negative \(a\) - dependent on both previous M marks
So, using the second positive \(t\) (if correct this is 1.25) is M0

A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\)
Must see a correct expression for \(v\) before given answer

Alternative energy approach to 13(a) (Appendix)

Scheme	Marks
\(R(\text{perp. plane}): R = mg\cos 36.86\ldots\) or \(mg \times 0.8\)	B1
\(F = 0.25 \times mg\cos 36.86\ldots\) or \(0.25 \times mg \times 0.8\)	M1
Work done by gravity/PE term \(= \pm mg \times (1.375 \times 0.6)\) or \(\pm mg \times (1.375 \times \sin 36.86\ldots)\) OR Work done by friction \(\pm 1.375 \times (0.25 \times mg \times 0.8)\) or \(\pm 1.375 \times (0.25 \times mg \times \cos 36.86\ldots)\)	B1
\(\frac{1}{2}mv^2 - \frac{1}{2}m \times 6^2 = -mg(1.375 \times 0.6) - 1.375 \times (0.25 \times mg \times 0.8)\)	M1
Correct W-E equation	A1
\(v^2 = 14.44 \Rightarrow v = 3.8\ (\mathrm{m\,s^{-1}})\)	A1

B1: Allow using an angle of 37 or better
soi (possibly in Work-Energy equation)

M1: Use of \(F = 0.25R\) with their \(R\) (with \(R \ne mg\)) – using an angle of 37 or better – allow sin/cos mix
\(R\) must be a component of weight not mass
soi (possibly in W-E equation)

B1: Correct expression for either WD by gravity or friction – using an angle of 37 or better
soi (possibly in W-E equation)

M1: Attempt at W-E principle – equation with the correct number of relevant terms, dimensionally correct, allow sign errors and sin/cos mix,
M0 if PE term is \(\pm 1.375mg\)

A1: Correct W-E equation

A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\)
Must see either \(v^2 = 14.44\) or \(v^2 = 36 - 21.56\) or \(v^2 = 6^2 - 21.56\) or a correct expression for \(v\) before given answer

SC – if \(m\) absent throughout (so implying \(m = 1\)) or \(m\) given a numerical value then at most B0 M1 B1 M1 A1 A0 can be awarded.

Mark scheme (b)

Scheme	Marks	AO
Time from \(A\) to \(B\) is \(t_1\) where \(3.8 = 6 + (-7.84)t_1\)	M1	3.4
\([t_1 =]\ 0.281\) (s)	A1	1.1
[Height of \(P\) at \(B\) is] 0.825 (m)	B1	3.1b
Equation for time \(t_2\) (from \(B\) to \(C\))	M1	3.1b
\(-0.825 = (3.8 \times 0.6)t_2 + 0.5(-g)t_2^{\,2}\) or \(-0.825 = (3.8 \times \sin 36.86\ldots)t_2 + 0.5(-g)t_2^{\,2}\) \(\left(4.9t_2^{\,2} - 2.28t_2 - 0.825 = 0\right)\)	A1	1.1
\(T\,[= 0.280\ldots + 0.704\ldots] = 0.985\)	A1	2.2a
	[6]

Notes

M1: Use of \(v = u + at\) with correct \(u\) and \(v\), and their negative \(a\) from (a) (or any other complete method e.g. \(1.375 = \left(\dfrac{6 + 3.8}{2}\right)t_1\))
Give BOD if \(6 = 3.8 + 7.84t_1\) seen but if \(v = u + at\) stated before then M0

A1: awrt 0.281 or accept 0.28 www. If exact then expect \(\frac{55}{196}\)
0.2806122...

B1: Must be exact e.g. \(\frac{33}{40}\) – allow un-simplified e.g. \(1.375 \times 0.6\)
Allow negative

M1: M1 for an equation for time \(t_2\) (from \(B\) to \(C\)) with awrt \(\pm 0.83\) for \(s\), \(a = \pm g\) and either \(u = 3.8 \times \sin\theta\) or \(u = 3.8 \times \cos\theta\) with \(\theta = 37\) (or better) substituted (or corresponding exact value) – condone sign errors

A1: A1 for a correct (un-simplified) equation for \(t_2\) - must be using 0.825 but allow \(\theta = 37\) or better
For reference: \(t_2 = 0.704346599\ldots\)

A1: awrt 0.985 only – dependent on all previous marks in this part
0.984958…

June 2024 Paper 3 Q11

OCR ACurrent spec7 marksMoments Resolving Forces

Rectangle ABCD leaning with corner A on horizontal ground and corner D against a vertical wall; AD makes an angle of 50 degrees with the ground, AB = 2 m and BC = 6 m

A uniform rectangular lamina \(ABCD\) has a mass of 0.5 kg. The length of \(AB\) is 2 m, and the length of \(BC\) is 6 m. The lamina is in limiting equilibrium with corner \(A\) in contact with rough horizontal ground and corner \(D\) in contact with a smooth vertical wall. The lamina rests in a vertical plane that is perpendicular to the wall, with \(AD\) inclined at 50° to the horizontal (see diagram).

(a) By taking moments, show that the magnitude of the normal contact force between the lamina and the wall is 1.24 N, correct to 3 significant figures. [4]

(b) Determine the coefficient of friction between the lamina and the ground. [3]

Mark scheme (a)

Scheme	Marks	AO
\(R_D \times k_1\sin 50 = 0.5g \times k_2\) or \(R_D \times k_1\cos 50 = 0.5g \times k_2\) (see notes)	M1	3.3
\(6(R_D\sin 50) = \ldots\)	A1	1.1
\(\ldots 3(0.5g\cos 50) - 1(0.5g\sin 50)\)	A1	3.1b
\(R_D = \dfrac{1.5g\cos 50 - 0.5g\sin 50}{6\sin 50} = 1.24\) (N)	A1	2.2a
	[4]

Notes

M1: Attempt at moments about \(A\) – at least two terms (one term for the weight and one term with a component of \(R_D\)) – must be written as an equation to score any marks in this part (see Answer column for the only acceptable forms)
with \(k_1, k_2 \gt 0\) but not \(= \pm 1\)

A1: Correct moment for the contact force at \(D\)

A1: Correct moment for the weight of the lamina (See Appendix) (e.g. \(0.5g \times (3 - \tan 50)\cos 50\) or \(0.5g \times \cos 50 \times 1.808\ldots\) etc. e.g. \(0.5g \times \sqrt{10}\sin\left(40 - \arctan\left(\tfrac{1}{3}\right)\right)\) or \(0.5g \times \sqrt{10}\sin(21.565\ldots)\) etc.) – if value stated with no or unclear working, then must see at least 5.6953… to award this mark - if moment stated as 5.6993… then A0 (and A0 for the next mark)
For reference 5.695360…

A1: AG – as this answer is given then working must be checked carefully – all previous marks must have been awarded
1.23912743…
A fully correct equation followed by stating 1.24 can score full marks

Additional guidance for 11(a) (Appendix)

There are many ways that candidates are correctly approaching this part:
For example,
\(6(R_D\sin 50) = 3(0.5g\cos 50) - (0.5g\sin 50)\)
\(6(R_D\sin 50) = 0.5g \times (3 - \tan 50)\cos 50\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\sin\left(40 - \arctan\left(\tfrac{1}{3}\right)\right)\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos(140 - \arctan(3))\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos\left(50 + \arctan\left(\tfrac{1}{3}\right)\right)\)

Any of these followed by the correct answer of 1.24 would score full marks (we do not need to see any intermediate working). If these exact expressions are not seen then sufficient working with values that are correct to at least 3 significant figures should be awarded full marks (and those that have 2 significant figures with sufficient working should get partial credit), e.g.

• \(6(R_D\sin 50) = 9.448977\ldots - 3.7536177\ldots\) so seeing \(6(R_D\sin 50) = 9.45 - 3.75\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 9.4 - 3.8\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.22 not 1.24)
• \(6(R_D\sin 50) = 5.7(0)\) is no marks (assume working backwards from the AG and no indication that the RHS contains a weight component)
• \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.808246\ldots\) so seeing \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.81\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.80\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.23 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 0.643 \times 1.81\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times 0.64 \times 1.8\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.23 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 1.16\) followed by 1.24 is M1 A1 A0 A0 (assume working backwards from the given answer on the RHS)
• \(6(R_D\sin 50) = 0.5g \times \sqrt{10}\sin(40 - 18.434948\ldots)\) so seeing \(6(R_D\sin 50) = 0.5g \times 3.16 \times \sin(21.6)\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times 3.16 \times \sin(22)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.26 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 3.2 \times \sin(22)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.28 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos(68.4349\ldots)\) so seeing \(6(R_D\sin 50) = 0.5g \times \sqrt{10} \times \cos(68)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.26 not 1.24)

Mark scheme (b)

Scheme	Marks	AO
\(R(\uparrow): R_A = 0.5g\) \(R(\rightarrow): 1.24 = F_A\)	B1*	3.3
\(1.24 = 0.5g\mu\)	M1dep*	3.4
\(\mu = 0.253\)	A1	2.2a
	[3]

Notes

B1*: Resolving horizontally and vertically (possibly implied by later working) – must be using given 1.24 (or a more accurate correct value) only (so not their incorrect value from part (a))
If taking moments about another point e.g. \(D\), then the corresponding equation(s) must be correct

M1dep*: Use of \(F = \mu R\) or \(F \leqslant \mu R\) with \(0.5g\) and 1.24 (or better) for \(F\)

A1: awrt 0.253
If final answer is \(\mu \geqslant 0.253\) then A0
Using the exact value or 1.24 for \(R_D\) leads to the same answer to 3 significant figures

June 2024 Paper 3 Q10

OCR ACurrent spec6 marksResolving Forces

Block of mass m kg on horizontal ground; a rope from its top runs at 50 degrees to the horizontal up to a small pulley, from which a 5 kg object hangs vertically; a force X N pulls the block on the other side at 20 degrees above the horizontal

A block of mass \(m\) kg is on smooth horizontal ground with one end of a light inextensible rope attached to its upper surface. The other end of the rope is attached to an object of mass 5 kg. The rope passes over a small smooth pulley, and the object hangs vertically below the pulley. The part of the rope between the block and the pulley makes an angle of 50° with the horizontal. A force of magnitude \(X\) N acts on the block at an angle of 20° above the horizontal in the vertical plane containing the rope (see diagram).

You are given that the block is in equilibrium.

(a) Determine the value of \(X\). [3]

You are also given that the magnitude of the contact force exerted by the ground on the block is 147 N.

(b) Determine the value of \(m\). [3]

Mark scheme (a)

Scheme	Marks	AO
\(R(\uparrow \text{object}):\quad T = 5g\)	B1	1.1
\(R(\rightarrow \text{block}):\quad T\cos 50 = X\cos 20\)	M1	3.3
\(X = 33.5\)	A1	1.1
	[3]

Notes

B1: Possibly implied by later working
Allow \(T - 5g = 0\)

M1: Resolving horizontally to form an equation – correct number of terms (possibly with \(T\) replaced with \(5g\)) – allow sin/cos mix but must be using correct angles
Must be components of both \(T\) and \(X\)

A1: awrt 33.5 – condone including of ‘N’
33.51797…

Mark scheme (b)

Scheme	Marks	AO
\(X\sin 20 + 147 + T\sin 50 = mg\)	M1	3.3
\((\text{‘33.5…’})\sin 20 + 147 + 5g\sin 50 = mg\)	A1FT	1.1
\(m = 20\)	A1	1.1
	[3]

Notes

M1: Resolving vertically to form an equation for the block with the correct number of relevant terms – allow sign errors and sin/cos mix but must be using correct angles
Allow \(X\), \(T\) or their value(s) for \(X\) and \(T\) – relevant means both dimensionally correct (so must be using \(mg\)) and that we must see components of \(T\) and \(X\) but no component of the weight or the 147

A1FT: Correct equation for \(m\) FT their value of \(X\) only (so not just \(X\))

A1: Accept either 20 or awrt 20.0; condone inclusion of ‘kg’
www

June 2024 Paper 3 Q9

OCR ACurrent spec8 marksResolving Forces

Two forces acting at a point O: 12 N at 20 degrees to the west of a dashed North line, and 17 N at 50 degrees to the east of North

Two horizontal forces of magnitudes 17 N and 12 N act at a point \(O\) along bearings of 050° and 340° respectively (see diagram).

(a) Determine the magnitude and bearing of the resultant force. [6]

A third horizontal force \(\mathbf{F}\) is now applied at \(O\). The three forces are in equilibrium.

(b) State the magnitude of \(\mathbf{F}\) and give the bearing along which it acts. [2]

Mark scheme (a)

Scheme	Marks	AO
Resolve vertically or horizontally	M1*	2.1
\(R(\uparrow): \pm(12\cos 20 + 17\cos 50)\)	A1	1.1
\(R(\rightarrow): \pm(17\sin 50 - 12\sin 20)\)	A1	1.1
Magnitude is \(\sqrt{(12\cos 20 + 17\cos 50)^2 + (17\sin 50 - 12\sin 20)^2}\)	M1dep*	3.4
\(\tan\theta = \dfrac{17\sin 50 - 12\sin 20}{12\cos 20 + 17\cos 50}\)	M1dep*	3.4
Resultant force is 23.9 (N) on a bearing of 022(°)	A1	2.2a
	[6]

Notes

M1*: Resolve vertically or horizontally – correct number of relevant terms. Allow sign errors and sin/cos mix but must be using correct angles

A1: Need not be simplified
\(\pm 22.20370081\ldots\)

A1: Need not be simplified
\(\pm 8.918513813\ldots\)

M1dep*: Correct method to calculate magnitude from expressions with the correct number of relevant terms

M1dep*: Correct method to calculate a relevant angle (so allow reciprocal) from expressions with the correct number of relevant terms

A1: awrt 23.9 and awrt 022 (so 021.9 is okay but 21.9 is A0)
23.927896…
21.883744…

Alternative method

Scheme	Marks
\(R^2 = 17^2 + 12^2 - 2(17)(12)\cos\alpha\)	M1*
\(R^2 = 17^2 + 12^2 - 2(17)(12)\cos 110\)	B1 A1
\(\dfrac{\sin\theta}{12} = \dfrac{\sin\alpha}{R}\) or \(\cos\theta = \dfrac{R^2 + 17^2 - 12^2}{2 \times R \times 17}\)	M1dep*
Bearing \(= 50 - \theta\)	M1dep*
Resultant force is 23.9 (N) on a bearing of 022(°)	A1

M1*: Correct cosine rule with any angle \(\alpha\)

B1 A1: B1 for correct angle of 110 seen, A1 for correct expression for \(R\) or \(R^2\)
110 may be seen on a diagram

M1dep*: Correct sine rule with their \(\alpha\) or correct cosine rule
If correct \(\theta = 28.11625\ldots\)

M1dep*: Correct calculation for bearing
Dep. both M marks

A1: awrt 23.9 and awrt 022 (so 021.9 is okay but 21.9 is A0)

Mark scheme (b)

Scheme	Marks	AO
Magnitude of \(\mathbf{F}\) is 23.9 (N)	B1FT	1.2
On a bearing of 202(°)	B1FT	3.1b
	[2]

Notes

B1FT: Follow through their resultant in (a)

B1FT: Follow through the answer to (a) – if \(0 \lt \theta \lt 180\) then FT is \(180 + \theta\) and if \(180 \lt \theta \lt 360\) then FT is \(\theta - 180\)
As in part (a) their bearing must be at least three figures (with leading zeros if necessary) but allow greater accuracy

June 2023 Paper 3 Q13

OCR ACurrent spec12 marksConnected Particles Resolving Forces

A block B of mass 2 kg on a horizontal surface is connected by a string over a pulley at the top edge to a particle P of mass 4 kg on a plane inclined at 60 degrees to the horizontal

The diagram shows a small block \(B\), of mass \(2\,\mathrm{kg}\), and a particle \(P\), of mass \(4\,\mathrm{kg}\), which are attached to the ends of a light inextensible string. The string is taut and passes over a small smooth pulley fixed at the intersection of a horizontal surface and an inclined plane. The particle can move on the inclined plane, which is rough, and which makes an angle of \(60^\circ\) with the horizontal. The block can move on the horizontal surface, which is also rough.

The system is released from rest, and in the subsequent motion \(P\) moves down the plane and \(B\) does not reach the pulley.

It is given that the coefficient of friction between \(P\) and the inclined plane is twice the coefficient of friction between \(B\) and the horizontal surface.

(a) Determine, in terms of \(g\), the tension in the string. [7]

When \(P\) is moving at \(2\,\mathrm{m\,s^{-1}}\) the string breaks. In the 0.5 seconds after the string breaks \(P\) moves \(1.9\,\mathrm{m}\) down the plane.

(b) Determine the deceleration of \(B\) after the string breaks. Give your answer correct to 3 significant figures. [5]

Mark scheme (a)

See appendix at the end of the MS for important information regarding this part

Scheme	Marks	AO
	M1*	2.1
\(4g\sin 60 - F_P - T = 4a\)	A1	1.1
\(R_P = 4g\cos 60\)	B1	3.3
\(4g\sin 60 - \mu_P(4g\cos 60) - T = 4a\) \(\Rightarrow 2\sqrt{3}g - 2\mu_P g - T = 4a\)	M1dep*	3.4
\(T - \mu_B(2g) = 2a\)	M1	3.3
\(2\sqrt{3}g - 2\mu_P g - T = 4a\) and \(2T - 4\mu_B g = 4a\) with \(\mu_P = 2\mu_B\) gives \(2\sqrt{3}g - T = 2T\)	A1	3.3
\(2\sqrt{3}g - T = 2T \Rightarrow T = \dfrac{2\sqrt{3}}{3}g\)	A1	2.2a
	[7]

Notes

For the first five marks condone: \(\mu\) used as the coefficient of friction for both \(P\) and \(B\), or implying that \(\mu_B = 2\mu_P\) rather than the correct \(\mu_P = 2\mu_B\)

M1*: Applying N2L parallel to the plane for \(P\) – correct number of terms and weight component resolved – allow sign errors and sin/cos confusion. Allow \(g\) missing but M0 if \(4ga\) in N2L

A1: NB \(T + F_P - 4g\sin 60 = 4a\) is A1 (taking up the plane as +ve dir.)
Where \(F_P\) is the frictional force for \(P\)

B1: Resolving correctly perpendicular to the plane for \(P\)
Where \(R_P\) is the normal contact force for \(P\)

M1dep*: Use of \(F = \mu R\) in the attempt at N2L for \(P\) with their \(R\) which must be a component of \(4g\) only
Where \(\mu_P\) is the coefficient of friction between \(P\) and the plane

M1: Applying N2L parallel to the surface for \(B\) – correct number of terms – allow sign errors but note that \(\mu_B(2g) - T = 2a\) is consistent with \(T + F_P - 4g\sin 60 = 4a\) and therefore gives the correct answer (and is not incorrect working)
Where \(\mu_B\) is the coefficient of friction between \(B\) and the plane
Allow \(g\) missing but M0 if \(2ga\) in N2L or for \(T - F_B = 2a\) only

A1: Solving simultaneously with \(\mu_P = 2\mu_B\) (soi) to obtain a correct equation in \(T\) only

A1: Must be seen in terms of \(g\)
Accept awrt \(1.15g\)

Appendix: exemplar responses for Q13(a)

See main MS for Guidance for the requirements/conditions to award each of these marks.

Case 1 (max 7): candidates who immediately set \(\mu\) as the coefficient of friction for \(B\) and \(2\mu\) as the coefficient of friction for \(P\) (correct)

Response	Marks
\(4g\sin 60 - F_P - T = 4a\)	M1 A1
\(R_P = 4g\cos 60\)	B1
\(4g\sin 60 - 2\mu(4g\cos 60) - T = 4a\)	M1
\(T - \mu(2g) = 2a\)	M1
\(2\sqrt{3}g - T = 2T\)	A1
\(T = \dfrac{2\sqrt{3}}{3}g\)	A1

Or equivalent e.g. replace \(2\mu\) above with \(\mu\) and \(\mu\) above with \(0.5\mu\)

Case 2 (max 5): candidates who immediately set \(2\mu\) as the coefficient of friction for \(B\) and \(\mu\) as the coefficient of friction for \(P\) or set both coefficients of frictions equal to \(\mu\) (both of these are incorrect)

Response (\(2\mu\) for \(B\), \(\mu\) for \(P\))	Marks
\(4g\sin 60 - F_P - T = 4a\)	M1 A1
\(R_P = 4g\cos 60\)	B1
\(4g\sin 60 - \mu(4g\cos 60) - T = 4a\)	M1
\(T - 2\mu(2g) = 2a\)	M1
\(4g\sin 60 - 4g\mu\cos 60 - T = 2T - 8\mu g\)	A0 A0

Response (\(\mu\) for both)	Marks
\(4g\sin 60 - F_P - T = 4a\)	M1 A1
\(R_P = 4g\cos 60\)	B1
\(4g\sin 60 - \mu(4g\cos 60) - T = 4a\)	M1
\(T - \mu(2g) = 2a\)	M1
\(4g\sin 60 - 4g\mu\cos 60 - T = 2T - 4\mu g\)	A0 A0

Case 3 (max 3): assuming that as \(\mu_P = 2\mu_B\) then this implies that \(F_P = 2F_B\) without justification (e.g. no calculation of the normal contact forces for \(P\) and \(B\) considered – so while it turns out to be correct in this case it is not true in general) – if this is assumed then

Response	Marks
\(4g\sin 60 - 2F - T = 4a\)	M1 A1
\(T - F = 2a\)	M1

(so now allow this M mark even though in the main MS this would have been M0)
leading to \(2g\sqrt{3} - 2(T - 2a) - T = 4a\) and therefore \(T = \dfrac{2\sqrt{3}}{3}g\) no further marks (3 marks max. also if assuming that \(2F_P = F_B\))

Case 4 (max 1): if assuming that \(a = 0\) or \(g\) or any other value then they can score the B mark only for \(R_P = 4g\cos 60\)

Mark scheme (b)

Scheme	Marks	AO
\(1.9 = 2(0.5) + \frac{1}{2}a_P(0.5)^2\)	B1	3.4
\(a_P = 7.2\)	B1	1.1
\(4g\sin 60 - \mu_P(4g\cos 60) = 4a_P\)	M1	3.1b
\(\mu_P = 0.26266\ldots \Rightarrow \mu_B = 0.13(1\ldots)\)	A1	3.4
\(a_B = -\mu_B g \Rightarrow\) deceleration is 1.29 \((\mathrm{m\,s^{-2}})\)	A1	3.2a
	[5]

Notes

B1: Applying \(s = ut + \frac{1}{2}at^2\) correctly to find \(a_P\)

M1: Set \(T = 0\) (or apply N2L) to obtain an expression for the acceleration of \(P\) when the string breaks
Correct number of terms, dimensionally correct (so \(g\) not missing), must be using the correct mass of 4 – allow sin/cos mix and sign errors only

A1: Using their acceleration of \(P\) to correctly calculate the coefficient of friction for \(B\). Can be implied from a correct deceleration of \(B\). Accept 0.13 (so 2 sf) or better
For reference: exact value is \(\dfrac{-72 + 49\sqrt{3}}{98}\) (which scores A1)

A1: Accept awrt 1.29 or \(-1.29\)
\(1.28704895\ldots\) or, for reference, exact value is \(\dfrac{-72 + 49\sqrt{3}}{10}\) (which scores A1)

June 2023 Paper 3 Q11

OCR ACurrent spec8 marksMoments Resolving Forces

A rod AB with end A on horizontal ground, inclined at 30 degrees to the horizontal; end B rests against a wall that makes an angle of 55 degrees with the ground

A uniform rod \(AB\), of weight \(20\,\mathrm{N}\) and length \(2.8\,\mathrm{m}\), rests in equilibrium with the end \(A\) in contact with rough horizontal ground and the end \(B\) resting against a smooth wall inclined at \(55^\circ\) to the horizontal. The rod, which rests in a vertical plane that is perpendicular to the wall, is inclined at \(30^\circ\) to the horizontal (see diagram).

(a) Show that the magnitude of the force acting on the rod at \(B\) is \(9.56\,\mathrm{N}\), correct to 3 significant figures. [3]

(b) Determine the magnitude of the contact force between the rod and the ground. Give your answer correct to 3 significant figures. [5]

Mark scheme (a)

Scheme	Marks	AO
	M1	3.1b
\(1.4(20\cos 30) = R_B(2.8\cos 25)\)	M1	1.1
\(R_B = \dfrac{28\cos 30}{2.8\cos 25} = 9.5555330\ldots\) \(= 9.56\) (N) (correct to 3 significant figures)	A1	1.1
	[3]

Notes

M1: Moments about \(A\) to form an equation with the correct number of terms and both the weight and contact force at \(B\) resolved - condone \(20g\) for the weight of the rod. All relevant values must have been substituted. May take moments about another point (and resolve) but must end up after elimination with an equation in \(R_B\) only
Resolved for the weight means either sin or cos with an angle of either 30 or 60 only, and for the contact force at \(B\) either sin or cos with an angle of 25 or 35 or 55 or 65 only

M1: One of the two moment terms correct
Must be as part of a two-term moment equation (so dimensionally correct so M0 if \(20g\) for the weight) but the other term need not be resolved (or resolved correctly) e.g. \(1.4(20\cos 30) = 2.8R_B\) scores M0 M1

A1: AG - correct equation followed by 9.56 scores all 3 marks
Allow awrt 9.56

For reference in parts (a) and (b):
Moments about \(B\): \(R_A(2.8\cos 30) = F_A(2.8\sin 30) + 20(1.4\cos 30)\)
Moments about the mid-point: \(R_A(1.4\cos 30) = F_A(1.4\sin 30) + R_B(1.4\cos 25)\)

Mark scheme (b)

Scheme	Marks	AO
	M1*	3.3
\(R_A + 9.56\sin 35 = 20\) \(F_A = 9.56\cos 35\)	A1 A1	1.1 1.1
\(\sqrt{(9.56\cos 35)^2 + (20 - 9.56\sin 35)^2}\)	M1dep*	1.1
16.5 (N)	A1	2.2a
	[5]

Notes

M1*: Attempt at resolving vertically or horizontally – correct number of terms with the contact force at \(B\) resolved (angle must be one of 25, 35, 55 or 65 only) – or taking moments again (see below) (same conditions for taking moments as in part (a)). M0 if using \(W\) or \(mg\) only for the weight of the rod (must substitute in the given value before awarding this mark – see second guidance column if using \(20g\))
Allow sign errors, sin/cos confusion and \(20g\) used as the weight for the M mark. Allow \(R_B\) instead of the given value of 9.56 for the M mark

A1: Correct expression/equation for \(R_A\)
\(R_A = 20 - 5.48\ldots = 14.5\ldots\)

A1: Correct expression/equation for \(F_A\)
\(F_A = 7.83\ldots\)

M1dep*: Correct method for calculating the magnitude of the contact force at \(A\) where \(R_A\) and \(F_A\) are a linear combination of the correct number of terms with 9.56 resolved in both equations (angle must be one of 25, 35, 55 or 65 only)

A1: www awrt 16.5. For reference if using 9.56 for \(R_B\) then \(16.494179\ldots\)
If using more accurate value for \(R_B\) then \(16.494698\ldots\)

June 2023 Paper 3 Q9

OCR ACurrent spec6 marksResolving Forces

A block B on a plane inclined at angle theta to the horizontal, rising to the right; a horizontal force of 2 N acts on B, pointing to the left

A block \(B\) of weight \(10\,\mathrm{N}\) lies at rest in equilibrium on a rough plane inclined at \(\theta\) to the horizontal. A horizontal force of magnitude \(2\,\mathrm{N}\), acting above a line of greatest slope, is applied to \(B\) (see diagram).

(a) Complete the diagram in the Printed Answer Booklet to show all the forces acting on \(B\). [1]

It is given that \(B\) remains at rest and the coefficient of friction between \(B\) and the plane is 0.8.

(b) Determine the greatest possible value of \(\tan\theta\). [5]

The diagram completed with the normal reaction R perpendicular to the plane, friction F up the slope, the weight 10 N vertically downwards, and the 2 N horizontal force to the left — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
	M1*	3.3
\(R + 2\sin\theta = 10\cos\theta\) \(F = 2\cos\theta + 10\sin\theta\)	A1 A1	1.1 1.1
\(2\cos\theta + 10\sin\theta \leqslant 0.8(10\cos\theta - 2\sin\theta)\)	M1dep*	3.4
\(10\cos\theta + 50\sin\theta \leqslant 40\cos\theta - 8\sin\theta\) \(\left(\tan\theta \leqslant \frac{30}{58} \Rightarrow \text{greatest value of}\right)\ \tan\theta\) is \(\frac{15}{29}\)	A1	2.2a
	[5]

Notes

M1*: Resolving parallel or perpendicular to the plane – correct number of terms - must be using 10 for the weight so no marks until this value used/applied (however, see SC in the final A mark)
Allow sin/cos confusion and sign errors – but the 2 and 10 must be resolved and the \(R\) and \(F\) should not (if considering \(\perp\) & \(\parallel\))

A1 A1: Condone \(\leqslant\) or \(\geqslant\) with \(F\) for this mark but withhold the final A mark
Where \(R\) is the normal contact force and \(F\) is the frictional contact force

M1dep*: Applying \(F \leqslant \mu R\) or \(F = \mu R\) to obtain an equation/inequality in \(\theta\) only – where \(R\) and \(F\) are a linear combination of the correct number of relevant resolved terms

A1: oe e.g. \(\frac{30}{58}\) but A0 for \(\frac{6}{11.6}\) as a final answer. Allow awrt 0.517. Isw if candidates go on to work out \(\theta\). Allow \(\tan\theta \leqslant \frac{15}{29}\) (oe) as a final answer
Can use equals throughout (no justification required) But not \(\tan\theta \lt \frac{15}{29}\)
If using \(10g\) for the weight, then SC M1* A1 A0 M1dep* A0 max (where the first A mark is for both equations correct using \(10g\) rather than 10)

Alternative for first three marks

Scheme	Marks
Resolving vertically and horizontally	M1*
\(R\cos\theta + F\sin\theta = 10\)	A1
\(F\cos\theta = 2 + R\sin\theta\)	A1

M1*: M1 conditions as above

A1: Final M1dep* and A marks as above

June 2022 Paper 3 Q11

OCR ACurrent spec7 marksMoments Resolving Forces

Rod AB, 3 m long, with A on horizontal ground, making 60° with the ground, with a particle at B; a string from point C on the rod, x m from A, goes to point D vertically above A, with angle 60° between DC and the vertical

A uniform rod \(AB\) of mass 4 kg and length 3 m rests in a vertical plane with \(A\) on rough horizontal ground.

A particle of mass 1 kg is attached to the rod at \(B\). The rod makes an angle of 60° with the horizontal and is held in limiting equilibrium by a light inextensible string \(CD\). \(D\) is a fixed point vertically above \(A\) and \(CD\) makes an angle of 60° with the vertical. The distance \(AC\) is \(x\) m (see diagram).

(a) Find, in terms of \(g\) and \(x\), the tension in the string. [3]

The coefficient of friction between the rod and the ground is \(\dfrac{9\sqrt{3}}{35}\).

(b) Determine the value of \(x\). [4]

Mark scheme (a)

Scheme	Marks	AO
Taking moments about \(A\) – correct number of terms, allow sin/cos confusion and sign errors – may take moments about another point (and resolve) but must end up after elimination with an equation in \(T\) and \(x\) only	M1	3.1b
\(4g\left(\frac{3}{2}\cos 60\right) + g(3\cos 60) = xT\)	A1	1.1
\(T = \dfrac{9g}{2x}\) (N)	A1	2.2a
	[3]

Notes

M1: Dimensionally correct.
Must be \(xT\) and not \(xT\cos(\ldots)\) or \(xT\sin(\ldots)\)

A1: Correct equation in \(g\), \(x\) and \(T\) only – condone \(g\) replaced by 9.8
\(T\) is the tension in the string

A1: An answer of \(\frac{44.1}{x}\) (or with trigonometric terms) is A0 unless correct answer in terms of \(g\) seen
oe exact answers in terms of \(g\) and \(x\) (condone correct triple decker fractions)

For reference for parts (a) and (b):
Moments about \(C\): \(R_A(x\cos 60) + g(3 - x)\cos 60 = 4g(x - 1.5)\cos 60 + F_A(x\sin 60)\)
Moments about \(B\): \(T(3 - x) + R_A(3\cos 60) = 4g(1.5\cos 60) + F_A(3\sin 60)\)
Moments about midpoint of \(AB\): \(R_A(1.5\cos 60) + g(1.5\cos 60) = T(x - 1.5) + F_A(1.5\sin 60)\)
Resolving perpendicular to \(AB\): \(T + R_A\cos 60 = 4g\cos 60 + g\cos 60 + F_A\sin 60\)
Resolving parallel to \(AB\): \(R_A\sin 60 + F_A\cos 60 = 4g\sin 60 + g\sin 60\)

Mark scheme (b)

Scheme	Marks	AO
Resolve vertically or horizontally – correct number of terms with the tension at \(C\) in terms of cos/sin, condone sign errors, allow sin/cos confusion but forces that require resolving must be (and correspondingly those that don’t require resolving e.g. the weights if resolving vertically, should not be resolved)	M1	3.3
\(T\cos 60 + R_A = 4g + g\) \(\left(\Rightarrow R_A = 5g - \dfrac{9g}{4x}\right)\) \(F_A = T\sin 60\) \(\left(\Rightarrow F_A = \dfrac{9\sqrt{3}g}{4x}\right)\)	A1	1.1
\(\dfrac{9\sqrt{3}g}{4x} = \dfrac{9\sqrt{3}}{35}\left(5g - \dfrac{9g}{4x}\right)\)	M1dep*	3.4
\(x = 2.2\)	A1	2.2a
	[4]

Notes

M1: Or obtain an equation in \(F_A\) and/or \(R_A\) in terms of \(T\) (or their \(T\)) only (see list of equations below)

A1: Both correct (unsimplified) – allow with \(T\) or their (possibly incorrect) \(T\) (oe eg two valid equations in \(R_A\) and \(F_A\))
\(R_A\) is the normal contact force at \(A\)
\(F_A\) is the frictional contact force at \(A\)

M1dep*: Use of \(F = \mu R\) with correct \(\mu\) to form an equation in \(x\) only – no forces missing from their \(R_A\) and \(F_A\) and all required forces resolved accordingly or not e.g. if resolving vertically the two weights should not contain sin/cos

A1: awrt 2.2
www

June 2022 Paper 3 Q10

OCR ACurrent spec8 marksConnected Particles Resolving Forces

Block B on a horizontal surface with particle P on its upper surface; a horizontal string from P passes over a pulley at the edge of the surface to particle Q hanging vertically

A rectangular block \(B\) is at rest on a horizontal surface. A particle \(P\) of mass 2.5 kg is placed on the upper surface of \(B\). The particle \(P\) is attached to one end of a light inextensible string which passes over a smooth fixed pulley. A particle \(Q\) of mass 3 kg is attached to the other end of the string and hangs freely below the pulley. The part of the string between \(P\) and the pulley is horizontal (see diagram).

The particles are released from rest with the string taut. It is given that \(B\) remains in equilibrium while \(P\) moves on the upper surface of \(B\). The tension in the string while \(P\) moves on \(B\) is 16.8 N.

(a) Find the acceleration of \(Q\) while \(P\) and \(B\) are in contact. [2]

(b) Determine the coefficient of friction between \(P\) and \(B\). [3]

(c) Given that the coefficient of friction between \(B\) and the horizontal surface is \(\frac{5}{49}\), determine the least possible value for the mass of \(B\). [3]

Mark scheme (a)

Scheme	Marks	AO
\(3g - 16.8 = 3a\)	M1	3.3
\(a = \dfrac{3g - 16.8}{3} = 4.2\) (\(\mathrm{m\,s^{-2}}\))	A1	1.1
	[2]

Notes

M1: N2L for \(Q\) – correct number of terms with the correct mass. Condone sign errors
M0 if using \(3g\) for the mass but allow \(g\) missing from the net force

Mark scheme (b)

Scheme	Marks	AO
\(16.8 - F_P = 2.5(4.2)\)	M1*	3.3
\(16.8 - 2.5(4.2) = \mu(2.5g)\)	M1dep*	3.4
\(\mu = 0.257\)	A1	1.1
	[3]

Notes

M1*: N2L for \(P\) horizontally using \(T = 16.8\) and their \(a\) (but not \(\pm 9.8\)) from (a) – allow sign errors but must have correct number of terms and correct mass (\(\ne 3\)) – if correct \(F_P = 6.3\)
\(F_P\) is the friction between \(P\) and \(B\)

M1dep*: Use of \(F = \mu R\) for \(P\) with \(R = 2.5g\)

A1: awrt 0.257 – condone working with \(F \leqslant \mu R\) provided that the value of \(\mu\) is explicitly stated (and not left in an inequality)
0.257142857…
allow \(\dfrac{9}{35}\)

Mark scheme (c)

Scheme	Marks	AO
\(R_B = 2.5g + Mg\)	M1*	3.1b
\(6.3 \leqslant \dfrac{5}{49}(2.5g + Mg)\)	M1dep*	3.4
\(M \geqslant 3.8\) so least possible value for the mass of \(B\) is 3.8 (kg)	A1	2.2a
	[3]

Notes

M1*: Resolving vertically for \(B\) – correct number of terms, allow sign errors and condone \(g\)’s missing
Where \(M\) is the mass of \(B\)

M1dep*: Use of \(F \leqslant \mu R\) or \(F = \mu R\) with correct \(R\) and \(\mu = \frac{5}{49}\) with \(F\) being their \(F_P\) from (b) where \(F_P \ne 16.8\)
No \(g\)’s missing for this mark

A1: 3.8 - allow use of ‘=’ throughout this part
No justification required

June 2022 Paper 3 Q8

OCR ACurrent spec2 marksResolving Forces

A sledge on horizontal ground with a rope attached to its front, rising at 15° above the horizontal

A child attempts to drag a sledge along horizontal ground by means of a rope attached to the sledge. The rope makes an angle of 15° with the horizontal (see diagram).

Given that the sledge remains at rest and that the frictional force acting on the sledge is 60 N, find the tension in the rope. [2]

Mark scheme

Scheme	Marks	AO
\(T\cos 15 = 60\)	M1	1.1
\(T = 62.1\) (N)	A1	1.1
	[2]

Notes

M1: Resolve horizontally, correct number of terms, allow sin/cos confusion (e.g. \(T\sin 15 = 60\) scores M1 but \(T = 60\cos 15\) scores M0)
Where \(T\) is the tension in the rope

A1: awrt 62.1 – allow the exact answer of \(60(\sqrt{6} - \sqrt{2})\)
62.11657082…

October 2021 Paper 3 Q14

OCR ACurrent spec11 marksConnected Particles Resolving Forces

Particle A of mass 2 kg on a plane inclined at 30 degrees, joined by a string over a pulley at the top of the plane to particle B of mass 3 kg on a horizontal surface; B is joined by a second string over another pulley to particle C of mass 4 kg on a plane, labelled Pi, inclined at 60 degrees

One end of a light inextensible string is attached to a particle \(A\) of mass \(2\,\mathrm{kg}\). The other end of the string is attached to a second particle \(B\) of mass \(3\,\mathrm{kg}\). Particle \(A\) is in contact with a smooth plane inclined at \(30^\circ\) to the horizontal and particle \(B\) is in contact with a rough horizontal plane.

A second light inextensible string is attached to \(B\). The other end of this second string is attached to a third particle \(C\) of mass \(4\,\mathrm{kg}\). Particle \(C\) is in contact with a smooth plane \(\mathit{\Pi}\) inclined at an angle of \(60^\circ\) to the horizontal.

Both strings are taut and pass over small smooth pulleys that are at the tops of the inclined planes. The parts of the strings from \(A\) to the pulley, and from \(C\) to the pulley, are parallel to lines of greatest slope of the corresponding planes (see diagram).

The coefficient of friction between \(B\) and the horizontal plane is \(\mu\). The system is released from rest and in the subsequent motion \(C\) moves down \(\mathit{\Pi}\) with acceleration \(a\,\mathrm{m\,s^{-2}}\).

(a) By considering an equation involving \(\mu\), \(a\) and \(g\) show that \(a \lt \frac{1}{9}g\left(2\sqrt{3} - 1\right)\). [7]

(b) Given that \(a = \frac{1}{9}g\), determine the magnitude of the contact force between \(B\) and the horizontal plane. Give your answer correct to 3 significant figures. [4]

Mark scheme (a)

Scheme	Marks	AO
\(T_{AB} - 2g\sin 30 = 2a\)	M1*	3.3
\(4g\sin 60 - T_{BC} = 4a\)	M1*	3.3
\(T_{BC} - T_{AB} - F_B = 3a\)	M1*	3.3
\(4g\sin 60 - 4a - 2g\sin 30 - 2a - F_B = 3a\)	M1dep*	2.1
\(9a = g(4\sin 60 - 2\sin 30 - 3\mu)\)	A1	3.3
\((\mu =)\,\dfrac{1}{3}\left(2\sqrt{3} - 1 - 9\dfrac{a}{g}\right) \gt 0\)	M1	3.1b
\(a \lt \dfrac{1}{9}g\left(2\sqrt{3} - 1\right)\)	A1	2.2a
	[7]

Notes

M1*: N2L parallel to plane for \(A\) – correct number of terms, allow cos/sin confusion
Dimensionally consistent equations for M marks

M1*: N2L parallel to plane for \(C\) – correct number of terms, allow cos/sin confusion
M1M0M0 if \(T\) used in both equations

M1*: N2L parallel to plane for \(B\)

M1dep*: Eliminates both tensions
Allow in terms of \(F_B\)

A1: Use of \(F_B = \mu(3g)\) to get a correct equation in \(a\) and \(\mu\)
\(9a = g\left(2\sqrt{3} - 1 - 3\mu\right)\)

M1: Explicitly uses \(\mu \gt 0\) to get a strict inequality in \(a\) and \(g\) only. If \(a = \frac{1}{9}g\left(2\sqrt{3} - 1 - 3\mu\right) \Rightarrow a \lt \frac{1}{9}g\left(2\sqrt{3} - 1\right)\) without justification is M0
Dependent on all previous M marks

A1: AG – must follow from a correct equation involving \(\mu, a\) and \(g\)
SC considering whole system with no friction B2 only for deriving \(9a = g\left(2\sqrt{3} - 1\right)\)

Mark scheme (b)

Scheme	Marks	AO
\(a = \dfrac{1}{9}g \Rightarrow \mu = \dfrac{2}{3}\left(\sqrt{3} - 1\right)\)	B1	1.1
\(F_B = 2\left(\sqrt{3} - 1\right)g\)	B1	3.4
\(\sqrt{(3g)^2 + \left(2\left(\sqrt{3} - 1\right)g\right)^2}\)	M1	3.1a
32.7 (N)	A1	2.2a
	[4]

Notes

B1: Correct value of \(\mu\) (oe) using given \(a\) (soi)
\(\mu = 0.488033\ldots\)

B1: Correct value for \(F_B\)
\(F_B = 14.34819\ldots\)

M1: \(\sqrt{(3g)^2 + F_B^{\,2}}\) allow any value for \(F_B\) or even just the expression \(F_B\)

A1: (For reference: 32.71438099…)

October 2021 Paper 3 Q10

OCR ACurrent spec6 marksResolving Forces

Block D resting on a horizontal surface; a force of 15 N pushes down on the top of the block at angle theta to the horizontal, directed down and to the right

A block \(D\) of weight \(50\,\mathrm{N}\) lies at rest in equilibrium on a fixed rough horizontal surface. A force of magnitude \(15\,\mathrm{N}\) is applied to \(D\) at an angle \(\theta\) to the horizontal (see diagram).

(a) Complete the diagram in the Printed Answer Booklet showing all the forces acting on \(D\). [1]

It is given that \(D\) remains at rest and the coefficient of friction between \(D\) and the surface is 0.2.

(b) Show that \[15\cos\theta - 3\sin\theta \leqslant 10.\] [5]

Mark scheme (b)

Scheme	Marks	AO
\(F = 15\cos\theta\)	B1	3.3
\(R = 15\sin\theta + 50\)	M1 A1	3.3 1.1
\(15\cos\theta \leqslant \frac{1}{5}(15\sin\theta + 50)\)	M1	3.4
\(15\cos\theta - 3\sin\theta \leqslant 10\)	A1	2.2a
	[5]

Notes

B1: Not for just seeing the expression \(15\cos\theta\)

M1: M1 for resolving vertically – allow sign errors and sin/cos confusion (three terms)
Condone inclusion of \(g\) with the 50 for the M mark only

M1: Use of \(F \leqslant \frac{1}{5}R\) with their \(F\) and \(R \neq 50\)
Allow \(F = \frac{1}{5}R\) for the M mark only

A1: AG

June 2025 Paper 1 Q14

OCR MEICurrent spec15 marksResolving Forces SUVAT

14 In this question the \(\mathbf{i}\) and \(\mathbf{j}\) vectors are horizontal and vertically upward respectively.

A particle of mass 5 kg is at rest on a rough horizontal shelf. The coefficient of friction between the particle and the shelf is \(\mu\).

(a) A force \(\mathbf{P} = 9\mathbf{i} + 20\mathbf{j}\) N acts on the particle. The particle is on the point of sliding along the shelf.

Determine the value of \(\mu\). [5]

The force \(\mathbf{P}\) is removed. One end of the shelf is lifted so that it is inclined at \(\alpha^\circ\) to the horizontal.

(b) The particle is on the point of sliding down the shelf.

Show that \(\alpha = 17.2\) to 3 significant figures. [4]

(c) The particle is projected up the shelf with an initial speed of \(5\,\mathrm{m\,s^{-1}}\).

Given that the particle remains in contact with the shelf, determine the time after projection at which the particle first comes to rest. [6]

Mark scheme (a)

Scheme	Marks	AO
\([-5g\mathbf{j} + R\mathbf{j} \pm F\mathbf{i} + 9\mathbf{i} + 20\mathbf{j} = \mathbf{0}]\)
Horizontal component \(F = 9\)	B1	3.1b
Vertical direction \(R - 5g + 20 = 0\)	M1	3.1b
\(R = 5g - 20\)	A1	1.1
On the point of sliding \(F = \mu R\) so	M1	3.4
So \(\mu = \dfrac{9}{29} = [0.310\) to 3 sf\(]\)	A1FT	1.1
	[5]

Notes

Allow wrong or missing vector equation

B1: Soi. Allow \(\boldsymbol{F} = -9\mathbf{i}\) or \(F = -9\)

M1: Soi Vertical equation – weight and 20 involved and not 9.
Allow sign errors

A1: Correct equation soi

M1: Uses their \(F\) and \(R\) to evaluate \(\mu\).
Where \(F = \mu\times 5g\) is seen it must be explicit that \(5g\) is their reaction and not weight

A1FT: FT their \(R\) and \(F\).
Final answer must follow from their working and be positive
Ignore unsupported \(\mu = 0.310\) seen.

Force diagram: particle on a slope at angle alpha, friction F1 up the slope, normal reaction R1, weight 5g — Mark scheme (b)

Mark scheme (c)

Scheme	Marks	AO
N2L up the slope \(-\left(5g\sin 17.2^\circ + \dfrac{9}{29}5g\cos 17.2^\circ\right) = 5a\)	B1 B1FT M1	3.3 3.4 3.1b
\(a = -5.80\ldots\)	A1	1.1
\(v = u + at\) with \(v = 0,\ u = 5\) and their \(a\)	M1	3.4
\(t = 0.862\) s	A1FT	1.1
	[6]

Notes

B1: Correct weight term seen (Allow \(\pm 14.5\) oe)

B1FT: Friction term seen FT their \(\mu\) (Allow \(\pm 14.5\) oe)

M1: Formulate equation of motion with their weight. Allow wrong or missing friction. Allow sign errors
No extra forces.

A1: Soi
Allow 5.8 if down the slope used as positive direction clear (eg on diagram, or \(u = -5\) or \(a = -5.8\) used subsequently)

M1: suvat equation(s) leading to a value for \(t\). Allow sign errors

A1FT: FT their acceleration \(\neq \pm g\)
Using \(\alpha = \tan^{-1}\frac{9}{29}\) gives 0.86067

June 2024 Paper 1 Q16

OCR MEICurrent spec7 marksResolving Forces

16 A block of mass \(m\) kg rests on rough horizontal ground. The coefficient of friction between the block and the ground is \(\mu\). A force of magnitude \(T\) N is applied at an angle \(\theta\) radians above the horizontal as shown in the diagram and the block slides without tilting or lifting.

Block on horizontal ground with a force T N applied at its top right corner at angle θ above the dashed horizontal

(a) Show that the acceleration of the block is given by \(\dfrac{T}{m}\cos\theta - \mu g + \dfrac{T}{m}\mu\sin\theta\). [4]

For a fixed value of \(T\), the acceleration of the block depends on the value of \(\theta\). The acceleration has its greatest value when \(\theta = \alpha\).

(b) Find an expression for \(\alpha\) in terms of \(\mu\). [3]

Force diagram of the block: R N upwards, mg N downwards, F N horizontally to the left and T N at angle θ above the horizontal — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}a}{\mathrm{d}\theta} = -\dfrac{T}{m}\sin\theta + \dfrac{T}{m}\mu\cos\theta\)	M1	3.1a
\(-\dfrac{T}{m}\sin\alpha + \dfrac{T}{m}\mu\cos\alpha = 0\)	M1	1.1a
\(\dfrac{\sin\alpha}{\cos\alpha} = \mu\) so \(\alpha = \tan^{-1}\mu\)	A1	1.1
	[3]

Notes

M1: Attempt to differentiate wrt \(\theta\)

M1: Equate their derivative to 0 and attempt to rearrange using a trig identity.
Condone using \(\theta\) not \(\alpha\)

A1: Must be \(\alpha =\)
Also allow for \(\alpha = \frac{\pi}{2} - \tan^{-1}\frac{1}{\mu}\)

Alternative solution

Scheme	Marks	AO
Maximum \(a\) when \(\frac{T}{m}(\cos\theta + \mu\sin\theta)\) is max
Acceleration is \((R\cos(\theta - \beta))\) where \(\beta = \tan^{-1}\mu\)	M1 M1
Max acceleration when \((\alpha - \beta) = 0\)
Giving \(\alpha = \tan^{-1}\mu\)	A1

M1: Uses trig identity

M1: Attempt to find the value of \(\beta\)

A1: Must be \(\alpha =\)
Also allow for \(\alpha = \frac{\pi}{2} - \tan^{-1}\frac{1}{\mu}\)

June 2024 Paper 1 Q3

OCR MEICurrent spec5 marksResolving Forces

3 A particle hangs at the end of a string. A horizontal force of magnitude \(F\) N acting on the particle holds it in equilibrium so that the string makes an angle of \(20^\circ\) with the vertical, as shown in the diagram. The tension in the string is 12 N.

Particle hanging on a string at 20° to the dashed vertical, with a horizontal force F N acting on the particle

(a) Find the value of \(F\). [2]

(b) Find the mass of the particle. [3]

Mark scheme (a)

Scheme	Marks	AO
Resolve horizontally \(F = 12\sin 20\)	M1	1.1a
\(F = 4.10\)	A1	1.1
	[2]

Notes

M1: Resolving horizontally – allow sin/cos interchange.
Allow if \(F = T\sin 20\) or similar seen

A1: www

Mark scheme (b)

Scheme	Marks	AO
Resolve vertically \([mg =]\,12\cos 20\)	M1	3.1b
\(m = \dfrac{12\cos 20^\circ}{g} = 1.15\ \text{kg}\)	M1 A1	3.4 1.1
	[3]

Notes

M1: Resolve to find vertical component of tension. Allow sin/cos interchange if consistent with (a)
If triangle of forces used, allow attempt to find by Pythagoras

M1: Equating weight to their component of tension and dividing their weight by \(g\) soi

A1: cao

June 2023 Paper 1 Q13

OCR MEICurrent spec12 marksConnected Particles Resolving Forces

13 A block of mass 8 kg is placed on a rough plane inclined at \(15^\circ\) to the horizontal. The coefficient of friction between the block and the plane is 0.3.

One end of a light rope is attached to the block. The rope passes over a smooth pulley fixed at the top of the plane, and a sphere of mass 5 kg, attached to the other end of the rope, hangs vertically below the pulley. The part of the rope between the block and the pulley is parallel to the plane. The system is released from rest, and as the sphere falls the block moves directly up the plane with acceleration \(a\,\mathrm{m\,s^{-2}}\).

Block on a plane inclined at 15 degrees, connected by a rope parallel to the plane over a pulley at the top to a sphere hanging vertically

(a) On the diagram in the Printed Answer Booklet, show all the forces acting on the block and on the sphere. [4]

(b) Write down the equation of motion for the sphere. [2]

Force diagram: on the block, normal reaction R perpendicular to the plane, tension T up the plane, friction F down the plane, weight 8g N downwards; on the sphere, tension T upwards and weight 5g N downwards — Mark scheme (a)

Mark scheme (c)

Scheme	Marks	AO
\(R = 8g\cos 15^\circ\)	B1	3.1b
\(F = 0.3R = 2.4g\cos 15^\circ\)	M1	3.4
Newton’s second law for the block \(T - 8g\sin 15^\circ - F = 8a\)	M1 A1	1.1a 1.1b
Add equations of motion \(5g - 8g\sin 15^\circ - 2.4g\cos 15^\circ = 13a\)	M1	1.1a
\(a = 0.461\)	A1	1.1b
	[6]

Notes

M1: FT their \(R\)

M1: Allow one missing or incorrect force, must be dimensionally correct. FT their \(F\)

A1: All correct FT their \(F\)

M1: Attempt to eliminate \(T\)
This equation with one missing or incorrect force implies the previous M1A0M1

June 2022 Paper 1 Q13

OCR MEICurrent spec12 marksConnected Particles Resolving Forces

13 A toy train consists of an engine of mass 0.5 kg pulling a coach of mass 0.4 kg. The coupling between the engine and the coach is light and inextensible. The train is pulled along with a string attached to the front of the engine.

At first, the train is pulled from rest along a horizontal carpet where there is a resistance to motion of 0.8 N on each part of the train. The string is horizontal, and the tension in the string is 5 N.

(a) Determine the velocity of the train after 1.5 s. [4]

The train is then pulled up a track inclined at \(20^\circ\) to the horizontal. The string is parallel to the track and the tension in the string is \(P\) N. The resistance on each part of the train along the track is \(R\) N.

(b) Draw a diagram showing all the forces acting on the train modelled as two connected particles. [3]

(d) The acceleration of the train when \(P = 5.5\) is double the acceleration when \(P = 5\).
Calculate the value of \(R\). [3]

Mark scheme (a)

Scheme	Marks	AO
Newton’s second law for the train \(5 - 2 \times 0.8 = (0.5 + 0.4)a\)	M1	3.1b
giving \(a = \frac{34}{9} = 3.78\ \mathrm{m\,s^{-2}}\).	A1	1.1b
Using \(v = u + at\) with \(u = 0,\ t = 1.5\)	M1	3.1b
\(v = \frac{34}{9} \times 1.5 = \frac{17}{3} = 5.67\ \mathrm{m\,s^{-1}}\) (3sf)	A1	1.1b
	[4]

Notes

M1: N2L for whole train with correct mass and all forces present
Alternative: \(5 - 0.8 - T = 0.5a\), \(T - 0.8 = 0.4a\). Also allow for 2 equations where both have correct mass and all forces present in each

M1: using suvat equation(s) with \(u = 0\) and their \(a \neq g\) leading to a value for \(v\)

A1: FT their \(a\). Any form

Force diagram on a slope at 20°: coach with weight 0.4g N, normal reaction N_C and resistance R down the slope; engine with weight 0.5g N, normal reaction N_E and resistance R down the slope; tension T in the coupling acting on both, and tension in the string up the slope — Mark scheme (b)

Mark scheme (c)

Scheme	Marks	AO
Newton’s second law \(P - 2R - 0.9g\sin 20^\circ = 0.9a\)	M1 A1	1.1b 3.3
	[2]

Notes

M1: Newton’s law with \(m = 0.9\). Allow for incorrect weight term(s) or \(R\) used instead of \(2R\)

A1: Fully correct
Any form

Mark scheme (d)

Scheme	Marks	AO
When \(P = 5\) the equation gives \(5 - 2R - 0.9g\sin 20^\circ = 0.9a\)	M1	3.1b
When \(P = 5.5\) the equation gives \(5.5 - 2R - 0.9g\sin 20^\circ = 0.9 \times 2a\)	M1	3.1b
Solve simultaneously giving \(R = 0.742\) \(\left[a = \frac{5}{9}\right]\)	A1	1.1b
	[3]

Notes

M1: establishes one equation linking \(R\) and \(a\). FT their (c)

M1: establishes another equation linking \(R\) and \(a\). Consistent with their first equation

A1: method need not be seen BC
correct value for \(R\) (\(a\) is not required)

Alternative method

Scheme	Marks
When \(P = 5\) \(a = \dfrac{5 - 2R - 0.9g\sin 20^\circ}{0.9}\) When \(P = 5.5\) \(a_1 = \dfrac{5.5 - 2R - 0.9g\sin 20^\circ}{0.9}\)	M1
So \(\dfrac{5.5 - 2R - 0.9g\sin 20^\circ}{0.9} = 2\left(\dfrac{5 - 2R - 0.9g\sin 20^\circ}{0.9}\right)\)	M1
giving \(R = 0.742\) \(\left[a = \frac{5}{9}\right]\)	A1

M1: Finds expression for \(a\) when \(P = 5\) or \(P = 5.5\). Soi

M1: Links corresponding acceleration for the other value of \(P\)
Do not allow factor of 2 on the wrong side

A1: correct value for \(R\) (\(a\) is not required)

June 2022 Paper 1 Q5

OCR MEICurrent spec5 marksResolving Forces

5 A sphere of mass 3 kg hangs on a string. A horizontal force of magnitude \(F\) N acts on the sphere so that it hangs in equilibrium with the string making an angle of \(25^\circ\) to the vertical. The force diagram for the sphere is shown below.

Force diagram: the sphere with Weight acting vertically down, a horizontal force F N to the left, and Tension acting up and to the right along the string at 25° to the vertical

(a) Sketch the triangle of forces for these forces. [2]

(b) Hence or otherwise determine each of the following:

the tension in the string
the value of \(F\).

[3]

Two acceptable triangles of forces: Weight (vertical), F N (horizontal) and Tension forming a closed loop, with the 25° angle between Weight and Tension — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
Using the triangle of forces \(F = 3g\tan 25^\circ\) or Tension \(= \dfrac{3g}{\cos 25^\circ}\)	M1	1.1a
\(F = 13.7\)	A1	1.1b
Tension \(= 32.4\) N	A1	1.1b
	[3]

Notes

M1: Allow sin/cos or \(25^\circ\)/\(65^\circ\) interchange to find \(F\) or \(T\)

A1: cao

Alternative method

Scheme	Marks
Resolve vertically \(T\cos 25^\circ = 3g\)	M1
\(T = 32.4\) N	A1
Resolve horizontally \(F = T\sin 25^\circ = 13.7\)	A1

M1: Allow sin/cos interchange or \(25^\circ\)/\(65^\circ\) interchange

A1: cao

October 2021 Paper 1 Q12

OCR MEICurrent spec7 marksResolving Forces SUVAT

12 A box of mass \(m\) kg slides down a rough slope inclined at \(15^\circ\) to the horizontal. The coefficient of friction between the box and the slope is 0.4. The box has an initial velocity of \(1.2\,\text{m}\,\text{s}^{-1}\) down the slope.

Calculate the distance the box travels before coming to rest. [7]

Mark scheme

Scheme	Marks	AO
Normal reaction \(mg\cos 15^\circ\)	B1	3.1b
Max friction \(\mu N = 0.4mg\cos 15^\circ\)	M1	1.1b
Resolve down the slope \(mg\sin 15^\circ - F = ma\)	B1	1.1b
\(mg\sin 15^\circ - 0.4mg\cos 15^\circ = ma\) giving \(a = -1.25\,\text{m}\,\text{s}^{-2}\)	M1 A1	3.1b 1.1b
Using \(v^2 = u^2 + 2as\) \(0^2 = 1.2^2 + 2 \times (-1.25)s\)	M1	3.1b
giving \(s = 0.576\) m	A1	1.1b
	[7]

Notes

B1: Correct normal reaction

M1: Attempt to evaluate friction FT their normal reaction.
Only allow \(0.4mg\) if it is clear that \(mg\) is their normal reaction and not just weight

B1: Correct component of weight seen \((2.536m)\)

M1: All terms present; allow sign errors, sin/cos interchange for weight and their \(F\)

A1: Correct equation (\(a\) need not be explicitly evaluated here)

M1: Use of suvat equation(s) leading to a value for \(s\) using \(v = 0\)

A1: FT their negative \(a\)

October 2020 Paper 1 Q15

OCR MEICurrent spec9 marksResolving Forces Vectors

15 Fig. 15 shows a particle of mass \(m\) kg on a smooth plane inclined at \(30^\circ\) to the horizontal. Unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are parallel and perpendicular to the plane, in the directions shown.

Fig. 15: particle on a plane inclined at 30 degrees; i points up the plane, j perpendicular to the plane away from it — Fig. 15

(a) Express the weight \(\mathbf{W}\) of the particle in terms of \(m\), \(g\), \(\mathbf{i}\) and \(\mathbf{j}\). [2]

The particle is held in equilibrium by a force \(\mathbf{F}\), and the normal reaction of the plane on the particle is denoted by \(\mathbf{R}\). The units for both \(\mathbf{F}\) and \(\mathbf{R}\) are newtons.

(b) Write down an equation relating \(\mathbf{W}\), \(\mathbf{R}\) and \(\mathbf{F}\). [1]

show that \(m = 1.22\) correct to 3 significant figures,
find the magnitude of \(\mathbf{R}\).

[6]

Mark scheme (a)

Scheme	Marks	AO
\(\mathbf{W} = (-mg\sin 30^\circ)\mathbf{i} + (-mg\cos 30^\circ)\mathbf{j}\)	M1	3.1b
\(\left[\mathbf{W} = \left(-\dfrac{1}{2}mg\right)\mathbf{i} + \left(-\dfrac{\sqrt{3}}{2}mg\right)\mathbf{j}\right]\)	A1	2.5
	[2]

Notes

M1: Attempting to resolve the weight. Allow sin/cos interchange and sign errors for the method mark.
\(mg\) must be seen for the method mark.

A1: All correct in this vector form

Mark scheme (b)

Scheme	Marks	AO
\(\mathbf{W} + \mathbf{R} + \mathbf{F} = \mathbf{0}\)	B1	2.5
	[1]

Notes

B1: Allow any rearrangement of this.
Allow if their expression for \(\mathbf{W}\) is used instead of \(\mathbf{W}\)

Mark scheme (c)

Scheme	Marks	AO
\(\mathbf{R} = R\mathbf{j}\) \(\left[(-mg\sin 30^\circ)\mathbf{i} + (-mg\cos 30^\circ)\mathbf{j} + (6\mathbf{i} + 8\mathbf{j}) + R\mathbf{j} = \mathbf{0}\right]\)	B1	3.1b
i component \(-mg\sin 30^\circ + 6 = 0\)	M1	3.1a
giving \(m = 1.22\) to 3 sf	A1 A1	1.1 1.1
j component: \(-12\cos 30 + 8 + R = 0\)	M1	3.1a
\(R = 2.39\) so magnitude is 2.39 N	A1	3.2a
	[6]

Notes

B1: Allow for any clear indication that \(\mathbf{R}\) is a multiple of \(\mathbf{j}\) or that it has no component in the \(\mathbf{i}\) direction
May be implied with an equation for the i direction with two terms and an equation in the j direction with three terms

M1: Forming equation from their \(\mathbf{i}\) terms, or equivalent by resolving parallel to the plane. FT their \(\mathbf{W}\)

A1: Correct equation in \(\mathbf{i}\) direction

A1: AG

M1: Equation from the \(\mathbf{j}\) terms (must include all three terms), oe, and using value of \(m\)
12 is the value for \(mg\). \(1.22 \times 9.8 = 11.956\)

A1: Accept arwt 2.4

October 2020 Paper 1 Q11

OCR MEICurrent spec11 marksConnected Particles Resolving Forces

11 A block of mass 2 kg is placed on a rough horizontal table. A light inextensible string attached to the block passes over a smooth pulley attached to the edge of the table. The other end of the string is attached to a sphere of mass 0.8 kg which hangs freely.

The part of the string between the block and the pulley is horizontal. The coefficient of friction between the table and the block is 0.35. The system is released from rest.

(a) Draw a force diagram showing all the forces on the block and the sphere. [3]

(b) Write down the equations of motion for the block and the sphere. [2]

(d) Calculate the time for the block to slide the first 0.5 m. Assume the block does not reach the pulley. [2]

Force diagram: block with R up, 2g N down, friction F to the left and tension T to the right towards the pulley; sphere with T up and 0.8g N down — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
\(T - F = 2a\)	B1	1.1a
\(0.8g - T = 0.8a\)	B1	1.1a
	[2]

Notes

B1: Allow any expression for F

B1: Allow distinct tensions if consistent with diagram

For \(F\), allow \(0.35R,\ 0.35 \times 2g\) \(0.7g\) or 6.86 N

Mark scheme (c)

Scheme	Marks	AO
Vertically for the block \(R = 2g\)	M1	3.1b
Friction \(F = 0.35R = 0.7g\)	A1	2.1
Add equations \(0.8g - F = 2.8a\)	M1	2.1
\(0.8g - 0.7g = 2.8a\) \(a = 0.35\,\mathrm{m\,s^{-2}}\)	A1	2.1
	[4]

Notes

M1: Attempt to use \(\mu\) to evaluate friction

A1: Correct value for F

M1: Eliminate \(T\) from their equations

A1: AG must follow from correct work

Some of this work may already have been seen in previous part

Mark scheme (d)

Scheme	Marks	AO
Use \(s = 0.5,\quad u = 0,\quad a = 0.35\) \(0.5 = \dfrac{1}{2} \times 0.35 \times t^2\)	M1	1.1a
\(t = 1.69\) s	A1	1.1
	[2]

Notes

M1: Using suvat equation(s) leading to a value for \(t\)

A1: Do not allow \(\pm 1.69\)