Connected Particles

Topic

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

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 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]

Scheme	Marks	AO
States that the resultant force is \(80g - T\) and states \(F = ma\)	E1	3.4
	(1)

Scheme	Marks	AO
States \(50g\cos 60 = 25g\) OE	B1	3.1b
	(1)

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

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

Scheme	Marks	AO
States any one of the following valid assumptions Rope is light rope inextensible pulley is fixed No air resistance	E1	3.5b
	(1)
	(9 marks)

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)

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

Scheme	Marks	AO
States one appropriate modelling assumption about the rod Accept Rod is rigid OE	E1	3.5b
	(1)
	(12 marks)

June 2025 Paper 3 Q8

OCR ACurrent spec4 marksConnected Particles

Particle A of mass 0.6 kg above particle B of mass 0.8 kg, joined by a vertical string; upward air resistance R N acts on A and 0.3 N acts on B

Two particles \(A\) and \(B\) of masses 0.6 kg and 0.8 kg respectively, are attached to the ends of a light inextensible string. Particle \(A\) is held at rest at a fixed point and \(B\) hangs vertically below \(A\).

Particle \(A\) is now released. As the particles fall the air resistance acting on \(A\) has a constant magnitude of \(R\) N and the air resistance acting on \(B\) has a constant magnitude of 0.3 N. The string remains taut throughout the subsequent motion (see diagram).

The downward acceleration of each of the particles is 9.2 m s⁻² and the tension in the string is \(T\) N.

(a) By applying Newton’s second law to \(B\), find the value of \(T\). [2]

(b) Hence, or otherwise, find the value of \(R\). [2]

Mark scheme (a)

Scheme	Marks	AO
\(0.8g - 0.3 - T = 0.8(9.2)\)	M1	3.3
\(T = 0.18\)	A1	1.1
	[2]

Notes

M1: Applying N2L for \(B\) with correct number of dimensionally correct terms (so allow sign errors only) to obtain an equation in \(T\) only

This mark is not awarded until all values have been correct substituted into N2L so, for example, must see 0.8 for the mass (and not just \(m\))

A1: cao – must be positive oe e.g. \(\frac{9}{50}\)

Mark scheme (b)

Scheme	Marks	AO
\(0.6g + T - R = 0.6(9.2)\)	M1	3.3
\(R = 0.54\)	A1	1.1
	[2]

Notes

M1: Applying N2L for \(A\) with correct number of dimensionally correct terms (so allow sign errors only) to obtain an equation in \(R\) (and possibly \(T\)) only. Alternative is to consider N2L for the system which if correct is \(0.8g - 0.3 + 0.6g - R = 1.4(9.2)\) (if considering this equation then allow sign errors only)

Allow the letter \(T\) or their (possibly incorrect) value of \(T\)

A1: cao – must be positive oe e.g. \(\frac{27}{50}\)

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 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

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…)

June 2025 Paper 1 Q6

OCR MEICurrent spec8 marksConnected Particles Time-Series Graphs

6 A car is travelling along a straight horizontal road. The car’s velocity \(v\,\mathrm{m\,s^{-1}}\) at time \(t\) s is shown in the velocity-time graph below. The points (0, 3) and (5, 3) are joined with a line segment, and the points (5, 3) and (15, \(-2\)) are joined with another line segment.

Velocity-time graph: v = 3 from t = 0 to t = 5, then a straight line down to v = -2 at t = 15

(a) Calculate the total distance travelled by the car in the first 15 s. [3]

The car is then attached to a caravan by means of a light inextensible horizontal tow bar and continues travelling along the road. You are given the following information.

The car and caravan accelerate at \(1.5\,\mathrm{m\,s^{-2}}\).
The mass of the car is 1400 kg and the mass of the caravan is 900 kg.
The driving force acting on the car is \(D\) N and the tension in the tow bar is \(T\) N.
The resistances to motion acting on the car and caravan are 400 N and 450 N respectively.

(b) Write down the equations of motion for the car and the caravan separately. [2]

Mark scheme (a)

Scheme	Marks	AO
\(v=0\) when \(t=11\) s	B1	3.1b
Distance \(= 5\times 3 + \frac{1}{2}\times 6\times 3 + \frac{1}{2}\times 4\times 2\)	M1	1.1
\([=15+9+4] = 28\) [m]	A1	1.1
	[3]

Notes

B1: soi. May be implied by \(t=6\) used in suvat

M1: Attempt to find the area between the graph and the \(x\)-axis
Allow \(\pm 2\) used for height of the triangle below the axis
soi
Trapezium plus triangle gives \(\frac{1}{2}\times 3(11+5)\) instead of the first two terms

A1: cao

Alternative method for M1A1

Scheme	Marks
Distance \(= 5\times 3 + \left(6\times 3 + \frac{1}{2}\left(-\frac{1}{2}\right)\times 6^2\right) + \left\|\frac{1}{2}\left(-\frac{1}{2}\right)\times 4^2\right\|\)	M1
\([=15+9+4] = 28\) [m]	A1

M1: Using suvat equation(s) with \(a = -\frac{5}{10}\). Award M1 even if \(t=10\) used

Mark scheme (b)

Scheme	Marks	AO
Car \(D - T - 400 = 1400\times 1.5\)	B1	1.1a
Caravan \(T - 450 = 900\times 1.5\)	B1	1.1a
	[2]

Notes

B1: No extra forces
Allow \(a\) used for 1.5 Any form
RHS may be 2100

B1: No extra forces
Allow \(a\) used for 1.5 Any form
RHS may be 1350

Mark scheme (c)

Scheme	Marks	AO
Caravan \(T-450 = 900\times 1.5\) gives \(T = 1800\)	B1	1.1
Add the equations \(D - 850 = 2300\times 1.5\)	M1	1.1
\(D = 4300\)	A1	1.1
	[3]

Notes

B1: cao

M1: Solving their equations leading to a value for \(D\) soi
May see use of \(T = 1800\) directly.
Allow for whole system equation soi

A1: cao

June 2024 Paper 1 Q2

OCR MEICurrent spec5 marksConnected Particles Forces & Newton's Laws

2 A car of mass 1400 kg pulls a trailer of mass 400 kg along a straight horizontal road. The engine of the car produces a driving force of 6000 N. A resistance of 800 N acts on the car. A resistance of 300 N acts on the trailer. The tow-bar between the car and the trailer is light and horizontal.

(a) Draw a force diagram showing all the horizontal forces on the car and the trailer. [2]

(b) Calculate the acceleration of the car and trailer. [3]

Force diagram: trailer and car on a horizontal line; 300 N to the left on the trailer, 800 N to the left on the car, 6000 N to the right on the car, and tension T acting in both directions in the tow-bar — Mark scheme (a)

Scheme	Marks	AO
	B1 B1	1.1 1.1
	[2]

Mark scheme (b)

Scheme	Marks	AO
N2L for system \(6000 - 800 - 300 = 1800a\)	M1 M1	1.1a 1.1a
\(a = 2.72\ \text{m s}^{-2}\) (to 3 s.f.)	A1	1.1
	[3]

Notes

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

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

A1: cao

Alternative solution

Scheme	Marks	AO
N2L for car \(6000 - 800 - T = 1400a\) N2L for trailer \(T - 300 = 400a\)	M1
Add equations to give	M1
\(a = 2.72\ \text{m s}^{-2}\) (to 3 s.f.)	A1

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

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

A1: cao

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)

Scheme	Marks	AO
	B1 B1 B1 B1	1.1b 1.1b 1.1b 1.1b
	[4]

Scheme	Marks	AO
Sphere (downwards positive) \(5g - T = 5a\)	M1 A1	1.1a 1.1b
	[2]

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)

Scheme	Marks	AO
	B1 B1 B1	1.1b 1.1b 1.1b
	[3]

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)

October 2021 Paper 1 Q9

OCR MEICurrent spec8 marksConnected Particles Forces & Newton's Laws

9 The diagram shows a toy caterpillar consisting of a head and three body sections each connected by a light inextensible ribbon. The head has a mass of 120 g and the body sections each have a mass of 90 g.

The toy is pulled on level ground using a horizontal string attached to the head. The tension in the string is 12 N. There are resistances to motion of 2.5 N for the head and each section of the body.

Toy caterpillar on level ground: three body sections on the left joined in a line to the head on the right, with a horizontal string attached to the head

(a)

(i) State the equation of motion for the toy caterpillar modelled as a single particle. [2]

(ii) Calculate the acceleration of the toy caterpillar. [1]

(b) Draw a diagram showing all the forces acting on the head of the toy caterpillar. [3]

Mark scheme (a)

Scheme	Marks	AO
(i) Total mass 390 g (0.390 kg) \(12 - 4 \times 2.5 = 0.390a\)	M1 A1	3.3 3.3
	[2]
(ii) \(5.13\ \text{m}\,\text{s}^{-2}\)	B1	1.1b
	[1]

Notes

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

A1: Fully correct equation need not be simplified

(ii) B1: cao

Force diagram for the head: R upwards, 0.12g N downwards, 12 N to the right, and to the left the tension [in the ribbon] and 2.5 N — Mark scheme (b)

Scheme	Marks	AO
	B1 B1 B1	1.1b 1.1b 1.1b
	[3]

Mark scheme (c)

Scheme	Marks	AO
N2L for the head only \(12 - 2.5 - T = 0.12a\) OR N2L for the three body sections together \(T - 3 \times 2.5 = 0.270a\)	M1	1.1b
\(T = 8.88\) N	A1	1.1b
	[2]

Notes

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

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

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)

Scheme	Marks	AO
	B1 B1 B1	1.1a 1.1a 1.1a
	[3]

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 \(\pm 1.69\)