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Newton's third law says that when two bodies interact they exert equal and opposite forces on each other. You use it to model connected particles: a car towing a trailer, a person in a lift, and particles joined by a string over a pulley. These are common multi-part questions on Paper 2, often with a second stage after a string breaks or a particle hits the ground.
Key facts
Newton's third law: equal and opposite forces on different bodies
Whole system: external forces = total mass × \(a\)
Each particle: \(F = ma\) with the same \(a\) and the same \(T\)
Smooth pulley and light string: same tension on both sides
Inextensible string: same magnitude of acceleration
Person in a lift: \(R - mg = ma\) (\(a\) upwards)
Force on a pulley from two vertical strings: \(2T\)
String slack: \(T = 0\)
Notes
Newton's third law
If body \(A\) exerts a force on body \(B\), then \(B\) exerts a force on \(A\) that is equal in magnitude and opposite in direction.
The two forces act on different bodies, so they never both appear on the force diagram for one body.
Examples: a person pushes down on a lift floor and the floor pushes up on the person; a tow bar pulls the trailer forwards and pulls the car backwards with a force of the same size.
Connected particles moving together
diagram
Objects joined by a rigid tow bar, or by a taut inextensible string, move with the same speed and the same magnitude of acceleration.
Whole system: treat the objects as one particle of total mass, using only the external forces. The tension is internal, so it cancels. This gives \(a\) quickly.
Separate parts: apply \(F = ma\) to one object to find the tension (or reaction) between them.
Tension T acts forwards on the trailer and backwards on the car: it cancels for the whole system
A tow bar can be in tension (pulling) or in thrust (compression, pushing), for example when the car brakes and the trailer pushes on it. If you assume a tension and get a negative answer, the force is a thrust.
Lift: for a person of mass \(m\) standing on the floor of a lift with upward acceleration \(a\),\(R - mg = ma\). For a lift of mass \(M\) and the person together, \(T - (M + m)g = (M + m)a\)
Pulleys
diagram
For a light inextensible string over a smooth pulley, the tension is the same on both sides and both particles have the same magnitude of acceleration.
Same tension on both sides; write F = ma for each particle in its own direction
Write \(F = ma\) for each particle in its own direction of motion, then add the equations to eliminate \(T\)
The force of the string on the pulley is the resultant of the two tensions: \(2T\) for two vertical strings, or \(T\sqrt{2}\) at 45° when one part is horizontal and the other vertical.
When one particle hits the ground the string goes slack (\(T = 0\)). The other particle keeps the speed it had at that moment and then moves freely under gravity (or under friction, on a rough table).
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Car and trailer: acceleration and force in the tow bar
6 to 10 marksGrade D
Whole system: driving force − total resistance = total mass × \(a\)
One part (usually the trailer): \(T\) − its resistance = its mass × \(a\)
If the driving force is removed, find the new (negative) acceleration from the whole system first.
A negative tension means the tow bar is in thrust.
Example. A car of mass 900 kg tows a trailer of mass 300 kg along a straight horizontal road by means of a light rigid tow bar. The resistances to motion are 400 N on the car and 50 N on the trailer. The engine of the car produces a driving force of 1650 N. (a) Find the acceleration of the car and trailer. (b) Find the tension in the tow bar. The car's engine is now switched off. The resistances are unchanged. (c) Find the magnitude of the force in the tow bar, and state whether it is a tension or a thrust.
(c) Whole system: \(-450 = 1200a\), so \(a = -0.375\) m s−2
(M1 A1)
Trailer: \(T - 50 = 300 \times (-0.375)\) (M1),\(T = -62.5\), so the force has magnitude 62.5 N (A1) and is a thrust
(A1)
Lift problems and Newton's third law
4 to 6 marksGrade C
For the lift and passenger together: \(T - (M + m)g = (M + m)a\)
For the passenger alone: \(R - mg = ma\)
Newton's third law: the passenger pushes down on the floor with a force equal in size to \(R\)
Example. A lift of mass 500 kg carries a passenger of mass 70 kg. The lift is raised by a vertical cable and moves upwards with constant acceleration 1.2 m s−2. (a) Find the tension in the cable. (b) Find the magnitude of the force exerted by the passenger on the floor of the lift.
(b) Passenger: \(R - 70g = 70 \times 1.2\) (M1), \(R = 770\) N (A1); by Newton's third law the passenger exerts a force of 770 N on the floor
(B1)
Particles hanging over a smooth pulley
6 to 10 marksGrade B
Write \(F = ma\) for each particle: the heavier one accelerates down, the lighter one up.
Add the equations to eliminate \(T\) and find \(a\); substitute back to find \(T\)
Force on the pulley: \(2T\) for vertical strings.
After the heavier particle hits the ground, the lighter one moves freely under gravity with the speed it had at that moment.
Example. Particles \(A\) and \(B\), of masses 5 kg and 3 kg, are attached to the ends of a light inextensible string that passes over a small smooth fixed pulley. The particles hang with the string taut and the hanging parts vertical. They are released from rest with \(A\) 1.5 m above the ground. (a) Find the acceleration of the particles and the tension in the string. (b) Find the magnitude of the force exerted on the pulley by the string. (c) \(A\) hits the ground and does not rebound, and \(B\) does not reach the pulley. Find the further distance that \(B\) rises after \(A\) hits the ground.
Show the model answerHide the model answer
(a) \(A\):\(5g - T = 5a\);\(B\):\(T - 3g = 3a\) (M1 A1 A1); adding, \(2g = 8a\), so \(a = 2.45\) m s−2 (A1); \(T = 3(9.8 + 2.45) = 36.75\), so 37 N (2 s.f.) or 36.8 N (3 s.f.)
(A1)
(b) \(2T = 73.5\) N
(B1)
(c) Speed when \(A\) lands: \(v^2 = 2(2.45)(1.5) = 7.35\) (M1); then \(B\) moves freely under gravity: \(0 = 7.35 - 2(9.8)s\) (M1),\(s = 0.375\), so 0.38 m (2 s.f.) or 0.375 m (3 s.f.)
(A1)
Particle on a table joined to a hanging particle
5 to 8 marksGrade B
Particle on the smooth table: \(T = m_1a\) horizontally.
Hanging particle: \(m_2g - T = m_2a\)
Solve for \(a\) and \(T\)
The force on the pulley is the resultant of a horizontal tension and a vertical tension: magnitude \(T\sqrt{2}\), at 45° below the horizontal.
Example. A block \(P\) of mass 4 kg lies on a smooth horizontal table. It is attached to one end of a light inextensible string, which passes over a small smooth pulley fixed at the edge of the table. A particle \(Q\) of mass 1 kg hangs freely from the other end of the string. The system is released from rest with the string taut. (a) Find the acceleration of \(P\) and the tension in the string. (b) Find the magnitude of the force exerted on the pulley by the string.
Show the model answerHide the model answer
(a) \(P\):\(T = 4a\);\(Q\):\(g - T = 1 \times a\) (M1 A1 A1); adding, \(g = 5a\), so \(a = 1.96\) m s−2 (A1) and \(T = 7.84\) N
(A1)
(b) \(\sqrt{T^2 + T^2} = T\sqrt{2}\) (M1) = 11 N (2 s.f.) or 11.1 N (3 s.f.)
(A1)
Don’t lose marks
Including the tension in the whole-system equation.
Using different accelerations for particles joined by a taut inextensible string.
Using the total mass with only one particle's forces.
Assuming the tension equals the weight of a hanging particle that is accelerating.
Forgetting that after the string goes slack the other particle keeps moving: it does not stop at once.
In lift questions, leaving out the weight of the lift or of the passenger.
More tips
Memory tricks
Two hanging masses \(m_1 > m_2\):\(a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}\). Use it to check, but write the two equations in the exam.
The tension always lies between the two weights: \(m_2g < T < m_1g\)
Check a tow bar force using the other vehicle's equation: both should give the same value.
Car and trailer slowing down with no driving force: if the trailer's resistance per kilogram is less than the car's, the tow bar is in thrust.
Exam technique
Draw a separate force diagram for each particle and mark the direction of its acceleration.
Label each equation, e.g. 'For \(A\) (↓): \(5g - T = 5a\)'.
For 'how have you used the fact that the string is inextensible / the pulley is smooth?', answer: the particles have the same acceleration / the tension is the same on both sides of the pulley.
Say 'tension' or 'thrust' when asked for the force in a tow bar or rod.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
State Newton's third lawThe forces between two bodies are equal in size, opposite in direction and act on different bodies.
Grade D
Solve car and trailer problemsUse the whole system to find the acceleration, then one part to find the tension or thrust.
Grade C
Find the reaction force in a liftApply \(F = ma\) to the person; the force on the floor is equal and opposite (Newton's third law).
Grade C
Solve problems with particles over a pulleyApply \(F = ma\) to each particle, with the same acceleration and the same tension.
Grade B
Find the force exerted on a pulleyIt is the resultant of the two tensions that act on the pulley.
Grade A
Analyse motion after a string goes slackOnce a particle hits the ground the tension is zero, and the other particle continues with the speed it had.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy9 marks
A particle \(A\) of mass 3 kg rests on a smooth horizontal table. It is attached to one end of a string, which passes over a pulley fixed at the edge of the table. The other end of the string is attached to a particle \(B\) of mass 2 kg, which hangs freely below the pulley. The system is released from rest with the string taut and the part of the string between \(A\) and the pulley horizontal. The string is modelled as light and inextensible, and the pulley is modelled as small and smooth. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
[object Object]
(a) Find the acceleration of \(A\).[4]
(b) Find the tension in the string.[2]
(c) State how you have used the modelling assumption that the string is inextensible.[1]
(d) State how you have used the modelling assumption that the pulley is smooth.[1]
(e) State how you have used the modelling assumption that the string is light.[1]
Show the answer and mark scheme
(a)Answer: 3.92 m s−2 (accept 3.9 m s−2) m s−2
M1 for Newton’s second law for \(A\): \(T = 3a\)
M1 for Newton’s second law for \(B\): \(2g - T = 2a\)
A1 for both equations correct
A1 for 3.92 or 3.9 (m s−2)
Worked solution: \(T = 3a\) and \(2g - T = 2a\), so \(2g = 5a\) and \(a = 0.4g = 3.92\) m s−2.
(b)Answer: 11.8 N (accept 12 N) N
M1 for \(T = 3a\) with their \(a\)
A1 for 11.8 or 12 (N)
Worked solution: \(T = 3 \times 3.92 = 11.76\), so 11.8 N (3 s.f.).
(c)Answer: \(A\) and \(B\) have accelerations of the same magnitude.
B1 for the accelerations of \(A\) and \(B\) have the same magnitude
Worked solution: The string does not stretch, so both particles move with the same speed and acceleration.
(d)Answer: The tension is the same in both parts of the string, on either side of the pulley.
B1 for the tension in the string is the same on both sides of the pulley
Worked solution: No friction acts at the pulley, so the same tension \(T\) acts on \(A\) and on \(B\).
(e)Answer: The weight of the string is ignored, so the tension is the same throughout each part of the string (and the string’s mass is not included in the equations).
B1 for the tension is the same throughout the string / the mass of the string is not included
Worked solution: A light string has no mass, so no resultant force is needed to accelerate it: the tension is the same all along it.
Question 2Medium8 marks
Two particles, \(P\) of mass 5 kg and \(Q\) of mass 3 kg, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed above the ground. The particles are released from rest with the string taut and the hanging parts of the string vertical. Take \(g = 9.8\) m s−2.
(a) Find the acceleration of the particles and the tension in the string.[4]
(b) Explain how each of the following modelling assumptions has been used in part (a): (i) the string is light (ii) the string is inextensible (iii) the pulley is smooth.[3]
(c) In reality the pulley is not smooth. State, with a reason, how this would affect the acceleration of the particles.[1]
Show the answer and mark scheme
(a)Answer: \(a = 2.45\) m s−2, \(T = 36.75\) N (36.8 N)
M1 for the equation of motion of \(P\): \(5g - T = 5a\)
M1 for the equation of motion of \(Q\): \(T - 3g = 3a\)
A1 for \(a = 2.45\) (m s−2)
A1 for \(T = 36.75\) (N) (accept 36.8)
Worked solution: \(P\): \(5g - T = 5a\). \(Q\): \(T - 3g = 3a\). Adding: \(2g = 8a\), so \(a = \frac{9.8}{4} = 2.45\) m s−2. \(T = 3(9.8 + 2.45) = 36.75\) N (36.8 N to 3 s.f.).
(b)Answer: (i) The tension is the same throughout the string (its weight is ignored). (ii) Both particles have the same magnitude of acceleration. (iii) The tension is the same on both sides of the pulley.
B1 for (i): the string has no weight, so the tension is the same along its length
B1 for (ii): the particles move with the same speed and acceleration (same \(a\) in both equations)
B1 for (iii): no friction at the pulley, so the tension is the same on both sides (same \(T\) in both equations)
Worked solution: (i) A light string has no weight, so the tension is the same all along it. (ii) An inextensible string keeps a constant length, so \(P\) and \(Q\) have the same magnitude of acceleration \(a\). (iii) A smooth pulley exerts no friction on the string, so the tension on each side is the same \(T\).
(c)Answer: The acceleration would be smaller, because friction at the pulley opposes the motion (the tension on \(P\)’s side is then greater than on \(Q\)’s side).
B1 for a smaller acceleration, because friction at the pulley opposes the motion
Worked solution: Friction at the pulley opposes the motion, so the tensions on the two sides differ and the resultant force driving the system is reduced: the acceleration would be less than 2.45 m s−2.
Question 3Hard10 marks
Two particles, \(P\) of mass 2 kg and \(Q\) of mass 3 kg, are attached to the ends of a light inextensible string which passes over a small smooth pulley fixed above the ground. The system is released from rest with the string taut, the hanging parts of the string vertical, \(P\) on the ground and \(Q\) 1.5 m above the ground. In the subsequent motion \(P\) does not reach the pulley. Take \(g = 9.8\) m s−2.
(a) Find the acceleration of the particles and the tension in the string before \(Q\) hits the ground.[3]
(b) Find the speed of \(Q\) as it hits the ground.[2]
(c) \(Q\) does not rebound after it hits the ground. Find the greatest height of \(P\) above the ground in the subsequent motion.[3]
(d) Explain why the string becomes slack when \(Q\) hits the ground, and state one assumption that you have made in part (c).[2]
Show the answer and mark scheme
(a)Answer: \(a = 1.96\) m s−2, \(T = 23.52\) N (23.5 N)
M1 for two equations of motion, e.g. \(3g - T = 3a\) and \(T - 2g = 2a\)
A1 for \(a = 1.96\) (m s−2)
A1 for \(T = 23.52\) (N) (accept 23.5)
Worked solution: \(Q\): \(3g - T = 3a\). \(P\): \(T - 2g = 2a\). Adding: \(g = 5a\), \(a = 1.96\) m s−2. \(T = 2(9.8 + 1.96) = 23.52\) N.
(b)Answer: 2.42 m s−1
M1 for \(v^2 = 2 \times 1.96 \times 1.5\)
A1 for 2.42 (accept \(\sqrt{5.88}\))
Worked solution: \(v^2 = u^2 + 2as = 2(1.96)(1.5) = 5.88\), so \(v = 2.42\) m s−1.
(c)Answer: 1.8 m
M1 for \(P\) moving freely under gravity with initial speed \(\sqrt{5.88}\): \(0 = 5.88 - 2(9.8)s\)
M1 for adding the extra height (0.3 m) to 1.5 m
A1 for 1.8 (m)
Worked solution: When \(Q\) lands, \(P\) is 1.5 m above the ground, moving upwards at \(\sqrt{5.88}\) m s−1. The string goes slack and \(P\) moves freely under gravity: extra height \(= \dfrac{5.88}{2 \times 9.8} = 0.3\) m. Greatest height \(= 1.5 + 0.3 = 1.8\) m.
(d)Answer: \(Q\) stops but \(P\) keeps moving upwards, so the distance between them along the string decreases and the string can no longer be taut (tension zero). Assumption: e.g. no air resistance, or \(P\) does not hit the pulley.
B1 for \(Q\) stops while \(P\) continues to move up, so the string no longer has a pulling tension (it slackens)
B1 for a relevant assumption, e.g. no air resistance on \(P\); \(P\) does not reach the pulley; \(P\) is a particle
Worked solution: When \(Q\) hits the ground it stops, but \(P\) is still moving upwards. The string cannot push, so it goes slack and the tension becomes zero; \(P\) then moves freely under gravity. Part (c) assumes, for example, that there is no air resistance on \(P\).