M8.4Newton's third law, connected particles and pulleys
Edexcel A level Maths (9MA0) · Mechanics › Forces and Newton's laws
Practise Newton's third law, connected particles and pulleys. 5 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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 \(P\).