Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
Forces and Newton's laws
40 free, high-quality past-paper-style questions on Forces and Newton's laws for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 146 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U1-2-A-01Unit 1 · ASEasy1 mark
A space probe has a weight of 520 N at the surface of the Earth. The gravitational field strength at the surface of Mars is 3.71 N kg−1. g at the Earth's surface = 9.81 N kg−1. What is the weight of the probe at the surface of Mars?
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Answer
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Notes
Mass = 520 ÷ 9.81 = 53.0 kg, which is the same on Mars. Weight on Mars = 53.0 × 3.71 = 197 N.
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Answer A
Answer: A
Common error: Mass stays the same from planet to planet; weight changes with g. Find the mass first.
A climbing helmet rests on a shelf. Its weight is the gravitational pull of the Earth on the helmet. Which force forms a Newton's third law pair with the weight of the helmet?
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A third-law pair involves the same two bodies and the same type of force, with the bodies swapped. The weight is a gravitational pull of the Earth on the helmet, so its partner is the gravitational pull of the helmet on the Earth.
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Answer D
Answer: D
Common error: The normal contact force is equal and opposite to the weight here, but it is a contact force acting on the same body — it balances the weight because of the first law, not the third.
In which situation is the resultant force on the named body zero?
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Zero resultant force means zero acceleration: at rest or constant velocity. Only the lift moving upwards at constant speed has constant velocity (constant speed in a straight line).
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Answer C
Answer: C
Common error: Constant speed is not the same as constant velocity. On a curved path the direction changes, so there is an acceleration and a resultant force.
A uniform wooden beam rests horizontally on two supports, one 0.40 m from each end. The beam is 3.0 m long. Where should the weight of the beam be drawn on its free-body diagram?
Figure: uniform beam on two supports (not to scale)
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The weight of an extended body acts at its centre of gravity. For a uniform beam this is at its geometric centre, 1.5 m from each end.
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Answer C
Answer: C
Exam tip: On a free-body diagram of an extended body, draw the whole weight as one arrow from the centre of gravity.
A geologist on the Moon collects a rock sample of mass 3.6 kg. g at the Earth's surface = 9.81 N kg−1.
(a)
State one difference between mass and weight.
(1)
(b)
Calculate the weight of the sample on the Earth.
(1)
(c)
On the Moon, a newton-meter shows that the weight of the sample is 5.8 N. Calculate the gravitational field strength at the surface of the Moon.
(2)
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Part (a)
mass is the amount of matter / a scalar in kg and is the same everywhere, whereas weight is the gravitational force on the body, a vector in N, which depends on g
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Part (b)
3.6 × 9.81 = 35 N (35.3 N)
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Part (c)
g = F ÷ m = 5.8 ÷ 3.6
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= 1.6 N kg−1 (1.61)
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Exam tip: Give gravitational field strength in N kg−1. It is a force per unit mass, which is why g = F/m.
A lander carries a force sensor to the surface of a planet. Five test masses are hung from the sensor in turn. The graph shows the weight W of each test mass against its mass m. An instrument of mass 7.5 kg is taken to the same planet. What is its weight there?
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The gradient of the graph is g on the planet: 12.0 N ÷ 5.0 kg = 2.4 N kg−1. Weight = mg = 7.5 × 2.4 = 18 N.
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Answer B
Answer: B
Exam tip: On a W–m graph the gradient is the gravitational field strength. Read it from a point far along the line on gridline intersections.
The figure shows a child pushing a heavy crate across a floor at constant velocity. Arrows 1 to 4 show four of the horizontal forces. The solid navy arrows act on the crate and the dashed gold arrows act on the child. Which two forces form a Newton's third law pair?
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A third-law pair acts on two different bodies, is the same type of force, and has equal size and opposite direction. Force 1 (child pushes crate) and force 2 (crate pushes child) are both contact forces at the hands, act on different bodies and point in opposite directions: 1 and 2.
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Answer C
Answer: C
Common error: Forces 1 and 3 are equal and opposite because the crate moves at constant velocity, but they both act on the crate. That is Newton's first law, not the third.
A hockey puck has been struck and is now sliding to the right across horizontal ice, slowing down. Nothing is in contact with the puck except the ice. Which diagram is the correct free-body diagram for the puck?
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Once the puck has left the stick there is no forward push. The forces are its weight, the normal contact force from the ice and friction, which opposes the motion and so acts to the left. That is diagram 4; the leftward resultant force explains why the puck slows down.
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Answer D
Answer: D
Common error: Adding a forward ‘force of motion’. A moving body needs no forward force to keep moving; a free-body diagram shows only forces from objects that are touching it, plus its weight.
A student hangs a mineral sample from a newton-meter in a laboratory on Earth. Figure 1 shows the scale of the newton-meter. g = 9.81 N kg−1.
(a)
State the weight of the sample.
(1)
(b)
Calculate the mass of the sample.
(1)
(c)
The gravitational field strength at the surface of Venus is 8.87 N kg−1. Calculate the weight of the sample at the surface of Venus.
(1)
(d)
Explain why the mass of the sample would be the same on Venus but its weight would not.
(1)
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Part (a)
7.4 N (allow 7.3–7.5 N)
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Part (b)
m = W ÷ g = 7.4 ÷ 9.81 = 0.75 kg (0.754 kg)
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Part (c)
W = 0.754 × 8.87 = 6.7 N (6.69 N; allow ECF from (b))
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Part (d)
mass is the quantity of matter in the sample and does not depend on where it is, whereas weight = mg is the gravitational force, which depends on the local field strength
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Exam tip: Read a scale to the nearest division and check what one small division is worth before you read: here each small division is 0.2 N.
A probe on the surface of a small moon with no atmosphere drops a steel sphere from rest from five different heights and times each fall. Figure 1 shows the distance fallen s against the square of the fall time t2, with a line of best fit.
(a)
Show that the gradient of the line is equal to g/2, where g is the gravitational field strength at the surface of the moon.
(1)
(b)
Use Figure 1 to determine g at the surface of the moon.
(2)
(c)
Calculate the weight of a 12 kg instrument at the surface of the moon.
(1)
(d)
A feather released at the same moment as the sphere, from the same height, lands at the same time. Explain why.
(1)
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Part (a)
s = ut + ½at2 with u = 0 and a = g, so s = (g/2)t2 and the gradient is g/2
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Part (b)
gradient = 3.00 ÷ 5.0 = 0.60 m s−2
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g = 2 × 0.60 = 1.2 N kg−1 (allow 1.18–1.22)
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Part (c)
W = 12 × 1.2 = 14 N (14.4 N; allow ECF from (b))
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Part (d)
there is no air resistance (no atmosphere), so the only force on each is its weight and each has acceleration W/m = g, whatever its mass
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Exam tip: Units of g: N kg−1 and m s−2 are equivalent. Use N kg−1 when the question calls it gravitational field strength.
A rowing boat and its crew have a total mass of 260 kg. The oars give a forward force of 410 N and the water exerts a resistive force of 150 N. What is the acceleration of the boat?
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Resultant force = 410 − 150 = 260 N forwards. a = 260 ÷ 260 = 1.0 m s−2.
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Answer B
Answer: B
Exam tip: ΣF in ΣF = ma is the resultant force. Write the subtraction before dividing by the mass.
A glider on a frictionless air track is released from rest. The track is inclined at 12° to the horizontal. g = 9.81 m s−2. What is the acceleration of the glider along the track?
Figure: glider on the frictionless air track
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The only force with a component along the track is the weight: mg sin 12°. So a = g sin 12° = 2.0 m s−2 (2.04).
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Answer A
Answer: A
Exam tip: On a slope, resolve along and perpendicular to the slope. The component of weight along the slope is mg sin θ, where θ is the slope angle.
A locomotive of mass 24 000 kg pulls two wagons, each of mass 8000 kg, along a straight level track. Resistive forces are negligible. The train accelerates at 0.40 m s−2. What is the tension in the coupling between the locomotive and the first wagon?
Figure: the locomotive pulling two wagons (not to scale)
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Answer
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The coupling behind the locomotive accelerates both wagons: T = (8000 + 8000) × 0.40 = 6.4 kN.
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Answer B
Answer: B
Exam tip: To find the tension in a coupling, draw a free-body diagram of everything behind it. The tension is the only horizontal force on that part.
An electric bus of mass 12 500 kg pulls away from a stop along a straight, level road. At one instant the driving force is 9.00 kN and the total resistive force is 2.75 kN.
(a)
Calculate the acceleration of the bus at this instant.
(2)
(b)
Later the bus travels at a constant speed of 14 m s−1 along the same road. The driving force is now 4.10 kN. State the total resistive force on the bus and explain your answer.
(2)
(c)
The driving force on the bus comes from the road pushing forwards on the tyres. Describe the force that forms a Newton's third law pair with this force.
(2)
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Part (a)
resultant force = 9.00 − 2.75 = 6.25 kN
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a = 6250 ÷ 12 500 = 0.500 m s−2
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Part (b)
4.10 kN
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constant velocity so zero acceleration, so the resultant force is zero / forces balance (Newton's first law)
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Part (c)
the tyres push backwards on the road
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with a force of equal size (4.10 kN at constant speed), the same type (contact / friction), acting on the road
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Common error: Thinking a moving bus needs an unbalanced forward force. At constant velocity the driving force exactly balances the resistive forces.
A skier of mass 68 kg slides down a straight slope inclined at 14° to the horizontal. The total resistive force (friction and air resistance) on the skier is 95 N, acting up the slope. g = 9.81 m s−2.
Figure: forces on the skier on the slope (not to scale)
(a)
Calculate the component of the skier's weight that acts down the slope.
(1)
(b)
Calculate the normal contact force on the skier.
(1)
(c)
Calculate the acceleration of the skier.
(2)
(d)
The skier is moving at 2.0 m s−1 at one instant. Calculate the skier's speed 6.0 s later, assuming the forces do not change.
(2)
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Part (a)
68 × 9.81 × sin 14° = 161 N
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Part (b)
R = 68 × 9.81 × cos 14° = 647 N
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Part (c)
resultant force = 161 − 95 = 66.4 N
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a = 66.4 ÷ 68 = 0.98 m s−2 (0.976)
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Part (d)
v = u + at = 2.0 + 0.976 × 6.0
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= 7.9 m s−1 (accept 7.8 m s−1; allow ECF from (c))
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Common error: Using the full weight, 667 N, along the slope. Only the component mg sin θ acts along the slope.
A hospital lift car and its passengers have a total mass of 950 kg. The lift rises from the ground floor: it accelerates upwards at 0.80 m s−2, then moves at constant velocity, then decelerates at 0.80 m s−2 before stopping. g = 9.81 m s−2.
(a)
Calculate the tension in the lift cable while the lift accelerates upwards.
(2)
(b)
State the tension in the cable while the lift moves at constant velocity.
(1)
(c)
A patient of mass 55 kg stands on scales in the lift. The scales divide the force on them by 9.81 and display the result in kilograms. Calculate the reading while the lift decelerates.
(2)
(d)
Explain why the patient feels lighter during this stage.
(1)
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Answer
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Part (a)
T − 950 × 9.81 = 950 × 0.80
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T = 1.01 × 104 N (10 080 N)
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Part (b)
equal to the weight: 950 × 9.81 = 9.32 × 103 N
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Part (c)
R = 55 × (9.81 − 0.80) = 496 N
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reading = 496 ÷ 9.81 = 51 kg (50.5 kg)
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Part (d)
the normal contact force from the floor on the patient is less than their weight, because the resultant force must act downwards to give a downward acceleration
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Common error: Thinking the patient's mass or weight changes in the lift. Only the normal contact force — what the scales measure — changes.
A student drops a paper cake case from rest and films its fall against a vertical metre rule. From the video she reads the distance fallen, s, every 0.10 s: t / s: 0, 0.10, 0.20, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80 s / cm: 0, 4.4, 14.4, 25.9, 37.8, 49.8, 61.8, 73.8, 85.8
Figure: distance fallen against time, plotted from the student's readings
(a)
Describe how the data show that the cake case reaches a terminal velocity.
(2)
(b)
Determine the terminal velocity of the cake case.
(2)
(c)
Explain, in terms of the forces on the cake case, why its acceleration decreases to zero.
(3)
(d)
Suggest one way to reduce the uncertainty in the readings of s.
(1)
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Answer
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Part (a)
the distance fallen in each 0.10 s interval increases at first (4.4, 10.0, 11.5 cm)
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then becomes constant at 12.0 cm per 0.10 s from about 0.40 s, so the velocity is constant
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Part (b)
use the constant-velocity section: (85.8 − 37.8) cm ÷ (0.80 − 0.40) s
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= 1.2 m s−1
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Part (c)
as the speed increases, the air resistance increases (while the weight stays constant)
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so the resultant force (weight − air resistance) decreases, and so does the acceleration (ΣF = ma)
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when air resistance equals weight the resultant force is zero, so the acceleration is zero
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Part (d)
any one: place the rule close to the path of the case to reduce parallax / use a higher frame rate camera / read the same point (e.g. the bottom edge) of the case each time / repeat and average
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Exam tip: For terminal velocity from a table, use only the readings where the intervals are equal, and take the widest such range to reduce the effect of reading errors.
Two ice skaters stand face to face on smooth ice and push each other apart. Skater P has a mass of 72 kg and skater Q has a mass of 48 kg. While they are in contact, P pushes on Q with a force of 180 N.
(a)
State the size and direction of the force that Q exerts on P.
(2)
(b)
Calculate the acceleration of each skater while they are in contact.
(2)
(c)
Explain why the two forces in part (a) do not cancel each other.
(1)
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Part (a)
180 N
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in the opposite direction to the force on Q / away from Q
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Part (b)
P: 180 ÷ 72 = 2.5 m s−2
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Q: 180 ÷ 48 = 3.8 m s−2 (3.75)
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Part (c)
they act on different bodies (one on P, one on Q), so they cannot be added to give a resultant on either skater
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Common error: Thinking the heavier skater pushes harder. The forces are always equal in size; the lighter skater has the larger acceleration.
A student investigates how the acceleration a of a trolley depends on the horizontal pulling force F. A force sensor on the trolley measures F, and two light gates measure a. The runway is level. She plots a (y-axis) against F (x-axis). The points lie on a straight line that passes through (0.20 N, 0.15 m s−2) and (1.00 N, 1.15 m s−2) and meets the F-axis at a positive value.
Figure: the best-fit line of acceleration against pulling force (data points not shown)
(a)
Explain why the line does not pass through the origin.
(2)
(b)
Determine the mass of the trolley.
(2)
(c)
Determine the frictional force on the trolley.
(1)
(d)
Suggest how the student could adjust the apparatus so that the line passes through the origin.
(1)
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Answer
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Part (a)
a frictional force acts on the trolley, opposing its motion
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the resultant force is F − friction, so the trolley only accelerates once F exceeds the friction
The graph shows how the velocity v of a cyclist changes during a short ride along a straight, level road. The total mass of the cyclist and bicycle is 84 kg. What is the size of the resultant force on the cyclist and bicycle at t = 11 s?
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At t = 11 s the cyclist is in the final section: a = (0 − 6.0) ÷ (13 − 10) = −2.0 m s−2. Size of the resultant force = 84 × 2.0 = 168 N.
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Answer C
Answer: C
Exam tip: Find the acceleration from the gradient of the section that contains the time asked about. Use the change in time for that section only, not the time from the start.
The figure shows block P on a smooth horizontal table, connected to a hanging block Q by a light string over a frictionless pulley. The blocks are released from rest. g = 9.81 N kg−1. What is the tension in the string while the blocks move?
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Answer
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Whole system: a = 0.50 × 9.81 ÷ (2.0 + 0.50) = 1.96 m s−2. The tension is the only horizontal force on P: T = 2.0 × 1.96 = 3.9 N.
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Answer B
Answer: B
Exam tip: Find the acceleration from the whole system first, then apply ΣF = ma to one block to get the tension. The tension is less than the weight of Q, otherwise Q could not accelerate downwards.
A delivery drone lifts a parcel of mass 2.0 kg vertically upwards from rest, using a light cord. The graph shows the tension T in the cord against time t. Air resistance on the parcel is negligible. g = 9.81 N kg−1. What is the acceleration of the parcel during the first 2.0 s?
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Weight = 2.0 × 9.81 = 19.6 N. From the graph T = 24 N. Upwards: 24 − 19.6 = 2.0a, so a = 2.2 m s−2 (2.19).
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Answer A
Answer: A
Common error: Dividing the tension by the mass. The tension is not the resultant force: subtract the weight first.
The figure shows a sledge being pulled across level snow at constant velocity. The tension in the rope and the angle of the rope are shown. Air resistance is negligible. What is the frictional force on the sledge?
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Answer
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Constant velocity, so the horizontal forces balance. Friction = horizontal component of the tension = 60 cos 25° = 54 N (54.4).
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Answer B
Answer: B
Exam tip: The component adjacent to the angle uses cos. Here the angle is measured from the horizontal, so the horizontal component is T cos θ.
Trolley P collides with trolley Q, which is at rest. The graph shows the force exerted on Q by P during the collision; positive values are to the right. What is the force exerted on P by Q at t = 0.15 s?
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Answer
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By Newton's third law the force of Q on P is equal in size and opposite in direction to the force of P on Q at every instant. At 0.15 s the graph shows 12 N to the right on Q, so the force on P is 12 N to the left, whatever the masses.
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Answer D
Answer: D
Common error: Scaling the forces by the masses. The masses decide the accelerations, not the sizes of the two forces in an interaction.
Figure 1 is a free-body diagram for a delivery van of mass 1600 kg moving along a straight, level road. At this instant the speed of the van is 14 m s−1. g = 9.81 N kg−1.
(a)
Show that the weight in Figure 1 is consistent with the mass of the van.
(1)
(b)
Calculate the acceleration of the van at this instant.
(2)
(c)
State the value of R1 + R2 and give a reason for your answer.
(1)
(d)
The driving force and the rolling resistance stay constant. The air resistance is proportional to the square of the speed. Calculate the maximum speed the van can reach.
(3)
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Answer
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Notes
Part (a)
W = mg = 1600 × 9.81 = 15 696 N ≈ 15.7 kN, which matches the 15.7 kN in Figure 1
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Part (b)
resultant force = 3.2 − 1.1 − 0.50 = 1.6 kN
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a = 1600 ÷ 1600 = 1.0 m s−2
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Part (c)
15.7 kN, because the van has no vertical acceleration, so the vertical forces balance
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Part (d)
at maximum speed the resultant force is zero, so air resistance = 3.2 − 0.50 = 2.7 kN
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(v ÷ 14)2 = 2.7 ÷ 1.1
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v = 14 × √2.45 = 22 m s−1 (21.9)
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Exam tip: Maximum (terminal) speed means zero resultant force. Write the force balance first, then use the stated relationship between air resistance and speed.
A badminton shuttlecock of mass 5.0 g is dropped from rest from a high balcony. Figure 1 shows how its velocity v changes with time t. The air resistance on the shuttlecock is proportional to the square of its speed. g = 9.81 N kg−1.
(a)
Use Figure 1 to state the terminal velocity of the shuttlecock.
(1)
(b)
State the acceleration of the shuttlecock at t = 0 and explain your answer.
(2)
(c)
Calculate the air resistance on the shuttlecock when it falls at its terminal velocity.
(2)
(d)
Read the velocity at point P from Figure 1. Use it to calculate the acceleration of the shuttlecock at P.
(3)
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Answer
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Notes
Part (a)
7.0 m s−1 (allow 6.9–7.1 m s−1)
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Part (b)
9.81 m s−2 / g
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at t = 0 the speed is zero so there is no air resistance; the only force is the weight, so a = W/m = g
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Part (c)
at terminal velocity the resultant force is zero, so air resistance = weight
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= 0.0050 × 9.81 = 0.049 N (0.0491 N)
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Part (d)
v = 3.5 m s−1, half the terminal velocity
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air resistance ∝ v2, so air resistance = ¼ × 0.0491 = 0.0123 N
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a = (0.0491 − 0.0123) ÷ 0.0050 = 7.4 m s−2 (7.36; ¾g)
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Common error: Thinking the air resistance at half the terminal speed is half the weight. With air resistance ∝ v2, halving the speed quarters the force.
Two ring magnets are threaded on a light, vertical plastic rod that is fixed to magnet B. Magnet B rests on a top-pan balance. Magnet A floats at rest above B because the facing poles repel (Figure 1). Friction between A and the rod is negligible. The balance shows the force on its pan divided by 9.81, in grams. g = 9.81 N kg−1.
(a)
Name the two forces acting on magnet A and explain how their sizes compare.
(2)
(b)
Use the balance reading to determine the mass of magnet A. Explain your reasoning.
(3)
(c)
Describe the force that forms a Newton's third law pair with the magnetic force of B on A.
(1)
(d)
Explain why the weight of A and the magnetic force of B on A are not a Newton's third law pair.
(1)
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Answer
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Notes
Part (a)
the weight of A (downwards) and the magnetic force of B on A (upwards)
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A is at rest, so the resultant force is zero and the two forces are equal in size (Newton's first law)
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Part (b)
A pushes down on B with a magnetic force equal in size to the force of B on A (Newton's third law), which equals the weight of A
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so the balance supports the weight of both magnets: total force = 0.325 × 9.81 = 3.19 N
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weight of A = 3.19 − 0.200 × 9.81 = 1.23 N, so mass of A = 0.125 kg (125 g)
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Part (c)
the magnetic force of A on B: 1.23 N, acting on B, vertically downwards
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Part (d)
they act on the same body (A) and are different types of force (gravitational and magnetic)
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Common error: Thinking the balance reads only the mass of B because A is not touching it. The magnetic push of A on B is passed on to the pan, so the reading includes the weight of A.
A student investigates how the acceleration a of a trolley depends on its total mass m when the resultant force on it is kept constant. She changes m by adding masses to the trolley, tilts the runway to compensate for friction, and measures a with light gates. Figure 1 shows her results and a line of best fit.
(a)
Explain why she plots a against 1/m rather than against m.
(2)
(b)
Use the line of best fit to determine the resultant force on the trolley.
(2)
(c)
Calculate the acceleration of the trolley when its total mass is 1.5 kg.
(1)
(d)
Describe how she can check that the runway is tilted by the right amount.
(1)
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Answer
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Notes
Part (a)
Newton's second law gives a = F × (1/m), so a should be proportional to 1/m
✓ 1
a graph of a against 1/m should then be a straight line through the origin, which is easy to test, and its gradient is F
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Part (b)
gradient = 1.50 ÷ 2.50 (any two well-separated points on the line)
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F = 0.60 N (allow 0.58–0.62 N)
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Part (c)
a = 0.60 ÷ 1.5 = 0.40 m s−2 (allow ECF from (b))
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Part (d)
with no pulling force, give the trolley a small push: it should move down the runway at constant velocity, e.g. the same speed at both light gates
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Exam tip: To find a gradient, use a large triangle on the best-fit line, not two of the plotted points: the points include experimental scatter.
A go-kart and its driver, of total mass 600 kg, start from rest on a straight, level track. Figure 1 shows the resultant force F on the go-kart against time t.
(a)
Calculate the acceleration of the go-kart during the first 5.0 s.
(1)
(b)
Calculate the speed of the go-kart at t = 10 s.
(3)
(c)
After t = 10 s the driving force on the go-kart is 0.80 kN. State the total resistive force on the go-kart and describe its motion.
(2)
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Answer
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Notes
Part (a)
a = 1200 ÷ 600 = 2.0 m s−2
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Part (b)
speed at 5.0 s = 2.0 × 5.0 = 10 m s−1
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acceleration from 5.0 s to 10 s = 600 ÷ 600 = 1.0 m s−2
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v = 10 + 1.0 × 5.0 = 15 m s−1
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Part (c)
resistive force = 0.80 kN, because the resultant force is zero
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it moves at a constant velocity of 15 m s−1 in a straight line (Newton's first law)
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Common error: Thinking the go-kart slows down when the resultant force falls. A smaller positive resultant force still speeds it up, only at a smaller rate.
A hotel guest of mass 70.0 kg stands on scales that measure force in newtons. The lift is moving upwards and slowing down with a deceleration of 1.20 m s−2. g = 9.81 m s−2. What is the reading on the scales?
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Answer
Mark
Notes
Moving up and slowing down means the acceleration is 1.20 m s−2downwards. Upwards positive: R − mg = m(−1.20), so R = 70.0 × (9.81 − 1.20) = 603 N.
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Answer B
Answer: B
Common error: The direction of the velocity does not decide the answer — the direction of the acceleration does. A lift slowing on its way up is accelerating downwards.
A bird feeder of weight 24 N hangs at rest from the midpoint of a wire stretched between two posts. Each half of the wire makes an angle of 30° with the horizontal. What is the tension in the wire?
Figure: the bird feeder hanging from the midpoint of the wire (not to scale)
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Answer
Mark
Notes
Vertically: 2T sin 30° = 24, so T = 24 ÷ (2 × 0.5) = 24 N.
✓ 1
Answer C
Answer: C
Exam tip: Only the vertical components of the tensions support the weight. The flatter the wire, the larger the tension needed.
At an airport, a baggage tractor of mass 1800 kg pulls two loaded baggage carts in a line along level ground. Each cart has a mass of 600 kg. The driving force on the tractor is 1250 N. The resistive force on the tractor is 350 N and the resistive force on each cart is 120 N. The couplings are horizontal.
Figure: horizontal forces on the tractor and carts (not to scale)
(a)
Calculate the acceleration of the tractor and carts.
(2)
(b)
Calculate the tension in the coupling between the two carts.
(2)
(c)
Calculate the tension in the coupling between the tractor and the first cart.
(2)
(d)
Explain why the tension in the front coupling is greater than the tension in the rear coupling.
(1)
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Answer
Mark
Notes
Part (a)
resultant force = 1250 − 350 − 2 × 120 = 660 N
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a = 660 ÷ 3000 = 0.22 m s−2
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Part (b)
rear cart: T2 − 120 = 600 × 0.22
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T2 = 252 N
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Part (c)
both carts together: T1 − 240 = 1200 × 0.22
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T1 = 504 N
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Part (d)
the front coupling must accelerate both carts and overcome the resistive forces on both, the rear coupling only one
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Exam tip: For the tension in a coupling, take everything behind the coupling as one body. The tension is then the only forward force on it.
A stage spotlight of weight 180 N hangs at rest from two cables, A and B, attached to the same point on the spotlight. Cable A makes an angle of 35° with the horizontal; cable B makes an angle of 50° with the horizontal on the other side.
Figure: the spotlight hanging from cables A and B (not to scale)
(a)
Describe the free-body diagram of the spotlight: name each force and give its direction.
(2)
(b)
By resolving horizontally, show that TA = 0.785 TB.
(1)
(c)
Calculate the tension in each cable.
(3)
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Answer
Mark
Notes
Part (a)
weight, 180 N, vertically downwards
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tension in A along cable A (up and to one side at 35°) and tension in B along cable B (up and to the other side at 50°)
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Part (b)
TA cos 35° = TB cos 50°, so TA = (0.643 ÷ 0.819) TB = 0.7847 TB
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Part (c)
vertically: TA sin 35° + TB sin 50° = 180
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0.7847 × 0.574 TB + 0.766 TB = 180, so TB = 148 N
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TA = 0.7847 × 148 = 116 N
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Exam tip: Check with a sketch: the steeper cable carries more of the weight, so it should have the larger tension.
A helicopter carries a load of mass 1200 kg on a cable. The helicopter flies in a straight horizontal line, speeding up at a constant rate. The cable stays straight and makes a constant angle of 12° with the vertical. Air resistance on the load is negligible. g = 9.81 m s−2.
Figure: the load hangs on a straight cable at 12° to the vertical (not to scale)
(a)
State the two forces that act on the load.
(1)
(b)
Calculate the tension in the cable.
(2)
(c)
Calculate the acceleration of the load.
(2)
(d)
The helicopter later flies at a constant velocity. State and explain the angle the cable now makes with the vertical.
(1)
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Answer
Mark
Notes
Part (a)
weight (vertically down) and the tension in the cable (along the cable)
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Part (b)
the load has no vertical acceleration: T cos 12° = 1200 × 9.81
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T = 1.20 × 104 N (12 030 N)
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Part (c)
horizontally: T sin 12° = 1200a
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a = 9.81 × tan 12° = 2.1 m s−2 (2.09)
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Part (d)
0° / the cable hangs vertically, because the resultant force is zero so the tension has no horizontal component
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Exam tip: When an acceleration is horizontal, resolve vertically to use “no vertical acceleration”, and horizontally to use ΣF = ma.
A plastic toy of mass 0.060 kg is dropped from rest. The graph shows its velocity v against time t. A tangent to the curve is drawn at point X; it passes through the two labelled points. g = 9.81 N kg−1. What is the air resistance on the toy at X?
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Answer
Mark
Notes
Gradient of the tangent = (8.8 − 2.8) ÷ 1.50 = 4.0 m s−2, the acceleration at X. Downwards: mg − D = ma, so D = 0.060 × (9.81 − 4.0) = 0.35 N.
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Answer A
Answer: A
Exam tip: The acceleration at an instant is the gradient of the tangent at that point on a v–t graph. Then write ΣF = ma with the weight and the air resistance in opposite directions.
A sack of sand slides down a straight chute. The figure shows the mass of the sack, the angle of the chute and the frictional force on the sack. g = 9.81 N kg−1. What is the acceleration of the sack?
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Answer
Mark
Notes
Down the slope: mg sin 30° − 12 = 5.0 × 9.81 × 0.5 − 12 = 24.5 − 12 = 12.5 N. a = 12.5 ÷ 5.0 = 2.5 m s−2.
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Answer A
Answer: A
Exam tip: On a slope, resolve the weight into mg sin θ along the slope and mg cos θ into it. Friction acts along the slope, opposite to the motion.
Figure 1 shows block A on a smooth slope, connected by a light string over a frictionless pulley to block B, which hangs vertically. The string is parallel to the slope. The blocks are released from rest. g = 9.81 N kg−1.
(a)
Show that the acceleration of the blocks is about 1.8 m s−2.
(3)
(b)
Calculate the tension in the string.
(1)
(c)
Calculate the speed of B just before it reaches the floor.
(2)
(d)
When B reaches the floor the string goes slack. Calculate the further distance that A moves up the slope before it stops momentarily.
(2)
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Answer
Mark
Notes
Part (a)
B: 2.50 × 9.81 − T = 2.50a
✓ 1
A: T − 3.00 × 9.81 × sin 30° = 3.00a
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adding: a = (24.5 − 14.7) ÷ 5.50 = 1.78 m s−2
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Part (b)
T = 2.50 × (9.81 − 1.78) = 20.1 N
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Part (c)
v2 = 2 × 1.78 × 0.600
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v = 1.46 m s−1
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Part (d)
the only force along the slope is now mg sin 30°, so the deceleration of A = 9.81 × sin 30° = 4.905 m s−2
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s = 1.462 ÷ (2 × 4.905) = 0.218 m
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Exam tip: For two connected blocks, write ΣF = ma for each block separately with the same a and T, then add the equations to remove T.
A parent holds a child on a swing at rest by pulling the seat horizontally with a force F (Figure 1). The total mass of the child and seat is 28 kg. Treat the swing ropes as one light rope. g = 9.81 N kg−1.
(a)
Explain why the three forces in Figure 1 form a closed triangle when drawn tip to tail.
(1)
(b)
Calculate the tension in the rope.
(2)
(c)
Calculate F.
(2)
(d)
The parent increases the pull to 150 N and holds the child at rest again. Calculate the new angle between the rope and the vertical.
(2)
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Answer
Mark
Notes
Part (a)
the child and seat are at rest, so the resultant force is zero
✓ 1
Part (b)
vertically: T cos 20° = 28 × 9.81
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T = 292 N
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Part (c)
horizontally: F = T sin 20° (or F = mg tan 20°)
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F = 1.00 × 102 N (99.98 N)
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Part (d)
tan θ = 150 ÷ (28 × 9.81) = 150 ÷ 275
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θ = 29° (28.6°)
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Exam tip: Choose the resolving direction that contains only one unknown. Here resolving vertically removes F, so it gives T directly.
Two blocks, A and B, are pushed along a rough horizontal floor by a horizontal force applied to A (Figure 1). The blocks stay in contact. The frictional forces on the blocks are constant.
(a)
Calculate the acceleration of the blocks.
(2)
(b)
Calculate the contact force between A and B.
(2)
(c)
Describe the force that forms a Newton's third law pair with the contact force of A on B.
(2)
(d)
The 30.0 N push is now applied to B instead, so that the blocks move to the left. The frictional forces keep the same sizes. Calculate the new contact force between the blocks.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
resultant force = 30.0 − 6.00 − 3.00 = 21.0 N
✓ 1
a = 21.0 ÷ 6.00 = 3.50 m s−2
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Part (b)
for B: C − 3.00 = 2.00 × 3.50
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C = 10.0 N
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Part (c)
the contact force of B on A, 10.0 N
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acting on A, to the left (opposite direction), and the same type of force (contact)
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Part (d)
the resultant force and so the acceleration are unchanged, 3.50 m s−2
✓ 1
B now pushes A: C′ − 6.00 = 4.00 × 3.50, so C′ = 20.0 N
✓ 1
Exam tip: For the contact force, apply ΣF = ma to the block that is being pushed by the other one. It has only two horizontal forces: the contact force and its own friction.
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