ExaminerPrep › Question bank › IGCSE Physics (Edexcel) › 1 Forces and motion
International GCSE Physics · Pearson Edexcel 4PH1 · Topic 1
317 free, high-quality past-paper-style questions on Forces and motion for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (c) Forces, Movement, Shape and Momentum; (b) Movement and Position. For when you have finished the past papers, or want to practise one topic at a time.Notes 1(b) →Notes 1(c) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
A spring in a hospital traction device extends by 6.0 cm when a force of 30 N is applied. The spring obeys Hooke's law throughout.
What force is needed to give an extension of 9.0 cm?
| Marking point | Mark | Notes |
|---|---|---|
| When a spring obeys Hooke's law, extension is directly proportional to force. The extension increases by a factor of 9.0 ÷ 6.0 = 1.5, so the force must also increase by 1.5: 30 × 1.5 = 45 N. 20 N treats force as inversely proportional to extension. 39 N adds the extension to the force. 270 N multiplies force by extension. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a spring that obeys Hooke's law, find the scale factor between the two extensions. Then apply the same factor to the force.
A sycamore seed spins as it falls from a tree. After a short time it falls at a constant speed.
Which statement is correct while the seed falls at constant speed?
| Marking point | Mark | Notes |
|---|---|---|
| At constant speed (terminal velocity) there is no acceleration, so the forces are balanced. The upward air resistance equals the downward weight, and the resultant force is zero. The weight does not disappear, and forces do still act. If air resistance were greater than weight, the seed would slow down. | ✓ 1 | Answer B |
Answer: B
Common error: Constant speed does not mean that no forces act. It means the forces are balanced, so the resultant force is zero.
A planetary rover has a mass of 900 kg. It is transported from Earth to Mars.
Use g = 10 N/kg on Earth.
Describe the difference between mass and weight.
Calculate the weight of the rover on Earth.
The weight of the rover on Mars is 3330 N. Calculate the gravitational field strength on Mars.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Mass is the amount of matter in an object, measured in kg, and does not change from place to place ; weight is the gravitational force on the object, measured in N, and depends on the gravitational field strength | ✓ 2 | |
| Part (b) | ||
| W = m × g = 900 × 10 ; W = 9000 N | ✓ 2 | |
| Part (c) | ||
| g = W ÷ m = 3330 ÷ 900 ; g = 3.7 N/kg | ✓ 2 | |
Common error: The mass of the rover is still 900 kg on Mars. Only its weight changes.
A delivery drone of mass 1.8 kg hovers at a fixed height above a garden.
Use g = 10 N/kg.
A force can change the speed of an object. State two other effects that a force can have on an object.
State whether each of these quantities is a vector or a scalar: force, mass, momentum, energy.
Calculate the weight of the drone.
State the size of the upward force from the drone's rotors while it hovers, and explain your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Change its direction of motion ; change its shape / deform it . Also accept: make it turn / rotate. | ✓ 2 | |
| Part (b) | ||
| Force and momentum are vectors; mass and energy are scalars. Two or three correctly classified earns ; all four correctly classified earns a second mark . | ✓ 2 | |
| Part (c) | ||
| W = m × g = 1.8 × 10 ; W = 18 N | ✓ 2 | |
| Part (d) | ||
| 18 N, equal to the weight ; the drone is stationary, so the forces are balanced and the resultant force is zero | ✓ 2 | |
Common error: A hovering drone is not weightless. Its weight is balanced by an equal upward force from the rotors.
A capsule endoscope is a tiny camera that a patient swallows. It travels 2.4 m through the patient's intestine in 8.0 hours.
What is the average speed of the capsule?
| Marking point | Mark | Notes |
|---|---|---|
| Average speed = distance ÷ time. The time must be in seconds: 8.0 × 3600 = 28 800 s. Speed = 2.4 ÷ 28 800 = 8.3 ×10−5 m/s. Option B converts hours to minutes only (2.4 ÷ 480). Option C does not convert the time at all, so its unit is really m/h. Option D divides time by distance. | ✓ 1 | Answer A |
Answer: A
Common error: Convert hours to seconds by multiplying by 3600, not 60, before calculating a speed in m/s.
A GPS marker fixed to the surface of a glacier shows that the ice moves an average distance of 0.75 m each day.
State the equation linking average speed, distance moved and time taken.
Calculate the distance, in km, that the marker moves in 365 days.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| average speed = distance moved ÷ time taken | ✓ 1 | |
| Part (b) | ||
| distance = 0.75 × 365 = 274 m (accept 273.75 m) | ✓ 1 | |
| = 0.274 km (accept 0.27 km) | ✓ 1 | |
Common error: Check the unit asked for. Here the answer in metres must be divided by 1000 to give kilometres.
A sprinter leaves the starting blocks from rest. She reaches a velocity of 10.8 m/s after 2.4 s.
State the unit of acceleration.
State what the gradient of a velocity–time graph represents.
Calculate the average acceleration of the sprinter.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| m/s2 | ✓ 1 | |
| Part (b) | ||
| acceleration | ✓ 1 | |
| Part (c) | ||
| a = (v − u) ÷ t = (10.8 − 0) ÷ 2.4 | ✓ 1 | |
| a = 4.5 m/s2 | ✓ 1 | |
Exam tip: 'From rest' tells you that the initial velocity u is 0.
In one city, the speed limit for electric scooters is 25 km/h.
State the difference between speed and velocity.
Calculate the speed limit in m/s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Velocity is speed in a given direction (velocity is a vector, speed is a scalar) | ✓ 1 | |
| Part (b) | ||
| 25 × 1000 ÷ 3600 | ✓ 1 | |
| = 6.9 m/s | ✓ 1 | |
Exam tip: To convert from km/h to m/s, divide by 3.6.
A small spring inside a medical syringe pump obeys Hooke's law. A force of 6.0 N gives the spring an extension of 3.0 cm. The limit of proportionality is not exceeded.
What is the extension when the force is 9.0 N?
| Marking point | Mark | Notes |
|---|---|---|
| Hooke's law says extension is directly proportional to force. The force increases by a factor of 9.0 ÷ 6.0 = 1.5, so the extension also increases by 1.5 times: 3.0 × 1.5 = 4.5 cm. 2.0 cm comes from treating the relationship as inverse. 6.0 cm comes from adding 3 cm for an extra 3 N, as if every spring stretches 1 cm per newton. 27 cm comes from multiplying 9.0 by 3.0 without dividing by 6.0. | ✓ 1 | Answer B |
Answer: B
Exam tip: For a proportional relationship, find the scale factor between the two forces and apply the same factor to the extension.
An astronaut on a space station pushes on a wall with a force of 40 N towards the wall.
According to Newton's third law, which force forms a pair with this push?
| Marking point | Mark | Notes |
|---|---|---|
| A Newton's third law pair acts on two different objects. The two forces are equal in size, opposite in direction and of the same type. The astronaut pushes on the wall, so the wall pushes back on the astronaut with 40 N, away from the wall. Weight is a gravitational force between the astronaut and the Earth, so it is not part of this contact pair. A force in the same direction breaks the 'opposite' condition. The muscle force acts within the astronaut's own body, not between the astronaut and the wall. | ✓ 1 | Answer B |
Answer: B
Exam tip: Check the three conditions for a third-law pair: two different objects, equal and opposite forces, and the same type of force.
A physiotherapist gives a patient a rubber resistance band for exercises. The band obeys Hooke's law only for forces up to 40 N. A force of 24 N stretches the band by 0.16 m.
Describe what is meant by elastic behaviour.
Calculate the force needed to stretch the band by 0.10 m.
The patient pulls the band with a force of 60 N. Explain why the extension cannot be found by the same method.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The material returns to its original shape/length when the deforming force is removed | ✓ 1 | |
| Part (b) | ||
| Uses proportionality, e.g. F = 24 × 0.10 ÷ 0.16, OR k = 24 ÷ 0.16 = 150 N/m and F = 150 × 0.10 | ✓ 1 | |
| F = 15 N | ✓ 1 | |
| Part (c) | ||
| 60 N is greater than 40 N, so it is beyond the limit of proportionality / outside the range where Hooke's law applies | ✓ 1 | |
| so the extension is no longer directly proportional to the force / the spring constant is not constant (accept: the band may be beyond its elastic limit) | ✓ 1 | |
Common error: Proportional reasoning only works in the linear region. Do not scale up beyond the stated limit of Hooke's law.
A sprinter of mass 64 kg starts from rest in the starting blocks. She reaches a velocity of 9.0 m/s in 1.8 s, with roughly uniform acceleration.
State the equation linking acceleration, change in velocity and time taken.
Calculate the acceleration of the sprinter.
Calculate the resultant force on the sprinter during this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| acceleration = change in velocity ÷ time taken, a = (v − u) ÷ t | ✓ 1 | |
| Part (b) | ||
| a = (9.0 − 0) ÷ 1.8 | ✓ 1 | |
| a = 5.0 m/s2 | ✓ 1 | |
| Part (c) | ||
| F = m × a = 64 × 5.0 | ✓ 1 | |
| F = 320 N | ✓ 1 | |
Exam tip: Write the substituted equation before the final answer. If your acceleration is wrong, a correct substitution into F = m × a can still earn credit using that value.
During a squash rally, a player hits a moving ball with a racket. The ball squashes against the strings and then flies off towards the front wall along a new path.
Using this example, state three effects that a force can have on an object.
State which of these quantities are vectors: speed, mass, velocity, force, momentum.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Change of speed: the ball speeds up or slows down (accept 'changes speed' or 'accelerates'/'decelerates') | ✓ 1 | |
| Change of direction: the ball moves off along a new path | ✓ 1 | |
| Change of shape: the ball squashes against the strings | ✓ 1 | |
| Part (b) | ||
| Any two of velocity, force and momentum identified as vectors | ✓ 1 | |
| All three of velocity, force and momentum identified, with neither speed nor mass included | ✓ 1 | |
Common error: Speed is a scalar. Velocity is the vector form of speed, because it includes direction.
A capsule camera swallowed by a patient travels 7.2 m through the digestive system in 8.0 hours.
What is the average speed of the capsule?
| Marking point | Mark | Notes |
|---|---|---|
| Time must be in seconds: 8.0 × 3600 = 28 800 s. Average speed = distance ÷ time = 7.2 ÷ 28 800 = 2.5 ×10−4 m/s. The value 0.90 m/s leaves the time in hours, 0.015 m/s converts hours to minutes only, and 4000 m/s divides time by distance. | ✓ 1 | Answer C |
Answer: C
Common error: Always convert hours to seconds (× 3600) before using average speed = distance ÷ time if the answer is wanted in m/s.
A tennis player serves a ball. The ball travels 24 m from the racket to the far baseline in 0.48 s.
State the equation linking average speed, distance moved and time taken.
Calculate the average speed of the ball.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| average speed = distance moved ÷ time taken | ✓ 1 | |
| Part (b) | ||
| 24 ÷ 0.48 | ✓ 1 | |
| = 50 m/s | ✓ 1 | |
Exam tip: Write the substitution line before the answer; it can earn a mark even if the final value is slipped.
A warehouse delivery robot moves in a straight line along an aisle. The table shows its distance from the start at different times.
time / s: 0, 10, 20, 30, 40, 50
distance / m: 0, 12, 24, 24, 24, 39
State what the robot was doing between 20 s and 40 s.
Calculate the speed of the robot during the first 20 s.
The robot moved fastest between 40 s and 50 s. Calculate its speed in this time interval.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| stationary / stopped (distance does not change) | ✓ 1 | |
| Part (b) | ||
| 24 ÷ 20 | ✓ 1 | |
| = 1.2 m/s | ✓ 1 | |
| Part (c) | ||
| distance = 39 − 24 = 15 m in 10 s | ✓ 1 | |
| speed = 15 ÷ 10 = 1.5 m/s | ✓ 1 | |
Common error: For a later time interval, use the change in distance and the change in time, not the total values read from the table.
A maglev train accelerates uniformly from rest with an acceleration of 0.80 m/s2 until it reaches a speed of 96 m/s.
State the equation linking acceleration, change in velocity and time taken.
Calculate the time taken for the train to reach 96 m/s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| acceleration = change in velocity ÷ time taken / a = (v − u) ÷ t | ✓ 1 | |
| Part (b) | ||
| t = (96 − 0) ÷ 0.80 | ✓ 1 | |
| = 120 s | ✓ 1 | |
Exam tip: Rearrange the equation to make the unknown the subject before substituting; here t = (v − u) ÷ a.
A delivery robot travels along a pavement. It covers 240 m in 3.0 minutes.
State the relationship between average speed, distance moved and time taken.
Calculate the average speed of the robot in m/s.
Suggest why the robot's speed at some moments during the journey is greater than this average speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| average speed = distance moved ÷ time taken | ✓ 1 | |
| Part (b) | ||
| time = 3.0 × 60 = 180 s | ✓ 1 | |
| speed = 240 ÷ 180 = 1.3 m/s (accept 1.33 m/s) | ✓ 1 | |
| Part (c) | ||
| Its speed is not constant: it stops, slows down or accelerates at times (e.g. at crossings or obstacles), so it must travel faster than the average at other times | ✓ 1 | |
Common error: Leaving the time in minutes gives 80, which is in m/min, not m/s. Convert the time to seconds first.
A planetary rover has a mass of 180 kg. It is tested on Earth and then sent to Mars.
Gravitational field strength on Earth = 10 N/kg
Gravitational field strength on Mars = 3.7 N/kg
Describe the difference between mass and weight.
Calculate the weight of the rover on Earth.
Calculate the weight of the rover on Mars.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Mass is the amount of matter in an object, measured in kg, and it does not depend on location. | ✓ 1 | |
| Weight is the gravitational force on the object, measured in N, and it depends on the gravitational field strength. | ✓ 1 | |
| Part (b) | ||
| weight = mass × g = 180 × 10 | ✓ 1 | |
| = 1800 N | ✓ 1 | |
| Part (c) | ||
| weight = mass × g = 180 × 3.7 | ✓ 1 | |
| = 666 N (accept 670 N) | ✓ 1 | |
Common error: The mass is still 180 kg on Mars. Only the weight changes, because g is different.
A hand-held luggage scale contains a spring. When a load of 20 N hangs from the scale, the spring extends by 1.6 cm. The spring obeys Hooke's law over the range used.
State Hooke's law.
Calculate the extension of the spring when a suitcase of weight 45 N hangs from the scale.
State what is meant by saying the spring behaves elastically.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The extension of a spring is directly proportional to the force applied, provided the limit of proportionality is not exceeded. | ✓ 1 | |
| Part (b) | ||
| extension per newton = 1.6 ÷ 20 = 0.08 cm/N, or use the ratio 45 ÷ 20 | ✓ 1 | |
| extension = 0.08 × 45 = 3.6 cm | ✓ 1 | |
| Part (c) | ||
| It returns to its original length when the force is removed. | ✓ 1 | |
Exam tip: Direct proportion means that doubling the force doubles the extension. Make sure you use the extension, not the total length of the spring.
A rider on an electric scooter sets off along a level path. The rider and scooter have a combined mass of 80 kg. The driving force is 180 N and the total resistive force is 60 N.
Calculate the resultant force on the rider and scooter.
Calculate the acceleration.
As the scooter speeds up, the resistive force increases until it equals 180 N. State what happens to the motion of the scooter.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 180 − 60 = 120 N (forwards) | ✓ 1 | |
| Part (b) | ||
| a = F ÷ m = 120 ÷ 80 | ✓ 1 | |
| = 1.5 m/s2 | ✓ 1 | |
| Part (c) | ||
| It moves at a constant speed / zero acceleration (accept constant velocity or terminal velocity; do not accept 'it stops'). | ✓ 1 | |
Common error: Balanced forces do not mean the scooter stops. It keeps moving, but at a steady speed.
A trolley accelerates at 1.2 m/s2 when pulled by a constant resultant force. The same force is applied after loads are added so that the total mass of the trolley is three times as large. What is the new acceleration?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: a = F/m, so tripling the mass with the same force divides a by 3 giving 0.40 m/s2. 1.2 ignores the mass change. 3.6 multiplies instead of dividing. 0.13 divides by 9, treating a as inversely proportional to mass squared. | ✓ 1 | Answer A |
Answer: A
Common error: With constant force, more mass means less acceleration, not more.
A student investigates how the mass of a trolley affects its acceleration.
State the independent variable and one control variable in this investigation.
In one run the trolley has mass 0.80 kg and accelerates at 1.5 m/s2. Calculate the resultant force on the trolley.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| independent variable: mass of the trolley | ✓ 1 | |
| control variable: the pulling force (or the surface, or the length of track) | ✓ 1 | |
| Part (b) | ||
| F = ma (or 0.80 × 1.5) | ✓ 1 | |
| 1.2 N | ✓ 1 | |
Exam tip: Only one variable changes at a time; everything else that could affect acceleration is kept the same.
A student uses two light gates to investigate the acceleration of a trolley. The trolley passes the first gate at 0.40 m/s and the second gate at 1.00 m/s. The time between the gates is 0.50 s.
Calculate the acceleration of the trolley.
The resultant force on the trolley is 0.60 N. Calculate the mass of the trolley.
State what happens to the acceleration if the mass is doubled and the force stays the same.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a = change in speed ÷ time (1.00 − 0.40) ÷ 0.50 | ✓ 1 | |
| 1.2 m/s2 | ✓ 1 | |
| Part (b) | ||
| m = F ÷ a (0.60 ÷ 1.2) (ECF from (a)) | ✓ 1 | |
| 0.50 kg | ✓ 1 | |
| Part (c) | ||
| acceleration halves | ✓ 1 | |
Exam tip: Acceleration comes from change in speed over time, not from either speed on its own.
A spring in a library book-return slot has a natural length of 20 cm. When a force of 6.0 N stretches it, its length is 24 cm. The spring obeys Hooke's law. What is the length of the spring when it is stretched by a force of 9.0 N?
| Marking point | Mark | Notes |
|---|---|---|
| Extension at 6.0 N is 24 − 20 = 4.0 cm. Extension is proportional to force, so at 9.0 N the extension is 4.0 × 9.0/6.0 = 6.0 cm and the length is 20 + 6.0 = 26 cm. 6.0 cm is the extension, not the length. 36 cm treats the total length as proportional to force (24 × 1.5). 23 cm comes from dividing the wrong way round (extension 4.0 × 6.0/9.0 ≈ 2.7 cm). | ✓ 1 | Answer C |
Answer: C
Common error: Hooke's law links force to extension, not to total length. Subtract the natural length first, then add it back at the end.
A maintenance van travels along a road between wind turbines. It covers 18 km in 15 minutes at a constant speed.
Calculate the speed of the van in km/h.
Convert this speed to m/s.
Later the van starts from rest and reaches 20 m/s uniformly in 8.0 s. Calculate its acceleration.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 15 minutes = 0.25 h | ✓ 1 | |
| speed = distance / time = 18 / 0.25 = 72 km/h | ✓ 1 | |
| Part (b) | ||
| 72 × 1000 = 72 000 m in 3600 s | ✓ 1 | |
| 20 m/s | ✓ 1 | |
| Part (c) | ||
| acceleration = change in speed / time | ✓ 1 | |
| 20 / 8.0 = 2.5 m/s2 | ✓ 1 | |
Exam tip: To change km/h to m/s, divide by 3.6. Check that your answer has a sensible size for the situation.
A skydiver jumps from an aircraft and, before opening the parachute, falls at a constant terminal velocity. Which statement about the forces on the skydiver is correct?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct: constant velocity means zero resultant force, so air resistance equals weight. A is wrong because motion does not need an unbalanced force. B is wrong because air resistance is largest at terminal velocity, not zero. C is wrong because a larger air resistance would make the skydiver slow down. | ✓ 1 | Answer D |
Answer: D
Common error: Constant velocity does not mean there is no force; it means the forces are balanced.
A student hangs a 0.30 kg load from a rubber band. The unstretched length of the band is 12.0 cm and its length with the load is 15.6 cm. Use g = 10 N/kg. Which row gives the force on the band and its extension?
| Marking point | Mark | Notes |
|---|---|---|
| The force is the weight, 0.30 × 10 = 3.0 N, and the extension is 15.6 − 12.0 = 3.6 cm, so C is correct. A uses the mass in kg as if it were the force. B uses the stretched length as the extension. D adds the two lengths instead of subtracting. | ✓ 1 | Answer C |
Answer: C
Common error: Extension is the increase in length, not the total length, and a load in kg must be converted to a weight in N.
A student investigates a rubber band used to hold sheets of music together. The band hangs from a clamp and masses are added one at a time to a hanger on the band. The unstretched length of the band is 8.0 cm. Use g = 10 N/kg.
Mass added (kg): 0.10, 0.20, 0.30, 0.40, 0.50
Length of band (cm): 9.2, 10.4, 11.6, 12.8, 15.2
Name the instrument used to measure the length of the band and state one way to reduce parallax error when reading it.
Calculate the extension of the band when the 0.40 kg mass is added. Give your answer in cm.
A student says the band obeys Hooke's law for all of these loads. Use the data to decide whether this is correct.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Metre rule / ruler | ✓ 1 | |
| Any one valid way, e.g. eye level with the end of the band, or ruler held vertical next to the band with a pointer or set square | ✓ 1 | |
| Part (b) | ||
| extension = new length − original length = 12.8 − 8.0 | ✓ 1 | |
| = 4.8 cm | ✓ 1 | |
| Part (c) | ||
| Extension per newton is constant (1.2 cm/N) for loads up to 0.40 kg (4.0 N) | ✓ 1 | |
| At 0.50 kg (5.0 N) the extension is 7.2 cm, which is 1.44 cm/N, so the band does not obey Hooke's law for all loads, the statement is not correct | ✓ 1 | |
Exam tip: When testing for proportionality, calculate extension divided by force for every point and look for the point where it changes.
A music stand has a spring clip that holds pages open. The spring has a spring constant of 150 N/m. It obeys Hooke's law for forces up to 12 N.
Calculate the force needed to stretch the spring by 0.020 m.
A force of 15 N is applied to the spring. State one possible effect on the spring.
Calculate the extension of the spring when a force of 4.5 N is applied.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = kx | ✓ 1 | |
| = 150 × 0.020 = 3.0 N | ✓ 1 | |
| Part (b) | ||
| Extension is no longer proportional to force (greater than predicted) / spring may not return to its original length | ✓ 1 | |
| Part (c) | ||
| x = F ÷ k = 4.5 ÷ 150 | ✓ 1 | |
| = 0.030 m | ✓ 1 | |
Exam tip: Rearrange F = kx for the unknown first, then substitute, and give the unit of your final answer.
A swimmer completes 6 lengths of a 25 m pool in 100 s. What is her average speed?
| Marking point | Mark | Notes |
|---|---|---|
| Total distance = 6 × 25 = 150 m, so average speed = 150 / 100 = 1.5 m/s, which is D. A (0 m/s) is the average velocity, because an even number of lengths returns her to the start. B (0.25 m/s) uses the length of one pool only (25 / 100). C (0.67 m/s) inverts the equation (100 / 150). | ✓ 1 | Answer D |
Answer: D
Common error: Average speed uses the total distance travelled along the path, not the displacement.
Which pair contains one contact force and one non-contact force?
| Marking point | Mark | Notes |
|---|---|---|
| Air resistance acts only where the object touches the air (contact), whereas weight is the gravitational pull of the Earth, which acts at a distance (non-contact), so C is correct. A and D contain two contact forces. B contains two non-contact forces. | ✓ 1 | Answer C |
Answer: C
Exam tip: A non-contact force acts without the objects touching: gravitational, magnetic and electrostatic forces.
A library book of mass 1.8 kg rests on a horizontal shelf. Use g = 10 N/kg.
State two effects that a force can have on an object.
Calculate the weight of the book.
State the size and direction of the force the shelf exerts on the book.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| any one effect, e.g. changes the shape or size of the object | ✓ 1 | |
| any second different effect, e.g. changes its speed, changes its direction of motion | ✓ 1 | |
| Part (b) | ||
| M1 weight = mass × g (or 1.8 × 10) | ✓ 1 | |
| A1 18 N | ✓ 1 | |
| Part (c) | ||
| 18 N (ECF from (b)) | ✓ 1 | |
| upwards (vertically) | ✓ 1 | |
Common error: Weight is a force in newtons; mass is in kilograms. Do not give the weight in kg.
A technician drives a service vehicle along a straight track between wind turbines. She covers 2.7 km in 6.0 minutes and then a further 1.8 km in 9.0 minutes.
Calculate her average speed during the first stage, in m/s.
Calculate her average speed for the whole journey, in m/s.
Explain why the average speed for the whole journey is less than the speed in the first stage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 speed = distance / time with 2700 m and 360 s | ✓ 1 | |
| A1 7.5 m/s | ✓ 1 | |
| Part (b) | ||
| total distance = 4500 m | ✓ 1 | |
| total time = 900 s | ✓ 1 | |
| average speed = 5.0 m/s (accept 5) | ✓ 1 | |
| Part (c) | ||
| the second stage was slower (1800 m in 540 s is about 3.3 m/s), which lowers the overall average | ✓ 1 | |
Exam tip: Convert kilometres to metres and minutes to seconds before using speed = distance / time.
A skydiver has been falling for some time and is moving at terminal velocity. Which statement about the forces on the skydiver is correct?
| Marking point | Mark | Notes |
|---|---|---|
| At terminal velocity the speed is constant, so the resultant force is zero and weight equals air resistance, which is C. A wrongly assumes that motion needs a resultant force. B would make the skydiver slow down. D confuses zero acceleration with zero weight; the weight of the skydiver does not change. | ✓ 1 | Answer C |
Answer: C
Common error: A constant speed does not mean there is no force; it means the forces are balanced.
Which list contains only contact forces?
| Marking point | Mark | Notes |
|---|---|---|
| Tension, friction and air resistance all need the objects to touch, so B is correct. A includes weight, which acts through a gravitational field with no contact. C includes the magnetic force, a non-contact force. D includes the electrostatic force, also a non-contact force. | ✓ 1 | Answer B |
Answer: B
Exam tip: Ask whether the force would still act if the two objects were separated by a gap; if so, it is a non-contact force.
A light aircraft accelerates uniformly in a straight line along a runway. It starts from rest and reaches a speed of 40 m/s after 16 s.
Calculate the acceleration of the aircraft.
Calculate the distance the aircraft travels along the runway in this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| acceleration = change in speed ÷ time = (40 − 0) ÷ 16 | ✓ 1 | |
| acceleration = 2.5 m/s² | ✓ 1 | |
| Part (b) | ||
| average speed = (0 + 40) ÷ 2 = 20 m/s and distance = average speed × time = 20 × 16 | ✓ 1 | |
| distance = 320 m | ✓ 1 | |
Exam tip: With uniform acceleration from rest, the average speed is half the final speed.
A quad bike moves along a straight farm track. The table shows its distance from a gate at different times.
Time (s): 0, 10, 20, 30, 40, 50, 60
Distance from gate (m): 0, 30, 60, 60, 60, 90, 120
State what the data show about the motion of the quad bike between 20 s and 40 s.
Calculate the speed of the quad bike during the first 20 s.
Calculate the average speed of the quad bike for the whole 60 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the quad bike is stationary / not moving (distance stays at 60 m) | ✓ 1 | |
| Part (b) | ||
| speed = distance ÷ time = 60 ÷ 20 | ✓ 1 | |
| speed = 3.0 m/s | ✓ 1 | |
| Part (c) | ||
| average speed = total distance ÷ total time = 120 ÷ 60 | ✓ 1 | |
| average speed = 2.0 m/s | ✓ 1 | |
Common error: Average speed uses the total time, including any time spent stationary.
A librarian pushes a loaded trolley along a level floor at a constant speed of 0.8 m/s. Which statement about the horizontal forces on the trolley is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Constant speed in a straight line means no acceleration, so the resultant force is zero and the push balances the resistive forces exactly. A push larger than friction would make the trolley accelerate. A zero push would let friction slow the trolley down. Weight acts vertically and is balanced by the floor, so it is not compared with the horizontal push. | ✓ 1 | Answer C |
Answer: C
Common error: Constant speed does not mean no force is needed; it means the forces are balanced.
A swimmer completes 8 lengths of a 25 m pool in 125 s. The swimmer has a mass of 60 kg.
Calculate the average speed of the swimmer.
At the start of one length, the swimmer pushes off the wall and accelerates from rest to 1.6 m/s in 2.0 s. Calculate the acceleration.
Calculate the resultant force on the swimmer during this acceleration.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance = 8 × 25 = 200 m, speed = distance / time = 200 / 125 | ✓ 1 | |
| A1 1.6 m/s | ✓ 1 | |
| Part (b) | ||
| M1 a = (1.6 − 0) / 2.0 | ✓ 1 | |
| A1 0.80 m/s² | ✓ 1 | |
| Part (c) | ||
| M1 F = ma = 60 × 0.80 | ✓ 1 | |
| A1 48 N | ✓ 1 | |
Common error: The distance for the average speed is the total distance swum, not the length of the pool.
A librarian pushes a trolley of books with a horizontal force of 90 N. The total mass of the trolley and books is 40 kg. A frictional force of 50 N acts on the trolley.
State two effects that a force can have on an object.
State one non-contact force acting on the trolley.
Calculate the resultant horizontal force on the trolley.
Calculate the acceleration of the trolley.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| first effect, e.g. changes its speed / starts or stops it moving | ✓ 1 | |
| a different second effect, e.g. changes its direction of motion / changes its shape or size | ✓ 1 | |
| Part (b) | ||
| weight / gravitational force | ✓ 1 | |
| Part (c) | ||
| 90 − 50 = 40 N | ✓ 1 | |
| Part (d) | ||
| M1 a = F / m = 40 / 40 (ECF from (c)) | ✓ 1 | |
| A1 1.0 m/s² | ✓ 1 | |
Exam tip: Find the resultant force by subtracting opposing forces before using F = ma.
A stone is released from rest at the top of a cliff and takes 3.0 s to reach the sea. Ignore air resistance and take the acceleration of free fall as 10 m/s2. What is the height of the cliff?
| Marking point | Mark | Notes |
|---|---|---|
| The speed on reaching the sea is v = 10 × 3.0 = 30 m/s. Using v2 = u2 + 2as with u = 0 gives s = 302 ÷ (2 × 10) = 45 m, which is the average speed (15 m/s) multiplied by the time. 30 m is wrong because it is the final speed, not a distance. 90 m is wrong because it multiplies the final speed by the time, which would only be correct for constant speed. 15 m is wrong because it is just half of the final speed in m/s, with the time ignored. | ✓ 1 | Answer B |
Answer: B
Common error: A falling object does not travel at its final speed for the whole fall, so distance is not final speed × time.
A diver steps off a platform that is 5.0 m above the surface of a pool. The diver starts from rest. Ignore air resistance and take the acceleration of free fall as 10 m/s2.
Calculate the speed of the diver when she reaches the water. Use v2 = u2 + 2as.
Calculate the time taken for the diver to fall to the water.
Suggest why the real speed of the diver on reaching the water is slightly less than your answer to (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| substitution into v2 = 0 + 2 × 10 × 5.0 = 100 | ✓ 1 | |
| v = 10 m/s | ✓ 1 | |
| Part (b) | ||
| use of a = (v − u) ÷ t, or equivalent rearrangement | ✓ 1 | |
| t = 10 ÷ 10 = 1.0 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| air resistance acts upwards on the diver, reducing the acceleration | ✓ 1 | |
Exam tip: Write down u, a and s before choosing the equation; if the object starts from rest, u = 0 and the equation simplifies.
A railway wagon A of mass 15 000 kg rolls at 1.6 m/s along a straight horizontal track and collides with a stationary wagon B of mass 10 000 kg. The two wagons couple together.
Calculate the momentum of wagon A before the collision.
Calculate the speed of the coupled wagons immediately after the collision.
State the direction in which the coupled wagons move.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity = 15 000 × 1.6 | ✓ 1 | |
| 24 000 kg m/s | ✓ 1 | |
| Part (b) | ||
| total momentum after = total momentum before = 24 000 kg m/s (ECF from (a)) | ✓ 1 | |
| total mass = 25 000 kg and speed = momentum ÷ mass | ✓ 1 | |
| 0.96 m/s | ✓ 1 | |
| Part (c) | ||
| in the same direction as wagon A was moving before the collision | ✓ 1 | |
Exam tip: Check your final speed is smaller than the speed of the moving wagon: adding mass to a system at constant momentum must reduce its speed.
A tower crane lifts a concrete beam of mass 600 kg. Take g = 10 N/kg.
Calculate the weight of the beam.
The beam rises at a constant speed. State the tension in the cable and explain your answer.
The beam then accelerates upwards at 0.50 m/s2. Calculate the tension in the cable.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight = mass × gravitational field strength = 600 × 10 | ✓ 1 | |
| 6000 N | ✓ 1 | |
| Part (b) | ||
| tension = 6000 N | ✓ 1 | |
| forces are balanced because there is no acceleration (resultant force is zero) | ✓ 1 | |
| Part (c) | ||
| resultant force = ma = 600 × 0.50 = 300 N | ✓ 1 | |
| tension = 6000 + 300 = 6300 N | ✓ 1 | |
Common error: Constant speed does not mean no force is acting; it means the forces are balanced.
A forager bee flies 1.8 km from its hive to a field of flowers in 10 minutes. What is its average speed?
| Marking point | Mark | Notes |
|---|---|---|
| Convert first: 1.8 km = 1800 m and 10 min = 600 s, so speed = 1800 ÷ 600 = 3.0 m/s, so this is correct. 0.18 comes from dividing 1.8 by 10 without converting units. 10.8 is the speed in km/h, not m/s. 180 comes from dividing 1800 m by 10 without converting minutes to seconds. | ✓ 1 | Answer B |
Answer: B
Common error: Convert distance to metres and time to seconds before calculating a speed in m/s.
A hospital sends a portable scanner of mass 6.0 kg from Earth, where g = 10 N/kg, to a planet where g = 4.0 N/kg. Which row gives the scanner's mass and weight on the planet?
| Marking point | Mark | Notes |
|---|---|---|
| Mass does not depend on location, so it stays 6.0 kg. Weight = mass × g = 6.0 × 4.0 = 24 N, so this is correct. A weight of 60 N is the weight on Earth, which would be wrong because g has changed. A mass of 2.4 kg wrongly makes mass change with g. A mass of 1.5 kg comes from dividing a weight of 6.0 N by g, mixing up mass and weight. | ✓ 1 | Answer D |
Answer: D
Common error: Mass stays the same everywhere; only weight changes with the gravitational field strength.
A Formula 1 car completes laps of a circuit of length 4.8 km. A lap takes 75 s.
Calculate the average speed of the car in m/s.
Calculate this speed in km/h. Give your answer to 3 significant figures.
The race is 52 laps long. Each lap takes 75 s. Calculate the time taken for the race in minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 speed = distance ÷ time = 4800 ÷ 75 | ✓ 1 | |
| A1 64 m/s | ✓ 1 | |
| Part (b) | ||
| M1 64 × 3.6 = 230.4 | ✓ 1 | |
| A1 230 km/h | ✓ 1 | |
| Part (c) | ||
| M1 52 × 75 = 3900 s | ✓ 1 | |
| A1 3900 ÷ 60 = 65 minutes | ✓ 1 | |
Exam tip: Write the equation in words first, then substitute values in the same units.
A hospital has a portable X-ray unit that weighs 650 N on Earth, where g = 10 N/kg.
Calculate the mass of the unit.
The unit is taken to a planet where the gravitational field strength is 4.0 N/kg. Calculate its weight on the planet.
State what happens to the mass of the unit when it is taken to the planet.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 mass = weight ÷ g = 650 ÷ 10 | ✓ 1 | |
| A1 65 kg | ✓ 1 | |
| Part (b) | ||
| M1 weight = 65 × 4.0 | ✓ 1 | |
| A1 260 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| it stays the same | ✓ 1 | |
Common error: Do not call weight and mass by the same name: weight is a force in newtons, mass is in kilograms.
A bee flies at an average speed of 6.0 m/s while it is foraging. One foraging trip, out to the flowers and back, lasts 15 minutes of flying time.
Calculate the distance flown on one trip. Give your answer in km.
Calculate the greatest distance from the hive that the bee could reach if it flew straight out and straight back.
The bee makes 8 trips in one day. Calculate the total distance flown in the day in km.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance = 6.0 × 15 × 60 = 5400 m | ✓ 1 | |
| A1 5.4 km | ✓ 1 | |
| Part (b) | ||
| 5.4 ÷ 2 = 2.7 km | ✓ 1 | |
| Part (c) | ||
| M1 8 × 5.4 | ✓ 1 | |
| A1 43.2 km | ✓ 1 | |
Exam tip: Convert minutes to seconds before using a speed in m/s.
A hospital porter pushes a trolley of total mass 80 kg along a corridor with a constant horizontal force of 120 N. Friction and other resistive forces total 40 N. What is the acceleration of the trolley?
| Marking point | Mark | Notes |
|---|---|---|
| The resultant force is 120 − 40 = 80 N, so a = F/m = 80 ÷ 80 = 1.0 m/s². 1.5 m/s² ignores the resistive force (120 ÷ 80). 2.0 m/s² adds the resistive force to the push instead of subtracting it. 0.50 m/s² uses only the resistive force (40 ÷ 80). | ✓ 1 | Answer C |
Answer: C
Common error: Always find the resultant force first: a push is reduced by friction, not increased by it.
A light aircraft of mass 1200 kg accelerates uniformly from rest along a runway. After 16 s it reaches its take-off speed of 40 m/s.
Calculate the acceleration of the aircraft.
Calculate the length of the take-off run.
Calculate the resultant force acting on the aircraft during the take-off run.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 acceleration = change in speed ÷ time, or 40 ÷ 16 | ✓ 1 | |
| A1 2.5 m/s2 | ✓ 1 | |
| Part (b) | ||
| M1 distance = average speed × time = 20 × 16 (or use s = ½(u + v)t) | ✓ 1 | |
| A1 320 m | ✓ 1 | |
| Part (c) | ||
| M1 F = ma = 1200 × 2.5 (ECF from (a)) | ✓ 1 | |
| A1 3000 N | ✓ 1 | |
Exam tip: Write the equation, substitute with units, then give the answer: this earns the method mark even if the arithmetic slips.
A tile falls from the roof of a hospital and hits the ground at 30 m/s. Ignore air resistance. Use g = 10 m/s2.
Calculate the height from which the tile fell.
Calculate the time taken for the tile to fall.
In reality the tile hits the ground at a speed lower than 30 m/s. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v2 = 2gs, so 302 = 2 × 10 × s | ✓ 1 | |
| A1 45 m | ✓ 1 | |
| Part (b) | ||
| M1 t = v ÷ g = 30 ÷ 10 | ✓ 1 | |
| A1 3.0 s | ✓ 1 | |
| Part (c) | ||
| air resistance (drag) acts upwards on the tile | ✓ 1 | |
| so the resultant force is smaller and the acceleration (and final speed) is less than calculated | ✓ 1 | |
Exam tip: If a question says to ignore air resistance, the only force is weight and the acceleration is g.
Which list contains only vector quantities?
| Marking point | Mark | Notes |
|---|---|---|
| A vector has both magnitude and direction. Acceleration, force and momentum all have direction, so this list contains only vectors. The first list contains only scalars. The second list includes speed, which is a scalar. The last list includes energy, which is a scalar. | ✓ 1 | Answer C |
Answer: C
Common error: Speed and velocity are not the same; speed is a scalar and velocity is a vector.
A volcanologist walks 800 m due east to a volcanic vent and then 600 m due north to a camp. The whole walk takes 1250 s.
State the difference between distance and displacement.
State the total distance walked.
Calculate the magnitude of the displacement from the start to the camp.
Calculate the average speed for the whole walk.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Distance is the total length of the path travelled and is a scalar (has magnitude only) | ✓ 1 | |
| Displacement is the straight-line distance from the start to the end in a given direction and is a vector | ✓ 1 | |
| Part (b) | ||
| 1400 m | ✓ 1 | |
| Part (c) | ||
| M1 use of Pythagoras: √(8002 + 6002) | ✓ 1 | |
| A1 1000 m | ✓ 1 | |
| Part (d) | ||
| M1 average speed = total distance ÷ time | ✓ 1 | |
| A1 1400 ÷ 1250 = 1.12 m/s (ECF from (b)) | ✓ 1 | |
Common error: Average speed uses the total distance travelled, not the displacement.
A light aircraft accelerates uniformly from rest along a runway at 2.5 m/s2 for 20 s before take-off.
Calculate the speed of the aircraft at take-off.
Calculate the distance travelled along the runway during the 20 s.
The runway is 600 m long. State whether it is long enough for take-off.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v = at, with a = 2.5 and t = 20 | ✓ 1 | |
| A1 50 m/s | ✓ 1 | |
| Part (b) | ||
| M1 distance = average speed × time = (50 ÷ 2) × 20 (ECF from (a)) | ✓ 1 | |
| A1 500 m | ✓ 1 | |
| Part (c) | ||
| Yes, because 500 m is less than 600 m (ECF from (b)) | ✓ 1 | |
Exam tip: Check units and use the correct form of the speed equation, such as v = at from rest.
The table shows the thinking distance and braking distance for a delivery van at three speeds.
| Speed / m/s | Thinking distance / m | Braking distance / m |
|---|---|---|
| 8 | 6 | 8 |
| 16 | 12 | 32 |
| 24 | 18 | 72 |
State what is meant by thinking distance.
Use the table to determine the reaction time of the driver.
Describe how the braking distance changes when the speed doubles from 8 m/s to 16 m/s. Use data from the table.
Calculate the stopping distance at 24 m/s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The distance travelled between the driver seeing a hazard and applying the brakes | ✓ 1 | |
| Part (b) | ||
| M1 time = distance ÷ speed, such as 12 ÷ 16 | ✓ 1 | |
| A1 0.75 s | ✓ 1 | |
| Part (c) | ||
| The braking distance increases by a factor of four | ✓ 1 | |
| from 8 m to 32 m | ✓ 1 | |
| Part (d) | ||
| 18 + 72 = 90 m | ✓ 1 | |
Exam tip: Stopping distance is the sum of thinking distance and braking distance.
A steel ball is released at the top of a tall column of oil and eventually falls at a constant speed. The upthrust on the ball is negligible. Which statement is correct while the ball falls at this constant speed?
| Marking point | Mark | Notes |
|---|---|---|
| At constant speed the acceleration is zero, so the resultant force is zero and the upward drag force must equal the downward weight. Drag larger than weight would slow the ball down. Drag is not zero: it has increased with speed until it balances the weight. The weight of an object depends only on its mass and the gravitational field strength, so it does not decrease. | ✓ 1 | Answer B |
Answer: B
Common error: A body moving at constant speed is not 'free of forces': the forces are balanced. Do not say that weight or drag has disappeared.
A car accelerates uniformly from rest to 20 m/s in 8.0 s, travels at a constant 20 m/s for 12 s, then decelerates uniformly to rest in 5.0 s. What is the average speed of the car for the whole journey?
| Marking point | Mark | Notes |
|---|---|---|
| Distance = ½ × 20 × 8.0 + 20 × 12 + ½ × 20 × 5.0 = 80 + 240 + 50 = 370 m in a total time of 25 s, so the average speed = 370 ÷ 25 = 14.8 m/s. 20 m/s is only the maximum speed. 10 m/s is the average of the starting and maximum speeds and ignores the long constant-speed stage. 18.5 m/s comes from dividing the distance by 20 s, leaving out the final 5.0 s. | ✓ 1 | Answer C |
Answer: C
Exam tip: Average speed = total distance ÷ total time. For a velocity–time graph, find each area separately and add them.
A parachute lowers an aid crate of mass 40 kg towards the ground. Take g = 10 N/kg.
The crate descends at a constant speed of 5.0 m/s. State the size of the resultant force on the crate.
Calculate the upward force from the air on the crate and parachute while it descends at constant speed.
The crate is released from the parachute. At the instant of release the crate is still moving at 5.0 m/s and the air resistance on the crate alone is 60 N. Calculate the acceleration of the crate at this instant.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0 N (zero) | ✓ 1 | |
| Part (b) | ||
| weight = mass × gravitational field strength = 40 × 10 | ✓ 1 | |
| upward force = 400 N (equal to the weight) | ✓ 1 | |
| Part (c) | ||
| resultant force = 400 − 60 = 340 N | ✓ 1 | |
| a = F ÷ m = 340 ÷ 40 | ✓ 1 | |
| 8.5 m/s2 (downwards) | ✓ 1 | |
Common error: The resultant force is the difference between weight and air resistance, not their sum. Always subtract before using F = ma.
A warehouse robot trolley starts from rest and accelerates uniformly to a speed of 1.5 m/s in 6.0 s. It then travels at a constant 1.5 m/s for a further 20 s.
Calculate the acceleration of the trolley during the first 6.0 s.
Calculate the distance travelled in the first 6.0 s.
Calculate the average speed of the trolley over the whole 26 s. Give your answer to 3 significant figures. Use the exact value from (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a = (v − u) ÷ t = 1.5 ÷ 6.0 | ✓ 1 | |
| 0.25 m/s2 | ✓ 1 | |
| Part (b) | ||
| distance = area under velocity–time graph = ½ × 6.0 × 1.5 (or average speed 0.75 × 6.0) | ✓ 1 | |
| 4.5 m | ✓ 1 | |
| Part (c) | ||
| distance in constant-speed stage = 1.5 × 20 = 30 m, so total distance = 34.5 m | ✓ 1 | |
| average speed = total distance ÷ total time = 34.5 ÷ 26 | ✓ 1 | |
| 1.33 m/s | ✓ 1 | |
Exam tip: On a velocity–time graph, gradient gives acceleration and area under the line gives distance. Split trapezium-shaped graphs into triangles and rectangles.
A shop assistant of mass 60 kg stands on the floor of a lift. The lift accelerates upwards at 1.5 m/s2. The gravitational field strength is 10 N/kg. What is the force exerted by the floor on the assistant?
| Marking point | Mark | Notes |
|---|---|---|
| The resultant upward force is ma = 60 × 1.5 = 90 N. The floor force must also support the weight of 600 N, so the floor force is 600 + 90 = 690 N. 600 N is the force only if the lift moves at constant velocity. 90 N is only the resultant force, not the force from the floor. 510 N subtracts the resultant instead of adding it, which would be correct for a downward acceleration. | ✓ 1 | Answer D |
Answer: D
Common error: Do not give the resultant force as the force from the floor. The floor force is the weight plus the extra force needed to accelerate upwards.
A delivery robot of mass 45 kg stands on the floor of a lift. The gravitational field strength is 10 N/kg.
Calculate the weight of the robot.
The lift moves upwards at a constant speed. State the force from the floor on the robot and explain your answer.
The lift now accelerates upwards at 0.80 m/s2. Calculate the resultant force on the robot.
Calculate the force from the floor on the robot while the lift accelerates upwards.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 W = m g | ✓ 1 | |
| A1 45 × 10 = 450 N | ✓ 1 | |
| Part (b) | ||
| B1 450 N | ✓ 1 | |
| B1 acceleration is zero so the resultant force is zero, and the force from the floor equals the weight | ✓ 1 | |
| Part (c) | ||
| M1 F = m a | ✓ 1 | |
| A1 45 × 0.80 = 36 N | ✓ 1 | |
| Part (d) | ||
| M1 force from floor = weight + resultant force = 450 + 36 (ECF from (a) and (c)) | ✓ 1 | |
| A1 486 N | ✓ 1 | |
Common error: The force from the floor is not the resultant force. The resultant is the difference between the force from the floor and the weight.
A fashion designer tests an elastic cord. Weights are added to the cord and its length is measured. The cord has a length of 20.0 cm with no load.
Load / N: 0, 3.0, 6.0, 9.0, 12.0
Length / cm: 20.0, 24.0, 28.0, 32.0, 36.0
Explain how the data show that the cord obeys Hooke's law.
Calculate the spring constant of the cord in N/m.
A load of 21 N is hung on the cord. Assume it is still within its limit of proportionality. Calculate the length of the cord.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 extensions are 4.0, 8.0, 12.0 and 16.0 cm | ✓ 1 | |
| B1 extension doubles when the load doubles, so extension is proportional to load | ✓ 1 | |
| Part (b) | ||
| M1 k = F ÷ x | ✓ 1 | |
| M1 convert extension to metres, e.g. 3.0 N and 0.040 m | ✓ 1 | |
| A1 75 N/m | ✓ 1 | |
| Part (c) | ||
| M1 extension = F ÷ k = 21 ÷ 75 (ECF from (b)) | ✓ 1 | |
| A1 0.28 m or 28 cm | ✓ 1 | |
| A1 length = 20.0 + 28 = 48 cm | ✓ 1 | |
Common error: The extension is the new length minus the original length. Do not use the total length in F = kx.
A spring in a model bridge obeys Hooke's law and has a spring constant of 400 N/m. What is its extension when the force applied to it is 30 N?
| Marking point | Mark | Notes |
|---|---|---|
| Extension = force ÷ spring constant = 30 ÷ 400 = 0.075 m, so 0.075 m is correct. 13 m comes from dividing the spring constant by the force. 12 000 m comes from multiplying the two quantities. 0.75 m is a power-of-ten slip in the division. | ✓ 1 | Answer C |
Answer: C
Common error: Rearranging F = kx the wrong way up gives an answer with the wrong size and the wrong units; check that your answer has the units of length.
A sprinter starts from rest. Her speed is recorded every second during the first part of a race.
Time / s: 0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0
Speed / m/s: 0, 4.0, 7.0, 9.0, 10.0, 10.0, 10.0
State the time from which she runs at constant speed.
Calculate her average acceleration during the first 2.0 s.
Estimate the distance she travels in the first 4.0 s. Assume her speed changes at a steady rate between readings.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.0 s | ✓ 1 | |
| Part (b) | ||
| acceleration = change in speed ÷ time | ✓ 1 | |
| 7.0 ÷ 2.0 = 3.5 m/s2 | ✓ 1 | |
| Part (c) | ||
| distance = area under the speed–time graph, or average speed × time for each second | ✓ 1 | |
| 2.0 + 5.5 + 8.0 + 9.5 | ✓ 1 | |
| 25 m | ✓ 1 | |
Exam tip: Distance on a speed–time graph is the area under the line; splitting the area into trapezia or strips and adding them is a reliable method for estimates.
A trolley loaded with roasted coffee beans has a total mass of 80 kg. A worker pushes it along a level floor with a force of 140 N. Friction between the wheels and floor is 60 N.
Calculate the resultant force on the trolley.
Calculate the acceleration of the trolley.
The worker stops pushing while the trolley is moving. Describe what happens to the motion of the trolley and explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 140 − 60 = 80 N | ✓ 1 | |
| Part (b) | ||
| acceleration = resultant force ÷ mass | ✓ 1 | |
| 80 ÷ 80 = 1.0 m/s2 | ✓ 1 | |
| Part (c) | ||
| the trolley slows down and comes to rest | ✓ 1 | |
| friction now acts alone, opposite to the motion, giving a resultant force backwards | ✓ 1 | |
Exam tip: Always find the resultant force first; the force in F = ma is the unbalanced force, not the push alone.
A mountaineer records several quantities during a climb. Which list contains only vector quantities?
| Marking point | Mark | Notes |
|---|---|---|
| Vectors have both magnitude and direction. Displacement, acceleration and weight (a force) all have direction, so every item in that list is a vector. Distance, speed and energy have magnitude only. Mass is a scalar although velocity and force are vectors. Speed and temperature are scalars although momentum is a vector. | ✓ 1 | Answer B |
Answer: B
Common error: Do not confuse mass (scalar) with weight (vector), or speed (scalar) with velocity (vector).
A telecoms relay drone flies a square patrol path of side 50 m, starting and finishing at the same point. The flight takes 40 s.
Calculate the average speed of the drone.
State the average velocity of the drone and explain your answer.
State which of speed and velocity is a vector.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance ÷ time, 200 ÷ 40 | ✓ 1 | |
| A1 5.0 m/s | ✓ 1 | |
| Part (b) | ||
| 0 m/s | ✓ 1 | |
| The displacement is zero because it returned to its starting point | ✓ 1 | |
| Part (c) | ||
| Velocity | ✓ 1 | |
Common error: Average velocity uses displacement, so a round trip gives zero.
A student hangs masses from a helical spring. A mass of 150 g produces an extension of 3.0 cm. The spring is not stretched beyond its limit of proportionality. The gravitational field strength is 10 N/kg.
What extension would a mass of 400 g produce?
| Marking point | Mark | Notes |
|---|---|---|
| Within the limit of proportionality extension is directly proportional to load, so the extension is 3.0 × 400/150 = 8.0 cm. 5.0 cm is only the extra extension caused by the additional 250 g, not the total. 12 cm comes from multiplying 3.0 cm by 4, treating 400 g as four times the original mass. 1.1 cm comes from inverting the ratio of the masses. | ✓ 1 | Answer B |
Answer: B
Exam tip: Within the limit of proportionality a simple ratio of loads gives the new extension, so you do not need to calculate the spring constant first.
A student investigates a helical spring hung from a clamp. She adds weights to the lower end and measures the length of the spring each time.
Load / N: 0, 2.0, 4.0, 6.0, 8.0, 10.0
Length / cm: 8.0, 10.5, 13.0, 15.5, 18.0, 23.5
Describe how she should measure the length of the spring accurately.
Calculate the extension of the spring for a load of 6.0 N.
Calculate the spring constant of the spring, in N/m, using the data for a load of 4.0 N.
One of the loads stretches the spring beyond its limit of proportionality. Identify this load and give a reason for your choice.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Ruler held vertically next to the spring (and zero of the ruler level with the top of the coils or a fixed mark) | ✓ 1 | |
| Read the scale with the eye level with the lower end of the spring or a pointer, to avoid parallax error | ✓ 1 | |
| Part (b) | ||
| 15.5 − 8.0 = 7.5 cm | ✓ 1 | |
| Part (c) | ||
| M1 extension converted to metres (0.050 m) and k = F/x = 4.0 / 0.050 | ✓ 1 | |
| A1 80 N/m | ✓ 1 | |
| Part (d) | ||
| 10.0 N | ✓ 1 | |
| The increase in extension is much larger than for the other 2.0 N steps (15.5 cm instead of about 12.5 cm), so extension is no longer proportional to load | ✓ 1 | |
Exam tip: Always subtract the original length to find the extension, and convert centimetres to metres before using k = F/x with an answer in N/m.
A cyclist travelling at 8.0 m/s sees a hazard ahead. Her reaction time is 0.70 s. Her braking distance at this speed is 7.2 m.
Calculate the thinking distance.
Calculate the total stopping distance.
State two factors, other than speed, that would increase the braking distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance = speed × time = 8.0 × 0.70 | ✓ 1 | |
| A1 5.6 m | ✓ 1 | |
| Part (b) | ||
| A1 5.6 + 7.2 = 12.8 m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Any two from: wet or icy road; worn tyres; worn or poorly adjusted brakes; larger mass; downhill slope | ✓ 2 | |
Exam tip: Stopping distance is thinking distance plus braking distance; a tired or distracted rider changes the thinking distance, not the braking distance.
A cyclist and bicycle have a combined mass of 80 kg. Starting from rest, the cyclist applies a driving force of 150 N while a constant resistive force of 30 N acts against the motion.
Calculate the resultant force on the cyclist.
Calculate the acceleration of the cyclist.
Calculate the time taken to reach a speed of 9.0 m/s.
Calculate the distance travelled in this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 150 − 30 = 120 N | ✓ 1 | |
| Part (b) | ||
| M1 a = F/m = 120/80 | ✓ 1 | |
| A1 1.5 m/s² (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 t = change in speed / a = 9.0 / 1.5 | ✓ 1 | |
| A1 6.0 s | ✓ 1 | |
| Part (d) | ||
| M1 average speed = (0 + 9.0)/2 = 4.5 m/s, distance = 4.5 × 6.0 | ✓ 1 | |
| A1 27 m | ✓ 1 | |
Common error: Use the resultant force, not the driving force, in F = ma.
A farmer balances a uniform 2.0 m bar on a pivot at its centre. A 30 N sack hangs 0.80 m from the pivot on one side. A second load hangs 0.50 m from the pivot on the other side. What weight must the second load have for the bar to balance?
| Marking point | Mark | Notes |
|---|---|---|
| The bar's own weight acts at the pivot, so it has no moment. Anticlockwise moment = clockwise moment: 30 × 0.80 = W × 0.50, so W = 24 ÷ 0.50 = 48 N. 19 N comes from inverting the distances (30 × 0.50 ÷ 0.80). 24 N is the moment of the sack, not yet divided by the distance of the second load. 30 N would only balance if both distances were equal. | ✓ 1 | Answer D |
Answer: D
Common error: Dividing the moment by the wrong distance. Write moment = force × distance for each side before rearranging.
A tractor moves 120 m in the first 40 s of a journey. It then stops for 20 s. It then moves a further 240 m in the next 60 s. What is the average speed of the tractor over the whole journey?
| Marking point | Mark | Notes |
|---|---|---|
| Average speed = total distance ÷ total time = (120 + 240) ÷ (40 + 20 + 60) = 360 ÷ 120 = 3.0 m/s. 3.5 m/s is the mean of the two moving speeds (3.0 and 4.0 m/s). 2.3 m/s is the mean of the three speeds 3.0, 0 and 4.0 m/s. 3.6 m/s leaves out the 20 s stop from the total time. | ✓ 1 | Answer A |
Answer: A
Common error: Average speed uses total time, including any time spent stationary. Do not average the speeds of the separate sections.
A crop-spraying drone flies in a straight line along the edge of a field. Its distance from the starting point is recorded every 10 s.
Time / s: 0, 10, 20, 30, 40, 50, 60
Distance / m: 0, 40, 80, 80, 80, 140, 200
State how the data show that the drone was stationary for part of the journey.
Calculate the speed of the drone during the first 20 s.
Describe how the line on a distance–time graph for the period 40 s to 60 s would differ from the line for the first 20 s.
Calculate the average speed of the drone over the whole 60 s. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the distance stays at 80 m between 20 s and 40 s (does not change) | ✓ 1 | |
| Part (b) | ||
| speed = distance ÷ time = 80 ÷ 20 | ✓ 1 | |
| 4.0 m/s | ✓ 1 | |
| Part (c) | ||
| both are straight lines sloping upwards (constant speed) | ✓ 1 | |
| the line from 40 s to 60 s is steeper because the speed (6.0 m/s) is greater than 4.0 m/s | ✓ 1 | |
| Part (d) | ||
| average speed = total distance ÷ total time = 200 ÷ 60 | ✓ 1 | |
| 3.3 m/s | ✓ 1 | |
Exam tip: The gradient of a distance–time graph is speed. A horizontal line means the object is stationary.
A farm scale uses a spring to weigh sacks. The spring obeys Hooke's law up to an extension of 0.30 m and has a spring constant of 250 N/m. Use g = 10 N/kg.
Calculate the weight of a sack of seed of mass 6.0 kg.
A farmer hangs a 9.0 kg sack on the spring. Explain why the scale should not be used for this sack.
Calculate the extension of the spring for the 6.0 kg sack, using your answer to (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight = 6.0 × 10 = 60 N | ✓ 1 | |
| Part (b) | ||
| weight = 9.0 × 10 = 90 N and extension would be 90 ÷ 250 = 0.36 m | ✓ 1 | |
| this is beyond the 0.30 m limit, so the spring no longer obeys Hooke's law | ✓ 1 | |
| so the reading would not be reliable (spring may be permanently stretched) | ✓ 1 | |
| Part (c) | ||
| extension = force ÷ spring constant = 60 ÷ 250 (ECF from (a)) | ✓ 1 | |
| 0.24 m | ✓ 1 | |
Exam tip: Use F = kx with extension in metres. Check that the extension is within the spring's Hooke's law range before using the result.
A pharmacist weighs a tablet and records its mass as 250 mg. What is this mass in kilograms?
| Marking point | Mark | Notes |
|---|---|---|
| 1 mg = 10−6 kg, so 250 mg = 250 × 10−6 kg = 2.5 × 10−4 kg. 2.5 × 10−3 kg uses 1 mg = 10−5 kg, a slip of one power of ten. 2.5 × 10−1 kg treats the value as 250 g and converts it to kilograms, ignoring the milli prefix. 2.5 × 10−7 kg uses 1 mg = 10−9 kg, which is far too small a factor. | ✓ 1 | Answer B |
Answer: B
Common error: Write the prefix as a power of ten (milli = 10⁻³) and then convert grams to kilograms in a second step.
In a water-treatment tank, a small particle of grit sinks through still water at a constant speed. Which statement about the forces on the particle is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Constant speed in a straight line means zero acceleration, so by Newton's first law the resultant force is zero: the downward weight is balanced by the upward forces from the water. The statement that weight is greater would give a resultant downward force and an acceleration. Saying there is no upward force confuses 'no resultant force' with 'no forces'. The idea that a resultant force is needed to keep an object moving is the common misconception that force is needed to maintain motion. | ✓ 1 | Answer A |
Answer: A
Common error: Constant velocity never means that no forces act; it means that the forces balance.
The speed of a sprinter is recorded from the start of a race. From t = 0 to t = 4.0 s the speed increases at a constant rate from 0 to 10.0 m/s. From t = 4.0 s to t = 10.0 s the speed stays constant at 10.0 m/s.
Calculate the acceleration of the sprinter during the first 4.0 s.
Explain, in terms of forces, why the speed is constant between 4.0 s and 10.0 s.
Calculate the average speed of the sprinter over the first 10.0 s. Use the area under the speed–time graph to find the total distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 acceleration = change in speed ÷ time | ✓ 1 | |
| A1 10.0 ÷ 4.0 = 2.5 m/s2 | ✓ 1 | |
| Part (b) | ||
| the resultant force on the sprinter is zero | ✓ 1 | |
| the forward (driving) force equals the backward resistive force (air resistance / friction) | ✓ 1 | |
| Part (c) | ||
| M1 distance in first 4.0 s = ½ × 4.0 × 10.0 = 20 m | ✓ 1 | |
| M1 total distance = 20 + 10.0 × 6.0 = 80 m | ✓ 1 | |
| A1 average speed = 80 ÷ 10.0 = 8.0 m/s | ✓ 1 | |
Exam tip: The area under a speed–time graph is the distance travelled; split a graph into triangles and rectangles.
A student investigates the motion of a marble rolling down a straight ramp of length 1.10 m. She times the marble with a stopwatch from release to the bottom of the ramp. Her four timings are 2.1 s, 2.3 s, 2.2 s and 2.2 s.
Calculate the mean time.
Calculate the average speed of the marble. Use the mean time from (a). Give your answer to 2 significant figures.
The student says that 'this is the speed of the marble at the bottom of the ramp'. Explain why she is incorrect.
Suggest why light gates and an electronic timer would give more reliable timings than a stopwatch.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 (2.1 + 2.3 + 2.2 + 2.2) ÷ 4 = 2.2 s | ✓ 1 | |
| Part (b) | ||
| M1 average speed = distance ÷ time = 1.10 ÷ 2.2 | ✓ 1 | |
| A1 0.50 m/s | ✓ 1 | |
| Part (c) | ||
| the value calculated is the average speed over the whole journey | ✓ 1 | |
| the marble speeds up as it rolls down, so its speed at the bottom is greater than the average | ✓ 1 | |
| Part (d) | ||
| a person's reaction time causes a delay when starting and stopping the stopwatch | ✓ 1 | |
| the electronic timer is triggered automatically and measures to a much smaller time interval | ✓ 1 | |
Exam tip: Repeating a measurement and taking a mean reduces the effect of random errors but does not remove an error that is the same every time.
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Simple statements, e.g. 'add masses and measure the spring'. How the extension is found, or how the results are used, is unclear. |
| Level 2 (3–4) | Workable method: masses added in steps, extension found from initial and new pointer readings, with either a valid accuracy/safety point or a correct use of a graph. |
| Level 3 (5–6) | Detailed, logically ordered method that would give valid results: pointer and fixed rule, equal steps, force from W = mg, extension calculated, at least one accuracy measure and one safety precaution, and a correct statement of how a straight line through the origin shows Hooke's law. |
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated points about the method, e.g. adds masses to the trolley or uses light gates, with no clear way of finding acceleration or of keeping the test fair. |
| Level 2 (3–4) | A partly logical method: varies the trolley mass and measures something from which acceleration could be found, with at least one control (e.g. same hanging mass). How results are processed or used may be missing. |
| Level 3 (5–6) | A logical, workable method: mass measured and varied over a suitable range, force kept constant by the same hanging mass, acceleration correctly determined from light-gate data, repeats, and a graph or comparison used to reach a conclusion. |
A physiotherapist tests a resistance band used for exercising an injured shoulder. She hangs weights from the band and measures its extension.
Force / N: 0, 10, 20, 30, 40
Extension / cm: 0, 4.0, 8.0, 12.0, 17.5
State what is meant by elastic behaviour.
Describe how the data show that the band obeys Hooke's law for forces up to 30 N.
Suggest what has happened to the band by the time the force reaches 40 N.
Calculate the extension of the band when a force of 25 N is applied.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The object returns to its original shape/length when the force is removed | ✓ 1 | |
| Part (b) | ||
| Each 10 N increase in force gives the same 4.0 cm increase in extension / extension doubles when force doubles ; so extension is directly proportional to force from 0 to 30 N | ✓ 2 | |
| Part (c) | ||
| The limit of proportionality has been exceeded, because the extension is 17.5 cm rather than the expected 16.0 cm | ✓ 1 | |
| Part (d) | ||
| Extension per newton = 4.0 ÷ 10 = 0.40 cm/N, or 25 ÷ 10 × 4.0 ; extension = 10 cm (10.0 cm) | ✓ 2 | |
Exam tip: To show proportionality from a table, quote equal increases or a doubling pattern using the actual numbers. Saying only 'it goes up' does not show proportionality.
A rider on an electric scooter sets off from rest along a level road. The total mass of the rider and scooter is 95 kg. At one moment the motor provides a forward driving force of 260 N. The total resistive force at that moment is 70 N.
Calculate the resultant force on the rider and scooter, and state its direction.
Calculate the acceleration of the rider and scooter at this moment.
The driving force stays constant. Explain why the acceleration decreases as the scooter's speed increases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 260 − 70 = 190 N ; in the forward direction / direction of motion | ✓ 2 | |
| Part (b) | ||
| F = m × a, or a = F ÷ m ; a = 190 ÷ 95 ; a = 2.0 m/s2 | ✓ 3 | |
| Part (c) | ||
| Air resistance/drag increases as speed increases ; so the resultant force decreases, and acceleration is proportional to resultant force | ✓ 2 | |
Exam tip: Always use the resultant force in F = m × a, not the driving force alone.
A communications satellite moves in a circular orbit around the Earth at a constant speed.
Which row correctly describes the satellite's motion?
| Marking point | Mark | Notes |
|---|---|---|
| Velocity is speed in a given direction. The satellite's direction changes all the time as it moves round its orbit, so its velocity changes even though its speed stays the same. A change in velocity means the satellite is accelerating. Option A treats speed and velocity as the same quantity. Options C and D wrongly say the velocity is constant. | ✓ 1 | Answer B |
Answer: B
Exam tip: Velocity is a vector. A change of direction alone is enough to change it, and so to produce an acceleration.
A high-speed lift in a skyscraper starts from rest.
• Stage 1: it accelerates uniformly to 6.0 m/s in 4.0 s.
• Stage 2: it travels at a constant 6.0 m/s for 20 s.
• Stage 3: it decelerates uniformly to rest in 3.0 s.
Calculate the acceleration of the lift during stage 1.
Calculate the total distance travelled by the lift.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a = 6.0 ÷ 4.0 | ✓ 1 | |
| a = 1.5 m/s2 | ✓ 1 | |
| Part (b) | ||
| Distance in stage 1 or stage 3 found as the area of a triangle, e.g. ½ × 4.0 × 6.0 = 12 m or ½ × 3.0 × 6.0 = 9 m | ✓ 1 | |
| Distance in stage 2 = 6.0 × 20 = 120 m | ✓ 1 | |
| Total distance = 12 + 120 + 9 = 141 m | ✓ 1 | |
Exam tip: Sketch the velocity–time graph roughly for yourself. The total distance is the total area under the line.
A tracker on a migrating goose records the total distance the goose has flown from its starting point.
time (h): 0, 2, 4, 6, 8
distance (km): 0, 90, 180, 180, 300
Describe the motion of the goose between 4 h and 6 h.
Calculate the average speed of the goose over the whole 8 hours. Give your answer in m/s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The goose is stationary (resting / speed is zero) because the distance does not change | ✓ 1 | |
| Part (b) | ||
| Distance = 300 000 m or time = 28 800 s (or average speed = 37.5 km/h) | ✓ 1 | |
| speed = 300 000 ÷ 28 800 (or 37.5 ÷ 3.6) | ✓ 1 | |
| = 10.4 m/s (accept 10 m/s) | ✓ 1 | |
Common error: An average speed must include any time spent stopped. Use the total time of 8 h, not just the time spent flying.
In curling, a player releases a stone across the ice at 2.8 m/s. The stone slides 28 m in a straight line before it stops.
Team-mates sweep the ice in front of the stone. Suggest why this makes the stone travel further.
Calculate the deceleration of the stone. Assume the deceleration is uniform.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Sweeping reduces the friction between the stone and the ice, so the stone decelerates less | ✓ 1 | |
| Part (b) | ||
| v2 = u2 + 2as, so 0 = 2.82 + 2 × a × 28 | ✓ 1 | |
| a = −7.84 ÷ 56 | ✓ 1 | |
| deceleration = 0.14 m/s2 | ✓ 1 | |
Exam tip: A negative value of a means a deceleration. Quote the deceleration as a positive number.
A small rocket launches vertically from rest. It accelerates at 25 m/s2 for 12 s.
State one assumption you need to make to use the equations of uniform motion for this rocket.
Calculate the velocity of the rocket after 12 s.
Calculate the height the rocket reaches in this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The acceleration of the rocket stays constant (uniform) for the whole 12 s | ✓ 1 | |
| Part (b) | ||
| v = u + at = 0 + 25 × 12 | ✓ 1 | |
| v = 300 m/s | ✓ 1 | |
| Part (c) | ||
| Correct substitution into any valid equation, e.g. 3002 = 02 + 2 × 25 × s, or s = ½ × (0 + 300) × 12, or s = ½ × 25 × 122; allow their v from the previous part | ✓ 1 | |
| s = 1800 m | ✓ 1 | |
Common error: Do not use distance = speed × time with the final speed. The rocket is speeding up from rest, so its average speed is only half of 300 m/s.
A dragster accelerates uniformly from rest along a straight 400 m track. It takes 5.0 s to reach the finish line.
Calculate the average speed of the dragster.
For uniform acceleration from rest, the final speed is twice the average speed. State the final speed of the dragster.
Calculate the acceleration of the dragster.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| average speed = distance ÷ time = 400 ÷ 5.0 | ✓ 1 | |
| = 80 m/s | ✓ 1 | |
| Part (b) | ||
| 160 m/s (allow 2 × their answer to part 1) | ✓ 1 | |
| Part (c) | ||
| a = (v − u) ÷ t = (160 − 0) ÷ 5.0, or a = 2s ÷ t2 = 800 ÷ 25 (allow their final speed from part 2) | ✓ 1 | |
| a = 32 m/s2 | ✓ 1 | |
Common error: Acceleration is the change in velocity divided by time. Using the average speed (80 m/s) instead of the final speed gives half the correct acceleration.
An electric scooter and its rider have a total mass of 90 kg. As the scooter moves off, the motor provides a driving force of 210 N. The total resistive force at that moment is 75 N.
State the resultant force on the scooter and rider.
Calculate the acceleration of the scooter and rider.
The driving force stays the same, but the acceleration decreases as the scooter speeds up. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 210 − 75 = 135 N (forwards) | ✓ 1 | |
| Part (b) | ||
| Uses force = mass × acceleration | ✓ 1 | |
| a = 135 ÷ 90 | ✓ 1 | |
| a = 1.5 m/s2 | ✓ 1 | |
| Part (c) | ||
| Air resistance (drag) increases as speed increases | ✓ 1 | |
| so the resultant force decreases, and the acceleration decreases | ✓ 1 | |
Exam tip: Use the resultant force in F = m × a, not the driving force alone.
A car park barrier is a uniform pole of weight 150 N and length 4.0 m. It pivots at a point 0.50 m from one end. A counterweight is fixed to the short side of the pole, 0.40 m from the pivot. The pole is held horizontal.
State what is meant by the centre of gravity of an object.
Calculate the moment of the pole's weight about the pivot.
Calculate the weight of the counterweight needed to balance the pole horizontally.
Suggest why the designer uses a counterweight slightly smaller than this value.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The point through which the whole weight of the object may be considered to act | ✓ 1 | |
| Part (b) | ||
| The weight acts at the midpoint, 2.0 m from the end, so it is 2.0 − 0.50 = 1.5 m from the pivot; moment = 150 × 1.5 | ✓ 1 | |
| = 225 N m | ✓ 1 | |
| Part (c) | ||
| Principle of moments: clockwise moment = anticlockwise moment (or sum of moments about the pivot = 0) | ✓ 1 | |
| W × 0.40 = 225 | ✓ 1 | |
| W = 562.5 N ≈ 560 N | ✓ 1 | |
| Part (d) | ||
| The long side then has the larger moment, so the barrier swings down and rests closed when it is not being lifted (accept: it fails safe, returning to the closed position if the power fails) | ✓ 1 | |
Common error: Measure the distance from the pivot to the centre of the pole, not from the end of the pole.
A delivery drone rises vertically from rest. Its velocity–time graph is a straight line from 0 m/s at time 0 s to 3.0 m/s at time 2.0 s.
How far does the drone rise in the first 2.0 s?
| Marking point | Mark | Notes |
|---|---|---|
| Distance travelled is the area under a velocity–time graph. Here the area is a triangle: ½ × 2.0 s × 3.0 m/s = 3.0 m. 1.5 is the gradient (the acceleration in m/s²), 6.0 m forgets the ½ for a triangle, and 0.67 m divides time by velocity. | ✓ 1 | Answer B |
Answer: B
Exam tip: On a velocity–time graph the gradient gives acceleration and the area gives distance. For uniform acceleration from rest, the area is a triangle, so include the ½.
A student investigates a toy car rolling down a ramp. A card of length 5.0 cm is fixed on top of the car. The card passes through two light gates placed on the ramp. Each light gate records how long the card blocks its beam.
Suggest why light gates give more accurate times than a hand-operated stopwatch in this experiment.
The card blocks the first light gate for 0.025 s.
Calculate the speed of the car as it passes the first light gate.
The card blocks the second light gate for 0.020 s. The car takes 0.40 s to travel from the first light gate to the second.
Calculate the acceleration of the car.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| removes human reaction time (which is large compared with the very short times measured) | ✓ 1 | |
| Part (b) | ||
| 5.0 cm = 0.050 m | ✓ 1 | |
| speed = 0.050 ÷ 0.025 | ✓ 1 | |
| = 2.0 m/s | ✓ 1 | |
| Part (c) | ||
| speed at second gate = 0.050 ÷ 0.020 = 2.5 m/s | ✓ 1 | |
| a = (2.5 − 2.0) ÷ 0.40 | ✓ 1 | |
| = 1.3 m/s2 (accept 1.25 m/s2) | ✓ 1 | |
Common error: The time the card blocks a gate gives the speed at that gate; the time between gates is the time to use in a = (v − u) ÷ t. Do not mix them up.
A sprinter starts from rest. She accelerates uniformly to a speed of 10 m/s in 4.0 s, then runs at a constant 10 m/s for a further 6.0 s.
Calculate her acceleration during the first 4.0 s.
Calculate the total distance she runs in the 10 s.
Calculate her average speed over the 10 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a = (10 − 0) ÷ 4.0 | ✓ 1 | |
| = 2.5 m/s2 | ✓ 1 | |
| Part (b) | ||
| distance while accelerating = ½ × 10 × 4.0 = 20 m | ✓ 1 | |
| distance at constant speed = 10 × 6.0 = 60 m | ✓ 1 | |
| total = 80 m | ✓ 1 | |
| Part (c) | ||
| average speed = 80 ÷ 10 | ✓ 1 | |
| = 8.0 m/s | ✓ 1 | |
Common error: Average speed is total distance ÷ total time. It is not the mean of the starting and top speeds unless the acceleration lasts the whole time.
After a collision, a road-traffic investigator measures skid marks 24 m long left by a car that braked to a stop. Tests show the car decelerated uniformly at 7.5 m/s2 on that road surface.
State the equation linking final speed, initial speed, acceleration and distance moved.
Calculate the speed of the car when the skid began.
The speed limit on the road was 50 km/h.
Use your answer to the previous part to decide whether the car was exceeding the speed limit. Show your working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v2 = u2 + 2as | ✓ 1 | |
| Part (b) | ||
| Substitution: 0 = u2 + 2 × (−7.5) × 24, or u2 = 2 × 7.5 × 24 | ✓ 1 | |
| u2 = 360 (m/s)2 | ✓ 1 | |
| u = 19 m/s | ✓ 1 | |
| Part (c) | ||
| 19 m/s × 3.6 = 68 km/h, or 50 km/h ÷ 3.6 = 14 m/s | ✓ 1 | |
| 68 km/h > 50 km/h (or 19 m/s > 14 m/s), so the car was exceeding the limit | ✓ 1 | |
Exam tip: A deceleration is a negative acceleration in v² = u² + 2as. To convert m/s to km/h, multiply by 3.6.
An astronaut on the Moon drops a hammer from rest at a height of 1.6 m. The acceleration of a falling object on the Moon is 1.6 m/s2. There is no air on the Moon.
Calculate the speed of the hammer just before it hits the ground.
Calculate the time taken for the hammer to fall.
The same hammer is dropped from a height of 1.6 m on Earth. Air resistance on the hammer can be ignored. Explain why it takes less time to fall than on the Moon.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v2 = u2 + 2as = 0 + 2 × 1.6 × 1.6 | ✓ 1 | |
| v2 = 5.12 | ✓ 1 | |
| v = 2.3 m/s | ✓ 1 | |
| Part (b) | ||
| t = (v − u) ÷ a = 2.26 ÷ 1.6, or t = √(2s ÷ a) = √(2 × 1.6 ÷ 1.6) | ✓ 1 | |
| t = 1.4 s | ✓ 1 | |
| Part (c) | ||
| The acceleration of free fall on Earth (about 10 m/s2) is greater, so the hammer covers the 1.6 m in less time | ✓ 1 | |
Exam tip: Find the final speed first with v² = u² + 2as, then use a = (v − u) ÷ t to find the time. Using s = ½at² also works. Keep unrounded values between steps.
A student wants to investigate how the angle of a ramp affects the average speed of a trolley rolling down it. She has a ramp, a trolley, a stopwatch, a metre rule, a protractor and a stack of books.
Describe how the student could carry out this investigation.
You should include how she would make it a fair test.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mark a fixed distance on the ramp (e.g. 1.00 m) using the metre rule | ✓ 1 | |
| set the angle using books and measure it with the protractor | ✓ 1 | |
| release the trolley from rest at the same start mark without pushing | ✓ 1 | |
| time how long the trolley takes to travel the marked distance with the stopwatch | ✓ 1 | |
| repeat each angle and find a mean time, then calculate average speed = distance ÷ time | ✓ 1 | |
| change the angle over a range of at least five values, keeping the same trolley, distance and ramp surface | ✓ 1 | |
Exam tip: For a 6-mark method, cover what you measure, how you measure it, how you calculate the result, how you change the independent variable and what you keep the same.
A survey drone of mass 1.2 kg accelerates vertically upwards at 2.0 m/s2. Air resistance is negligible. Take g = 10 N/kg.
What upward thrust do its rotors provide?
| Marking point | Mark | Notes |
|---|---|---|
| The resultant force needed is F = ma = 1.2 × 2.0 = 2.4 N upwards. The thrust must also support the drone's weight, which is 1.2 × 10 = 12 N. So thrust = 12 + 2.4 = 14.4 N. 2.4 N is the resultant force only. 12 N would only keep the drone hovering or moving at constant speed. 9.6 N subtracts the resultant force instead of adding it. | ✓ 1 | Answer D |
Answer: D
Exam tip: F = ma uses the resultant force. Identify the forces first, then write resultant = thrust − weight.
A city tram is travelling at 15 m/s when the driver sees a cyclist fall onto the track ahead. The driver's reaction time is 0.80 s. Once the brakes are applied, the tram decelerates uniformly at 2.5 m/s2.
State two factors that would increase the driver's thinking distance.
Calculate the thinking distance.
Calculate the total stopping distance of the tram.
Use the equation v2 = u2 + 2as.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Any two of: greater speed / driver tiredness / alcohol or drugs / distractions (e.g. a phone) | ✓ 2 | |
| Part (b) | ||
| distance = speed × time = 15 × 0.80 | ✓ 1 | |
| = 12 m | ✓ 1 | |
| Part (c) | ||
| 0 = 152 − 2 × 2.5 × s, so braking distance s = 225 ÷ 5.0 | ✓ 1 | |
| braking distance = 45 m | ✓ 1 | |
| stopping distance = 12 + 45 = 57 m | ✓ 1 | |
Common error: Stopping distance = thinking distance + braking distance. Giving only the braking distance from the equation loses the final mark.
A nutcracker consists of two arms hinged at one end. A person squeezes the arms with a force of 15 N, applied 14 cm from the hinge. The nut is held 3.5 cm from the hinge. All forces act at right angles to the arms, and the arm is in equilibrium.
State the relationship between the moment of a force, the force and the perpendicular distance from the pivot.
Calculate the moment of the 15 N force about the hinge. Give your answer in N m.
Use the principle of moments to calculate the force exerted on the nut by the arm.
Explain why squeezing the arms closer to the hinge makes it harder to crack the nut.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moment = force × perpendicular distance from the pivot | ✓ 1 | |
| Part (b) | ||
| moment = 15 × 0.14 | ✓ 1 | |
| = 2.1 N m | ✓ 1 | |
| Part (c) | ||
| F × 0.035 = 2.1, or F × 3.5 = 15 × 14 | ✓ 1 | |
| F = 60 N | ✓ 1 | |
| Part (d) | ||
| Smaller perpendicular distance gives a smaller moment for the same squeezing force, so the force on the nut is smaller (or a larger squeezing force is needed to crack it) | ✓ 1 | |
Common error: A moment calculated in N cm is not the same number as one in N m. Convert 14 cm to 0.14 m when the answer is asked for in N m.
A horizontal beam in an architect's model is pivoted at its centre. A downward force of 300 N acts 1.5 m to the left of the pivot. A downward force F acts 2.5 m to the right of the pivot. The beam is balanced. What is F?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: balance needs 300 × 1.5 = F × 2.5, so F = 180 N. 300 N assumes equal forces balance regardless of distance. 500 N inverts the ratio of distances. 450 N is the left moment without dividing by 2.5. | ✓ 1 | Answer A |
Answer: A
Exam tip: Moments balance when clockwise moment equals anticlockwise moment, each taken about the same pivot.
In a warehouse, a 200 kg cart moving at 1.5 m/s collides with a stationary 100 kg cart. The carts couple together and move off. What is their speed immediately afterwards?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: momentum is conserved, 200 × 1.5 = 300 kg m/s = (300 kg) × v, so v = 1.0 m/s. 1.5 m/s forgets the extra mass. 3.0 m/s divides by only the stationary cart's mass. 0.75 m/s averages 1.5 and 0. | ✓ 1 | Answer A |
Answer: A
Common error: After a coupling collision, divide the total momentum by the combined mass.
A student plans to find how acceleration depends on the mass of a trolley when the pulling force is constant.
Describe a method the student could use. Include how the acceleration is measured and how the force is kept constant.
With mass 0.30 kg the trolley accelerates at 2.0 m/s2. Calculate the acceleration when the total mass is 0.80 kg and the force is unchanged.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| trolley pulled along a level (friction-compensated) track by a string over a pulley with a hanging weight | ✓ 1 | |
| the hanging weight (pulling force) kept the same for every run | ✓ 1 | |
| masses added to the trolley to change the total mass, with the hanging weight unchanged | ✓ 1 | |
| two light gates (or a ticker timer / video) measure speeds or times along the track | ✓ 1 | |
| acceleration calculated from change in speed ÷ time between the gates | ✓ 1 | |
| repeat each mass and take a mean, then compare acceleration for different masses | ✓ 1 | |
| Part (b) | ||
| F = 0.30 × 2.0 = 0.60 N | ✓ 1 | |
| a = 0.60 ÷ 0.80 = 0.75 m/s2 | ✓ 1 | |
Common error: Hanging masses taken from the trolley and put on the hanger keep the total mass constant, which is a different investigation.
In a distribution centre a parcel of mass 4.0 kg lands on the floor at 5.0 m/s and stops without bouncing. On a hard floor the parcel stops in 0.020 s. On a foam mat it stops in 0.10 s.
Calculate the change in momentum of the parcel.
Calculate the average force on the parcel from the hard floor.
Explain why the foam mat reduces the chance of damage to the parcel.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum before = 4.0 × 5.0 | ✓ 1 | |
| 20 kg m/s | ✓ 1 | |
| Part (b) | ||
| force = change in momentum ÷ time (20 ÷ 0.020) | ✓ 1 | |
| 1000 N | ✓ 1 | |
| Part (c) | ||
| the change in momentum is the same on both surfaces | ✓ 1 | |
| the time to stop is longer on foam | ✓ 1 | |
| force = change in momentum ÷ time, so the force on the parcel is smaller | ✓ 1 | |
Exam tip: Explanations about cushioning should link longer stopping time to smaller force for the same change in momentum.
A pool hoist has a horizontal arm that is pivoted at P. A chair carrying a swimmer, total weight 600 N, hangs 2.0 m to the left of P. A counterweight hangs 0.50 m to the right of P and the arm is balanced. Ignore the weight of the arm. What is the weight of the counterweight?
| Marking point | Mark | Notes |
|---|---|---|
| Balanced, so clockwise moment = anticlockwise moment: 600 × 2.0 = W × 0.50, giving W = 2400 N. 150 N comes from inverting the ratio (600 × 0.50 / 2.0). 600 N assumes equal forces balance whatever the distances. 480 N uses the total distance 2.5 m instead of 0.50 m. | ✓ 1 | Answer A |
Answer: A
Common error: The smaller the distance from the pivot, the larger the force needed to give the same moment.
A maintenance drone climbs beside a wind turbine tower. Its speed increases uniformly from 0 to 1.6 m/s in 8.0 s. It then climbs at a constant 1.6 m/s for a further 12 s. What is its average speed over the whole 20 s?
| Marking point | Mark | Notes |
|---|---|---|
| Distance = area under the speed–time graph = ½ × 8.0 × 1.6 + 12 × 1.6 = 6.4 + 19.2 = 25.6 m. Average speed = 25.6/20 = 1.28 m/s. 0.80 m/s is the average over the first 8.0 s only. 1.20 m/s averages the two stage averages (0.80 and 1.60) as if they lasted equally long. 1.60 m/s is the maximum speed. | ✓ 1 | Answer D |
Answer: D
Common error: Average speed is total distance divided by total time. Do not average the averages of stages that last different times.
The jib of a tower crane is a horizontal beam pivoted at the top of the tower. A concrete counterweight of weight 30 000 N hangs 8.0 m from the tower on one side. A load hangs from a trolley on the other side. Ignore the weight of the jib.
State the principle of moments.
The load hangs 12 m from the tower and the jib is exactly balanced. Calculate the weight of the load.
The gravitational field strength is 10 N/kg. Calculate the mass of the load.
The trolley is moved so that the same load now hangs 15 m from the tower. Explain what happens to the jib.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| When an object is in equilibrium (balanced) | ✓ 1 | |
| the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about the same point | ✓ 1 | |
| Part (b) | ||
| moment of counterweight = 30 000 × 8.0 = 240 000 N m | ✓ 1 | |
| 240 000 = W × 12 | ✓ 1 | |
| W = 20 000 N | ✓ 1 | |
| Part (c) | ||
| mass = 20 000 / 10 = 2000 kg (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| moment of the load is 20 000 × 15 = 300 000 N m, which is greater than the counterweight moment of 240 000 N m | ✓ 1 | |
| so the jib turns (tips) down on the load side | ✓ 1 | |
Common error: Moments must be taken about the same pivot, and the distance is the perpendicular distance from the pivot to the line of the force.
A student investigates how the mass of a trolley affects its acceleration. Each time, the resultant force on the trolley is 1.2 N. The results are shown.
Mass of trolley / kg: 0.60, 1.20, 2.40, 3.00
Acceleration / m/s2: 2.00, 1.00, 0.50, 0.40
State one variable that the student must keep constant during the investigation.
Describe the relationship between the mass of the trolley and its acceleration shown by the results.
Use the equation F = ma to predict the acceleration of a 4.0 kg trolley acted on by the same resultant force.
Describe how the student could use two light gates and a data logger to measure the acceleration of the trolley.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the resultant force on the trolley | ✓ 1 | |
| Part (b) | ||
| as the mass increases the acceleration decreases | ✓ 1 | |
| the acceleration is inversely proportional to the mass (the product of mass and acceleration is constant at 1.2) | ✓ 1 | |
| Part (c) | ||
| a = F / m | ✓ 1 | |
| 1.2 / 4.0 = 0.30 m/s2 | ✓ 1 | |
| Part (d) | ||
| The data logger records the speed of the trolley at each light gate | ✓ 1 | |
| and the time taken to travel between the gates | ✓ 1 | |
| acceleration = (final speed − initial speed) / time taken | ✓ 1 | |
Exam tip: In a practical about two variables, name the one you change, the one you measure and the ones you keep the same.
A swimmer swims one length of a 50 m pool. Her speed increases uniformly from rest to 2.0 m/s in 5.0 s, stays constant at 2.0 m/s for 20 s, and then decreases uniformly to rest in 5.0 s as she reaches the wall.
Calculate her acceleration during the first 5.0 s.
Show that the distance she swims is 50 m.
Calculate her average speed for the whole length. Give your answer to 3 significant figures.
Explain why her average speed is less than 2.0 m/s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| acceleration = change in speed / time | ✓ 1 | |
| 2.0 / 5.0 = 0.40 m/s2 | ✓ 1 | |
| Part (b) | ||
| distance = area under the speed–time graph | ✓ 1 | |
| first stage ½ × 2.0 × 5.0 = 5.0 m and last stage 5.0 m | ✓ 1 | |
| constant stage 2.0 × 20 = 40 m, so total 5.0 + 40 + 5.0 = 50 m | ✓ 1 | |
| Part (c) | ||
| average speed = total distance / total time | ✓ 1 | |
| total time = 30 s, so 50 / 30 = 1.67 m/s | ✓ 1 | |
| Part (d) | ||
| She spends part of the time speeding up and slowing down, when her speed is below 2.0 m/s | ✓ 1 | |
Exam tip: For a speed–time graph made of straight lines, split the area into triangles and rectangles.
A van drives along an icy mountain road to a wind farm at 20 m/s. The driver's reaction time is 0.60 s. When the brakes are applied the van decelerates uniformly at 5.0 m/s2.
Calculate the thinking distance.
Use the equation v2 = u2 + 2as to calculate the braking distance.
Calculate the total stopping distance.
Explain why the braking distance is longer on an icy road than on a dry road.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distance = speed × time | ✓ 1 | |
| 20 × 0.60 = 12 m | ✓ 1 | |
| Part (b) | ||
| 0 = 202 + 2 × (−5.0) × s | ✓ 1 | |
| 10 s = 400 | ✓ 1 | |
| s = 40 m | ✓ 1 | |
| Part (c) | ||
| 12 + 40 = 52 m (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| There is less friction between the tyres and the road, so the braking force (deceleration) is smaller | ✓ 1 | |
| so a greater distance is needed to stop from the same speed | ✓ 1 | |
Common error: Thinking distance depends on reaction time; braking distance depends on speed and the braking force. Icy roads change only the second.
A spring is loaded with forces of 1.0 N, 2.0 N, 3.0 N, 4.0 N and 5.0 N. The extensions are 2.0 cm, 4.0 cm, 6.0 cm, 8.0 cm and 11.0 cm. The 5.0 N point is beyond the limit of proportionality. What is the spring constant of the spring?
| Marking point | Mark | Notes |
|---|---|---|
| Use a point in the proportional region: k = F/x = 1.0 N / 0.020 m = 50 N/m. B uses the extension in cm, giving N/cm, and labels it N/m. C uses the 5.0 N point, which is beyond the limit of proportionality. D inverts the ratio (x/F). | ✓ 1 | Answer A |
Answer: A
Exam tip: Convert extension to metres before using F = kx if the spring constant is needed in N/m.
A skydiver of mass 80 kg has been falling at terminal velocity. She then opens her parachute. At the instant the parachute opens, the air resistance on her is 2400 N. Use g = 10 N/kg.
State the size of the air resistance on the skydiver just before the parachute opens.
Calculate the deceleration of the skydiver at the instant the parachute opens.
Describe and explain how her speed changes after this instant until she reaches a new steady speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 800 N (equal to her weight) | ✓ 1 | |
| Part (b) | ||
| resultant force = 2400 − 800 = 1600 N | ✓ 1 | |
| a = F ÷ m = 1600 ÷ 80 = 20 m/s2 | ✓ 1 | |
| Part (c) | ||
| Her speed decreases | ✓ 1 | |
| As she slows down, the air resistance decreases | ✓ 1 | |
| Until air resistance equals her weight, the resultant force is zero and she falls at a new, lower, constant (terminal) speed | ✓ 1 | |
Common error: The air resistance at the instant the parachute opens is larger than the weight, so the resultant force is upwards even though she is still moving downwards.
A student tests a spring from a farm feed dispenser. The unstretched length of the spring is 20.0 cm. Different forces are applied and the length is measured.
Force (N): 2.0, 4.0, 6.0, 8.0, 10.0
Length of spring (cm): 22.5, 25.0, 27.5, 30.0, 34.0
Calculate the extension of the spring when the force is 6.0 N.
State between which two forces the limit of proportionality lies and give a reason for your answer.
Calculate the spring constant of the spring, in N/m. Use data from the region where the spring obeys Hooke's law.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 27.5 − 20.0 = 7.5 cm | ✓ 1 | |
| Part (b) | ||
| Between 8.0 N and 10.0 N | ✓ 1 | |
| Extension is proportional to force up to 8.0 N (10.0 cm), but at 10.0 N the extension is 14.0 cm, more than the 12.5 cm expected | ✓ 1 | |
| Part (c) | ||
| M1 k = F ÷ x with extension in metres, e.g. 4.0 ÷ 0.050 (any point from 2.0 N to 8.0 N) | ✓ 1 | |
| A1 k = 80 N/m | ✓ 1 | |
Exam tip: Use a point from the straight-line region of the data when finding the spring constant, and convert cm to m first.
A hailstone of mass 4.0 g falls from a cloud above an orchard. Use g = 10 N/kg.
The hailstone falls at terminal velocity. Calculate the air resistance on it.
At one moment during its fall the air resistance on the hailstone is 0.028 N. Calculate its acceleration at that moment.
A larger hailstone has a greater terminal velocity than a smaller one. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight = mg = 0.0040 × 10 = 0.040 N | ✓ 1 | |
| at terminal velocity air resistance equals weight, 0.040 N | ✓ 1 | |
| Part (b) | ||
| resultant force = 0.040 − 0.028 = 0.012 N | ✓ 1 | |
| a = F ÷ m = 0.012 ÷ 0.0040 = 3.0 m/s2 | ✓ 1 | |
| Part (c) | ||
| The larger hailstone has a greater weight, so a greater air resistance is needed to balance it | ✓ 1 | |
| Air resistance increases with speed, so a higher speed is needed before the forces balance | ✓ 1 | |
Common error: Convert grams to kilograms before using weight = mass × g or F = ma.
A student hangs a crate of fruit from a spring and investigates how the extension depends on the force. The force–extension graph for the spring is a straight line through the origin up to a force of 3.2 N. The point (0.050 m, 2.0 N) lies on this line. Use g = 10 N/kg.
Describe the shape of the force–extension graph from zero force to a force greater than 3.2 N.
State how the spring constant is found from the straight part of the graph.
The crate produces an extension of 0.070 m. Calculate the mass of the crate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A straight line through the origin up to 3.2 N | ✓ 1 | |
| Beyond 3.2 N the line curves, so extension increases more for each extra newton (the gradient decreases) | ✓ 1 | |
| Part (b) | ||
| Gradient of the straight-line section (force ÷ extension) | ✓ 1 | |
| Part (c) | ||
| M1 k = 2.0 ÷ 0.050 = 40 N/m | ✓ 1 | |
| M1 F = kx = 40 × 0.070 = 2.8 N | ✓ 1 | |
| A1 mass = 2.8 ÷ 10 = 0.28 kg | ✓ 1 | |
Exam tip: If a question gives one point on the straight section, use it to find the gradient, then check the force is below the limit of proportionality.
A gymnast lands on a thick crash mat and comes to rest. Which statement best explains why the mat reduces the risk of injury compared with landing on a hard floor?
| Marking point | Mark | Notes |
|---|---|---|
| The gymnast's momentum change is the same on either surface (from her landing momentum to zero). The mat makes the stopping time longer, so force = change in momentum / time is smaller, which is B. A is wrong because the momentum before landing is set by her fall, not the mat. C is wrong because the change in momentum is unchanged. D reverses the effect of the mat on the time. | ✓ 1 | Answer B |
Answer: B
Common error: The mat does not change the momentum change; it changes how long the change takes.
A student hangs weights from a rubber band whose unstretched length is 10.0 cm and measures its length each time.
Force / N: 1.0, 2.0, 3.0, 4.0, 5.0
Length / cm: 12.0, 14.0, 16.0, 19.0, 24.0
Calculate the extension of the band when the force is 4.0 N.
The student says the band obeys Hooke's law for forces up to 3.0 N. Use the data to justify this.
Calculate the spring constant of the band, in N/m, using the 3.0 N reading.
State one way to reduce the error in reading the length of the band.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 9.0 cm | ✓ 1 | |
| Part (b) | ||
| extensions of 2.0, 4.0 and 6.0 cm for 1.0, 2.0 and 3.0 N | ✓ 1 | |
| extension is proportional to force (the ratio extension / force is constant at 2.0 cm per N) | ✓ 1 | |
| Part (c) | ||
| convert 6.0 cm to 0.060 m | ✓ 1 | |
| M1 k = F / x | ✓ 1 | |
| A1 50 N/m | ✓ 1 | |
| Part (d) | ||
| any one of: view the ruler at eye level; use a set square or pointer against the scale; use a fixed vertical ruler | ✓ 1 | |
Common error: Use the extension (length minus unstretched length), not the length itself, when testing proportionality.
A swimmer pushes off from the end wall of a 25 m pool at a speed of 3.0 m/s. She glides and slows uniformly to 1.0 m/s in 5.0 s, then swims the rest of the length at a constant 1.0 m/s.
Calculate her deceleration during the glide.
Calculate the distance she travels during the glide.
Calculate the time she takes to swim the rest of the length.
Calculate her average speed for the whole length.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 acceleration = change in speed / time (2.0 / 5.0) | ✓ 1 | |
| A1 0.40 m/s² (accept −0.40 m/s²) | ✓ 1 | |
| Part (b) | ||
| M1 distance = average speed × time = 2.0 × 5.0 | ✓ 1 | |
| A1 10 m | ✓ 1 | |
| Part (c) | ||
| remaining distance = 25 − 10 = 15 m (ECF from (b)) | ✓ 1 | |
| time = 15 / 1.0 = 15 s | ✓ 1 | |
| Part (d) | ||
| M1 average speed = total distance / total time (25 / 20) | ✓ 1 | |
| A1 1.25 m/s | ✓ 1 | |
Exam tip: Average speed over several stages is total distance divided by total time, never the mean of the speeds.
A service vehicle of mass 1200 kg accelerates uniformly from rest along a wind farm track at 1.25 m/s². A resistive force of 600 N acts against its motion.
Calculate the resultant force on the vehicle.
Calculate the driving force provided by the engine.
Calculate the speed of the vehicle 8.0 s after starting from rest.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 F = ma (1200 × 1.25) | ✓ 1 | |
| A1 1500 N | ✓ 1 | |
| Part (b) | ||
| M1 driving force = resultant force + resistive force | ✓ 1 | |
| A1 2100 N | ✓ 1 | |
| Part (c) | ||
| M1 v = at (or v = u + at) | ✓ 1 | |
| A1 10 m/s | ✓ 1 | |
Common error: The driving force is not the resultant force: the resistive force must be added back on.
A spring-loaded stopper on a library trolley has a spring constant of 300 N/m.
State what is meant by the limit of proportionality of a spring.
The spring is stretched by a force of 18 N, within its limit of proportionality. Calculate the extension in cm.
A second spring has a spring constant of 450 N/m. Calculate the force needed to give it the same extension as the first spring.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the point beyond which the extension is no longer proportional to the force applied (Hooke's law no longer obeyed) | ✓ 1 | |
| Part (b) | ||
| M1 extension = F / k (18 / 300) | ✓ 1 | |
| A1 0.060 m | ✓ 1 | |
| A1 6.0 cm | ✓ 1 | |
| Part (c) | ||
| M1 F = kx = 450 × 0.060 (ECF from (b)) | ✓ 1 | |
| A1 27 N | ✓ 1 | |
Exam tip: Convert extensions to metres before using F = kx with a spring constant in N/m.
A van travels at 20 m/s along the access road to a wind farm. The driver's normal reaction time is 0.70 s, and the van's braking distance at this speed is 38 m.
State what is meant by thinking distance.
Calculate the thinking distance at 20 m/s.
The driver is tired and her reaction time increases to 1.1 s. The speed and the braking distance are unchanged. Calculate the new stopping distance.
State two factors, other than tiredness, that would increase the stopping distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the distance travelled between the driver seeing a hazard and applying the brakes (during the reaction time) | ✓ 1 | |
| Part (b) | ||
| M1 distance = speed × time (20 × 0.70) | ✓ 1 | |
| A1 14 m | ✓ 1 | |
| Part (c) | ||
| thinking distance = 20 × 1.1 = 22 m | ✓ 1 | |
| stopping distance = thinking distance + braking distance | ✓ 1 | |
| 60 m | ✓ 1 | |
| Part (d) | ||
| any one valid factor, e.g. wet or icy road, worn tyres, worn brakes, greater mass of vehicle, higher speed | ✓ 1 | |
| any second different valid factor | ✓ 1 | |
Exam tip: Stopping distance = thinking distance + braking distance; tiredness changes only the first of these.
A drone moves along a straight line. Its distance–time graph has three straight sections. In the first 8.0 s the distance from the start rises from 0 to 36 m. From 8.0 s to 20 s the distance stays at 36 m. From 20 s to 30 s the distance falls steadily from 36 m back to 0 m. What is the speed of the drone in the final section?
| Marking point | Mark | Notes |
|---|---|---|
| Speed is the gradient magnitude: 36 m ÷ 10 s = 3.6 m/s, which is C. A is the speed in the middle section (a horizontal line), not the final one. B is 36 ÷ 30, using the whole journey time. D is 36 ÷ 8.0, the speed in the first section. | ✓ 1 | Answer C |
Answer: C
Exam tip: Read the time interval of the section you need from the graph and divide the change in distance by that interval only.
A skydiver of mass 80 kg jumps from an aircraft. Use g = 10 N/kg.
Calculate the weight of the skydiver.
Describe how the forces on the skydiver cause the speed to change until terminal velocity is reached.
At one moment during the fall the air resistance is 560 N. Calculate the acceleration of the skydiver at this moment.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight = mass × g = 80 × 10 | ✓ 1 | |
| weight = 800 N | ✓ 1 | |
| Part (b) | ||
| as speed increases the air resistance (drag) increases | ✓ 1 | |
| so the resultant force (and acceleration) decreases | ✓ 1 | |
| at terminal velocity air resistance equals weight, the resultant force is zero and the speed is constant | ✓ 1 | |
| Part (c) | ||
| resultant force = 800 − 560 = 240 N | ✓ 1 | |
| a = F ÷ m = 240 ÷ 80 | ✓ 1 | |
| a = 3.0 m/s² | ✓ 1 | |
Common error: Use the resultant force, not the weight, when calculating acceleration with F = ma.
A farmer pushes a loaded wheelbarrow of total mass 60 kg along a flat path with a constant horizontal force of 150 N. The wheelbarrow speeds up. The horizontal resistive force on it is 90 N.
State two effects that a force can have on an object.
Name the type of force that mainly opposes the motion of the wheelbarrow along the path.
Calculate the resultant force on the wheelbarrow.
Calculate the acceleration of the wheelbarrow.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| changes its speed / starts or stops its motion | ✓ 1 | |
| changes its direction of motion or its shape (accept any two different valid effects) | ✓ 1 | |
| Part (b) | ||
| friction (accept rolling resistance) | ✓ 1 | |
| Part (c) | ||
| 150 − 90 = 60 N | ✓ 1 | |
| Part (d) | ||
| a = F ÷ m = 60 ÷ 60 | ✓ 1 | |
| a = 1.0 m/s² (ECF from (c)) | ✓ 1 | |
Exam tip: Subtract opposing forces along the same line to find the resultant before using F = ma.
A marching band walks along a straight parade route. It walks at a constant speed of 1.5 m/s for 360 s, stops for 90 s, then walks at a constant speed of 2.0 m/s for 150 s.
Calculate the distance the band walks in the first 360 s.
State the speed of the band between 360 s and 450 s, and describe how this part would look on a distance–time graph.
Calculate the average speed of the band over the whole 600 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distance = speed × time = 1.5 × 360 | ✓ 1 | |
| distance = 540 m | ✓ 1 | |
| Part (b) | ||
| speed = 0 m/s | ✓ 1 | |
| a horizontal (flat) line | ✓ 1 | |
| Part (c) | ||
| distance in last section = 2.0 × 150 = 300 m, so total distance = 540 + 300 = 840 m | ✓ 1 | |
| total time = 360 + 90 + 150 = 600 s | ✓ 1 | |
| average speed = 840 ÷ 600 = 1.4 m/s | ✓ 1 | |
Common error: Average speed is not the mean of the section speeds; divide the total distance by the total time.
A hailstone of mass 2.0 g falls through the air above a field of crops. Use g = 10 N/kg.
Calculate the weight of the hailstone.
The hailstone reaches terminal velocity. State the size of the air resistance on it and give a reason for your answer.
Earlier in the fall, the speed is 8 m/s and the air resistance is 0.012 N. Calculate the acceleration of the hailstone at this speed.
A larger hailstone of the same shape and mass 8.0 g has a higher terminal velocity. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mass = 0.0020 kg | ✓ 1 | |
| weight = 0.0020 × 10 = 0.020 N | ✓ 1 | |
| Part (b) | ||
| 0.020 N (accept equal to the weight) | ✓ 1 | |
| the forces are balanced / the resultant force is zero so the speed is constant | ✓ 1 | |
| Part (c) | ||
| resultant force = 0.020 − 0.012 = 0.008 N | ✓ 1 | |
| a = F ÷ m = 0.008 ÷ 0.0020 | ✓ 1 | |
| a = 4.0 m/s² | ✓ 1 | |
| Part (d) | ||
| the larger hailstone has a greater weight, so a greater air resistance is needed to balance it | ✓ 1 | |
| air resistance increases with speed, so it must fall faster before the forces balance | ✓ 1 | |
Common error: Convert grams to kilograms before using weight = mass × g or F = ma.
A helical spring has a natural length of 20 cm and obeys Hooke's law. A load of 4.0 N makes it 6.0 cm longer. What is the length of the spring with a load of 10 N, assuming the limit of proportionality is not exceeded?
| Marking point | Mark | Notes |
|---|---|---|
| Extension is proportional to force, so the extension for 10 N is 6.0 × 10/4.0 = 15 cm. The length is the natural length plus the extension: 20 + 15 = 35 cm. 15 cm is the extension only and leaves out the natural length. 26 cm adds the extension for 4.0 N, not 10 N. 50 cm wrongly scales the whole length by 10/4.0. | ✓ 1 | Answer A |
Answer: A
Exam tip: Hooke's law links force to extension, not to total length, so always subtract or add the natural length.
A swimmer of mass 45 kg stands at rest on a floating raft of mass 30 kg in still water. She dives horizontally off the raft at 2.0 m/s. Ignoring water resistance, what is the velocity of the raft immediately afterwards?
| Marking point | Mark | Notes |
|---|---|---|
| Total momentum is zero before and must be zero after. The swimmer has momentum 45 × 2.0 = 90 kg m/s forwards, so the raft has 90 kg m/s backwards: v = 90 / 30 = 3.0 m/s in the opposite direction. 2.0 m/s ignores the different masses. The same direction would give a total momentum that is not zero. 1.2 m/s divides by the combined mass of 75 kg, which is wrong because the raft alone moves backwards. | ✓ 1 | Answer D |
Answer: D
Common error: Momentum is a vector, so the two bodies moving apart must have momenta of opposite sign that add to zero.
A student hangs different loads from a spring and records its length. The original length of the spring is 12.0 cm.
| Load / N | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 |
| Length / cm | 12.0 | 14.0 | 16.0 | 18.0 | 20.0 | 22.0 | 26.0 |
Calculate the extension of the spring when the load is 4.0 N.
State the evidence from the data that the limit of proportionality has been exceeded by the time the load reaches 6.0 N.
Use the data for a load of 5.0 N to calculate the spring constant of the spring in N/m.
A load of 3.5 N is hung on the spring. Calculate the length of the spring.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 20.0 − 12.0 = 8.0 cm | ✓ 1 | |
| Part (b) | ||
| up to 5.0 N the extension increases by 2.0 cm for each 1.0 N | ✓ 1 | |
| from 5.0 N to 6.0 N it increases by 4.0 cm, more than 2.0 cm, so extension is no longer proportional to load | ✓ 1 | |
| Part (c) | ||
| M1 F = kx with extension converted to metres: 5.0 = k × 0.100 | ✓ 1 | |
| A1 k = 50 N/m | ✓ 1 | |
| Part (d) | ||
| M1 extension = 3.5 / 50 = 0.070 m = 7.0 cm | ✓ 1 | |
| A1 length = 12.0 + 7.0 = 19.0 cm (ECF from (c)) | ✓ 1 | |
Exam tip: Convert centimetres to metres before using F = kx with a spring constant in N/m.
A van carrying equipment to a wind farm is travelling at 20 m/s along a straight road. The driver sees a hazard, and after a reaction time of 0.65 s applies the brakes. The braking distance of the van is 40 m.
Calculate the thinking distance.
Calculate the total stopping distance.
State two factors that could increase the thinking distance.
Explain why the braking distance of the van is greater on a wet road.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance = speed × time = 20 × 0.65 | ✓ 1 | |
| A1 13 m | ✓ 1 | |
| Part (b) | ||
| 13 + 40 = 53 m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| one valid factor, e.g. tiredness / alcohol or drugs / distraction | ✓ 1 | |
| a second different valid factor, e.g. higher speed of the van | ✓ 1 | |
| Part (d) | ||
| less friction between the tyres and road, so the maximum braking force is smaller | ✓ 1 | |
| so the deceleration is smaller and the van takes longer to stop / travels further | ✓ 1 | |
Exam tip: Stopping distance is thinking distance plus braking distance, and only the first depends on reaction time.
A railway wagon of mass 12 000 kg moving at 2.0 m/s collides with a stationary wagon of mass 8000 kg. The wagons couple together and move off as one. What is their speed immediately after the collision?
| Marking point | Mark | Notes |
|---|---|---|
| Momentum is conserved: 12 000 × 2.0 = (12 000 + 8000) × v, so v = 24 000 ÷ 20 000 = 1.2 m/s. 2.0 m/s is wrong because it ignores the extra mass that must now be moved. 1.0 m/s is wrong because it takes the mean of the two speeds, 2.0 and 0. 0.80 m/s is wrong because it uses the momentum of the lighter wagon only in the calculation (8000 × 2.0 ÷ 20 000). | ✓ 1 | Answer D |
Answer: D
Common error: After coupling, use the total mass of both wagons when dividing the total momentum to find the common speed.
The load on the jib of a tower crane is moved along the jib towards the tower. The load is not lifted or lowered. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| The moment of a force about a pivot is force × perpendicular distance from the pivot. The weight of the load is unchanged but the distance to the tower is smaller, so the moment decreases. The first option is wrong because moving the load sideways does not change its weight. The second is wrong because the moment depends on distance as well as force. The fourth is wrong because the length of the jib is not what matters; the distance of the load from the pivot is. | ✓ 1 | Answer C |
Answer: C
Exam tip: Moment = force × perpendicular distance from the pivot; change either factor and the moment changes.
A footballer kicks a stationary ball of mass 0.45 kg. The ball leaves the boot at 20 m/s and the boot is in contact with the ball for 0.015 s. Assume the force is constant while they are in contact.
Calculate the change in momentum of the ball.
Calculate the average force exerted on the ball by the boot.
A softer boot increases the contact time to 0.025 s, but the ball again leaves at 20 m/s. Calculate the new average force on the ball.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| change in momentum = mass × change in velocity = 0.45 × 20 | ✓ 1 | |
| 9.0 kg m/s | ✓ 1 | |
| Part (b) | ||
| force = change in momentum ÷ time taken = 9.0 ÷ 0.015 | ✓ 1 | |
| 600 N | ✓ 1 | |
| Part (c) | ||
| force = 9.0 ÷ 0.025 | ✓ 1 | |
| 360 N | ✓ 1 | |
Exam tip: Use F = Δp ÷ Δt: for the same change in momentum, a longer contact time means a smaller average force.
A sports scientist throws a ball vertically downwards from a platform 20 m above the ground. The ball hits the ground at 25 m/s. Ignore air resistance and take the acceleration of free fall as 10 m/s2.
Calculate the speed at which the ball was thrown.
Calculate the time the ball takes to reach the ground.
Calculate the average speed of the ball during its fall.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| substitution into v2 = u2 + 2as: 252 = u2 + 2 × 10 × 20 | ✓ 1 | |
| u2 = 625 − 400 = 225 | ✓ 1 | |
| u = 15 m/s | ✓ 1 | |
| Part (b) | ||
| use of t = (v − u) ÷ a, or equivalent | ✓ 1 | |
| t = (25 − 15) ÷ 10 = 1.0 s | ✓ 1 | |
| Part (c) | ||
| average speed = distance ÷ time = 20 ÷ 1.0 (ECF from (b)) | ✓ 1 | |
| 20 m/s | ✓ 1 | |
Common error: In v² = u² + 2as, square the speeds before subtracting; do not take 25 − 20 − 15 or square the difference.
A tower crane has a horizontal jib. A counterweight of weight 45 kN is fixed 8.0 m from the tower on one side. A load hangs from a trolley on the other side of the tower. Ignore the weight of the jib.
Calculate the moment of the counterweight about the tower.
The crane lifts a load of weight 15 kN. Calculate the greatest distance of this load from the tower for which the crane does not tip.
The jib extends 30 m from the tower. Explain why the 15 kN load must not be carried to the end of the jib.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moment = force × perpendicular distance from the pivot = 45 × 8.0 | ✓ 1 | |
| 360 kN m | ✓ 1 | |
| Part (b) | ||
| for balance, moment of load = moment of counterweight: 15 × d = 360 (ECF from (a)) | ✓ 1 | |
| d = 24 m | ✓ 1 | |
| Part (c) | ||
| moment of the load there = 15 × 30 = 450 kN m, which is greater than 360 kN m | ✓ 1 | |
| so there is a resultant moment towards the load and the crane would tip forwards | ✓ 1 | |
Exam tip: For a balanced object, total clockwise moment = total anticlockwise moment about the same pivot.
A coach times a sprinter during a training run. The sprinter accelerates uniformly from rest to 10.0 m/s in 2.5 s, then runs at a constant 10.0 m/s for a further 7.5 s.
Calculate the acceleration of the sprinter during the first 2.5 s.
Use the area under a velocity–time graph to calculate the total distance run in the 10.0 s.
Calculate the average speed of the sprinter over the 10.0 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 acceleration = change in speed ÷ time = 10.0 ÷ 2.5 | ✓ 1 | |
| A1 4.0 m/s² | ✓ 1 | |
| Part (b) | ||
| distance while accelerating = ½ × 10.0 × 2.5 = 12.5 m | ✓ 1 | |
| distance at constant speed = 10.0 × 7.5 = 75 m | ✓ 1 | |
| total distance = 87.5 m | ✓ 1 | |
| Part (c) | ||
| M1 average speed = 87.5 ÷ 10.0 | ✓ 1 | |
| A1 8.75 m/s (ECF from (b)) | ✓ 1 | |
Exam tip: Average speed is total distance divided by total time, not the mean of the highest and lowest speeds.
A beekeeper knocks a hive tool off a platform. The tool falls freely from rest and hits the ground after 0.60 s. Ignore air resistance and use g = 10 m/s².
Calculate the speed of the tool just before it lands.
Calculate the height of the platform.
State one reason why the real tool would hit the ground slightly more slowly.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v = u + at = 0 + 10 × 0.60 | ✓ 1 | |
| A1 6.0 m/s | ✓ 1 | |
| Part (b) | ||
| M1 average speed = (0 + 6.0) ÷ 2 = 3.0 m/s | ✓ 1 | |
| M1 height = average speed × time = 3.0 × 0.60 | ✓ 1 | |
| A1 1.8 m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| air resistance acts upwards on the tool | ✓ 1 | |
Common error: Free fall from rest has a constant acceleration of 10 m/s², so the average speed is half the final speed.
A forager bee flies 1200 m from its hive to a field of clover with a following wind at 8.0 m/s. It returns along the same path against the wind at 4.0 m/s.
Calculate the time taken for the outward journey.
Calculate the time taken for the return journey.
Calculate the average speed of the bee for the whole round trip. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 time = distance ÷ speed = 1200 ÷ 8.0 | ✓ 1 | |
| A1 150 s | ✓ 1 | |
| Part (b) | ||
| 1200 ÷ 4.0 = 300 s | ✓ 1 | |
| Part (c) | ||
| M1 total distance = 2400 m | ✓ 1 | |
| M1 total time = 150 + 300 = 450 s | ✓ 1 | |
| A1 2400 ÷ 450 = 5.33 m/s (ECF from (a) and (b)) | ✓ 1 | |
Common error: The average speed for a round trip is not the mean of the two speeds, which would give 6.0 m/s here.
A sprinter's speed is measured every second at the start of a race.
Time = 0 s: speed 0 m/s
Time = 1 s: speed 5.0 m/s
Time = 2 s: speed 8.0 m/s
Time = 3 s: speed 9.5 m/s
Time = 4 s: speed 10.0 m/s
Time = 5 s: speed 10.0 m/s
Describe how the sprinter's speed changes between 0 and 5 s.
Calculate the average acceleration during the first 2.0 s.
The sprinter has a mass of 70 kg. Calculate the average resultant force on the sprinter during the first 2.0 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the speed increases, but the increase in each second gets smaller (the acceleration decreases) | ✓ 1 | |
| after about 4 s the speed is constant at 10.0 m/s | ✓ 1 | |
| Part (b) | ||
| M1 acceleration = change in speed ÷ time = 8.0 ÷ 2.0 | ✓ 1 | |
| A1 4.0 m/s² | ✓ 1 | |
| Part (c) | ||
| M1 F = ma = 70 × 4.0 | ✓ 1 | |
| A1 280 N (ECF from (b)) | ✓ 1 | |
Exam tip: On a velocity–time graph the gradient is the acceleration, so a curve that flattens means the acceleration is falling.
A car travelling at 12 m/s on a dry road has a thinking distance of 9 m and a braking distance of 14 m. The driver has the same reaction time and the same braking force, and the road is unchanged. The car now travels at 24 m/s. What is the total stopping distance?
| Marking point | Mark | Notes |
|---|---|---|
| Thinking distance is proportional to speed, so it doubles to 18 m. Braking distance is proportional to speed squared, so it becomes 4 × 14 = 56 m. The total is 18 + 56 = 74 m. 46 m doubles the whole 23 m stopping distance. 32 m doubles only the thinking distance but leaves the braking distance unchanged. 92 m multiplies the whole stopping distance by 4. | ✓ 1 | Answer A |
Answer: A
Exam tip: Treat thinking distance and braking distance separately: one scales with speed, the other with speed squared.
A water balloon is dropped from rest from a hospital window 45 m above the ground. Ignore air resistance. Use g = 10 m/s2. What is the speed of the balloon just before it hits the ground?
| Marking point | Mark | Notes |
|---|---|---|
| Using v² = u² + 2as with u = 0: v² = 2 × 10 × 45 = 900, so v = 30 m/s. 900 m/s forgets to take the square root. 450 m/s multiplies g by the height (v = gs). 21 m/s uses v = √(gs), leaving out the factor of 2. | ✓ 1 | Answer D |
Answer: D
Common error: v² = 2as gives v squared; remember to take the square root as the last step.
A beekeeper records the position of a bee as it leaves the hive and flies in a straight line. Assume the bee moves at a steady speed between readings.
Time / s: 0, 10, 20, 30, 40, 50
Distance from hive / m: 0, 25, 50, 50, 50, 110
Describe the motion of the bee between 20 s and 40 s.
Calculate the speed of the bee during the first 20 s.
Calculate the average speed of the bee for the whole 50 s.
A second bee flies away from the hive at a constant speed of 2.4 m/s for the same 50 s. Calculate how much further than the first bee the second bee travels.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the bee is stationary / at rest | ✓ 1 | |
| Part (b) | ||
| M1 speed = distance ÷ time = 50 ÷ 20 | ✓ 1 | |
| A1 2.5 m/s | ✓ 1 | |
| Part (c) | ||
| M1 average speed = total distance ÷ total time = 110 ÷ 50 | ✓ 1 | |
| A1 2.2 m/s | ✓ 1 | |
| Part (d) | ||
| M1 distance = 2.4 × 50 = 120 m | ✓ 1 | |
| A1 10 m | ✓ 1 | |
Common error: Average speed uses the total time, including any time spent stationary.
A Formula 1 circuit has a lap length of 5.4 km. A driver completes a lap in 1 minute 30 seconds at a steady average speed.
Write the lap length in metres and the lap time in seconds.
Calculate the average speed in m/s.
Convert this speed to km/h.
A race lasts 56 laps, each taking 90 s. Calculate the race time in minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 5400 m | ✓ 1 | |
| 90 s | ✓ 1 | |
| Part (b) | ||
| M1 speed = distance ÷ time = 5400 ÷ 90 | ✓ 1 | |
| A1 60 m/s | ✓ 1 | |
| Part (c) | ||
| M1 × 3600 ÷ 1000 (or × 3.6) | ✓ 1 | |
| A1 216 km/h | ✓ 1 | |
| Part (d) | ||
| M1 56 × 90 = 5040 s | ✓ 1 | |
| A1 84 minutes | ✓ 1 | |
Common error: To change m/s to km/h multiply by 3.6; do not multiply by 60 twice without dividing by 1000.
A ship sails 30 km due east and then 40 km due north. What is the magnitude of the ship's displacement from its starting point?
| Marking point | Mark | Notes |
|---|---|---|
| The two legs are at right angles, so the displacement magnitude is found by Pythagoras: √(30² + 40²) = 50 km. 70 km is the total distance travelled, found by adding the lengths. 35 km is the mean of the two lengths. 10 km is the difference of the lengths, which would apply only if the legs were in opposite directions. | ✓ 1 | Answer A |
Answer: A
Exam tip: Displacement is the straight-line distance from start to finish, so it is not usually the total path length.
A delivery van travelling at 10 m/s has a thinking distance of 7 m and a braking distance of 12 m. The driver reaction time and the braking force are unchanged. What is the stopping distance at 20 m/s?
| Marking point | Mark | Notes |
|---|---|---|
| At double the speed, the thinking distance doubles to 14 m because the reaction time is the same. The braking distance increases by a factor of four to 48 m because it depends on speed squared. Stopping distance = 14 + 48 = 62 m. 76 m multiplies the whole stopping distance at 10 m/s by four. 38 m doubles the whole stopping distance. 26 m doubles only the thinking distance and leaves the braking distance unchanged. | ✓ 1 | Answer B |
Answer: B
Common error: Thinking distance is proportional to speed but braking distance is proportional to speed squared.
An aircraft accelerates uniformly from rest to a take-off speed of 50 m/s in 25 s. What distance does it travel along the runway?
| Marking point | Mark | Notes |
|---|---|---|
| With uniform acceleration from rest the average speed is half the final speed, 25 m/s. Distance = average speed × time = 25 × 25 = 625 m. 1250 m uses the final speed instead of the average speed. 50 m comes from multiplying the acceleration of 2.0 m/s² by the time. 25 m comes from using ½at instead of ½at². | ✓ 1 | Answer D |
Answer: D
Exam tip: For uniform acceleration from rest, use average speed = ½ × final speed.
A cargo aircraft has a mass of 1.2 × 104 kg. During take-off the engines produce a forward force of 36 kN. The average resistive force is 6.0 kN. The aircraft starts from rest and reaches a take-off speed of 60 m/s.
Calculate the resultant force on the aircraft.
Calculate the acceleration of the aircraft. Assume it is constant.
Calculate the time taken to reach take-off speed.
Calculate the length of runway needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 36 − 6.0 = 30 kN | ✓ 1 | |
| Part (b) | ||
| M1 a = F ÷ m = 30 000 ÷ 12 000 (ECF from (a)) | ✓ 1 | |
| A1 2.5 m/s2 | ✓ 1 | |
| Part (c) | ||
| M1 t = v ÷ a = 60 ÷ 2.5 (ECF from (b)) | ✓ 1 | |
| A1 24 s | ✓ 1 | |
| Part (d) | ||
| M1 use of v2 = u2 + 2as, so s = 602 ÷ (2 × 2.5) (ECF from (b)) | ✓ 1 | |
| A1 720 m | ✓ 1 | |
Common error: Use the resultant force, not just the engine force, in F = ma.
A delivery van of mass 2400 kg is travelling at 15 m/s when the driver brakes. The van comes to rest in a braking distance of 30 m. Assume the deceleration is uniform.
Calculate the deceleration of the van.
Calculate the braking force.
Explain why the braking distance is greater on a wet road.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 use of v2 = u2 + 2as, so a = 152 ÷ (2 × 30) | ✓ 1 | |
| A1 3.75 m/s2 | ✓ 1 | |
| Part (b) | ||
| M1 use of F = ma | ✓ 1 | |
| A1 2400 × 3.75 = 9000 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| There is less friction between the tyres and the road so the maximum braking force is smaller | ✓ 1 | |
| so the deceleration is smaller and the van takes a longer distance to stop | ✓ 1 | |
Exam tip: Link a longer braking distance to a smaller braking force and smaller deceleration.
A driver travelling at 18 m/s in a van normally has a reaction time of 0.50 s. When tired the reaction time is 1.00 s. In both cases the braking distance is 27 m.
Calculate the thinking distance when the driver is (i) not tired and (ii) tired.
Calculate how much further the van travels during the thinking time when the driver is tired.
Calculate the stopping distance when the driver is tired.
Calculate the percentage increase in stopping distance caused by tiredness compared with the normal stopping distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (i) 18 × 0.50 = 9.0 m | ✓ 1 | |
| (ii) 18 × 1.00 = 18 m | ✓ 1 | |
| Part (b) | ||
| 18 − 9.0 = 9.0 m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 stopping distance = thinking distance + braking distance | ✓ 1 | |
| A1 18 + 27 = 45 m (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| M1 normal stopping distance = 9.0 + 27 = 36 m and percentage = 9.0 ÷ 36 × 100 (ECF) | ✓ 1 | |
| A1 25% | ✓ 1 | |
Common error: Percentage increase is the increase divided by the original value, not the new value.
An architect tests a model of a balanced walkway using a light horizontal rod resting on a pivot. A weight of 300 N hangs 2.0 m to the left of the pivot. A weight of 400 N is hung on the right so that the rod balances horizontally. How far from the pivot is the 400 N weight?
| Marking point | Mark | Notes |
|---|---|---|
| For balance the clockwise moment equals the anticlockwise moment: 300 N × 2.0 m = 600 N m, so 400 N × d = 600 N m and d = 1.5 m. The heavier weight must be closer to the pivot. 2.0 m assumes equal distances balance unequal weights. 2.7 m comes from inverting the ratio (2.0 × 400 ÷ 300). 0.75 m comes from dividing 300 by 400 and ignoring the 2.0 m distance. | ✓ 1 | Answer D |
Answer: D
Exam tip: Moment = force × perpendicular distance from the pivot. Work out one moment first, then use it to find the unknown.
A tower crane on a building site is balanced about its vertical tower. A concrete counterweight of weight 36 kN is fixed 5.0 m from the tower on one side. A load hangs from the jib on the other side, 15 m from the tower. The weight of the jib itself can be ignored. Take g = 10 N/kg.
Calculate the moment of the counterweight about the tower. Give your answer in kN m.
The crane is exactly balanced. State the principle of moments and use it to find the weight of the load.
Calculate the mass of the load. Use the weight from (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moment = force × perpendicular distance = 36 × 5.0 | ✓ 1 | |
| 180 kN m | ✓ 1 | |
| Part (b) | ||
| for a balanced object the sum of clockwise moments equals the sum of anticlockwise moments | ✓ 1 | |
| load × 15 = 180 (ECF from (a)) | ✓ 1 | |
| load weight = 12 kN | ✓ 1 | |
| Part (c) | ||
| mass = weight ÷ g = 12 000 ÷ 10 (ECF from (b)) | ✓ 1 | |
| 1200 kg | ✓ 1 | |
Exam tip: Keep units consistent: if the force is in kN and the distance in m, the moment is in kN m. Convert to N before using W = mg.
A marine biologist drops a waterproof camera pod from a cliff ledge into the sea. Ignore air resistance. The pod hits the water at 30 m/s. Take g = 10 m/s2.
Calculate the height of the ledge above the water.
Calculate the time the pod takes to fall. Use the value 30 m/s.
Explain why the real time of fall would be slightly longer than your answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v2 = u2 + 2as with u = 0 (so 302 = 2 × 10 × s) | ✓ 1 | |
| s = 900 ÷ 20 | ✓ 1 | |
| 45 m | ✓ 1 | |
| Part (b) | ||
| t = (v − u) ÷ a = 30 ÷ 10 | ✓ 1 | |
| 3.0 s | ✓ 1 | |
| Part (c) | ||
| air resistance acts upwards on the falling pod | ✓ 1 | |
| so the resultant force (and acceleration) is smaller than for free fall, giving a longer time | ✓ 1 | |
Exam tip: In free fall near the surface of the Earth every object has the same acceleration of 10 m/s². Air resistance is what makes real falls differ.
A lorry driver on a motorway is travelling at 20 m/s when a hazard appears ahead. The driver's reaction time is 0.80 s. After the brakes are applied the lorry decelerates uniformly at 4.0 m/s2 until it stops.
Calculate the thinking distance.
Calculate the braking distance.
Calculate the total stopping distance. Use your answers to (a) and (b).
State one factor, other than speed, that would increase the thinking distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distance = speed × time = 20 × 0.80 | ✓ 1 | |
| 16 m | ✓ 1 | |
| Part (b) | ||
| v2 = u2 + 2as (or equivalent) with final speed 0 and a = −4.0 m/s2 | ✓ 1 | |
| s = 202 ÷ (2 × 4.0) | ✓ 1 | |
| 50 m | ✓ 1 | |
| Part (c) | ||
| 16 + 50 = 66 m (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| driver tiredness, alcohol, drugs or distraction (any one) | ✓ 1 | |
Common error: Stopping distance is thinking distance plus braking distance. Doubling the speed doubles the thinking distance but quadruples the braking distance.
A marine biologist weighs a catch of fish in a net using a spring balance. The spring obeys Hooke's law. When a force of 24 N is applied the spring extends by 0.060 m. Take g = 10 N/kg.
Calculate the spring constant of the spring.
The spring now extends by 0.105 m when the net of fish hangs from it. Calculate the weight of the fish and net.
Calculate the mass of the fish and net. Use the value 42 N.
Explain what is meant by the limit of proportionality of a spring.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| k = F ÷ x = 24 ÷ 0.060 | ✓ 1 | |
| 400 N/m | ✓ 1 | |
| Part (b) | ||
| F = kx = 400 × 0.105 (ECF from (a)) | ✓ 1 | |
| 42 N | ✓ 1 | |
| Part (c) | ||
| m = W ÷ g = 42 ÷ 10 | ✓ 1 | |
| 4.2 kg | ✓ 1 | |
| Part (d) | ||
| the force beyond which the extension is no longer directly proportional to the force | ✓ 1 | |
Exam tip: Hooke's law only applies up to the limit of proportionality. Check that your extension is a change in length, not the total length.
A uniform beam of length 4.0 m and weight 120 N rests on a support at each end. A box of weight 80 N is placed on the beam 1.0 m from the left-hand end. What is the upward force from the left-hand support?
| Marking point | Mark | Notes |
|---|---|---|
| Taking moments about the right-hand end or the left-hand end gives the right-hand support force: 4.0 × R = 120 × 2.0 + 80 × 1.0 = 320, so R = 80 N. The upward forces must balance the total weight of 200 N, so the left-hand force is 200 − 80 = 120 N. 80 N is the right-hand support force. 100 N wrongly shares the weight equally, which ignores the off-centre box. 200 N is the total weight carried by both supports together. | ✓ 1 | Answer B |
Answer: B
Exam tip: Take moments about one support so that its unknown force has no moment, then use the fact that the upward forces equal the downward forces.
In an experiment a constant resultant force acts on a trolley. When the trolley has a mass of 0.50 kg its acceleration is 1.2 m/s2. The mass is increased to 2.0 kg and the resultant force is unchanged. What is the new acceleration?
| Marking point | Mark | Notes |
|---|---|---|
| The force is F = ma = 0.50 × 1.2 = 0.60 N. The new acceleration is a = F/m = 0.60 ÷ 2.0 = 0.30 m/s2. The mass is four times larger, so the acceleration is four times smaller. 4.8 m/s2 multiplies by four instead of dividing. 0.60 m/s2 halves the acceleration, as if the mass had only doubled. 0.075 m/s2 divides by 16, as if acceleration were inversely proportional to mass squared. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a constant force, acceleration is inversely proportional to mass: multiply the mass by a factor and divide the acceleration by the same factor.
A uniform walkway plank of length 5.0 m and weight 200 N rests on a support at each end, on a fishing pier. A fisherman of weight 700 N stands on the plank 2.0 m from the left-hand end.
Calculate the upward force from the right-hand support.
Calculate the upward force from the left-hand support.
The fisherman now walks to the right-hand end of the plank. Calculate the force from the left-hand support.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 take moments about the left-hand support (clockwise moments = anticlockwise moments) | ✓ 1 | |
| A1 moment of plank 200 × 2.5 = 500 N m and moment of fisherman 700 × 2.0 = 1400 N m | ✓ 1 | |
| A1 right-hand support force = 1900 ÷ 5.0 = 380 N | ✓ 1 | |
| Part (b) | ||
| M1 upward forces equal downward forces: 200 + 700 − 380 | ✓ 1 | |
| A1 520 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 moments about the left-hand end: right-hand force = (200 × 2.5 + 700 × 5.0) ÷ 5.0 = 800 N, then left-hand force = 900 − 800 | ✓ 1 | |
| A1 100 N | ✓ 1 | |
Exam tip: Choose a pivot at one support so that its unknown force drops out of the moments equation. Show the moment of each weight separately.
A student investigates how the mass of a trolley affects its acceleration. A constant resultant force of 0.60 N pulls the trolley along a bench.
State the independent variable and one control variable in this investigation.
The trolley has a mass of 0.80 kg. Calculate its acceleration.
The trolley starts from rest. Calculate the time taken to reach a speed of 3.0 m/s.
The mass of the trolley is doubled and the resultant force stays the same. State what happens to the acceleration and explain your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 independent variable is the mass of the trolley | ✓ 1 | |
| B1 any one control: resultant (pulling) force, same bench or friction compensated, same starting position | ✓ 1 | |
| Part (b) | ||
| M1 a = F ÷ m | ✓ 1 | |
| A1 0.60 ÷ 0.80 = 0.75 m/s2 | ✓ 1 | |
| Part (c) | ||
| M1 t = v ÷ a (or (v − u) ÷ a) | ✓ 1 | |
| A1 3.0 ÷ 0.75 = 4.0 s (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| B1 acceleration halves | ✓ 1 | |
| B1 a = F ÷ m, so acceleration is inversely proportional to mass for a constant force | ✓ 1 | |
Exam tip: When a question says a quantity is doubled, write what happens to the other quantity as a factor (halves, doubles) and link it to the equation.
A lorry carrying fish travels at 20 m/s along a harbour road. The driver has a reaction time of 0.70 s. After the brakes are applied the lorry slows with a constant deceleration of 5.0 m/s2.
Calculate the thinking distance.
Calculate the braking distance.
Calculate the total stopping distance.
State two factors that would increase the thinking distance of a driver.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance = speed × time | ✓ 1 | |
| A1 20 × 0.70 = 14 m | ✓ 1 | |
| Part (b) | ||
| M1 v2 = u2 + 2as with v = 0 and a = −5.0 | ✓ 1 | |
| M1 0 = 202 − 2 × 5.0 × s | ✓ 1 | |
| A1 s = 40 m | ✓ 1 | |
| Part (c) | ||
| A1 14 + 40 = 54 m (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| B1 any one from: tiredness, alcohol or drugs, distraction, slower reaction due to illness | ✓ 1 | |
| B1 any second different factor from the list above, or higher speed | ✓ 1 | |
Common error: Stopping distance is thinking distance plus braking distance. Do not use the reaction time in the braking calculation.
A space probe sits at rest on the surface of the Moon. Which pair of forces is an example of Newton's third law (an action–reaction pair)?
| Marking point | Mark | Notes |
|---|---|---|
| A third-law pair acts on two different objects, is the same type of force and has equal size in opposite directions: the Moon pulls the probe and the probe pulls the Moon, both gravitational. The weight and the contact force both act on the probe, so they only balance each other. The weight and the push on the surface are different types of force, one gravitational and one contact. The last pair mixes a contact force with a gravitational force, so they are not a pair. | ✓ 1 | Answer B |
Answer: B
Common error: Two equal and opposite forces that both act on the same object are balanced forces, not a third-law pair.
A parachutist falls with the parachute open at a steady speed of 5 m/s. Which statement about the forces on the parachutist is correct?
| Marking point | Mark | Notes |
|---|---|---|
| A constant velocity means there is no acceleration, so the resultant force is zero and the upward air resistance equals the downward weight. If the weight were larger the parachutist would still be speeding up. Air resistance is not zero, since it is exactly what cancels the weight. If the air resistance were larger the parachutist would be slowing down. | ✓ 1 | Answer D |
Answer: D
Common error: Motion in one direction does not need a resultant force in that direction; only a change in velocity does.
A student hangs masses from a helical spring and plots a graph of force against extension. The graph is a straight line through the origin at first and then begins to curve. Which statement describes the curved part of the graph?
| Marking point | Mark | Notes |
|---|---|---|
| Beyond the limit of proportionality the spring stretches more for each extra newton, so force divided by extension falls and the gradient of the force–extension graph decreases. The second option has the effect the wrong way round. A horizontal graph would mean the force changes with no extension change at all. If force were still proportional to extension the graph would remain straight. | ✓ 1 | Answer A |
Answer: A
Exam tip: On a force–extension graph the gradient equals the spring constant, so a curve bending towards the extension axis shows the spring has gone past its limit of proportionality.
A student investigates a helical spring hung from a clamp. Its natural length is 8.0 cm. She adds 100 g masses one at a time and measures the length of the spring each time. Use g = 10 N/kg.
Mass added / g: 0, 100, 200, 300, 400, 500
Length of spring / cm: 8.0, 10.0, 12.0, 14.0, 16.0, 18.5
State one way to make the length readings more accurate.
Show that the spring obeys Hooke's law for loads up to 400 g.
Identify the first load for which the spring is beyond its limit of proportionality. Give a reason.
Calculate the spring constant of the spring in N/m using the 400 g reading.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| any one from: read the scale at eye level to avoid parallax; use a set square to compare the end of the spring with the ruler; wait for the spring to stop moving before reading; fix a pointer to the bottom coil | ✓ 1 | |
| Part (b) | ||
| extension increases by the same amount, 2.0 cm, for each extra 100 g | ✓ 1 | |
| so extension is directly proportional to the load, for example 4.0 cm at 200 g and 8.0 cm at 400 g | ✓ 1 | |
| Part (c) | ||
| 500 g | ✓ 1 | |
| extension is 10.5 cm, which is more than the 10.0 cm expected from proportionality | ✓ 1 | |
| Part (d) | ||
| force = 0.400 × 10 = 4.0 N | ✓ 1 | |
| extension = 16.0 − 8.0 = 8.0 cm = 0.080 m | ✓ 1 | |
| k = F ÷ x = 4.0 ÷ 0.080 = 50 N/m | ✓ 1 | |
Common error: Extension is the stretched length minus the natural length; using the total length of the spring in F = kx gives a wrong spring constant.
A sprinter accelerates uniformly from rest to 8.0 m/s in 2.0 s and then runs at a constant speed of 8.0 m/s for the rest of a 100 m race.
Calculate her acceleration.
Calculate the distance she covers while accelerating.
Calculate the total time she takes to run the 100 m.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| acceleration = change in speed ÷ time | ✓ 1 | |
| 8.0 ÷ 2.0 = 4.0 m/s2 | ✓ 1 | |
| Part (b) | ||
| average speed = 4.0 m/s, so distance = 4.0 × 2.0, or area under graph = ½ × 2.0 × 8.0 | ✓ 1 | |
| 8.0 m | ✓ 1 | |
| Part (c) | ||
| remaining distance = 100 − 8.0 = 92 m (ECF from (b)) | ✓ 1 | |
| time at constant speed = 92 ÷ 8.0 = 11.5 s | ✓ 1 | |
| total time = 11.5 + 2.0 = 13.5 s | ✓ 1 | |
Common error: Do not use the constant-speed equation for the whole race; treat the accelerating stage and the steady stage separately and add the times.
A car travels at 24 m/s when the driver sees an obstacle ahead. The driver takes 0.75 s to react. The car then slows down uniformly and stops in a further 4.0 s.
Calculate the thinking distance.
State two factors, other than speed, that increase the thinking distance.
Calculate the braking distance.
Calculate the total stopping distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distance = speed × time | ✓ 1 | |
| 24 × 0.75 = 18 m | ✓ 1 | |
| Part (b) | ||
| any two from: tiredness; alcohol or drugs; distraction such as a phone; poor concentration | ✓ 2 | |
| Part (c) | ||
| average speed = 12 m/s, or area under speed–time graph | ✓ 1 | |
| 12 × 4.0 = 48 m | ✓ 1 | |
| Part (d) | ||
| 18 + 48 = 66 m (ECF from (a) and (c)) | ✓ 1 | |
Common error: Stopping distance is thinking distance plus braking distance; the car does not slow down at all during the reaction time.
A climber walks 3.0 km up a path in 2.0 hours, then returns down the same path in 1.0 hour. What is the average speed for the whole trip?
| Marking point | Mark | Notes |
|---|---|---|
| Average speed = total distance ÷ total time = 6.0 km ÷ 3.0 h = 2.0 km/h. Zero would be the average velocity, because the displacement is zero, not the average speed. Averaging the two speeds (1.5 and 3.0 km/h) gives 2.25 km/h, which is wrong because the two parts take different times. Adding the speeds gives 4.5 km/h, which has no meaning. | ✓ 1 | Answer D |
Answer: D
Exam tip: Average speed is always total distance divided by total time, never the mean of two speeds.
In a food-science experiment, a steel ball is dropped into a tall cylinder of syrup and quickly reaches a constant speed. Ignoring upthrust, which statement is correct while the ball moves at this constant speed?
| Marking point | Mark | Notes |
|---|---|---|
| A constant speed means zero acceleration, so the resultant force is zero: the upward drag force balances the downward weight. Drag larger than weight would slow the ball down. Weight does not vanish because the ball is moving steadily, and the forces have not stopped acting, they are balanced. | ✓ 1 | Answer A |
Answer: A
Common error: Constant speed does not mean no forces. It means balanced forces.
A telecoms engineer drives a van to a mast. The van travels 45 km in 0.50 h, then 60 km in 1.00 h. The engineer then stops for 0.30 h before the journey ends.
Calculate the average speed of the van during the first 45 km, in km/h.
Calculate the average speed for the whole journey, including the stop. Give your answer to 2 significant figures in km/h.
Convert a speed of 90 km/h to m/s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 speed = distance ÷ time, 45 ÷ 0.50 | ✓ 1 | |
| A1 90 km/h | ✓ 1 | |
| Part (b) | ||
| M1 total distance 105 km and total time 1.80 h | ✓ 1 | |
| M1 105 ÷ 1.80 | ✓ 1 | |
| A1 58 km/h | ✓ 1 | |
| Part (c) | ||
| M1 90 × 1000 ÷ 3600 | ✓ 1 | |
| A1 25 m/s | ✓ 1 | |
Exam tip: Include any time spent stopped when finding the average speed of a whole journey.
A mountaineering team pulls a sledge with equipment of total mass 40 kg across snow. The pulling force is 120 N and the friction force is 60 N.
Calculate the resultant force on the sledge.
Calculate the acceleration of the sledge.
The sledge starts from rest. Calculate its speed after 4.0 s.
Calculate the distance the sledge travels in the 4.0 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 60 N | ✓ 1 | |
| Part (b) | ||
| M1 a = F ÷ m, 60 ÷ 40 | ✓ 1 | |
| A1 1.5 m/s² | ✓ 1 | |
| Part (c) | ||
| M1 v = a t, 1.5 × 4.0 | ✓ 1 | |
| A1 6.0 m/s | ✓ 1 | |
| Part (d) | ||
| M1 distance = average speed × time, (0 + 6.0) ÷ 2 × 4.0 | ✓ 1 | |
| A1 12 m | ✓ 1 | |
Common error: Use the resultant force, not the pulling force, in F = m a.
A food-science student drops a steel ball bearing of mass 0.020 kg into a tall tube of glycerol. Ignore upthrust. Take g = 10 N/kg.
Explain why the ball bearing accelerates at first.
Explain why the ball bearing eventually reaches a constant speed.
Calculate the weight of the ball bearing.
State the size of the drag force on the ball bearing when it moves at constant speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| At the start the drag force is small (zero) | ✓ 1 | |
| Weight is greater than drag, so there is a downward resultant force | ✓ 1 | |
| Part (b) | ||
| Drag increases as speed increases | ✓ 1 | |
| Until the drag equals the weight | ✓ 1 | |
| The resultant force is zero so there is no more acceleration | ✓ 1 | |
| Part (c) | ||
| M1 W = m g, 0.020 × 10 | ✓ 1 | |
| A1 0.20 N | ✓ 1 | |
| Part (d) | ||
| 0.20 N (ECF from (c)) | ✓ 1 | |
Exam tip: For terminal velocity, say that the forces are balanced, so the resultant force is zero.
A food-delivery cart moves along a straight corridor. A student records its distance from the start every second.
Time / s: 0, 1, 2, 3, 4, 5
Distance / m: 0.0, 0.6, 1.4, 2.4, 3.6, 5.0
Calculate the average speed of the cart over the whole 5.0 s.
Describe the motion of the cart. Use the data to support your answer.
Calculate the average speed between 2.0 s and 4.0 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance ÷ time, 5.0 ÷ 5.0 | ✓ 1 | |
| A1 1.0 m/s | ✓ 1 | |
| Part (b) | ||
| The cart speeds up (accelerates) | ✓ 1 | |
| The distance travelled in each second increases, for example 0.6 m in the first second and 1.4 m in the fifth | ✓ 1 | |
| Part (c) | ||
| M1 (3.6 − 1.4) ÷ (4.0 − 2.0) | ✓ 1 | |
| A1 1.1 m/s | ✓ 1 | |
Exam tip: Average speed over an interval uses the change in distance, not the distance at the end.
A mountaineer drops a stone from a cliff edge. Ignore air resistance and take g = 10 N/kg (10 m/s² acceleration).
Calculate the speed of the stone after 3.0 s.
Calculate the distance the stone falls in 3.0 s.
A feather dropped from the same place falls much more slowly than the stone. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v = a t, 10 × 3.0 | ✓ 1 | |
| A1 30 m/s | ✓ 1 | |
| Part (b) | ||
| M1 distance = average speed × time, (0 + 30) ÷ 2 × 3.0 | ✓ 1 | |
| A1 45 m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Air resistance (drag) is significant compared with the weight of the feather | ✓ 1 | |
| The feather's resultant force and acceleration are therefore much smaller | ✓ 1 | |
Exam tip: Air resistance can be ignored only when it is small compared with the weight of the object.
A student adds masses to a rubber band one at a time and then removes them one at a time, measuring the extension each time.
Which observation shows that the rubber band does not obey Hooke's law?
| Marking point | Mark | Notes |
|---|---|---|
| Hooke's law requires extension to be directly proportional to load. Doubling the load from 2.0 N to 4.0 N should double the extension from 3.0 cm to 6.0 cm, but it only reaches 5.0 cm, so the law is not obeyed. Extension increasing with load happens for any stretching material and does not test proportionality. Returning to the original length shows elastic behaviour, not proportionality. Doubling the load from 1.0 N to 2.0 N doubles the extension, which is consistent with Hooke's law. | ✓ 1 | Answer D |
Answer: D
Common error: Many students say a material obeys Hooke's law because it springs back; springing back shows elasticity, but proportionality of extension to load is the test.
Two skaters stand at rest on smooth ice, facing each other. Skater P has a mass of 60 kg and skater Q has a mass of 45 kg. They push each other apart and P moves away at 1.5 m/s. Friction is negligible.
What is the speed of Q?
| Marking point | Mark | Notes |
|---|---|---|
| The total momentum is zero before and after, so the two momenta are equal in size: 60 × 1.5 = 45 × v, giving v = 2.0 m/s. 1.5 m/s assumes both skaters move at the same speed, which ignores their different masses. 1.1 m/s comes from dividing by the wrong mass, so the lighter skater ends up slower. 3.5 m/s comes from using the combined mass of both skaters in the momentum equation. | ✓ 1 | Answer A |
Answer: A
Exam tip: When two objects push apart from rest, their momenta are equal and opposite, so the lighter object moves faster.
A student investigates a rubber band. She adds weights to it and measures its extension, then removes the weights one at a time and measures the extension again.
Load / N: 0, 1.0, 2.0, 3.0, 4.0
Extension while loading / cm: 0, 2.0, 5.0, 9.0, 15.0
Extension while unloading / cm: 0, 3.5, 7.0, 11.0, 15.0
Calculate the increase in extension when the load is increased from 2.0 N to 3.0 N while loading.
State whether the rubber band obeys Hooke's law over this range of loads. Use the loading data to justify your answer.
Calculate how much larger the extension at a load of 2.0 N is while unloading than while loading.
State what the answer to (c) shows about the behaviour of the rubber band.
Suggest one change to the method that would make the results more reliable.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 9.0 − 5.0 = 4.0 cm | ✓ 1 | |
| Part (b) | ||
| No | ✓ 1 | |
| Extension is not directly proportional to load, for example 1.0 N gives 2.0 cm but 2.0 N does not give 4.0 cm (accept the increase in extension per newton is not constant) | ✓ 1 | |
| Part (c) | ||
| 7.0 − 5.0 = 2.0 cm | ✓ 1 | |
| Part (d) | ||
| The extension for a given load depends on whether the load is being added or removed (the band does not follow the same path / it does not return along the same curve) | ✓ 1 | |
| Part (e) | ||
| Repeat the experiment and average (or wait the same time before each reading, or use smaller steps of load) | ✓ 1 | |
Common error: Do not say a rubber band is not elastic just because it does not obey Hooke's law; the problem is that extension is not proportional to load.
Two ice skaters, Amara of mass 50 kg and Bao of mass 75 kg, stand at rest on smooth ice. They push each other and move apart in opposite directions. Amara moves at 3.0 m/s. Friction is negligible.
State the total momentum of the two skaters before they push.
Calculate the speed of Bao.
Calculate the total kinetic energy of the two skaters after the push.
State the energy store from which this kinetic energy was transferred.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Zero | ✓ 1 | |
| Part (b) | ||
| M1 total momentum after is also zero, so the two momenta are equal in size | ✓ 1 | |
| M1 50 × 3.0 = 75 × v | ✓ 1 | |
| A1 2.0 m/s | ✓ 1 | |
| Part (c) | ||
| M1 kinetic energy = ½ m v² used for either skater | ✓ 1 | |
| A1 Amara 225 J and Bao 150 J | ✓ 1 | |
| A1 total 375 J (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| Chemical energy store (in the skaters' muscles) | ✓ 1 | |
Common error: Momentum is conserved in the push, but kinetic energy is not: the skaters gain kinetic energy that was not there before.
In a data centre a technician balances a uniform metal bar, of length 2.0 m and weight 60 N, on a pivot placed 0.60 m from the left-hand end. A server component hangs from the left-hand end and the bar is balanced horizontally. The weight of the bar acts at its centre.
State the principle of moments.
Calculate the weight W of the server component.
Calculate the mass of the server component. Gravitational field strength = 10 N/kg.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| For an object in equilibrium, the sum of the clockwise moments equals the sum of the anticlockwise moments | ✓ 1 | |
| Both calculated about the same point | ✓ 1 | |
| Part (b) | ||
| M1 moment of the bar's weight about the pivot = 60 × 0.40 = 24 N m | ✓ 1 | |
| M1 W × 0.60 = 24 | ✓ 1 | |
| A1 40 N | ✓ 1 | |
| Part (c) | ||
| A1 40/10 = 4.0 kg (ECF from (b)) | ✓ 1 | |
Common error: The bar's weight acts at its centre, which is 1.0 m from the end; the distance in the moment is measured from the pivot, not from the end.
On an excavation, a loaded trolley of mass 40 kg moves along a level track at 1.5 m/s. It collides with a stationary empty trolley of mass 60 kg. The two trolleys couple together and move off together. Friction is negligible.
Calculate the speed of the trolleys immediately after the collision.
Calculate the kinetic energy transferred away in the collision.
Suggest where this energy goes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 momentum before = 40 × 1.5 = 60 kg m/s | ✓ 1 | |
| M1 60 = (40 + 60) × v | ✓ 1 | |
| A1 0.60 m/s | ✓ 1 | |
| Part (b) | ||
| M1 kinetic energy before = ½ × 40 × 1.5² = 45 J | ✓ 1 | |
| M1 kinetic energy after = ½ × 100 × 0.60² = 18 J | ✓ 1 | |
| A1 27 J (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Transferred to thermal energy stores (and sound) of the trolleys and surroundings | ✓ 1 | |
Exam tip: In a collision where objects stick together, add the masses on the right-hand side of the momentum equation.
A trolley of mass 0.40 kg is pulled along a bench by a constant resultant force of 0.50 N. Masses are then added so the trolley has a total mass of 1.0 kg. The resultant force stays the same. What is the new acceleration of the trolley?
| Marking point | Mark | Notes |
|---|---|---|
| Using a = F ÷ m, the new acceleration is 0.50 ÷ 1.0 = 0.50 m/s². 0.20 m/s² multiplies the force by the original mass instead of dividing. 0.65 m/s² subtracts the extra 0.60 kg from the original acceleration of 1.25 m/s². 3.1 m/s² assumes acceleration increases in proportion to mass. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a fixed resultant force, acceleration is inversely proportional to mass: a larger mass gives a smaller acceleration.
A student investigates how the mass of a trolley affects its acceleration. A constant horizontal force pulls the trolley along a bench, and friction has been compensated. The mass is changed by adding masses to the trolley.
Mass / kg: 0.50, 1.00, 1.50, 2.00
Acceleration / m/s²: 1.60, 0.80, 0.53, 0.40
State the variable that must be kept constant for the results to be valid.
Describe how the student could use two light gates to find the acceleration of the trolley.
Describe the relationship between acceleration and mass shown by the data.
Use the data to calculate the acceleration expected if the trolley mass is 3.0 kg. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the pulling (accelerating) force | ✓ 1 | |
| Part (b) | ||
| a card of known length on the trolley; speed at each gate = card length ÷ time to pass the gate | ✓ 1 | |
| acceleration = (final speed − initial speed) ÷ time taken between the two gates | ✓ 1 | |
| Part (c) | ||
| as mass increases the acceleration decreases | ✓ 1 | |
| acceleration is inversely proportional to mass (doubling the mass halves the acceleration) | ✓ 1 | |
| Part (d) | ||
| force = mass × acceleration = 0.50 × 1.60 = 0.80 N (any data pair) | ✓ 1 | |
| acceleration = force ÷ mass = 0.80 ÷ 3.0 | ✓ 1 | |
| 0.27 m/s² | ✓ 1 | |
Exam tip: When asked for a relationship from data, test it: for inverse proportion, mass × acceleration stays constant.
A van carrying musical instruments travels at 20 m/s. The driver's reaction time is 0.80 s. After the brakes are applied, the van decelerates uniformly at 5.0 m/s² until it stops.
Calculate the thinking distance.
Calculate the braking distance.
Calculate the total stopping distance (ECF from (a) and (b)).
The van is now fully loaded and the braking force is unchanged. Explain why the braking distance increases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distance = speed × time = 20 × 0.80 | ✓ 1 | |
| 16 m | ✓ 1 | |
| Part (b) | ||
| v² = u² + 2as gives 20² = 2 × 5.0 × s | ✓ 1 | |
| 40 m | ✓ 1 | |
| Part (c) | ||
| 16 + 40 = 56 m | ✓ 1 | |
| Part (d) | ||
| greater mass means a smaller deceleration for the same braking force (a = F ÷ m) | ✓ 1 | |
| so the van takes a longer distance to stop from the same speed | ✓ 1 | |
Common error: Stopping distance is thinking distance plus braking distance. Reaction time affects only the thinking distance.
A farmer pushes a wheelbarrow carrying 40 kg of grain. The weight of the grain acts 0.50 m horizontally from the axle of the wheel. The farmer lifts the handles with a vertical force at a horizontal distance of 1.30 m from the axle. Ignore the weight of the wheelbarrow. Use g = 10 N/kg.
State the principle of moments.
Calculate the vertical force the farmer applies to the handles. Give your answer to 3 significant figures.
Suggest one change that would reduce the force the farmer needs to apply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| for an object in equilibrium, the sum of the clockwise moments about any point | ✓ 1 | |
| equals the sum of the anticlockwise moments about the same point | ✓ 1 | |
| Part (b) | ||
| weight of grain = 40 × 10 = 400 N | ✓ 1 | |
| moment of grain about axle = 400 × 0.50 = 200 N m | ✓ 1 | |
| force = 200 ÷ 1.30 = 154 N | ✓ 1 | |
| Part (c) | ||
| move the grain closer to the axle (or hold the handles further from the axle) | ✓ 1 | |
Exam tip: Take moments about the pivot. Use the perpendicular distance from the pivot to the line of action of each force.
A driver's reaction time and the car's braking deceleration do not change. At 10 m/s the thinking distance is 6.0 m and the braking distance is 10 m. What is the total stopping distance at 20 m/s?
| Marking point | Mark | Notes |
|---|---|---|
| Thinking distance is proportional to speed, so it doubles to 12 m. Braking distance is proportional to speed squared, so it increases by a factor of 4 to 40 m. Total = 12 + 40 = 52 m. 22 m leaves the braking distance unchanged. 32 m doubles the total of 16 m, treating both parts as proportional to speed. 64 m multiplies the whole 16 m by 4, treating the thinking distance as proportional to speed squared. | ✓ 1 | Answer D |
Answer: D
Exam tip: Treat the two parts of stopping distance separately: thinking distance ∝ speed, braking distance ∝ speed².
A spring obeys Hooke's law and has a spring constant of 250 N/m. A force of 10 N stretches it. What is the extension of the spring?
| Marking point | Mark | Notes |
|---|---|---|
| Extension = force ÷ spring constant = 10 ÷ 250 = 0.040 m = 4.0 cm. 0.040 cm forgets to convert metres to centimetres. 25 cm comes from dividing the spring constant by the force, the wrong way up. 40 cm converts 0.040 m using a factor of 1000 instead of 100. | ✓ 1 | Answer C |
Answer: C
Common error: Check that the spring constant and the extension use the same length unit before you finish.
In a driving-simulator game, a virtual car travels at 15 m/s when a hazard appears. The car then brakes with a constant deceleration of 4.5 m/s2. The total stopping distance is 37 m.
Calculate the braking distance. Use v2 = u2 + 2as.
Calculate the reaction time programmed into the game.
The game now simulates a wet road, with a constant deceleration of 3.0 m/s2 from 15 m/s. Calculate the new braking distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 0 = 152 − 2 × 4.5 × s, so s = 152 ÷ (2 × 4.5) | ✓ 1 | |
| A1 25 m | ✓ 1 | |
| Part (b) | ||
| thinking distance = 37 − 25 = 12 m (ECF from (a)) | ✓ 1 | |
| M1 time = distance ÷ speed = 12 ÷ 15 | ✓ 1 | |
| A1 0.80 s | ✓ 1 | |
| Part (c) | ||
| M1 s = 152 ÷ (2 × 3.0) | ✓ 1 | |
| A1 37.5 m | ✓ 1 | |
Common error: Do not add the two distances before you have found each one separately: thinking distance uses constant speed, braking distance uses deceleration.
A pharmacy uses a spring balance to check the weight of small sample bags. A force of 1.5 N hung from the spring produces an extension of 3.0 cm. The spring obeys Hooke's law for this load. Use g = 10 N/kg.
Calculate the spring constant, in N/m.
State what is meant by the limit of proportionality of a spring.
A sample bag of mass 0.24 kg is hung from the spring. Calculate the extension of the spring in centimetres.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 k = F ÷ e = 1.5 ÷ 0.030 (extension converted to metres) | ✓ 1 | |
| A1 50 N/m | ✓ 1 | |
| Part (b) | ||
| the point beyond which the extension is no longer directly proportional to the force | ✓ 1 | |
| Part (c) | ||
| M1 weight = 0.24 × 10 = 2.4 N | ✓ 1 | |
| M1 e = F ÷ k = 2.4 ÷ 50 = 0.048 m | ✓ 1 | |
| A1 4.8 cm | ✓ 1 | |
Exam tip: Convert centimetres to metres before using k in N/m, and convert back at the end if the question asks for centimetres.
A water tanker is driven along a straight, level road. When the tank is empty the tanker has a mass of 6000 kg. The driving force is 7200 N and the resistive force is 1200 N.
Calculate the resultant force on the empty tanker.
Calculate the acceleration of the empty tanker.
When the tank is filled with water, the driving force and resistive force are the same but the acceleration is 0.60 m/s2. Calculate the mass of water in the tank.
The full tanker starts from rest and keeps accelerating at 0.60 m/s2. Calculate its speed after 12 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 7200 − 1200 = 6000 N | ✓ 1 | |
| Part (b) | ||
| M1 a = F ÷ m = 6000 ÷ 6000 | ✓ 1 | |
| A1 1.0 m/s2 | ✓ 1 | |
| Part (c) | ||
| M1 total mass = F ÷ a = 6000 ÷ 0.60 | ✓ 1 | |
| A1 total mass = 10 000 kg | ✓ 1 | |
| A1 mass of water = 10 000 − 6000 = 4000 kg | ✓ 1 | |
| Part (d) | ||
| M1 v = u + at = 0 + 0.60 × 12 | ✓ 1 | |
| A1 7.2 m/s | ✓ 1 | |
Exam tip: Find the resultant force first, then use F = ma: the mass in the equation is the whole moving mass.
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated points, e.g. names weight and drag, or states that the ball reaches a constant speed. Little or no link between resultant force and change of velocity. |
| Level 2 (3–4) | Describes the motion in both air and oil and links some changes in velocity to resultant force, but the chain is incomplete — typically misses that drag exceeds weight on entry, or treats the ball as speeding up to terminal velocity in the oil. |
| Level 3 (5–6) | Coherent, logically sequenced explanation of both stages: acceleration in air, deceleration on entry because drag > weight, drag falling with speed so the deceleration decreases, and forces balancing at terminal velocity. |
A lift in a tall office building has a total mass of 800 kg, including its passengers. It hangs from a steel cable.
Use g = 10 N/kg.
Name the two vertical forces acting on the lift.
As the lift starts moving upwards, the tension in the cable is 8800 N. Calculate the acceleration of the lift.
The lift then moves upwards at a constant speed. Explain the value of the tension in the cable during this stage.
Near the top floor, the lift slows down with a deceleration of 1.5 m/s2 while still moving upwards. Calculate the tension in the cable.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Tension in the cable acting upwards and weight acting downwards (both needed) | ✓ 1 | |
| Part (b) | ||
| Weight = 800 × 10 = 8000 N, so resultant force = 8800 − 8000 = 800 N ; a = F ÷ m = 800 ÷ 800 ; a = 1.0 m/s2 upwards | ✓ 3 | |
| Part (c) | ||
| At constant speed there is no acceleration, so the resultant force is zero ; so the tension equals the weight, 8000 N | ✓ 2 | |
| Part (d) | ||
| Resultant force needed = 800 × 1.5 = 1200 N downwards ; tension = weight − resultant = 8000 − 1200 ; tension = 6800 N | ✓ 3 | |
Exam tip: First decide which way the resultant force must point. For an upward-moving lift that is slowing down, it points downwards, so the tension is less than the weight.
A high-speed train of mass 4.0 × 105 kg is travelling at 60 m/s on level track. In an emergency it must stop within a distance of 2400 m. Assume the deceleration is constant.
Calculate the deceleration needed for the train to stop in exactly 2400 m.
Calculate the resultant force needed to produce this deceleration.
Air resistance and friction together provide a constant backward force of 4.0 × 104 N. Calculate the braking force needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Uses v2 = u2 + 2as with v = 0, u = 60 m/s and s = 2400 m: 0 = 602 + 2 × a × 2400 ; a = −3600 ÷ 4800 ; a = −0.75 m/s2, so the deceleration is 0.75 m/s2. Accept 0.75 m/s2 with or without the minus sign | ✓ 3 | |
| Part (b) | ||
| F = m × a = 4.0 × 105 × 0.75 ; F = 3.0 × 105 N | ✓ 2 | |
| Part (c) | ||
| Braking force = resultant force − other resistive forces = 3.0 × 105 − 4.0 × 104 ; braking force = 2.6 × 105 N | ✓ 2 | |
Exam tip: When finding a braking force, calculate the total resultant force first. Then subtract any forces that already act in the same backward direction.
A steam catapult on an aircraft carrier accelerates a jet uniformly from rest to 70 m/s over a distance of 90 m.
What is the acceleration of the jet?
| Marking point | Mark | Notes |
|---|---|---|
| Use v2 = u2 + 2as with u = 0. This gives a = v2 ÷ 2s = 4900 ÷ 180 = 27 m/s2. Option D leaves out the factor of 2. Option B divides velocity by distance, which is not an acceleration. Option A divides velocity by twice the distance. | ✓ 1 | Answer C |
Answer: C
Exam tip: If no time is given, use v² = u² + 2as. Remember to square the velocity.
A train driver makes an emergency stop on a straight, level track. The train decelerates at 1.2 m/s2 and travels 650 m before it stops. The speed limit on this section of track is 125 km/h.
Calculate the speed of the train when the brakes were applied.
Suggest why the speed calculated in the previous part might not be exactly the true speed.
Evaluate whether the train was exceeding the speed limit when the brakes were applied.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0 = u2 + 2 × (−1.2) × 650 | ✓ 1 | |
| u2 = 1560 | ✓ 1 | |
| u = 39.5 m/s | ✓ 1 | |
| Part (b) | ||
| The deceleration may not have been constant, e.g. because wheel grip or braking force changed | ✓ 1 | |
| Part (c) | ||
| 39.5 m/s = 39.5 × 3.6 = 142 km/h (or 125 km/h = 34.7 m/s) | ✓ 1 | |
| 142 km/h is greater than 125 km/h, so the train was exceeding the limit | ✓ 1 | |
Exam tip: When you evaluate, convert both values to the same unit, compare them, and finish with a clear conclusion.
In a quality-control test for engine oil, a steel ball is released from rest at the surface of oil in a tall glass tube. The ball falls to the bottom of the tube.
Explain how the forces on the ball change as it falls, and describe the effect on its motion. Refer to its velocity and acceleration in your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| At release, the velocity is zero and the drag is zero | ✓ 1 | |
| The weight is greater than the upward forces, so the resultant force is downwards and the ball accelerates | ✓ 1 | |
| As the velocity increases, the drag increases | ✓ 1 | |
| so the resultant force decreases and the acceleration decreases | ✓ 1 | |
| Eventually, drag (+ upthrust) equals the weight, so the resultant force is zero | ✓ 1 | |
| The acceleration is then zero and the ball falls at a constant (terminal) velocity | ✓ 1 | |
Exam tip: Build a chain: forces → resultant force → acceleration → velocity. Each link earns credit only if it is stated explicitly.
A high-speed lift in a tower block travels upwards from rest through a total height of 45 m and stops.
It accelerates uniformly at 1.2 m/s2 until it reaches 3.0 m/s, travels at a constant 3.0 m/s, and then decelerates uniformly at 1.2 m/s2 to rest.
Calculate the time the lift takes to reach 3.0 m/s.
Calculate the distance the lift travels while accelerating.
Calculate the total time for the journey.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| t = (3.0 − 0) ÷ 1.2 | ✓ 1 | |
| = 2.5 s | ✓ 1 | |
| Part (b) | ||
| distance = ½ × 3.0 × 2.5 (or average speed 1.5 m/s × 2.5 s) | ✓ 1 | |
| = 3.75 m | ✓ 1 | |
| Part (c) | ||
| distance at constant speed = 45 − 2 × 3.75 = 37.5 m | ✓ 1 | |
| time at constant speed = 37.5 ÷ 3.0 = 12.5 s | ✓ 1 | |
| total time = 2.5 + 12.5 + 2.5 = 17.5 s | ✓ 1 | |
Common error: The deceleration stage also covers distance; subtract both the speeding-up and slowing-down distances before finding the constant-speed time.
A car travelling at a constant 25 m/s passes a stationary police motorcyclist. At that instant the motorcyclist sets off in the same direction, accelerating uniformly at 2.5 m/s2 until reaching 30 m/s, and then continues at a constant 30 m/s. The car keeps a constant speed of 25 m/s.
Calculate the time taken for the motorcyclist to reach 30 m/s.
Calculate the distance travelled in this time by
(i) the motorcyclist
(ii) the car.
Calculate the total time, from when the car passes the motorcyclist, for the motorcyclist to catch up with the car.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| t = (30 − 0) ÷ 2.5 | ✓ 1 | |
| = 12 s | ✓ 1 | |
| Part (b) | ||
| (i) ½ × 30 × 12 | ✓ 1 | |
| = 180 m | ✓ 1 | |
| (ii) 25 × 12 = 300 m | ✓ 1 | |
| Part (c) | ||
| gap at 12 s = 300 − 180 = 120 m | ✓ 1 | |
| motorcyclist gains 30 − 25 = 5 m every second, so extra time = 120 ÷ 5 = 24 s | ✓ 1 | |
| total time = 12 + 24 = 36 s | ✓ 1 | |
Exam tip: When one object chases another, find the gap and divide it by the difference in their speeds to get the time needed to close it.
A tower crane has a uniform horizontal jib of length 24 m and weight 36 000 N. The jib is pivoted on the tower at a point 6.0 m from its left-hand end. A counterweight hangs from the left-hand end. A load of weight 20 000 N hangs from the right-hand end. The jib is balanced. The gravitational field strength is 10 N/kg.
Calculate the moment of the load about the pivot.
Calculate the weight of the counterweight.
Calculate the mass of the counterweight.
Explain why the crane still needs a counterweight when it carries no load.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moment = force × distance (18 m from the pivot) | ✓ 1 | |
| 20 000 × 18 = 360 000 N m | ✓ 1 | |
| Part (b) | ||
| The centre of the jib is 12 m from the left end, so 6.0 m from the pivot, giving a moment of 36 000 × 6.0 = 216 000 N m | ✓ 1 | |
| total clockwise moment = 216 000 + 360 000 = 576 000 N m (ECF from (a)) | ✓ 1 | |
| counterweight × 6.0 = 576 000 | ✓ 1 | |
| counterweight weight = 96 000 N | ✓ 1 | |
| Part (c) | ||
| 96 000 / 10 = 9600 kg (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| The weight of the jib acts to the right of the pivot and produces a clockwise moment | ✓ 1 | |
| so an anticlockwise moment from the counterweight is needed to balance it | ✓ 1 | |
Common error: A uniform beam's weight acts at its centre, so measure its distance from the pivot, not from the end of the beam.
A library uses a spring to test small weights. A student hangs weights from the spring and measures the extension. The results are shown.
Force / N: 0, 3.0, 6.0, 9.0, 12.0, 15.0
Extension / cm: 0, 1.5, 3.0, 4.5, 6.0, 8.4
State between which two forces the limit of proportionality is passed.
Calculate the spring constant of the spring in N/m.
A book hung on the spring produces an extension of 5.4 cm. Calculate the mass of the book. The gravitational field strength is 10 N/kg. Give your answer to 3 significant figures.
Explain why this spring should not be used to measure a weight of 15 N.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Up to 12 N the extension is directly proportional to the force | ✓ 1 | |
| The limit is passed between 12 N and 15 N, where 8.4 cm is more than the 7.5 cm expected | ✓ 1 | |
| Part (b) | ||
| spring constant = force / extension | ✓ 1 | |
| extension 1.5 cm = 0.015 m | ✓ 1 | |
| 3.0 / 0.015 = 200 N/m | ✓ 1 | |
| Part (c) | ||
| force = 200 × 0.054 = 10.8 N (ECF from (b)) | ✓ 1 | |
| weight = mass × g so mass = 10.8 / 10 | ✓ 1 | |
| mass = 1.08 kg | ✓ 1 | |
| Part (d) | ||
| 15 N is beyond the limit of proportionality, so extension is no longer proportional to force | ✓ 1 | |
| so the scale would give inaccurate readings (and the spring may be permanently stretched) | ✓ 1 | |
Exam tip: Always convert extensions from cm to m before using F = kx with k in N/m.
A small goods lift in a library carries books between floors. The lift and its load have a total mass of 120 kg. Starting from rest, the lift accelerates upwards uniformly at 0.50 m/s2 for 4.0 s. The gravitational field strength is 10 N/kg.
Calculate the speed of the lift after 4.0 s.
Calculate the distance the lift travels in the 4.0 s.
Calculate the tension in the lift cable during the 4.0 s.
After the 4.0 s the tension in the cable is 1200 N. Describe the motion of the lift.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v = u + at (or speed = acceleration × time) | ✓ 1 | |
| 0.50 × 4.0 = 2.0 m/s | ✓ 1 | |
| Part (b) | ||
| average speed = (0 + 2.0)/2 = 1.0 m/s (or s = ½at2) | ✓ 1 | |
| 1.0 × 4.0 = 4.0 m | ✓ 1 | |
| Part (c) | ||
| weight = 120 × 10 = 1200 N | ✓ 1 | |
| resultant force = ma = 120 × 0.50 = 60 N | ✓ 1 | |
| tension = 1200 + 60 = 1260 N | ✓ 1 | |
| Part (d) | ||
| The resultant force is zero | ✓ 1 | |
| so the lift moves at constant speed (2.0 m/s) upwards | ✓ 1 | |
Common error: The cable tension must be bigger than the weight for the lift to accelerate upwards: tension − weight = ma.
A spring is hung from a hook in a seed hopper. When a weight of 3.0 N hangs from it, its total length is 18.0 cm. When the weight is 5.0 N its total length is 22.0 cm. The spring obeys Hooke's law for both loads.
Calculate the spring constant, in N/m.
Calculate the unstretched length of the spring, in cm. Use the unrounded value of k.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 k = change in force ÷ change in length = 2.0 ÷ 0.040 | ✓ 1 | |
| A1 k = 50 N/m | ✓ 1 | |
| Part (b) | ||
| M1 extension at 3.0 N = F ÷ k = 3.0 ÷ 50 | ✓ 1 | |
| A1 = 0.060 m = 6.0 cm | ✓ 1 | |
| A1 unstretched length = 18.0 − 6.0 = 12.0 cm | ✓ 1 | |
Exam tip: Total length and extension are different: subtract the extension from the stretched length to get the unstretched length.
Two library trolleys with spring bumpers collide on a level floor. Trolley A (mass 18 kg) moves at 1.5 m/s towards trolley B (mass 12 kg), which is at rest. After the collision A continues in the same direction at 0.30 m/s.
Calculate the momentum of trolley A before the collision.
Calculate the speed of trolley B after the collision.
Calculate the kinetic energy of the trolleys before the collision.
Calculate the total kinetic energy after the collision, to 2 decimal places, and state whether the collision is elastic.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 momentum = mass × velocity | ✓ 1 | |
| A1 27 kg m/s | ✓ 1 | |
| Part (b) | ||
| momentum of A after = 18 × 0.30 = 5.4 kg m/s | ✓ 1 | |
| M1 momentum conserved: momentum of B = 27 − 5.4 = 21.6 kg m/s | ✓ 1 | |
| A1 speed of B = 21.6 / 12 = 1.8 m/s | ✓ 1 | |
| Part (c) | ||
| M1 KE = ½mv² = ½ × 18 × 1.5² | ✓ 1 | |
| A1 20.25 J | ✓ 1 | |
| Part (d) | ||
| KE of A = ½ × 18 × 0.30² = 0.81 J and KE of B = ½ × 12 × 1.8² = 19.44 J | ✓ 1 | |
| total KE = 20.25 J | ✓ 1 | |
| total KE is unchanged (20.25 J before and after), so the collision is elastic | ✓ 1 | |
Exam tip: Use 1.8 m/s exactly (not a rounded value) when finding the kinetic energy of B so the totals can be compared fairly.
Two tractors, A and B, travel in the same direction along a straight, level farm track. At time 0 tractor A is at the start of the track moving at a constant speed of 4.0 m/s. At time 0 tractor B is 150 m ahead of A, moving at a constant speed of 2.5 m/s.
Calculate the distance of tractor B from the start of the track after 20 s.
Calculate the speed at which the gap between the tractors decreases.
Calculate the time taken for tractor A to catch up with tractor B.
Calculate the distance from the start of the track at which A catches B.
Describe what would be seen on a distance–time graph of both tractors at the moment A catches B.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distance moved by B = 2.5 × 20 = 50 m | ✓ 1 | |
| 150 + 50 = 200 m | ✓ 1 | |
| Part (b) | ||
| 4.0 − 2.5 | ✓ 1 | |
| 1.5 m/s | ✓ 1 | |
| Part (c) | ||
| time = 150 ÷ 1.5 | ✓ 1 | |
| time = 100 s (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| 4.0 × 100 = 400 m (ECF from (c); accept 150 + 2.5 × 100) | ✓ 1 | |
| Part (e) | ||
| the two lines cross (intersect) at that time | ✓ 1 | |
Exam tip: Check a catching-up answer by finding the distance each vehicle has travelled when they meet; both must give the same position.
At a concert, a loudspeaker of mass 12 kg is lowered from the ceiling on a vertical rope. Use g = 10 N/kg.
Calculate the weight of the loudspeaker.
At one moment the loudspeaker accelerates downwards at 2.0 m/s². Calculate the resultant force on it.
Calculate the tension in the rope at this moment.
The loudspeaker is then lowered at a constant speed. State the tension in the rope and give a reason.
Later the loudspeaker is raised and accelerates upwards at 2.0 m/s². Calculate the tension in the rope.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight = mass × g = 12 × 10 | ✓ 1 | |
| weight = 120 N | ✓ 1 | |
| Part (b) | ||
| F = ma = 12 × 2.0 | ✓ 1 | |
| F = 24 N | ✓ 1 | |
| Part (c) | ||
| tension = weight − resultant force = 120 − 24 | ✓ 1 | |
| tension = 96 N (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| 120 N | ✓ 1 | |
| the resultant force is zero, so tension equals weight | ✓ 1 | |
| Part (e) | ||
| tension = weight + resultant force = 120 + 24 | ✓ 1 | |
| tension = 144 N | ✓ 1 | |
Common error: When the acceleration is downwards the tension is less than the weight; when it is upwards the tension is greater.
Two identical springs, each with spring constant 40 N/m, hang side by side from a fixed bar. A light rigid beam joins their lower ends and a 12 N weight hangs from its centre, so each spring carries an equal share of the load. Both springs stay within the limit of proportionality. What is the extension of each spring?
| Marking point | Mark | Notes |
|---|---|---|
| Each spring carries 12 / 2 = 6.0 N, so extension = F / k = 6.0 / 40 = 0.15 m. 0.30 m uses the whole 12 N on one spring and ignores the sharing. 0.075 m halves the load and also halves the extension again by using the combined stiffness of 80 N/m on the shared load. 0.60 m multiplies the single-spring answer by two instead of dividing the load. | ✓ 1 | Answer B |
Answer: B
Exam tip: Decide how much force acts on each spring first, then apply F = kx to one spring at a time.
A uniform diving board at a swimming pool is 3.0 m long and weighs 150 N. It rests on a support P that is 1.0 m from end A. End A is held down by a bolt. A swimmer of weight 500 N stands at the free end B. The board is in equilibrium.
State the equation used to calculate the moment of a force.
Calculate the moment of the swimmer's weight about P.
Calculate the moment of the weight of the board about P.
Calculate the downward force that the bolt exerts at A.
State what happens to the force at A if the swimmer moves towards the support P.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moment = force × perpendicular distance from the pivot | ✓ 1 | |
| Part (b) | ||
| M1 500 × 2.0 | ✓ 1 | |
| A1 1000 N m | ✓ 1 | |
| Part (c) | ||
| M1 the centre of mass is 1.5 m from A, which is 0.5 m from P, so 150 × 0.5 | ✓ 1 | |
| A1 75 N m | ✓ 1 | |
| Part (d) | ||
| M1 anticlockwise moment of bolt force = sum of clockwise moments: F × 1.0 = 1000 + 75 (ECF from (b) and (c)) | ✓ 1 | |
| A1 1075 N | ✓ 1 | |
| Part (e) | ||
| the force decreases because the swimmer's moment about P is smaller | ✓ 1 | |
Common error: Measure each distance from the pivot, and place the weight of a uniform board at its centre.
A gymnast of mass 50 kg drops from rest and falls freely through 3.2 m before landing on a thick mat. The mat compresses by 0.50 m while it brings her to rest. Ignore air resistance, assume a constant deceleration on the mat and take g = 10 N/kg (10 m/s2).
Calculate her speed when she first touches the mat.
Calculate her deceleration on the mat.
Calculate the average force exerted on her by the mat.
Explain why a thicker mat reduces the force on a gymnast who lands at the same speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v2 = 0 + 2 × 10 × 3.2 = 64 | ✓ 1 | |
| v = 8.0 m/s | ✓ 1 | |
| Part (b) | ||
| use of v2 = u2 + 2as with u = 8.0 and s = 0.50: a = 64 ÷ (2 × 0.50) | ✓ 1 | |
| 64 m/s2 | ✓ 1 | |
| Part (c) | ||
| resultant force = ma = 50 × 64 = 3200 N (ECF from (b)) | ✓ 1 | |
| the mat must also support her weight, 50 × 10 = 500 N | ✓ 1 | |
| force from mat = 3200 + 500 = 3700 N | ✓ 1 | |
| Part (d) | ||
| a greater stopping distance means a smaller deceleration (or longer stopping time) | ✓ 1 | |
| a smaller deceleration means a smaller resultant force, since F = ma | ✓ 1 | |
Common error: The force from the mat is larger than ma because the mat must also support the gymnast's weight.
A tower crane has a uniform horizontal jib of length 30 m and weight 60 kN. The jib is pivoted on the tower at a point 6.0 m from its left-hand end, so its centre of mass is 9.0 m to the right of the pivot. A counterweight is fixed at the left-hand end of the jib. A load of weight 20 kN hangs from the right-hand end of the jib.
Calculate the moment of the weight of the jib about the pivot.
Calculate the total clockwise moment about the pivot when the load is on the jib.
The jib is exactly balanced. Calculate the weight of the counterweight.
The load is removed. Calculate the resultant moment about the pivot and state which end of the jib tends to go down.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moment = force × perpendicular distance = 60 × 9.0 | ✓ 1 | |
| 540 kN m | ✓ 1 | |
| Part (b) | ||
| moment of load = 20 × 24 = 480 kN m (distance of right-hand end from pivot is 24 m) | ✓ 1 | |
| total = 540 + 480 = 1020 kN m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| anticlockwise moment = clockwise moment: W × 6.0 = 1020 (ECF from (b)) | ✓ 1 | |
| W = 170 kN | ✓ 1 | |
| Part (d) | ||
| anticlockwise moment = 170 × 6.0 = 1020 kN m and clockwise = 540 kN m, so resultant = 480 kN m (ECF from (c)) | ✓ 1 | |
| the left-hand (counterweight) end tends to go down | ✓ 1 | |
Exam tip: The weight of a uniform beam acts at its centre, so use the distance from the pivot to the centre of the beam, not to its end.
A hospital lift accelerates upwards at 2.0 m/s². A patient and trolley with a combined mass of 60 kg stand on its floor. Using g = 10 N/kg, what is the upward force from the floor on the trolley?
| Marking point | Mark | Notes |
|---|---|---|
| The resultant upward force is ma = 60 × 2.0 = 120 N. The floor force must also support the weight of 600 N, so force = 600 + 120 = 720 N, so this is correct. 120 N is the resultant force only. 480 N subtracts the resultant from the weight, as if the lift were accelerating downwards. 600 N is the force for constant speed, ignoring the acceleration. | ✓ 1 | Answer C |
Answer: C
Common error: When a lift accelerates upwards the floor force is larger than the weight, because it supplies the resultant force as well.
In a 100 m race a sprinter accelerates uniformly from rest for 3.0 s until reaching a top speed v. The sprinter then runs at constant speed v until the finish. The total race time is 10.5 s.
Determine the top speed v. Give your answer to 3 significant figures.
Calculate the acceleration during the first 3.0 s. Use the unrounded value of v and give your answer to 2 significant figures.
Calculate the average speed over the whole race. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distance while accelerating = ½ × v × 3.0 | ✓ 1 | |
| distance at constant speed = v × 7.5 | ✓ 1 | |
| total distance: 1.5v + 7.5v = 100, so 9v = 100 | ✓ 1 | |
| v = 11.1 m/s | ✓ 1 | |
| Part (b) | ||
| M1 a = 11.11 ÷ 3.0 | ✓ 1 | |
| A1 3.7 m/s² (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 average speed = 100 ÷ 10.5 | ✓ 1 | |
| A1 9.52 m/s | ✓ 1 | |
Common error: The top speed is greater than the average speed because the sprinter spends the first part of the race speeding up.
An ambulance of mass 3200 kg travels at 25 m/s on a dry road. The driver's reaction time is 0.80 s.
Calculate the thinking distance.
The ambulance then brakes with a constant deceleration of 5.0 m/s2 until it stops. Calculate the braking distance.
Calculate the total stopping distance.
On a wet road the maximum braking force is only 9600 N. Calculate the braking distance from 25 m/s on the wet road. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance = speed × time = 25 × 0.80 | ✓ 1 | |
| A1 20 m | ✓ 1 | |
| Part (b) | ||
| M1 v2 = 2as, so 252 = 2 × 5.0 × s | ✓ 1 | |
| A1 62.5 m | ✓ 1 | |
| Part (c) | ||
| A1 20 + 62.5 = 82.5 m (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| M1 deceleration = F ÷ m = 9600 ÷ 3200 = 3.0 m/s2 | ✓ 1 | |
| M1 s = v2 ÷ 2a = 625 ÷ 6.0 | ✓ 1 | |
| A1 104 m | ✓ 1 | |
Exam tip: Stopping distance = thinking distance + braking distance; a smaller braking force gives a smaller deceleration and a much longer braking distance.
A delivery van travels at 20 m/s. The driver has a reaction time of 0.80 s. After the brakes are applied the van decelerates uniformly at 5.0 m/s2. A pedestrian steps into the road 46 m ahead of the van when the driver first sees the pedestrian.
Calculate the braking distance if the van is brought to rest.
Calculate the thinking distance and the total stopping distance.
Explain why the van does not stop before reaching the pedestrian.
Determine the speed of the van when it reaches the pedestrian.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 use of v2 = u2 + 2as, with s = 202 ÷ (2 × 5.0) | ✓ 1 | |
| A1 40 m | ✓ 1 | |
| Part (b) | ||
| M1 thinking distance = 20 × 0.80 = 16 m | ✓ 1 | |
| A1 stopping distance = 16 + 40 = 56 m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| The stopping distance of 56 m is greater than 46 m (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 distance during braking = 46 − 16 = 30 m | ✓ 1 | |
| M1 v2 = 202 − 2 × 5.0 × 30 = 100 | ✓ 1 | |
| A1 10 m/s | ✓ 1 | |
Common error: The van travels at constant speed during the thinking time, so only the remaining distance is used when braking.
A steel ball bearing of mass 0.050 kg is released at the top of a tall tube of glycerine. The upthrust is negligible. The ball accelerates, then reaches terminal velocity. Take g = 10 N/kg.
Calculate the weight of the ball bearing.
At one instant during the fall the drag force on the ball is 0.20 N. Calculate the acceleration of the ball at this instant. Use the exact value from (a).
Explain why the acceleration of the ball is zero once it reaches terminal velocity although it is still moving.
A second ball of the same size has twice the mass. Explain why its terminal velocity is greater.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = mg = 0.050 × 10 | ✓ 1 | |
| 0.50 N | ✓ 1 | |
| Part (b) | ||
| resultant force = 0.50 − 0.20 = 0.30 N | ✓ 1 | |
| a = F ÷ m = 0.30 ÷ 0.050 | ✓ 1 | |
| 6.0 m/s2 | ✓ 1 | |
| Part (c) | ||
| the drag force has increased with speed until it equals the weight (0.50 N) | ✓ 1 | |
| the resultant force is zero, so there is no change in speed (constant velocity) | ✓ 1 | |
| Part (d) | ||
| its weight is larger, so a larger drag force is needed to balance it | ✓ 1 | |
| drag increases with speed, so a higher speed is needed to reach that drag force | ✓ 1 | |
Common error: Terminal velocity does not mean the forces have stopped acting. It means the forces are balanced so the resultant force is zero.
A tower crane has a counterweight of weight 60 kN fixed 10 m from the tower on one side. The uniform jib has a weight of 20 kN that can be assumed to act at a point 12 m from the tower on the other side. A load of weight 15 kN hangs from a trolley that can be moved along the jib. Give moments in kN m.
Calculate the moment of the counterweight about the tower.
Calculate the moment of the weight of the jib about the tower.
The crane is just balanced with the 15 kN load at its furthest safe distance. Calculate this distance.
A heavier load of weight 20 kN is now used. Calculate the greatest distance from the tower at which it can be safely hung.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moment = force × distance = 60 × 10 | ✓ 1 | |
| 600 kN m | ✓ 1 | |
| Part (b) | ||
| moment = 20 × 12 | ✓ 1 | |
| 240 kN m | ✓ 1 | |
| Part (c) | ||
| moment available for load = 600 − 240 = 360 kN m (ECF from (a) and (b)) | ✓ 1 | |
| distance = 360 ÷ 15 | ✓ 1 | |
| 24 m | ✓ 1 | |
| Part (d) | ||
| distance = 360 ÷ 20 (ECF from (c)) | ✓ 1 | |
| 18 m | ✓ 1 | |
Exam tip: For a balance problem, list every force that has a turning effect about the pivot, including the weight of the beam itself acting at its centre of mass.
A lift cabin and its load of boutique stock have a total mass of 440 kg. While the lift accelerates upwards from rest, the tension in the lift cable is 5280 N. The gravitational field strength is 10 N/kg.
Calculate the total weight of the cabin and load.
Calculate the resultant force on the cabin and load.
Calculate the acceleration of the lift.
The lift accelerates at this rate for 3.0 s from rest. Calculate the distance it travels in this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 W = m g | ✓ 1 | |
| A1 440 × 10 = 4400 N | ✓ 1 | |
| Part (b) | ||
| M1 resultant = tension − weight | ✓ 1 | |
| A1 5280 − 4400 = 880 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 a = F ÷ m | ✓ 1 | |
| A1 880 ÷ 440 = 2.0 m/s2 (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 use s = ut + ½at2 with u = 0, or find v = at and use an average speed | ✓ 1 | |
| M1 s = ½ × 2.0 × 3.02 (ECF from (c)) | ✓ 1 | |
| A1 9.0 m | ✓ 1 | |
Common error: The tension in the cable is not the resultant force. Subtract the weight from the tension before using F = ma.
A fashion show catwalk is a uniform horizontal beam of length 8.0 m and weight 300 N. It rests on a support P at its left-hand end and a support Q that is 6.0 m from P. A model of weight 450 N stands on the beam 7.0 m from P.
Calculate the upward force from support Q.
Calculate the upward force from support P.
The model walks further towards the right-hand end of the beam. Explain what happens to the force from support P.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 take moments about P | ✓ 1 | |
| A1 beam moment 300 × 4.0 = 1200 N m and model moment 450 × 7.0 = 3150 N m | ✓ 1 | |
| A1 Q = 4350 ÷ 6.0 = 725 N | ✓ 1 | |
| Part (b) | ||
| M1 upward forces equal downward forces: 300 + 450 − 725 (ECF from (a)) | ✓ 1 | |
| A1 25 N | ✓ 1 | |
| Part (c) | ||
| B1 the force from P decreases | ✓ 1 | |
| B1 the clockwise moment about Q from the model increases, and when it exceeds the anticlockwise moment of the beam's weight, P's force falls to zero and the beam starts to tip | ✓ 1 | |
Exam tip: If a support force comes out negative, the beam would need to be held down, so check that every support force is positive.
A uniform beam on a test rig is 6.0 m long and weighs 60 N. It rests on a pivot placed 1.0 m from its left-hand end. A load of 30 N hangs from the right-hand end. A second load W hangs from the left-hand end, and the beam is balanced horizontally.
State the principle of moments.
Calculate the load W.
Calculate the upward force exerted by the pivot on the beam.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| for a body in equilibrium the sum of the clockwise moments about a point | ✓ 1 | |
| equals the sum of the anticlockwise moments about the same point | ✓ 1 | |
| Part (b) | ||
| the centre of the beam is 2.0 m from the pivot, so the moment of its weight is 60 × 2.0 = 120 N m | ✓ 1 | |
| moment of the 30 N load = 30 × 5.0 = 150 N m | ✓ 1 | |
| W × 1.0 = 120 + 150 | ✓ 1 | |
| W = 270 N | ✓ 1 | |
| Part (c) | ||
| the beam is in equilibrium, so upward force = total downward force | ✓ 1 | |
| 270 + 60 + 30 = 360 N (ECF from (b)) | ✓ 1 | |
Common error: Measure each distance from the pivot, not from the end of the beam; the weight of a uniform beam acts at its centre.
A telecoms relay drone of mass 2.0 kg descends vertically. At one instant the upward drag force is 12 N. Take g = 10 N/kg. What is the acceleration of the drone at this instant?
| Marking point | Mark | Notes |
|---|---|---|
| Weight = 2.0 × 10 = 20 N down. Resultant = 20 − 12 = 8.0 N down, so a = 8.0 ÷ 2.0 = 4.0 m/s² downwards. Dividing the drag by the mass gives 6.0, ignoring the drag gives 10, and adding drag to weight gives 16. | ✓ 1 | Answer C |
Answer: C
Exam tip: Find the resultant force first, taking directions into account, then use F = m a.
An archaeologist uses a helical spring to find the mass of a small clay figure. A mass of 0.20 kg hung on the spring produces an extension of 2.4 cm. When the clay figure is hung from the same spring, the extension is 6.0 cm. The spring obeys Hooke's law for both loads. The gravitational field strength is 10 N/kg.
Calculate the spring constant of the spring, in N/m, to 2 significant figures.
Calculate the mass of the clay figure. Use the unrounded value of k from (a).
The archaeologist later hangs a much heavier object on the spring and finds that the spring no longer returns to its original length. Explain this observation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 weight = 0.20 × 10 = 2.0 N | ✓ 1 | |
| M1 k = F/x = 2.0 / 0.024 | ✓ 1 | |
| A1 83 N/m | ✓ 1 | |
| Part (b) | ||
| M1 F = kx with x = 0.060 m | ✓ 1 | |
| A1 weight = 5.0 N (accept 4.98 N from the rounded k) | ✓ 1 | |
| A1 mass = 0.50 kg (accept 0.498 kg) | ✓ 1 | |
| Part (c) | ||
| The load took the spring beyond its elastic limit | ✓ 1 | |
| The spring has been permanently deformed | ✓ 1 | |
Exam tip: If the spring constant is already known, a ratio of extensions gives the unknown mass directly: 0.20 × 6.0/2.4.
A tower crane has a horizontal jib that balances about the top of the tower. A concrete counterweight of mass 8000 kg hangs 6.0 m from the tower on one side. A load is lifted on the other side of the tower. Ignore the weight of the jib. Use g = 10 N/kg.
Calculate the moment of the counterweight about the tower.
Calculate the maximum mass of a load that can hang 24 m from the tower without the jib tipping towards the load.
The operator lifts the maximum load from (b) and moves it further from the tower. Explain what happens to the jib.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight of counterweight = 8000 × 10 = 80 000 N | ✓ 1 | |
| moment = 80 000 × 6.0 = 480 000 N m | ✓ 1 | |
| Part (b) | ||
| at balance, moment of load = moment of counterweight (480 000 N m, ECF from (a)) | ✓ 1 | |
| weight of load = 480 000 ÷ 24 = 20 000 N | ✓ 1 | |
| mass of load = 20 000 ÷ 10 = 2000 kg | ✓ 1 | |
| Part (c) | ||
| the moment of the load increases (same weight, greater distance) | ✓ 1 | |
| this is now greater than the moment of the counterweight, so the jib tips (rotates) towards the load | ✓ 1 | |
Exam tip: Moments are always taken about a stated pivot. For equilibrium, total clockwise moment equals total anticlockwise moment about that pivot.
Two sprinters, A and B, run a 100 m race. Each starts from rest and accelerates uniformly until reaching a top speed of 10.0 m/s, which each then holds until the finish. Sprinter A accelerates at 2.5 m/s2. Sprinter B accelerates at 2.0 m/s2.
Calculate the time taken by sprinter A to reach 10.0 m/s.
Calculate the distance covered by sprinter A while accelerating.
Show that sprinter A finishes the race in 12.0 s.
Calculate how far behind the finish line sprinter B is at the moment sprinter A finishes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = (v − u) ÷ a = 10.0 ÷ 2.5 | ✓ 1 | |
| A1 4.0 s | ✓ 1 | |
| Part (b) | ||
| M1 distance = average speed × time = ½ × 10.0 × 4.0 | ✓ 1 | |
| A1 20 m | ✓ 1 | |
| Part (c) | ||
| M1 remaining distance 100 − 20 = 80 m at 10.0 m/s takes 80 ÷ 10.0 = 8.0 s | ✓ 1 | |
| A1 total time = 4.0 + 8.0 = 12.0 s | ✓ 1 | |
| Part (d) | ||
| M1 time for B to reach top speed = 10.0 ÷ 2.0 = 5.0 s | ✓ 1 | |
| M1 distance covered by B while accelerating = ½ × 10.0 × 5.0 = 25 m | ✓ 1 | |
| M1 distance covered at top speed in the remaining 12.0 − 5.0 = 7.0 s = 70 m | ✓ 1 | |
| A1 B has covered 95 m, so is 5.0 m behind the line | ✓ 1 | |
Common error: Do not use the 12.0 s for the whole of B's journey at 10.0 m/s; B spends its first 5.0 s accelerating, so it covers less distance than 10.0 × 12.0.
A delivery van of mass 1200 kg carries medicines at 25 m/s. The driver sees a hazard 70 m ahead and, after a reaction time of 0.70 s, brakes with a constant braking force of 6000 N.
Calculate the deceleration of the van while braking.
Calculate the distance travelled by the van during the reaction time.
Calculate the braking distance needed to stop the van completely.
Show that the van cannot stop before reaching the hazard, and calculate the speed at which it reaches the hazard.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 a = F ÷ m = 6000 ÷ 1200 | ✓ 1 | |
| A1 5.0 m/s2 | ✓ 1 | |
| Part (b) | ||
| M1 distance = speed × time = 25 × 0.70 | ✓ 1 | |
| A1 17.5 m | ✓ 1 | |
| Part (c) | ||
| M1 s = v2 ÷ 2a = 252 ÷ (2 × 5.0) | ✓ 1 | |
| A1 62.5 m | ✓ 1 | |
| Part (d) | ||
| distance available for braking = 70 − 17.5 = 52.5 m, which is less than 62.5 m so the van cannot stop in time | ✓ 1 | |
| M1 v2 = 252 − 2 × 5.0 × 52.5 = 100 | ✓ 1 | |
| A1 v = 10 m/s | ✓ 1 | |
Exam tip: Subtract the thinking distance from the distance to the hazard before applying v² = u² + 2as, because the van does not decelerate during the reaction time.
A loaded library trolley of mass 40 kg moves at 1.5 m/s and collides with a stationary empty trolley of mass 20 kg. The two trolleys couple together. What is their speed immediately after the collision? Ignore friction.
| Marking point | Mark | Notes |
|---|---|---|
| Momentum is conserved: 40 × 1.5 = (40 + 20) × v, so v = 60/60 = 1.0 m/s. 1.5 m/s assumes speed is unchanged. 0.75 m/s halves the speed because there are two trolleys. 3.0 m/s divides the momentum by the mass of the stationary trolley only. | ✓ 1 | Answer B |
Answer: B
Common error: After a collision where objects stick together, divide the total momentum by the total mass of both objects.
A gymnast of mass 45 kg lands on a crash mat at a speed of 6.0 m/s. The mat brings her to rest in 0.30 s.
Calculate the momentum of the gymnast just before she touches the mat.
Calculate the average force exerted by the mat on the gymnast.
Explain why landing on a crash mat reduces the risk of injury compared with landing on a hard floor that stops her in 0.05 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity (or p = mv) | ✓ 1 | |
| 270 kg m/s | ✓ 1 | |
| Part (b) | ||
| force = change in momentum / time | ✓ 1 | |
| 270 / 0.30 = 900 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| The change in momentum is the same in both cases | ✓ 1 | |
| The mat increases the time taken to stop (force = change in momentum / time) | ✓ 1 | |
| so the force on her is smaller | ✓ 1 | |
Exam tip: Write the equation F = (mv − mu)/t first, then substitute. Stopping over a longer time with the same momentum change means a smaller force.
A fisherman releases a lead weight from rest on a bridge. It falls 45 m to the water. Ignore air resistance and use g = 10 m/s2. What is the speed of the weight just before it reaches the water?
| Marking point | Mark | Notes |
|---|---|---|
| With u = 0, v2 = 2as = 2 × 10 × 45 = 900, so v = 30 m/s. 900 m/s is the value of v2 and the square root was not taken. 21 m/s comes from leaving out the factor of 2 (v2 = 10 × 45 = 450). 90 m/s comes from multiplying 2 × 45 and forgetting both g and the square root. | ✓ 1 | Answer A |
Answer: A
Common error: v² = u² + 2as gives v squared. Always take the square root as the last step.
A mountaineer walks 8.0 km due east and then 6.0 km due north.
State the difference between distance and displacement.
State the total distance walked.
Calculate the magnitude of the mountaineer's displacement.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Distance is the total length of path travelled, a scalar with no direction | ✓ 1 | |
| Displacement is the straight-line distance from start to finish in a given direction, a vector | ✓ 1 | |
| Part (b) | ||
| 14 km | ✓ 1 | |
| Part (c) | ||
| M1 Pythagoras, √(8.0² + 6.0²) | ✓ 1 | |
| A1 10 km | ✓ 1 | |
Common error: Displacement is not the path length: the straight line from start to finish is shorter.
A tennis ball of mass 0.058 kg hits a practice wall at 20 m/s. It rebounds along the same line at 15 m/s.
What is the size of the change in momentum of the ball?
| Marking point | Mark | Notes |
|---|---|---|
| Momentum is a vector, so the direction reversal must be included. Taking towards the wall as positive, the initial momentum is 0.058 × 20 = +1.16 kg m/s. The final momentum is 0.058 × (−15) = −0.87 kg m/s. The change has size 1.16 + 0.87 = 2.03 kg m/s. 0.29 kg m/s ignores the change of direction and subtracts the speeds. 1.16 and 0.87 kg m/s are only the momentum before or after the collision. | ✓ 1 | Answer D |
Answer: D
Common error: When an object bounces back, the velocity changes sign. The sizes of the momenta before and after are added, not subtracted.
A gymnast of mass 48 kg drops from a high bar. She lands on a thick foam mat at a speed of 6.0 m/s. The mat brings her to rest in 0.30 s.
Calculate the momentum of the gymnast just before she lands.
Calculate the average force that the mat exerts to stop the gymnast. Ignore her weight.
Explain how the thick mat reduces the risk of injury compared with landing on a hard floor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = m × v = 48 × 6.0 ; p = 288 kg m/s (290 kg m/s) | ✓ 2 | |
| Part (b) | ||
| F = change in momentum ÷ time = 288 ÷ 0.30 ; F = 960 N | ✓ 2 | |
| Part (c) | ||
| The mat increases the time taken for the gymnast to stop ; her change in momentum is the same, so the rate of change of momentum is smaller ; so the force on her body is smaller | ✓ 3 | |
Exam tip: In 'safety feature' explanations, link all three ideas: longer time, same change in momentum, and so smaller force.
A robotic lander of mass 1200 kg touches down on the Moon with a vertical speed of 2.5 m/s. Its legs contain crushable aluminium sections. These bring the lander to rest in 0.40 s.
Calculate the momentum of the lander just before touchdown.
Calculate the average resultant force on the lander as it comes to rest.
Explain how the crushable sections protect the lander's instruments.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity = 1200 × 2.5 | ✓ 1 | |
| = 3000 kg m/s | ✓ 1 | |
| Part (b) | ||
| F = change in momentum ÷ time = 3000 ÷ 0.40 | ✓ 1 | |
| F = 7500 N | ✓ 1 | |
| Part (c) | ||
| They increase the time taken for the lander to stop | ✓ 1 | |
| The change in momentum is the same | ✓ 1 | |
| so the rate of change of momentum is smaller, and therefore the force is smaller | ✓ 1 | |
Exam tip: Safety explanations need three linked ideas: a longer time, the same change in momentum, and therefore a smaller force.
An astronaut of mass 90 kg is floating at rest outside a space station. She pushes a 6.0 kg toolbox away from her at 3.0 m/s.
What is the astronaut's speed just after the push?
| Marking point | Mark | Notes |
|---|---|---|
| The total momentum before the push is zero, so the total momentum afterwards is also zero. The toolbox's momentum is 6.0 × 3.0 = 18 kg m/s, so the astronaut has 18 kg m/s in the opposite direction. Her speed is 18 ÷ 90 = 0.20 m/s. 0.19 m/s divides by the combined mass of 96 kg. 3.0 m/s assumes both objects move at the same speed. 45 m/s uses the mass ratio the wrong way round. | ✓ 1 | Answer B |
Answer: B
Exam tip: In an explosion or push-apart problem, start from 'total momentum before = 0'. The two momenta afterwards are equal in size and opposite in direction.
A rock climber of mass 70 kg slips and falls. When the safety rope starts to stretch, she is moving downwards at 8.0 m/s. The rope brings her to rest in 0.50 s.
Calculate her momentum just as the rope starts to stretch.
Calculate the average resultant force on the climber while the rope stops her.
Climbing ropes are designed to stretch. Explain how this reduces the risk of injury to the climber.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity = 70 × 8.0 | ✓ 1 | |
| = 560 kg m/s (downwards) | ✓ 1 | |
| Part (b) | ||
| force = change in momentum ÷ time = 560 ÷ 0.50 | ✓ 1 | |
| = 1120 N (accept 1100 N); direction (upwards) not required | ✓ 1 | |
| Part (c) | ||
| Stretching increases the time taken to bring her to rest, for the same change in momentum (accept: smaller deceleration). | ✓ 1 | |
| So the rate of change of momentum, and therefore the force on her, is smaller. | ✓ 1 | |
Exam tip: For safety-feature questions, link three ideas: same change in momentum, longer time, smaller force.
Two skaters, A and B, stand together at rest on smooth ice and push against each other. Skater A has mass 60 kg and moves off at 1.5 m/s. Skater B has mass 45 kg.
State the total momentum of the two skaters before they push apart.
Calculate the speed of skater B after the push.
Calculate the total kinetic energy of the skaters after the push. Give the exact value.
State the energy transfer that provides this kinetic energy.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| zero | ✓ 1 | |
| Part (b) | ||
| momentum of A = 60 × 1.5 = 90 kg m/s | ✓ 1 | |
| momentum of B equals 90 kg m/s in the opposite direction (total momentum remains zero) | ✓ 1 | |
| speed = 90 ÷ 45 = 2.0 m/s | ✓ 1 | |
| Part (c) | ||
| KE = ½mv2 used for each skater: ½ × 60 × 1.52 = 67.5 J and ½ × 45 × 2.02 = 90 J | ✓ 1 | |
| 157.5 J (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| chemical energy store of the skaters' muscles is transferred to kinetic energy stores | ✓ 1 | |
Common error: Momentum has direction; the skaters move in opposite directions, so their momenta cancel to zero in total.
A research boat of mass 1200 kg is used to follow whales. It starts from rest and reaches a speed of 6.0 m/s in 15 s.
Calculate the change in momentum of the boat.
Calculate the average resultant force on the boat.
The same force acts for 15 s on a loaded boat of mass 2000 kg starting from rest. Calculate its final speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = mv (1200 × 6.0) | ✓ 1 | |
| 7200 kg m/s | ✓ 1 | |
| Part (b) | ||
| force = change in momentum ÷ time (7200 ÷ 15) (ECF from (a)) | ✓ 1 | |
| 480 N | ✓ 1 | |
| Part (c) | ||
| change in momentum = 480 × 15 = 7200 kg m/s | ✓ 1 | |
| v = 7200 ÷ 2000 = 3.6 m/s | ✓ 1 | |
Exam tip: Force equals change in momentum divided by time taken; this works even when the mass is different.
A lifeguard of mass 70 kg stands at rest on an inflatable raft of mass 30 kg floating on still water. She dives horizontally off the raft at a speed of 1.5 m/s relative to the water. Assume there is no resistance from the water.
Calculate the momentum of the lifeguard as she leaves the raft.
State the principle of conservation of momentum.
Calculate the speed of the raft immediately after she dives, and state its direction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity | ✓ 1 | |
| 70 × 1.5 = 105 kg m/s | ✓ 1 | |
| Part (b) | ||
| The total momentum before an interaction equals the total momentum after it | ✓ 1 | |
| provided no external force acts | ✓ 1 | |
| Part (c) | ||
| total momentum before = 0, so the raft has momentum 105 kg m/s (ECF from (a)) | ✓ 1 | |
| 30 × v = 105 | ✓ 1 | |
| v = 3.5 m/s in the opposite direction to the lifeguard | ✓ 1 | |
Common error: Momentum has direction. When the total before is zero, the two bodies must have equal and opposite momentum afterwards.
Grain falls vertically at 3.0 m/s onto a weighing platform. Each second 4.0 kg of grain arrives and comes to rest on the platform. What is the average force that the grain exerts on the platform because of this change in momentum?
| Marking point | Mark | Notes |
|---|---|---|
| Force equals the rate of change of momentum: each second the momentum lost is 4.0 × 3.0 = 12 kg m/s, so F = 12 N. 7.0 N comes from adding mass flow rate and speed. 0.75 N comes from dividing 3.0 by 4.0. 36 N comes from using mv2 (4.0 × 3.02). | ✓ 1 | Answer B |
Answer: B
Common error: Check the units of what you calculate: force is a momentum change divided by a time, not a product of mass and speed squared.
A light aircraft of mass 1.2 × 103 kg accelerates along a runway from 20 m/s to 38 m/s in 6.0 s. Assume the mass stays constant.
Calculate the change in momentum of the aircraft.
Calculate the resultant force on the aircraft.
The engine provides a forward thrust of 4200 N. Calculate the resistive force acting on the aircraft. Use the answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| change in momentum = m × Δv = 1200 × (38 − 20) | ✓ 1 | |
| = 21 600 kg m/s (2.16 × 104 kg m/s) | ✓ 1 | |
| Part (b) | ||
| force = change in momentum ÷ time = 21 600 ÷ 6.0 | ✓ 1 | |
| = 3600 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| resistive force = thrust − resultant force = 4200 − 3600 | ✓ 1 | |
| = 600 N (ECF from (b)) | ✓ 1 | |
Exam tip: The resultant force is the difference between the forces in opposite directions, not the engine force alone.
A gymnast of mass 40 kg lands on a crash mat at a speed of 6.0 m/s and comes to rest.
Calculate her momentum just before she touches the mat.
She is brought to rest in 0.50 s by the mat. Calculate the average force on her.
On a thin hard floor she would stop in 0.060 s. Calculate the average force in this case.
Explain how the mat reduces the risk of injury.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 momentum = mass × velocity | ✓ 1 | |
| A1 240 kg m/s | ✓ 1 | |
| Part (b) | ||
| M1 force = change in momentum / time (240 / 0.50) | ✓ 1 | |
| A1 480 N | ✓ 1 | |
| Part (c) | ||
| M1 force = 240 / 0.060 | ✓ 1 | |
| A1 4000 N | ✓ 1 | |
| Part (d) | ||
| the change in momentum is the same, but the mat makes the stopping time longer | ✓ 1 | |
| force = change in momentum / time, so a longer time gives a smaller force | ✓ 1 | |
Exam tip: Write force = change in momentum / time and substitute carefully; stopping from speed v means the change in momentum is mv.
A bale of hay is dropped from the door of a hayloft and hits the ground at a speed of 12 m/s. Air resistance is negligible and g = 10 N/kg. From what height was the bale dropped?
| Marking point | Mark | Notes |
|---|---|---|
| Using v² = u² + 2as with u = 0: 12² = 2 × 10 × h, so h = 144 ÷ 20 = 7.2 m, which is C. A (1.2 m) is v ÷ g, which is the fall time in seconds, not a height. B (3.6 m) comes from dividing by 4g instead of 2g. D (14.4 m) comes from forgetting the factor 2 and using h = v² ÷ g. | ✓ 1 | Answer C |
Answer: C
Common error: In v² = u² + 2as the factor 2 is part of the equation; leaving it out doubles the height.
During a concert, a metal tuning fork of mass 0.15 kg is knocked off a raised stage and falls 1.8 m to the floor. Assume air resistance is negligible and g = 10 N/kg.
Calculate the weight of the tuning fork.
Calculate the speed of the tuning fork just before it hits the floor.
Calculate the time taken for the tuning fork to fall. Use the speed from (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight = mass × g = 0.15 × 10 | ✓ 1 | |
| weight = 1.5 N | ✓ 1 | |
| Part (b) | ||
| v² = u² + 2as with u = 0 | ✓ 1 | |
| v² = 2 × 10 × 1.8 = 36 | ✓ 1 | |
| v = 6.0 m/s | ✓ 1 | |
| Part (c) | ||
| time = (v − u) ÷ a = 6.0 ÷ 10 | ✓ 1 | |
| time = 0.60 s (ECF from (b)) | ✓ 1 | |
Exam tip: State the equation, then substitute with u = 0 for an object that is dropped rather than thrown.
At a wind farm, air hits the blades of one turbine. Each second a mass of 3500 kg of air passes through the blades. The air arrives at a speed of 14 m/s and leaves at a speed of 9.0 m/s in the same direction.
State the equation that links force, change in momentum and time taken.
Calculate the change in momentum of the air each second.
Determine the size of the force that the air exerts on the blades, and state its direction.
Explain why the force on the blades would be smaller if the air left the turbine at a higher speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| force = change in momentum / time taken | ✓ 1 | |
| Part (b) | ||
| M1 3500 × (14 − 9.0) | ✓ 1 | |
| A1 17500 kg m/s | ✓ 1 | |
| Part (c) | ||
| 17500 N (ECF from (b)) | ✓ 1 | |
| in the direction of the wind, because the blades exert an equal and opposite force on the air | ✓ 1 | |
| Part (d) | ||
| the change in velocity / momentum of the air each second would be smaller | ✓ 1 | |
| so the rate of change of momentum, and hence the force, would be smaller | ✓ 1 | |
Exam tip: Force on one body from another is found from the rate of change of momentum of the body that is acted on, then use equal and opposite forces.
A ball of mass 0.40 kg hits a wall horizontally at 15 m/s and rebounds along the same line at 10 m/s. The ball is in contact with the wall for 0.020 s. What is the average force exerted on the ball by the wall?
| Marking point | Mark | Notes |
|---|---|---|
| Taking the rebound direction as positive, the change in momentum is 0.40 × (10 − (−15)) = 10 kg m/s. Force = change in momentum ÷ time = 10 ÷ 0.020 = 500 N. 100 N is wrong because it uses the difference in speeds (5 m/s) instead of adding them. 300 N is wrong because it treats the ball as just stopping (0.40 × 15 ÷ 0.020). 200 N is wrong because it only uses the rebound speed (0.40 × 10 ÷ 0.020). | ✓ 1 | Answer A |
Answer: A
Common error: When a ball rebounds, its velocity reverses, so the change in velocity is the sum of the two speeds.
A Formula 1 car brakes in a straight line from 60 m/s to 20 m/s over a distance of 160 m. Assuming constant deceleration, what is the size of the deceleration?
| Marking point | Mark | Notes |
|---|---|---|
| Use v² = u² + 2as: 20² = 60² + 2 × a × 160, so a = (400 − 3600) ÷ 320 = −10 m/s², a deceleration of 10 m/s², so this is correct. 0.25 comes from using speeds instead of squared speeds. 12.5 comes from adding 3600 and 400. 20 comes from forgetting the factor of 2 in the equation. | ✓ 1 | Answer A |
Answer: A
Exam tip: In v² = u² + 2as a negative acceleration means slowing down, so quote the size of the deceleration.
A Formula 1 car travelling at 70 m/s brakes in a straight line with a constant deceleration of 50 m/s² until its speed is 20 m/s. The driver has a mass of 75 kg.
Use v² = u² + 2as to calculate the distance travelled while braking.
Calculate the time taken to brake.
Calculate the resultant force on the driver during braking.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| substitutes into the equation: 20² = 70² + 2 × (−50) × s | ✓ 1 | |
| rearranges: 100s = 4900 − 400 | ✓ 1 | |
| s = 45 m | ✓ 1 | |
| Part (b) | ||
| M1 time = change in speed ÷ deceleration = (70 − 20) ÷ 50 | ✓ 1 | |
| A1 1.0 s | ✓ 1 | |
| Part (c) | ||
| M1 F = ma = 75 × 50 | ✓ 1 | |
| A1 3750 N | ✓ 1 | |
Exam tip: Use a negative value for the acceleration in v² = u² + 2as when the object is slowing down.
A squash ball of mass 0.50 kg hits a wall at 8.0 m/s and rebounds along the same line at 6.0 m/s. The contact with the wall lasts 0.020 s. What is the average force exerted by the wall on the ball?
| Marking point | Mark | Notes |
|---|---|---|
| The velocity reverses, so the change in velocity is 8.0 + 6.0 = 14 m/s. Change in momentum = 0.50 × 14 = 7.0 kg m/s and F = Δp ÷ t = 7.0 ÷ 0.020 = 350 N. 50 N subtracts the speeds (8.0 − 6.0). 200 N uses only the incoming speed. 150 N uses only the rebound speed. | ✓ 1 | Answer B |
Answer: B
Common error: When a ball rebounds, add the two speeds: the velocity changes direction, so the change in velocity is larger than either speed.
During a theatre performance a stage prop of mass 20 kg is on a smooth horizontal floor. Two ropes pull on it horizontally at right angles to each other. One rope pulls with a force of 40 N and the other with a force of 30 N. Assume no other horizontal forces act.
State what is meant by a vector quantity.
Calculate the magnitude of the resultant force on the prop.
Calculate the angle between the resultant force and the 40 N rope.
Calculate the acceleration of the prop.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A quantity that has both magnitude and direction | ✓ 1 | |
| Part (b) | ||
| M1 use of Pythagoras: √(402 + 302) | ✓ 1 | |
| A1 50 N | ✓ 1 | |
| Part (c) | ||
| M1 tan θ = 30 ÷ 40 (or cos θ = 40 ÷ 50 or sin θ = 30 ÷ 50) | ✓ 1 | |
| A1 37° (36.9°) | ✓ 1 | |
| Part (d) | ||
| M1 use of F = ma, rearranged to a = F ÷ m | ✓ 1 | |
| A1 50 ÷ 20 = 2.5 m/s2 (ECF from (b)) | ✓ 1 | |
Exam tip: Forces at right angles are combined using Pythagoras, not by adding their magnitudes.
A loaded trolley of mass 1200 kg moves along a track at 3.0 m/s and couples to a stationary trolley of mass 600 kg. The two trolleys move off together. Friction is negligible. What is their speed after coupling?
| Marking point | Mark | Notes |
|---|---|---|
| Momentum is conserved: 1200 × 3.0 = 3600 kg m/s before, and this equals (1200 + 600) × v after, so v = 3600 ÷ 1800 = 2.0 m/s. 1.5 m/s is the mean of 3.0 m/s and 0 m/s, which ignores the masses. 3.0 m/s assumes the speed does not change although the mass being moved has increased. 1.0 m/s comes from using the 600 kg mass for the initial momentum (600 × 3.0 ÷ 1800). | ✓ 1 | Answer A |
Answer: A
Common error: After a collision where objects stick together, use the total mass in the final momentum. Do not average the speeds.
During a rugby match an attacking player of mass 100 kg runs at 6.0 m/s. A defender of mass 80 kg runs directly towards him at 5.0 m/s. The two players collide, hold on to each other and move off together. Take the attacker's direction as positive.
Calculate the momentum of the attacker before the collision.
Calculate the velocity of the players immediately after the collision. Give your answer to 3 significant figures.
State how the force on the defender from the attacker compares with the force on the attacker from the defender. Explain why the two changes in momentum are equal in size.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity = 100 × 6.0 | ✓ 1 | |
| 600 kg m/s | ✓ 1 | |
| Part (b) | ||
| total momentum before = 600 − (80 × 5.0) = 200 kg m/s (momentum is conserved) | ✓ 1 | |
| total mass = 180 kg, so v = 200 ÷ 180 | ✓ 1 | |
| 1.11 m/s in the attacker's original direction | ✓ 1 | |
| Part (c) | ||
| the two forces are equal in size and opposite in direction (Newton's third law) | ✓ 1 | |
| the forces act for the same time, so the changes in momentum (force × time) are equal and opposite | ✓ 1 | |
Common error: Momentum is a vector. When two objects move towards each other, subtract their momenta using one direction as positive.
A fisherman on a pier throws a lead weight downwards at 3.0 m/s. The weight falls 8.0 m to the sea. Ignore air resistance and use g = 10 m/s2.
Calculate the speed of the weight just before it reaches the sea.
Calculate the time taken for the weight to reach the sea.
State and explain one effect that air resistance would have on the speed of the weight as it reaches the sea.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v2 = u2 + 2as | ✓ 1 | |
| M1 v2 = 3.02 + 2 × 10 × 8.0 = 169 | ✓ 1 | |
| A1 v = 13 m/s | ✓ 1 | |
| Part (b) | ||
| M1 t = (v − u) ÷ a (ECF from (a)) | ✓ 1 | |
| A1 (13 − 3.0) ÷ 10 = 1.0 s | ✓ 1 | |
| Part (c) | ||
| B1 the speed would be lower | ✓ 1 | |
| B1 air resistance acts upwards, opposing the motion, so the resultant force and the acceleration are smaller | ✓ 1 | |
Exam tip: Write down u, v, a, s and t before choosing an equation. Include the sign or direction of the acceleration.
A trawler winch cart of mass 800 kg rolls along a quay at 2.5 m/s. It collides with a stationary cart of mass 1200 kg and the two carts couple together and move off at the same velocity.
Calculate the momentum of the moving cart before the collision.
Calculate the velocity of the coupled carts after the collision.
The collision lasts 0.80 s. Calculate the average force on the stationary cart during the collision.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p = m v | ✓ 1 | |
| A1 800 × 2.5 = 2000 kg m/s | ✓ 1 | |
| Part (b) | ||
| M1 total momentum is conserved, so v = 2000 ÷ (800 + 1200) (ECF from (a)) | ✓ 1 | |
| A1 1.0 m/s | ✓ 1 | |
| Part (c) | ||
| M1 change in momentum of the stationary cart = 1200 × 1.0 = 1200 kg m/s (ECF from (b)) | ✓ 1 | |
| M1 F = change in momentum ÷ time | ✓ 1 | |
| A1 1200 ÷ 0.80 = 1500 N | ✓ 1 | |
Exam tip: In collision questions, state that total momentum before equals total momentum after, then use the combined mass for coupled objects.
A bridge-inspection drone hovers 45 m above a river and releases a bolt from rest. Ignore air resistance. Use g = 10 N/kg.
Calculate the speed of the bolt as it reaches the water.
Calculate the time taken for the bolt to fall.
In practice the bolt reaches the water more slowly than your answer to (a). Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v2 = u2 + 2as with u = 0 | ✓ 1 | |
| v2 = 2 × 10 × 45 = 900 | ✓ 1 | |
| v = 30 m/s | ✓ 1 | |
| Part (b) | ||
| time = change in speed ÷ acceleration, or t = v ÷ g | ✓ 1 | |
| 30 ÷ 10 = 3.0 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| air resistance acts upwards on the bolt | ✓ 1 | |
| so the resultant force and the acceleration are smaller than with weight alone | ✓ 1 | |
Common error: Use the equation v² = u² + 2as with v² on its own; taking the square root only at the end avoids carrying rounding errors forward.
Archaeologists estimate the depth of a flooded shaft by dropping a stone from the rim. The stone is released from rest and falls 45 m to the surface of the water. Ignore air resistance. The acceleration of free fall is 10 m/s².
Calculate the speed of the stone when it reaches the water.
Calculate the time taken for the stone to fall.
In practice the stone would reach the water at a lower speed than the value calculated in (a). Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v² = u² + 2as | ✓ 1 | |
| M1 v² = 0 + 2 × 10 × 45 = 900 | ✓ 1 | |
| A1 30 m/s | ✓ 1 | |
| Part (b) | ||
| M1 t = (v − u)/a = 30/10 | ✓ 1 | |
| A1 3.0 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Air resistance acts upwards, opposing the motion | ✓ 1 | |
| The resultant force (and so the acceleration) is smaller than the weight alone, so the final speed is smaller | ✓ 1 | |
Exam tip: Write the equation, substitute the values with their units, then give the answer; this earns the method marks even if the arithmetic slips.
An airport baggage tug of mass 1200 kg moves at 3.0 m/s and hits a stationary line of trolleys of total mass 800 kg. The tug and trolleys couple together and move off as one. What is their speed immediately after the collision?
| Marking point | Mark | Notes |
|---|---|---|
| Momentum is conserved: 1200 × 3.0 = (1200 + 800) × v, so v = 3600 ÷ 2000 = 1.8 m/s. 1.5 m/s is the average of 3.0 m/s and 0 m/s, which ignores the different masses. 3.0 m/s assumes the speed does not change when mass is added. 4.5 m/s comes from dividing by the trolley mass alone (3600 ÷ 800). | ✓ 1 | Answer B |
Answer: B
Common error: After a coupling collision, divide the total momentum by the combined mass, not by one object's mass.
Two skaters in an ice show stand still together on smooth ice. As the music starts they push apart. Skater A has mass 60 kg and moves away at 1.5 m/s. Skater B has mass 45 kg and moves in the opposite direction. Friction between the skates and the ice is negligible.
State the total momentum of the two skaters before they push apart.
Calculate the speed of skater B immediately after the push.
The push lasts 0.30 s. Use v = 2.0 m/s to calculate the average force on skater B during the push.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| zero (0 kg m/s) | ✓ 1 | |
| Part (b) | ||
| total momentum after the push is also zero (momentum is conserved) | ✓ 1 | |
| 45 × v = 60 × 1.5 | ✓ 1 | |
| v = 2.0 m/s | ✓ 1 | |
| Part (c) | ||
| force = change in momentum ÷ time = (45 × 2.0) ÷ 0.30 | ✓ 1 | |
| 300 N | ✓ 1 | |
Common error: Momentum is a vector. When two objects move apart from rest, their momenta are equal in size and opposite in direction, so the total stays zero.
A drummer's stick of mass 0.050 kg hits a drum skin at 4.0 m/s and rebounds in the opposite direction at 1.0 m/s. The stick is in contact with the drum skin for 0.010 s.
Calculate the momentum of the stick just before it hits the drum skin.
Calculate the change in momentum of the stick.
Calculate the average force the drum skin exerts on the stick.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = 0.050 × 4.0 = 0.20 kg m/s | ✓ 1 | |
| Part (b) | ||
| taking the initial direction as positive, momentum after = −0.050 × 1.0 = −0.050 kg m/s | ✓ 1 | |
| change = 0.20 − (−0.050) | ✓ 1 | |
| 0.25 kg m/s | ✓ 1 | |
| Part (c) | ||
| force = change in momentum ÷ time = 0.25 ÷ 0.010 (ECF from (b)) | ✓ 1 | |
| 25 N | ✓ 1 | |
Common error: A rebound reverses the direction of the velocity, so the change in momentum is the sum of the two magnitudes, not their difference.
A uniform pipe section of length 6.0 m and mass 80 kg rests horizontally on two supports. Support P is at the left end of the pipe. Support Q is 4.0 m from P. A worker of mass 60 kg stands on the pipe 2.0 m from P. Use g = 10 N/kg.
Calculate the upward force exerted by support Q. Take moments about P.
Calculate the upward force exerted by support P.
The worker walks slowly towards support Q. Explain what happens to the force exerted by support P.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the weight of the pipe, 800 N, acts at its centre, 3.0 m from P, and the worker's weight is 600 N | ✓ 1 | |
| M1 anticlockwise moment of Q = clockwise moments: Q × 4.0 = 800 × 3.0 + 600 × 2.0 | ✓ 1 | |
| A1 Q = 3600 ÷ 4.0 = 900 N | ✓ 1 | |
| Part (b) | ||
| M1 total upward force = total downward force: P + Q = 800 + 600 = 1400 N | ✓ 1 | |
| A1 P = 1400 − 900 = 500 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| the force at P decreases | ✓ 1 | |
| the worker's weight is now closer to Q, so Q must provide a larger share of the total weight (and P a smaller share) | ✓ 1 | |
Exam tip: Take moments about one of the unknown forces so that it drops out of the equation, then use 'upward = downward' for the other.
A gymnast of mass 55 kg lands on a crash mat with a downward speed of 6.0 m/s and comes to rest in 0.50 s.
Calculate the momentum of the gymnast just before she touches the mat.
Calculate the average resultant force on the gymnast while she is stopping.
Landing on a thin, hard floor instead stops the gymnast in a much shorter time. Explain why the force on her would be larger.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 momentum = mass × velocity = 55 × 6.0 | ✓ 1 | |
| A1 330 kg m/s | ✓ 1 | |
| Part (b) | ||
| M1 force = change in momentum ÷ time = 330 ÷ 0.50 | ✓ 1 | |
| A1 660 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| the change in momentum is the same | ✓ 1 | |
| force = rate of change of momentum, so a shorter time gives a larger force | ✓ 1 | |
Exam tip: Write force = change in momentum ÷ time, then use the same change in momentum to compare different stopping times.
A uniform lighting truss in a theatre is 4.0 m long and weighs 240 N. It hangs horizontally from two vertical chains, one at each end. A spotlight of weight 160 N is clamped to the truss 1.0 m from the left-hand end.
State the principle of moments.
State the total upward force provided by the two chains.
Calculate the tension in the right-hand chain.
Calculate the tension in the left-hand chain.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| For an object in equilibrium, the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about that point. | ✓ 1 | |
| Part (b) | ||
| 240 + 160 = 400 N | ✓ 1 | |
| Part (c) | ||
| Take moments about the left-hand end: the truss's weight acts at its centre, 2.0 m from the left end. | ✓ 1 | |
| TR × 4.0 = (240 × 2.0) + (160 × 1.0) = 640 N m | ✓ 1 | |
| TR = 160 N | ✓ 1 | |
| Part (d) | ||
| TL = 400 − 160 = 240 N | ✓ 1 | |
Exam tip: Take moments about one of the supports. The unknown force at that support then has no moment, leaving only one unknown.
An astronaut is floating at rest outside a space station. The astronaut and spacesuit have a total mass of 120 kg. The astronaut throws a 2.4 kg wrench away from their body. The wrench leaves the astronaut's hand at 5.0 m/s relative to the space station. The throw lasts 0.40 s.
Calculate the average force the astronaut exerts on the wrench.
State the size and direction of the force the wrench exerts on the astronaut, and name the law you used.
Calculate the velocity of the astronaut immediately after the throw.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = (mv − mu) ÷ t ; F = (2.4 × 5.0 − 0) ÷ 0.40 ; F = 30 N | ✓ 3 | |
| Part (b) | ||
| 30 N in the opposite direction to the throw ; Newton's third law | ✓ 2 | |
| Part (c) | ||
| Total momentum before = 0, so momentum of astronaut = momentum of wrench ; 120 × v = 2.4 × 5.0 = 12 kg m/s ; v = 0.10 m/s in the opposite direction to the wrench | ✓ 3 | |
Exam tip: When the starting momentum is zero, the two objects move apart with equal and opposite momenta. State the direction as well as the speed.
A uniform footbridge of weight 40 000 N and length 12 m rests on two supports, A and B, one at each end. A maintenance vehicle of weight 20 000 N is parked on the bridge, 3.0 m from support A.
Calculate the upward force from support B.
Calculate the upward force from support A.
The vehicle drives slowly from A to B. Describe how the forces from the two supports change.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Taking moments about A: (40 000 × 6.0) + (20 000 × 3.0) = 300 000 N m | ✓ 1 | |
| RB × 12 = 300 000 | ✓ 1 | |
| RB = 25 000 N | ✓ 1 | |
| Part (b) | ||
| Total upward force = total downward force: RA + RB = 60 000 N | ✓ 1 | |
| RA = 35 000 N | ✓ 1 | |
| Part (c) | ||
| The force from A decreases and the force from B increases | ✓ 1 | |
| The sum of the two forces stays constant at 60 000 N | ✓ 1 | |
Exam tip: Take moments about one support so that support's unknown force has no moment. Then use 'upward forces = downward forces' to find the other force.
A cargo probe of mass 2400 kg approaches a stationary storage module of mass 9600 kg in space at 0.50 m/s. The probe docks and locks onto the module, and they move off together. External forces are negligible.
State the principle of conservation of momentum.
Calculate the momentum of the probe before docking.
Calculate the velocity of the probe and module together after docking.
The docking takes 4.0 s. Calculate the average force exerted on the module during docking.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| When no external force acts, the total momentum before an interaction equals the total momentum after it. | ✓ 1 | |
| Part (b) | ||
| momentum = 2400 × 0.50 | ✓ 1 | |
| = 1200 kg m/s | ✓ 1 | |
| Part (c) | ||
| total momentum after = 1200 kg m/s | ✓ 1 | |
| combined mass = 2400 + 9600 = 12 000 kg | ✓ 1 | |
| v = 1200 ÷ 12 000 = 0.10 m/s (in the probe's original direction) | ✓ 1 | |
| Part (d) | ||
| change in momentum of module = 9600 × 0.10 = 960 kg m/s | ✓ 1 | |
| force = 960 ÷ 4.0 = 240 N | ✓ 1 | |
Common error: After docking, divide by the combined mass, not by the probe's mass alone.
A 0.50 kg ball hits a wall at 8.0 m/s and rebounds along the same line at 6.0 m/s. It is in contact with the wall for 0.020 s. What is the average force on the ball?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: the velocity changes from +8.0 to −6.0 m/s, a change of 14 m/s, so F = 0.50 × 14 / 0.020 = 350 N. 50 N subtracts the speeds (2.0). 200 N ignores the rebound. 150 N uses only the rebound speed. | ✓ 1 | Answer A |
Answer: A
Common error: A rebound reverses direction, so the speeds add when finding the change in velocity.
Two ice skaters stand together at rest and push apart. Skater B has mass 72 kg and moves off at 1.5 m/s. Skater A moves off in the opposite direction at 2.25 m/s.
Calculate the mass of skater A.
Explain why the lighter skater has the greater speed.
Calculate the ratio kinetic energy of A : kinetic energy of B.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum of B = 72 × 1.5 = 108 kg m/s | ✓ 1 | |
| momentum of A has the same size in the opposite direction | ✓ 1 | |
| mass = 108 ÷ 2.25 = 48 kg | ✓ 1 | |
| Part (b) | ||
| the two momenta are equal in size | ✓ 1 | |
| for the same momentum a smaller mass needs a larger speed | ✓ 1 | |
| Part (c) | ||
| KE of A = ½ × 48 × 2.252 = 121.5 J and KE of B = ½ × 72 × 1.52 = 81 J | ✓ 1 | |
| ratio = 121.5 ÷ 81 = 1.5 (ECF from (a)) | ✓ 1 | |
Common error: Equal and opposite momenta do not mean equal kinetic energies.
An irrigation hose directs a horizontal jet of water at a vertical wall. In each second, 0.80 kg of water leaves the hose at 12 m/s and hits the wall.
Assume that the water stops when it hits the wall. Calculate the force exerted by the water on the wall.
In practice the water splashes back from the wall. Explain why the force on the wall is greater than your answer to (a).
The water rebounds from the wall at 3.0 m/s. Calculate the force exerted on the wall.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum lost per second = 0.80 × 12 = 9.6 kg m/s | ✓ 1 | |
| force = change in momentum ÷ time = 9.6 N | ✓ 1 | |
| Part (b) | ||
| The water's velocity changes direction, so its change in momentum each second is larger | ✓ 1 | |
| Force = rate of change of momentum, so the force is larger | ✓ 1 | |
| Part (c) | ||
| M1 change in velocity = 12 + 3.0 = 15 m/s (water reverses direction) | ✓ 1 | |
| A1 force = 0.80 × 15 = 12 N | ✓ 1 | |
Common error: When an object bounces back, the change in velocity is the sum of the speeds, not the difference.
A uniform metre rule of weight 3.0 N is pivoted at its 0.20 m mark. A weight W hung from the 0 m mark balances the rule horizontally. What is W?
| Marking point | Mark | Notes |
|---|---|---|
| The rule's weight acts at its centre (0.50 m mark), 0.30 m from the pivot, giving a moment of 3.0 × 0.30 = 0.90 N m. W acts 0.20 m from the pivot, so W × 0.20 = 0.90 and W = 4.5 N, which is D. A gives the moment instead of the force. B swaps the two distances (3.0 × 0.20 / 0.30). C wrongly assumes equal weights balance. | ✓ 1 | Answer D |
Answer: D
Exam tip: Measure each distance from the pivot, and remember that a uniform object's weight acts at its centre.
A uniform plank of length 4.0 m and weight 200 N is pivoted at a point 1.0 m from its left-hand end. An adult of weight 500 N sits at the left-hand end. A bag is placed at the right-hand end so that the plank balances horizontally. Use g = 10 N/kg.
State the principle of moments.
Calculate the anticlockwise moment of the adult about the pivot.
Calculate the weight of the bag.
Calculate the mass of the bag.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| for a body in equilibrium, the sum of the clockwise moments about any point | ✓ 1 | |
| equals the sum of the anticlockwise moments about that point | ✓ 1 | |
| Part (b) | ||
| M1 moment = force × perpendicular distance (500 × 1.0) | ✓ 1 | |
| A1 500 N m | ✓ 1 | |
| Part (c) | ||
| weight of plank acts at its centre, 1.0 m to the right of the pivot, giving a moment of 200 N m | ✓ 1 | |
| M1 clockwise moments equal anticlockwise: 200 + W × 3.0 = 500 | ✓ 1 | |
| A1 W = 100 N | ✓ 1 | |
| Part (d) | ||
| M1 mass = weight / g (100 / 10) | ✓ 1 | |
| A1 10 kg (ECF from (c)) | ✓ 1 | |
Common error: Take the plank's own weight as acting at its centre, which here is not at the pivot.
A passenger aircraft touches down on a runway at a speed of 60 m/s and decelerates uniformly to rest in a distance of 900 m. The mass of the aircraft is 1.5 × 104 kg.
Calculate the deceleration of the aircraft.
Calculate the resultant braking force on the aircraft. Use a deceleration of 2.0 m/s².
Calculate the time taken for the aircraft to stop.
On another occasion the aircraft touches down at 66 m/s and has the same deceleration of 2.0 m/s². Calculate its stopping distance.
The runway is 1000 m long. State whether the aircraft in (d) can stop on the runway, with a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0 = 60² + 2 × a × 900 (use of v² = u² + 2as) | ✓ 1 | |
| a = −3600 ÷ 1800 | ✓ 1 | |
| deceleration = 2.0 m/s² | ✓ 1 | |
| Part (b) | ||
| F = ma = 1.5 × 104 × 2.0 | ✓ 1 | |
| F = 3.0 × 104 N | ✓ 1 | |
| Part (c) | ||
| t = (v − u) ÷ a = 60 ÷ 2.0 | ✓ 1 | |
| t = 30 s | ✓ 1 | |
| Part (d) | ||
| 0 = 66² − 2 × 2.0 × s, so s = 66² ÷ (2 × 2.0) | ✓ 1 | |
| s = 1089 m | ✓ 1 | |
| Part (e) | ||
| no, because 1089 m is greater than 1000 m (ECF from (d)) | ✓ 1 | |
Exam tip: Stopping distance rises with the square of the speed, so a small increase in speed gives a larger increase in distance.
Two skaters move towards each other along a straight line on smooth ice. Skater A has a mass of 48 kg and moves east at 5.0 m/s. Skater B has a mass of 72 kg and moves west at 3.0 m/s. They collide, hold on to each other and move off together. Take east as positive.
State the equation for momentum.
Calculate the total momentum of the two skaters before the collision.
Calculate the velocity of the skaters immediately after the collision, giving the magnitude and the direction.
Explain why the total momentum is the same before and after the collision.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity | ✓ 1 | |
| Part (b) | ||
| M1 momentum of A = 48 × 5.0 = +240 kg m/s | ✓ 1 | |
| M1 momentum of B = 72 × (−3.0) = −216 kg m/s | ✓ 1 | |
| A1 total = +24 kg m/s | ✓ 1 | |
| Part (c) | ||
| M1 total momentum after = (48 + 72) × v with value 24 (ECF from (b)) | ✓ 1 | |
| A1 v = 0.20 m/s | ✓ 1 | |
| A1 east / direction in which A was moving | ✓ 1 | |
| Part (d) | ||
| there is no resultant external force on the skaters (friction negligible), so momentum is conserved | ✓ 1 | |
Common error: Choose a positive direction first and give any momentum in the opposite direction a negative sign.
A spacecraft in deep space fires a thruster that ejects 0.040 kg of gas backwards every second. The gas leaves at a speed of 3000 m/s relative to the spacecraft. At the start of the burn the spacecraft has mass 800 kg. Ignore the change in the mass of the spacecraft in parts (a) to (c).
Calculate the momentum given to the gas each second.
Explain why the gas exerts a forward force on the spacecraft, and state the size of this force.
Calculate the initial acceleration of the spacecraft.
The force stays the same, but the acceleration gradually increases during the burn. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum = mass × velocity = 0.040 × 3000 | ✓ 1 | |
| 120 kg m/s | ✓ 1 | |
| Part (b) | ||
| the spacecraft exerts a backward force on the gas, so by Newton's third law the gas exerts an equal and opposite force on the spacecraft | ✓ 1 | |
| force = rate of change of momentum = 120 ÷ 1 = 120 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| use of a = F ÷ m = 120 ÷ 800 (ECF from (b)) | ✓ 1 | |
| 0.15 m/s2 | ✓ 1 | |
| Part (d) | ||
| the spacecraft loses mass as gas is ejected | ✓ 1 | |
| a = F ÷ m, so the same force gives a larger acceleration | ✓ 1 | |
Exam tip: Force equals the rate of change of momentum: ejecting more mass per second, or at a higher speed, gives a bigger force.
A squeegee slips from the hand of a window cleaner working on the wall of a hospital. It falls vertically from rest and hits the ground at 12 m/s. Ignore air resistance and use g = 10 m/s².
Calculate the height from which the squeegee fell.
Calculate the time taken to fall.
A second identical squeegee is thrown vertically downwards with an initial speed of 4.0 m/s from the same height. Calculate its speed when it hits the ground. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| uses v² = u² + 2as with u = 0 | ✓ 1 | |
| 144 = 2 × 10 × s | ✓ 1 | |
| s = 7.2 m | ✓ 1 | |
| Part (b) | ||
| M1 t = (v − u) ÷ a = 12 ÷ 10 | ✓ 1 | |
| A1 1.2 s | ✓ 1 | |
| Part (c) | ||
| M1 v² = 4.0² + 2 × 10 × 7.2 = 160 (ECF from (a)) | ✓ 1 | |
| A1 v = 12.6 m/s | ✓ 1 | |
Common error: Add the squares of speeds in v² = u² + 2as, not the speeds themselves: 4.0 + 12 would not give the right answer.
Two trolleys move towards each other along a straight hospital corridor. Trolley A has mass 80 kg and moves to the right at 2.0 m/s. Trolley B has mass 120 kg and moves to the left at 1.0 m/s. They collide and couple together. Take movement to the right as positive.
Calculate the momentum of each trolley before the collision.
State the total momentum before the collision.
Calculate the velocity of the coupled trolleys after the collision.
The collision lasts 0.50 s. Calculate the average force on trolley B during the collision.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A: 80 × 2.0 = +160 kg m/s | ✓ 1 | |
| B: 120 × (−1.0) = −120 kg m/s | ✓ 1 | |
| Part (b) | ||
| +40 kg m/s (to the right) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 total momentum is conserved, so v = 40 ÷ (80 + 120) | ✓ 1 | |
| A1 0.20 m/s to the right | ✓ 1 | |
| Part (d) | ||
| M1 change in velocity of B = 0.20 − (−1.0) = 1.2 m/s | ✓ 1 | |
| M1 F = m × Δv ÷ t = 120 × 1.2 ÷ 0.50 | ✓ 1 | |
| A1 288 N | ✓ 1 | |
Common error: Momentum has direction: give velocities going the other way a negative sign before adding.
The table gives the speed of a cargo aircraft of mass 8000 kg during part of its take-off run on a runway.
| Time / s | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| Speed / m/s | 0 | 15 | 27 | 36 | 42 |
Calculate the acceleration of the aircraft during the first 5 s.
Calculate the resultant force on the aircraft between 15 s and 20 s.
Explain why the acceleration of the aircraft decreases during the take-off run.
Use the data to estimate the distance travelled in the 20 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 a = change in speed ÷ time = 15 ÷ 5 | ✓ 1 | |
| A1 3.0 m/s2 | ✓ 1 | |
| Part (b) | ||
| M1 acceleration = (42 − 36) ÷ 5 | ✓ 1 | |
| A1 1.2 m/s2 | ✓ 1 | |
| A1 F = ma = 8000 × 1.2 = 9600 N | ✓ 1 | |
| Part (c) | ||
| Resistive forces (such as air resistance) increase as speed increases while the engine force stays about the same | ✓ 1 | |
| so the resultant force decreases and therefore the acceleration decreases | ✓ 1 | |
| Part (d) | ||
| M1 average speed in each 5 s interval multiplied by 5 s (or area under the speed–time graph) | ✓ 1 | |
| M1 (0 + 15) ÷ 2 × 5 + (15 + 27) ÷ 2 × 5 + (27 + 36) ÷ 2 × 5 + (36 + 42) ÷ 2 × 5 | ✓ 1 | |
| A1 495 m | ✓ 1 | |
Exam tip: For non-uniform acceleration, split the motion into short time intervals and find the average speed for each.
A container ship of mass 6.0 × 107 kg is moving at 8.0 m/s when its engines are switched off. The water exerts a constant resistive force of 3.0 × 105 N on the ship. Assume no other horizontal forces act.
Calculate the deceleration of the ship.
Calculate the time taken for the ship to stop.
Calculate the distance travelled by the ship while stopping.
A tug now also pulls back on the ship with a constant force of 1.0 × 105 N, so the total resisting force is 4.0 × 105 N. Calculate the new stopping distance from 8.0 m/s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 a = F ÷ m = 3.0 × 105 ÷ 6.0 × 107 | ✓ 1 | |
| A1 0.0050 m/s2 | ✓ 1 | |
| Part (b) | ||
| M1 t = v ÷ a = 8.0 ÷ 0.0050 (ECF from (a)) | ✓ 1 | |
| A1 1600 s | ✓ 1 | |
| Part (c) | ||
| M1 distance = average speed × time = 4.0 × 1600 (ECF from (b)) | ✓ 1 | |
| A1 6400 m | ✓ 1 | |
| Part (d) | ||
| M1 new deceleration = 4.0 × 105 ÷ 6.0 × 107 = 0.00667 m/s2 | ✓ 1 | |
| M1 use of v2 = 2as, so s = 8.02 ÷ (2 × 0.00667) | ✓ 1 | |
| A1 4800 m (4.8 km) | ✓ 1 | |
Exam tip: A larger resisting force gives a larger deceleration and a shorter stopping distance from the same speed.
In a training session a rugby player of mass 90 kg running at 8.0 m/s is brought to rest by a tackle. The player slows down at a constant rate and the tackle lasts 0.60 s.
Calculate the change in momentum of the player.
Calculate the average force exerted on the player by the tackler.
Use Newton's third law to describe the force the player exerts on the tackler during the tackle.
Calculate the distance the player travels during the tackle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| change in momentum = mass × change in velocity = 90 × 8.0 | ✓ 1 | |
| 720 kg m/s | ✓ 1 | |
| Part (b) | ||
| force = change in momentum ÷ time = 720 ÷ 0.60 (ECF from (a)) | ✓ 1 | |
| 1200 N | ✓ 1 | |
| Part (c) | ||
| the force is also 1200 N (equal in size) | ✓ 1 | |
| opposite in direction (acting on the tackler) | ✓ 1 | |
| Part (d) | ||
| average speed = (8.0 + 0) ÷ 2 = 4.0 m/s (constant deceleration) | ✓ 1 | |
| distance = average speed × time = 4.0 × 0.60 | ✓ 1 | |
| 2.4 m | ✓ 1 | |
Exam tip: Force = change in momentum ÷ time. A longer contact time for the same change in momentum means a smaller average force, which is why padded surfaces reduce injury.
A robotic arm throws a ball vertically upwards at 12 m/s. Ignore air resistance and use g = 10 m/s2.
State the velocity of the ball at its highest point.
Calculate the maximum height reached above the point of release.
Calculate the time taken for the ball to reach its highest point.
Calculate the total time for the ball to return to the height from which it was released.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 zero | ✓ 1 | |
| Part (b) | ||
| M1 v2 = u2 + 2as with v = 0 and a = −10 | ✓ 1 | |
| M1 0 = 122 − 2 × 10 × s | ✓ 1 | |
| A1 s = 7.2 m | ✓ 1 | |
| Part (c) | ||
| M1 t = (v − u) ÷ a = (0 − 12) ÷ (−10) | ✓ 1 | |
| A1 1.2 s | ✓ 1 | |
| Part (d) | ||
| M1 time down equals time up, so double the time to the top (ECF from (c)) | ✓ 1 | |
| A1 2 × 1.2 = 2.4 s | ✓ 1 | |
Common error: At the highest point the velocity is zero but the acceleration is still 10 m/s² downwards.
A sensor package is thrown vertically upwards at 10 m/s from the deck of a bridge. The river is 15 m below the deck and the package falls into it on the way down. Ignore air resistance. Use g = 10 N/kg.
Calculate the maximum height reached by the package above the deck.
Calculate the speed of the package as it reaches the river.
A second package is thrown vertically downwards from the deck at 10 m/s. Calculate its speed as it reaches the river.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0 = 102 − 2 × 10 × h, or v2 = u2 + 2as with v = 0 | ✓ 1 | |
| h = 5.0 m | ✓ 1 | |
| Part (b) | ||
| on returning to deck level the speed is 10 m/s downwards | ✓ 1 | |
| v2 = 102 + 2 × 10 × 15 = 400 | ✓ 1 | |
| v = 20 m/s | ✓ 1 | |
| (accept any valid method, for example a fall of 20 m from rest at the top giving v2 = 2 × 10 × 20 = 400 after adding 5.0 m, ECF from (a)) | — | |
| Part (c) | ||
| v2 = 102 + 2 × 10 × 15 = 400 | ✓ 1 | |
| v = 20 m/s | ✓ 1 | |
Exam tip: A ball thrown upwards passes the launch point on the way down at the same speed it was thrown, so the second half of the motion can be treated as a fall that starts at that speed.
A telecoms trolley of mass 12 kg carrying cable drums moves at 2.0 m/s and couples to a stationary trolley of mass 8.0 kg. The two trolleys move off together. Assume there are no external horizontal forces.
Calculate the momentum of the 12 kg trolley before the collision.
Calculate the speed of the trolleys after they couple.
The coupling takes 0.30 s. Calculate the average force on the 8.0 kg trolley during this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p = m v, 12 × 2.0 | ✓ 1 | |
| A1 24 kg m/s | ✓ 1 | |
| Part (b) | ||
| Momentum is conserved so total momentum after is 24 kg m/s (ECF from (a)) | ✓ 1 | |
| M1 24 = (12 + 8.0) × v | ✓ 1 | |
| A1 1.2 m/s | ✓ 1 | |
| Part (c) | ||
| M1 change in momentum = 8.0 × 1.2 = 9.6 kg m/s (ECF from (b)) | ✓ 1 | |
| M1 force = change in momentum ÷ time, 9.6 ÷ 0.30 | ✓ 1 | |
| A1 32 N | ✓ 1 | |
Common error: After the trolleys couple, use the combined mass in p = m v.
A BASE jumper of total mass 80 kg falls from a cliff. Take g = 10 N/kg.
Calculate the weight of the jumper.
Later, the drag force on the jumper is 1400 N upwards. Calculate the resultant force and state its direction.
Calculate the acceleration of the jumper at this instant and state its direction.
Explain how the jumper's speed changes after this instant until it becomes constant.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 W = m g, 80 × 10 | ✓ 1 | |
| A1 800 N | ✓ 1 | |
| Part (b) | ||
| M1 1400 − 800 | ✓ 1 | |
| A1 600 N upwards (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 a = F ÷ m, 600 ÷ 80 | ✓ 1 | |
| A1 7.5 m/s² upwards (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| The jumper slows down because the resultant force is upwards | ✓ 1 | |
| As speed decreases, drag decreases | ✓ 1 | |
| Until drag equals weight (800 N), when the speed becomes constant | ✓ 1 | |
Exam tip: A resultant force acting against the direction of motion causes deceleration, which can be shown as a negative acceleration.
A cyclist and bicycle have a combined mass of 80 kg. On a level track the cyclist brakes with a constant resultant force, slowing from 12 m/s to 4.0 m/s over a distance of 32 m.
What is the size of the resultant force?
| Marking point | Mark | Notes |
|---|---|---|
| From v² = u² + 2as, 4.0² = 12² + 2 × a × 32 gives a deceleration of 2.0 m/s². Then F = ma = 80 × 2.0 = 160 N. 20 N comes from using (12 − 4.0)/32 as the acceleration, which is not the correct equation. 80 N comes from using (12 − 4.0)² = 64 in place of the difference of the squared speeds, giving 1.0 m/s². 320 N comes from leaving out the factor of 2 in 2as, giving 4.0 m/s². | ✓ 1 | Answer C |
Answer: C
Common error: The equation v² = u² + 2as needs the squares of the speeds subtracted, not the squared difference and not the difference of the speeds.
A technician on a gantry in a data centre throws a small cable clip vertically upwards at 6.0 m/s. It leaves her hand 8.0 m above the floor and later lands on the floor. Ignore air resistance. The acceleration of free fall is 10 m/s².
Calculate the maximum height reached by the clip above the point of release.
Calculate the speed of the clip when it hits the floor.
A second clip is thrown vertically downwards from the same point at 6.0 m/s. State whether it hits the floor at a speed greater than, equal to or less than 14 m/s, and justify your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 0 = 6.0² − 2 × 10 × s (or s = v²/2g) | ✓ 1 | |
| A1 1.8 m | ✓ 1 | |
| Part (b) | ||
| M1 v² = u² + 2as | ✓ 1 | |
| M1 v² = 6.0² + 2 × 10 × 8.0 = 196 | ✓ 1 | |
| A1 14 m/s | ✓ 1 | |
| Part (c) | ||
| Equal to 14 m/s | ✓ 1 | |
| The first clip returns to the release point moving downwards at 6.0 m/s, so its motion from there on is the same as the second clip's | ✓ 1 | |
Common error: When using v² = u² + 2as, take the direction of the acceleration into account: a clip thrown upwards has acceleration −10 m/s² until it turns round.
A uniform plank of length 5.0 m and mass 30 kg rests horizontally on two supports. The left support is 0.50 m from the left end of the plank. The right support is 1.0 m from the right end. A person of mass 50 kg walks from the middle of the plank towards the right end. Use g = 10 N/kg.
State the distance from the left end of the plank to the right support.
Calculate the moment of the weight of the plank about the right support.
The plank is about to tip over the right support. At that moment the force from the left support is zero. Calculate the distance of the person from the right support.
Determine whether the person can reach the right end of the plank without the plank tipping. Use your answer to (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 4.0 m | ✓ 1 | |
| Part (b) | ||
| M1 weight = 30 × 10 = 300 N acting at the centre, 1.5 m from the right support | ✓ 1 | |
| A1 moment = 300 × 1.5 = 450 N m | ✓ 1 | |
| Part (c) | ||
| M1 taking moments about the right support, moment of person = moment of plank | ✓ 1 | |
| M1 500 × d = 450 (ECF from (b)) | ✓ 1 | |
| A1 d = 0.90 m beyond the right support | ✓ 1 | |
| Part (d) | ||
| the end of the plank is 1.0 m beyond the right support | ✓ 1 | |
| 1.0 m is greater than 0.90 m, so the plank tips before the person reaches the end (ECF from (c)) | ✓ 1 | |
Exam tip: At the point of tipping the force at the support being left behind is zero, so only two moments about the other support remain.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy