International GCSE Physics · Pearson Edexcel 4PH1 · Topic 1
Forces and motion
Original exam-style questions with mark schemes. Questions marked “Paper 2 only” use content assessed only on Paper 2. Notes 1(b) →Notes 1(c) →
40questions
248marks
Freefull mark schemes
Original questions in the style of each exam boardEvery answer re-solved independently before publishingCreated by Evan Kotronis with AI assistanceUpdated 13 October 2026 · new questions added regularlyHow every question is checked →
1IGCSE-PHY-1-001Scalars and vectorsPapers 1 & 2Grade 4–51 mark
Which of these quantities is a vector?
Show mark scheme
D (weight);
A is incorrect because distance has size only (it is the scalar version of displacement)
B is incorrect because mass has size only; it is weight, not mass, that acts downwards
C is incorrect because speed has size only (velocity is the vector)
State the independent variable in this investigation.
(1)
(b)
Hooke's law states that extension is directly proportional to the applied force. Use the data to explain whether the rubber band obeys Hooke's law.
(2)
(c)
When the weights are removed, the rubber band returns to its original length. State the word used to describe this behaviour.
(1)
(d)
State one safety precaution the student should take.
(1)
Show mark scheme
(a)
force / weight (added to the band);
Allow
allow 'number of weights' / 'mass added'
Do not accept
'extension'
(b)
does not obey Hooke's law / extension is not (directly) proportional to force;
use of data e.g. doubling force from 1.0 N to 2.0 N does not double extension (3.0 → 5.2 cm) OR increases in extension get smaller (3.0, 2.2, 1.6, 1.2, 0.9 cm);
Allow
allow 'graph would not be a straight line through the origin' for MP2 if linked to data
Do not accept
'it obeys Hooke's law because extension increases' scores 0
(c)
elastic;
Allow
allow 'elasticity'
Do not accept
'plastic'
'stretchy'
(d)
any one from:
wear eye protection / goggles (in case the band snaps);
keep feet away from under the weights / place a tray of sand or a cushion beneath;
clamp the stand to the bench / use a counterweight so it does not topple;
7IGCSE-PHY-1-021Distance–time graph: comparing speedsPapers 1 & 2Grade 4–51 mark
A robot vacuum cleaner moves along a straight hallway. The graph shows how the distance travelled by the robot changes with time. The graph is divided into four sections, P, Q, R and S.
In which section is the robot moving fastest?Show mark scheme
B (section R);
A is incorrect because the robot travels the greatest distance in P (10 m) but takes 20 s, so its speed is only 0.5 m/s; the steepest gradient, not the largest distance, shows the greatest speed
C is incorrect because the line in Q is horizontal, so the robot is stationary (speed 0 m/s)
D is incorrect because S is highest on the graph (greatest total distance) but its gradient is the smallest of the moving sections (0.3 m/s)
8IGCSE-PHY-1-022Forces on a sliding ice hockey puckPapers 1 & 2Grade 4–58 marks
An ice hockey player hits a puck with a stick. The puck slides in a straight line across the ice and gradually slows down. While it slides, three forces act on the puck: its weight, the upward force from the ice and friction. The puck then hits a wooden board at the edge of the rink and bounces straight back.
(a)
Which of the three forces acting on the sliding puck is a non-contact force?
(1)
(b)
The weight of the puck is 1.7 N. State the size of the upward force from the ice on the sliding puck. Give a reason for your answer.
(2)
(c)
A student says: 'Nothing is pushing the puck backwards once it leaves the stick, so the resultant force on the sliding puck is zero.' Explain why the student is wrong.
(2)
(d)
The puck hits the board at a speed of 12 m/s and bounces straight back at a speed of 12 m/s. A student says: 'The velocity of the puck has not changed, because its speed is still 12 m/s.' Explain why the student is wrong.
(2)
(e)
During the collision the puck is squashed slightly. State which effect of a force this shows.
(1)
Show mark scheme
(a)
weight;
Allow
allow 'gravity' / 'gravitational force'
Do not accept
'friction'
'normal force'
(b)
1.7 (N);
vertical forces are balanced / puck does not accelerate up or down / no resultant vertical force;
Allow
allow 'equal to the weight' for MP1
allow 'the puck stays on the surface of the ice' for MP2
Do not accept
MP2 without a value for MP1 or 'equal to the weight'
'Newton's third law pair' as the reason
(c)
the puck is slowing down / decelerating (so there must be a resultant force);
friction acts (backwards) opposite to the motion / friction is the (unbalanced) resultant force;
Allow
allow 'its speed changes so the forces are not balanced' for MP1
allow 'friction opposes motion' for MP2
Do not accept
'there is a forward force from the stick still acting'
'gravity slows it down'
(d)
velocity is a vector / velocity has (size and) direction;
the direction (of motion) has changed / reversed (so the velocity has changed);
Allow
allow 'speed is a scalar but velocity is a vector' for MP1
allow 'velocity changes from +12 m/s to −12 m/s' for MP2
10IGCSE-PHY-1-024Velocity–time graph of a sprinterPapers 1 & 2Grade 4–57 marks
The graph shows how the velocity of a sprinter changes during the first 8.0 s of a race.
(a)
Describe the motion of the sprinter between 4.0 s and 8.0 s.
(1)
(b)
Determine the acceleration of the sprinter in the first 4.0 s.
(2)
(c)
Determine the distance the sprinter runs in the first 4.0 s.
(2)
(d)
A second sprinter reaches the same top velocity in a shorter time. Describe how the velocity–time graph for the second sprinter would differ from the graph shown during the first part of the race.
(2)
Show mark scheme
(a)
constant velocity / constant speed (of 10 m/s);
Allow
allow 'steady speed'
allow 'zero acceleration'
Do not accept
'stationary' / 'not moving'
(b)
use of gradient / substitution into a = (v − u)/t;
evaluation;
e.g. a = (10 − 0) / 4.0
(a =) 2.5 (m/s²)
Allow
correct answer with no working scores 2
Do not accept
4.0 / 10 = 0.4 (m/s²) scores 0
(c)
use of area under the graph / ½ × 4.0 × 10;
evaluation;
(distance =) 20 (m)
Allow
correct answer with no working scores 2
Do not accept
10 × 4.0 = 40 (m) scores 0
(d)
(first section of graph is) steeper / has a greater gradient;
reaches 10 m/s / becomes horizontal at an earlier time (less than 4.0 s);
11IGCSE-PHY-1-025Practical: analysing force–extension data for a springPapers 1 & 2Grade 4–55 marks
A student investigates how the extension of a steel spring from a kitchen scale varies with the force applied to it. The table shows her results. The graph shows her results plotted as points.
12IGCSE-PHY-1-005Practical: motion of everyday objectsPapers 1 & 2Grade 6–79 marks
A student investigates the motion of a toy car rolling down a ramp. The student releases the car from rest at a start line. The car rolls 1.50 m down the ramp to a finish line. The student times how long the car takes to reach the finish line. She repeats this three times.
Time / s: 1.84, 1.78, 1.90
(a)
Name two measuring instruments the student needs.
(2)
(b(i))
Determine the mean time.
(1)
(b(ii))
Calculate the average speed of the toy car between the start line and the finish line.
(2)
(c)
Suggest one way to improve the accuracy of the time measurements. Explain why your suggestion is an improvement.
(2)
(d)
The student wants to find out if the toy car accelerates as it moves down the ramp. Suggest what additional measurements she should take and how she would use them.
(2)
Show mark scheme
(a)
metre rule / tape measure;
stopwatch / stopclock;
Allow
allow timer
allow light gate(s) (and data logger) for the second mark
Do not accept
'ruler' alone (unless 'metre ruler')
'clock'
(b(i))
1.84 (s);
Allow
allow 1.8 (s)
(b(ii))
substitution into speed = distance / time;
evaluation;
e.g. speed = 1.50 / 1.84
(speed =) 0.82 (m/s)
Allow
allow 0.815, 0.8152… (m/s)
ECF from (b)(i)
allow 82 (cm/s)
(c)
use light gates (connected to a timer / data logger) OR film the car and use video playback with timer;
idea that this removes (human) reaction time (in starting/stopping the stopwatch);
Allow
allow 'use a longer ramp so that reaction time is a smaller fraction of the time measured' with matching explanation
Do not accept
'repeat more times' (improves reliability/precision of mean, not accuracy of each reading) — scores 0
'be more careful'
(d)
measure the time taken to reach a halfway point (e.g. 0.75 m) as well as the finish line;
calculate the average speed for each half and compare: if the second half has a greater average speed the car is accelerating;
Allow
allow light gates at two points to measure two speeds and compare
allow measure time for equal successive distances and show the times decrease
Do not accept
'measure the speed at the end' without a second speed for comparison
13IGCSE-PHY-1-006Velocity–time graphs: area and average speedPapers 1 & 2Grade 6–71 mark
An electric scooter is moving at 4.0 m/s. It accelerates uniformly to a velocity of 12 m/s in 5.0 s. It then travels at a constant velocity of 12 m/s for a further 5.0 s. What is the average speed of the scooter for the whole 10.0 s?
Show mark scheme
C (10 m/s);
A is incorrect because it is the mean of the initial and final velocities of the accelerating section, not total distance ÷ total time
B is incorrect because it treats the accelerating section as a triangle starting from rest (½ × 5.0 × 12 = 30 m), ignoring the initial velocity of 4.0 m/s
D is incorrect because it is the maximum speed, not the average
15IGCSE-PHY-1-009Stopping distance: thinking and braking distancePapers 1 & 2Grade 6–71 mark
A car is travelling along a straight road at 20 m/s. The driver sees a hazard and brakes. The driver's reaction time is 0.70 s. The braking distance of the car is 32 m. What is the stopping distance of the car?
Show mark scheme
C (46 m);
A is incorrect because it is the thinking distance only (20 × 0.70); the braking distance must be added
B is incorrect because it is the braking distance only; the car also travels during the driver's reaction time
D is incorrect because the thinking distance has been calculated as speed ÷ reaction time (20 ÷ 0.70 = 29 m) instead of speed × reaction time
16IGCSE-PHY-1-011Resultant force in a liftPapers 1 & 2Grade 6–78 marks
A lift and its passengers have a total mass of 800 kg. The lift starts from rest and accelerates upwards. After 3.0 s its velocity is 2.4 m/s. The lift is pulled upwards by a steel cable. Ignore friction. Use g = 10 N/kg.
(a)
State the formula linking acceleration, change in velocity and time taken.
(1)
(b)
Calculate the acceleration of the lift.
(2)
(c)
Calculate the resultant force on the lift.
(2)
(d)
Calculate the force exerted on the lift by the cable while the lift accelerates.
(3)
Show mark scheme
(a)
acceleration = change in velocity / time taken;
Allow
standard symbols and rearrangements e.g. a = (v − u)/t
Do not accept
a = v/t unless v defined as change in velocity
(b)
substitution;
evaluation;
e.g. a = (2.4 − 0) / 3.0
(a =) 0.80 (m/s²)
Allow
allow 0.8 (m/s²)
(c)
substitution into F = m × a;
evaluation;
e.g. F = 800 × 0.80
(F =) 640 (N)
Allow
ECF from (b)
(d)
weight = 8000 (N);
idea that cable force − weight = resultant force OR cable force = weight + resultant force;
17IGCSE-PHY-1-014Practical: extension of a helical springPapers 1 & 2Grade 6–76 marks
A student wants to find out whether a steel helical spring obeys Hooke's law. She has a clamp stand, the spring, a 100 g mass hanger with eight 100 g slotted masses, a metre rule and a pointer that can be attached to the bottom of the spring.
Describe how the student could carry out this investigation and how she should use her results to decide whether the spring obeys Hooke's law. You should include how to obtain accurate results and how to work safely.
(6)
Show mark scheme
Indicative content:
MP1. hang the spring from the clamp (stand) with the metre rule fixed vertically beside it;
MP2. attach the pointer to the bottom of the spring and record its initial reading on the rule;
MP3. add masses in equal steps / one 100 g mass at a time;
MP4. force calculated using W = m × g (e.g. 0.10 kg × 10 N/kg = 1.0 N per mass);
MP5. record the new pointer reading after each mass is added;
MP6. extension = new reading − initial reading;
MP7. read the rule at eye level / perpendicular to the scale to avoid parallax;
MP8. wait for the spring to stop oscillating before taking a reading;
MP9. repeat (e.g. while removing the masses) and calculate a mean extension for each force;
MP10. safety: wear eye protection in case the spring or masses fly off;
MP11. safety: clamp the stand to the bench / use a counterweight, and keep feet clear of the masses;
MP12. plot a graph of extension against force (or force against extension);
MP13. obeys Hooke's law where the graph is a straight line through the origin / extension directly proportional to force;
Allow
allow 'pointer' to be a marker / tape flag
allow use of a set square to read the pointer for MP7
Point-based alternative matching 4PH1 practice: award 1 mark per MP to a maximum of 6; a method with no way of determining extension cannot score above 3
Do not accept
'measure the length of the spring' with no initial reading for MP2/MP6
'be careful' for safety
'repeat to make it accurate' without a mean for MP9
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.
18IGCSE-PHY-1-026Resultant force and acceleration: sailing dinghyPapers 1 & 2Grade 6–71 mark
A sailing dinghy and its sailor have a total mass of 150 kg. At one instant, the wind exerts a forward force of 480 N on the sail and the water exerts a backward resistive force of 300 N on the dinghy.
What is the acceleration of the dinghy at this instant?
Show mark scheme
A (1.2 m/s²);
B is incorrect because it uses only the resistive force: 300 ÷ 150 = 2.0
C is incorrect because it uses only the forward force: 480 ÷ 150 = 3.2, ignoring the resistive force
D is incorrect because it adds forces acting in opposite directions: (480 + 300) ÷ 150 = 5.2
19IGCSE-PHY-1-027Aircraft take-off runPapers 1 & 2Grade 6–77 marks
An airliner starts from rest at one end of a runway and accelerates uniformly along it. The airliner leaves the ground when its speed reaches 72 m/s. The acceleration of the airliner is 2.0 m/s².
(a)
Calculate the minimum length of runway needed for the airliner to take off.
(3)
(b)
Calculate the time taken for the airliner to reach take-off speed.
(2)
(c)
The mass of the airliner is 65 000 kg. Calculate the resultant force on the airliner during its take-off run.
20IGCSE-PHY-1-028Practical: timing a ball rolling along a corridorPapers 1 & 2Grade 6–79 marks
Some students investigate the motion of a tennis ball rolling along a school corridor. They mark lines on the floor every 2.0 m. A student stands at each line with a stopwatch. The ball is pushed and all the stopwatches are started as the ball passes the 0 m line. Each student stops their stopwatch as the ball passes their line.
Name a suitable measuring instrument for marking out the lines every 2.0 m.
(1)
(b)
Calculate the average speed of the ball over the 10.0 m.
(2)
(c)
Describe how the speed of the ball changes as it rolls along the corridor. Use data from the table to support your answer.
(2)
(d)
Name the force that causes the change in speed of the ball.
(1)
(e)
The students plot a distance–time graph of their results. Describe the shape of the graph.
(1)
(f)
Each time in the table has an uncertainty of about 0.2 s because of the students' reaction times. Explain why the average speed calculated for the first 2.0 m is much less reliable than the average speed calculated for the whole 10.0 m.
(2)
Show mark scheme
(a)
tape measure / metre rule / trundle wheel;
Allow
allow 'measuring tape'
allow 'metre ruler'
Do not accept
'stopwatch'
'ruler' alone
(b)
substitution into average speed = distance / time;
evaluation;
e.g. average speed = 10.0 / 5.6
(average speed =) 1.8 (m/s)
Allow
allow 1.79 (m/s)
correct answer with no working scores 2
Do not accept
5.6 / 10.0 = 0.56 (m/s) scores 0
(c)
speed decreases / ball slows down;
use of data e.g. time to travel each 2.0 m increases (0.8 s, 0.9 s, 1.1 s, 1.3 s, 1.5 s) OR speed falls from 2.5 m/s to 1.3 m/s;
Allow
allow 'decelerates'
allow any two correct interval times or speeds compared for MP2
Do not accept
'time increases so speed increases'
(d)
friction / air resistance;
Allow
allow 'drag' / 'rolling resistance'
Do not accept
'gravity'
'weight'
(e)
curve with decreasing gradient / curve that levels off;
Allow
allow 'curve getting less steep'
Do not accept
'straight line'
'curve getting steeper'
(f)
0.2 s is a large fraction of the time for the first 2.0 m (0.8 s) / about 25%;
but only a small fraction of the time for 10.0 m (5.6 s) / about 4% (so the percentage uncertainty in the speed is much smaller);
Allow
allow 'the uncertainty is the same size but the time measured is much longer' for MP2 if linked to the 10.0 m time
allow percentages in the range 20–30% and 3–5%
Do not accept
'the ball is slower at the end so it is easier to time' with no reference to the size of the uncertainty compared with the time
'repeat the experiment' (does not answer the question)
Use data from the table to calculate the reaction time of the driver.
(2)
(c)
On another journey on a dry road, the driver is travelling at 20 m/s when she sees a fallen tree 50 m ahead. Determine whether the van stops before it reaches the tree.
(2)
(d)
Compare how the thinking distance and the braking distance change when the speed doubles from 15 m/s to 30 m/s.
(2)
(e)
The van is now fully loaded with heavy parcels and is driven on a wet road. Explain how these two changes affect the stopping distance of the van at a given speed.
(3)
Show mark scheme
(a)
distance travelled (at constant speed) between the driver seeing a hazard and applying the brakes / during the driver's reaction time;
Allow
allow 'distance travelled while the driver reacts'
Do not accept
'distance travelled while braking'
(b)
substitution into time = distance / speed using any matching pair;
evaluation;
e.g. time = 6.0 / 10
(reaction time =) 0.60 (s)
Allow
allow 0.6 (s)
correct answer with no working scores 2
Do not accept
10 / 6.0 = 1.7 (s) scores 0
(c)
(stopping distance =) 12.0 + 30.0 = 42 (m);
42 m is less than 50 m so the van stops (8 m) before the tree;
Allow
ECF for MP2 from an incorrect stopping distance
allow 'yes, with 8 m to spare' if 42 m seen
Do not accept
comparison using braking distance only (30 m) scores 0
comparison using thinking distance only scores 0
(d)
thinking distance doubles / is directly proportional to speed (9.0 m → 18.0 m);
braking distance increases by (a factor of) four / quadruples (16.9 m → 67.5 m);
Allow
allow 'braking distance more than doubles' for MP2
22IGCSE-PHY-1-030Practical: how mass affects the acceleration of a trolleyPapers 1 & 2Grade 6–76 marks
A student wants to investigate how the mass of a trolley affects its acceleration when the same force acts on it. The diagram shows some of the apparatus she can use. A string attached to the trolley passes over a pulley and holds a hanging mass. A card is fixed to the top of the trolley so it can pass through two light gates connected to a data logger. She also has a top-pan balance, a set of 0.50 kg masses that can be stacked on the trolley and a metre rule.
Describe how the student could use this apparatus to investigate how the mass of the trolley affects its acceleration. You should include how she would make the investigation a fair test and how she would use her results.
(6)
Show mark scheme
Indicative content:
MP1. measure the mass of the trolley (and any added masses) using the balance;
MP2. keep the hanging mass the same so that the (accelerating) force is constant;
MP3. release the trolley from rest from the same starting position each time;
MP4. measure the length of the card (with the rule) and use the light gates/data logger to find the speed at each gate (length of card ÷ time for the card to pass through the gate);
MP5. measure the time to travel between the gates and calculate acceleration = (v − u)/t;
MP6. add 0.50 kg masses to the trolley one at a time and repeat for at least five different masses;
MP7. repeat each measurement and calculate a mean acceleration;
MP8. other control variables e.g. same runway/surface, same distance between gates, same card;
MP9. plot a graph of acceleration against mass (or against 1/mass) and describe the relationship, e.g. acceleration decreases as mass increases;
MP10. safety e.g. catch box/soft landing under the hanging mass or a block at the end of the runway;
Allow
award 1 mark per MP to a maximum of 6 if a points-based approach is used
allow use of v² = u² + 2as with the distance between the gates for MP5
allow 'tilt the runway slightly to compensate for friction' as a control for MP8
allow 'weight' for 'hanging mass' in MP2
allow 1/(total mass) or 1/(mass of trolley + hanging mass) for MP9
Do not accept
'change the hanging mass' as the independent variable (this investigates force, not mass)
'use a stopwatch to time the trolley' with no light gates for MP4
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.
23IGCSE-PHY-1-034Terminal velocity of falling paper casesPapers 1 & 2Grade 6–78 marks
A student drops paper cupcake cases from a height of 2 m. They quickly reach a terminal velocity. The diagram shows the two vertical forces, X and Y, acting on a single case as it falls. She then drops a stack of four cases fitted inside each other. The stack has the same shape and size as a single case.
(a)
Name forces X and Y.
(2)
(b)
The mass of a single case is 0.60 g. Calculate the size of force X when the single case falls at its terminal velocity.
(3)
(c)
The stack of four cases has a greater terminal velocity than a single case. Explain why.
(3)
Show mark scheme
(a)
X: air resistance / drag;
Y: weight;
Allow
allow 'gravity' / 'gravitational force' for Y
allow 'friction' for X
Do not accept
'upthrust' for X
'mass' for Y
(b)
X = weight (at terminal velocity);
conversion of 0.60 g to 0.00060 kg and substitution into W = m × g;
evaluation;
e.g. W = 0.00060 × 10
(X =) 0.0060 (N)
Allow
allow 6.0 × 10⁻³ (N)
allow 0.0059 (N) using g = 9.8
correct answer with no working scores 3
0.60 g used as 0.60 kg giving 6.0 (N) scores 2 (MP1 and MP2; POT error)
(c)
stack has a greater weight / four times the weight;
at terminal velocity air resistance = weight / resultant force is zero, so the stack needs a greater air resistance;
air resistance increases with speed (and the shape is the same) so the stack must fall faster before the forces balance;
Allow
allow 'drag' for air resistance throughout
'heavier objects fall faster because gravity pulls them more' scores MP1 only
Calculate the acceleration of the train in the first 20 s.
(2)
(b)
Describe how the acceleration of the train changes between 0 s and 60 s. Use data from the table to support your answer.
(2)
(c)
A student plots the data on a velocity–time graph and joins the points with straight lines. She uses the area under her graph to estimate the distance travelled by the train in the first 60 s. Calculate her estimate.
(3)
(d)
The actual velocity–time graph of the train is a smooth curve through the four points. Explain whether the student's estimate is greater than or less than the actual distance travelled.
(2)
Show mark scheme
(a)
substitution into a = (v − u) / t;
evaluation;
e.g. a = (14 − 0) / 20
(a =) 0.70 (m/s²)
Allow
allow 0.7 (m/s²)
(b)
(acceleration) decreases;
use of data e.g. change in velocity in each 20 s decreases (14, 10, 6 m/s) OR accelerations of 0.70, 0.50 and 0.30 m/s² quoted;
Allow
allow 'velocity increases by less each 20 s' for MP2
allow 'train speeds up more slowly'
Do not accept
'train slows down' / 'decelerates' scores 0 for MP1
(c)
any one area correct (140, 380 or 540 m);
adding three areas (trapezia/triangle);
evaluation;
e.g. ½ × 20 × 14 + ½ × 20 × (14 + 24) + ½ × 20 × (24 + 30)
= 140 + 380 + 540
(distance =) 1060 (m)
Allow
allow 1100 (m)
allow one triangle and rectangles method if equivalent
Do not accept
30 × 60 = 1800 (m) or 15 × 60 = 900 (m) scores 0
(d)
(estimate is) less than the actual distance;
idea that the curve lies above the straight lines (because acceleration/gradient decreases), so the area under the curve is greater;
25IGCSE-PHY-1-012Falling objects and terminal velocityPapers 1 & 2Grade 8–96 marks
A small steel ball is held 50 cm above a tall glass tube filled with thick oil. The ball is released from rest. It falls through the air, enters the oil and falls to the bottom of the tube. When the ball enters the oil, it is moving faster than the terminal velocity it would have in the oil.
Explain how the velocity of the ball changes from the moment it is released until it reaches the bottom of the tube. Your answer should refer to the forces acting on the ball.
(6)
Show mark scheme
Indicative content:
In the air:
MP1. weight acts downwards (and is constant);
MP2. air resistance is small / less than weight so resultant force is downwards;
MP3. so the ball accelerates / speeds up (in the air);
In the oil:
MP4. drag / resistive force from the oil acts upwards / opposes motion;
MP5. on entry, drag (+ upthrust) is greater than weight;
MP6. so resultant force is upwards and the ball decelerates / slows down;
MP7. as the ball slows, the drag decreases;
MP8. so the resultant force and the deceleration decrease;
MP9. eventually drag (+ upthrust) equals weight / forces become balanced;
MP10. resultant force is zero so the ball moves at a constant (terminal) velocity (lower than its speed on entering the oil);
Allow
allow 'friction' for air resistance or drag
allow 'gravity' for weight
upthrust is not required but credit correct reference to it
Point-based alternative matching 4PH1 practice: award 1 mark per MP to a maximum of 6, with at least one MP from each of 'in the air' and 'in the oil' for full marks
Do not accept
'the ball stops / speed becomes zero in the oil'
'weight increases / decreases'
'terminal velocity is when the ball reaches maximum speed' (it is reached from above here)
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.
26IGCSE-PHY-1-031Velocity–time graphs: police car catching a speeding carPapers 1 & 2Grade 8–99 marks
Car X travels along a straight motorway at a constant velocity of 30 m/s. At time t = 0, car X passes a stationary police car. At the same instant, the police car starts to accelerate along the motorway after car X. The graph shows how the velocity of each car changes with time.
(a)
Determine the acceleration of the police car in the first 10 s.
(2)
(b)
The two lines on the graph cross at t = 7.5 s. A student says this is the time when the police car catches up with car X. Explain why the student is wrong.
(1)
(c)
Determine the time at which the police car catches up with car X. Show your working clearly.
(5)
(d)
Determine the distance travelled by the police car from t = 0 until it catches up with car X.
(1)
Show mark scheme
(a)
gradient / substitution into a = (v − u)/t;
evaluation;
e.g. a = (40 − 0) / 10
(a =) 4.0 (m/s²)
Allow
correct answer with no working scores 2
allow 30 / 7.5 = 4.0 (m/s²) using the crossing point (same gradient)
(b)
(at 7.5 s) the cars have the same velocity, not the same position / car X has travelled further (the area under its line is greater);
Allow
allow 'the police car is still behind because it was slower until then'
Do not accept
'the graph shows distance'
(c)
distance travelled by police car in first 10 s = 200 (m);
distance travelled by car X in first 10 s = 300 (m) / gap of 100 (m) at 10 s;
(after 10 s) police car closes the gap at 40 − 30 = 10 (m/s) / equation e.g. 200 + 40(t − 10) = 30t;
time to close the gap = 100 / 10 = 10 (s) / solving the equation;
(t =) 20 (s);
Allow
allow equating areas under the two graphs as a valid method
correct answer with no working scores 1 only
ECF from incorrect distances in MP1/MP2 for MP3 and MP4
Do not accept
t = 7.5 (s) scores 0
t = 10 (s) (when police car reaches 40 m/s) scores 0 for MP5
(d)
600 (m);
Allow
allow 30 × candidate's (c) answer
ECF from (c)
Do not accept
800 (m) (= 40 × 20, ignoring the acceleration phase)
27IGCSE-PHY-1-032Forces on a lunar lander during its final descentPapers 1 & 2Grade 8–98 marks
A lunar lander of mass 15 000 kg is descending vertically towards the surface of the Moon. The gravitational field strength on the Moon is 1.6 N/kg. The Moon has no atmosphere.
(a)
Explain why the lander would not reach a terminal velocity if its engine were switched off.
(2)
(b)
The lander is moving downwards at 12 m/s. Its engine is then switched on and provides a constant upward force of 45 000 N, so the lander decelerates uniformly. Calculate the minimum height above the surface at which the engine must be switched on so that the velocity of the lander is zero when it reaches the surface.
(6)
Show mark scheme
(a)
no atmosphere so no air resistance/drag (acts on the lander);
so resultant force (= weight) is never zero / forces never balance so it continues to accelerate;
Allow
allow 'only force is weight/gravity' for MP1 if 'no air resistance' is implied
Do not accept
'there is no gravity on the Moon' scores 0
'it is too light to reach terminal velocity'
(b)
weight = 15 000 × 1.6 = 24 000 (N);
resultant force = 45 000 − 24 000 = 21 000 (N);
deceleration = 21 000 / 15 000 = 1.4 (m/s²);
substitution into v² = u² + 2as;
rearrangement;
evaluation;
e.g. 0 = 12² + 2 × (−1.4) × s
s = 12² / (2 × 1.4)
(s =) 51 (m)
Allow
allow 51.4 (m)
ECF from an incorrect weight, resultant force or deceleration for later marks
rearrangement and substitution in either order
weight ignored: a = 45 000 / 15 000 = 3.0 (m/s²), s = 24 (m) scores 3 (MP4–MP6)
28IGCSE-PHY-1-033Practical: loading and unloading a polythene stripPapers 1 & 2Grade 8–910 marks
A student investigates a strip cut from a polythene carrier bag. She hangs the strip from a clamp and adds 1.0 N weights one at a time up to 5.0 N, measuring the extension each time. She then removes the weights one at a time and measures the extension again.
Use the loading results to describe how the extension of the strip depends on the force. Refer to Hooke's law in your answer.
(3)
(b)
Explain whether the strip shows elastic behaviour over the whole range of forces used.
(2)
(c)
The student thinks the strip may behave elastically for forces up to 3.0 N. Describe how she could use a new strip and the same apparatus to test this idea.
(2)
(d)
The student wants to use a strip like this to hang a bag of mass 0.35 kg. Evaluate whether she can use her results to predict the extension of the strip, and whether the strip would return to its original length when the bag is removed.
(3)
Show mark scheme
(a)
up to 3.0 N extension is (directly) proportional to force / extension increases by equal amounts (about 2 mm per 1.0 N);
so it obeys Hooke's law (in this region);
above 3.0 N extension increases by much more for each 1.0 N (6.4 mm then 13.5 mm) so it no longer obeys Hooke's law;
Allow
allow 'graph would be a straight line through the origin up to 3 N' for MP1
allow limit stated anywhere between 3.0 N and 4.0 N
Do not accept
'obeys Hooke's law because extension increases with force' scores 0 for MP2
(b)
no / not elastic (over the whole range);
extension is not zero when all the weights are removed / 19.5 mm permanent extension / does not return to original length;
Allow
MP1 can be scored from a correct explanation in MP2
Do not accept
'yes, because the extension decreases when weights are removed'
(c)
add weights up to 3.0 N only then remove them all;
(measure the length/extension with no load and) check whether the extension returns to zero / the strip returns to its original length;
Allow
allow 'load and unload in steps up to 3.0 N and compare loading and unloading extensions'
Do not accept
'repeat the whole experiment up to 5.0 N'
(d)
weight of bag = 3.5 (N);
3.5 N is beyond the Hooke's law/linear region (above 3.0 N) so the extension cannot be found by proportion / can only say it lies between 6.1 mm and 12.5 mm;
her results cannot show whether the strip returns to its original length after 3.5 N because she only unloaded after loading to 5.0 N / she would need to load to 3.5 N and then unload to find out;
Allow
allow 'extension would be about 7 mm by proportion but this is unreliable' for MP2
allow 'it may not return to its original length, since it did not after 5.0 N, but the results do not show what happens after 3.5 N' for MP3
Do not accept
7.0 mm stated as a reliable prediction scores 0 for MP2
'0.35 N' for the weight (mass not converted) scores 0 for MP1
'it will be permanently stretched because it is beyond the linear region' scores 0 for MP3 (the end of the linear region is not where elastic behaviour stops)
29IGCSE-PHY-1-035Momentum and Newton's third law in a rugby tacklePaper 2 onlyGrade 4–56 marks
A rugby player, A, of mass 90 kg runs in a straight line at 8.0 m/s.
(a)
State the unit of momentum.
(1)
(b)
Calculate the momentum of player A.
(2)
(c)
Player A tackles player B. Player A says: 'I pushed player B with a force of 900 N, but he pushed back on me with a smaller force because he is lighter than me.' Explain why player A is wrong.
(2)
(d)
Player C has a mass of 110 kg and has the same momentum as player A. Determine the speed of player C.
(1)
Show mark scheme
(a)
kg m/s;
Allow
allow N s
allow kilogram metre per second
Do not accept
kg/m/s
kg m/s²
(b)
substitution into p = m × v;
evaluation;
e.g. p = 90 × 8.0
(p =) 720 (kg m/s)
Allow
correct answer with no working scores 2
Do not accept
90 / 8.0 = 11 scores 0
(c)
(Newton's third law) the forces are equal in size / B also pushes on A with 900 N;
(forces are) in opposite directions (and act on different bodies), whatever the masses;
Allow
allow 'action and reaction are equal and opposite' for MP1 and MP2 if 900 N or 'equal' stated
Do not accept
'the forces cancel out'
'B pushes back with less force because he has less momentum'
30IGCSE-PHY-1-015Newton's third lawPaper 2 onlyGrade 6–71 mark
A book rests on a table. The weight of the book is the gravitational force of the Earth on the book. Which force forms a Newton's third law pair with the weight of the book?
Show mark scheme
A (the gravitational force of the book on the Earth);
B is incorrect because this is a contact force; it pairs with the upward force of the table on the book
C is incorrect because this force acts on the same object as the weight (it balances the weight; it is not its third-law pair)
D is incorrect because it acts on a different pair of objects (Earth and table)
31IGCSE-PHY-1-016Conservation of momentum: skaters pushing apartPaper 2 onlyGrade 6–77 marks
Two ice skaters, X and Y, stand still facing each other on smooth ice. They push against each other and move apart. Skater X has a mass of 72 kg. Skater Y has a mass of 52 kg. Just after the push, skater X moves at 1.4 m/s.
(a)
State the total momentum of the two skaters before the push. Give a reason for your answer.
(2)
(b)
Calculate the speed of skater Y just after the push.
(3)
(c)
Explain why the two skaters move in opposite directions.
(2)
Show mark scheme
(a)
zero;
(because) both skaters are stationary / have zero velocity;
Allow
allow 0 kg m/s
Do not accept
'zero because they have no mass'
(b)
momentum of X = 72 × 1.4 = 100.8 (kg m/s);
idea that momentum of Y = momentum of X / 52 × v = 100.8;
32IGCSE-PHY-1-017Force and rate of change of momentumPaper 2 onlyGrade 6–711 marks
A carpenter uses a hammer to drive a nail into wood. The head of the hammer has a mass of 0.60 kg. It hits the nail at a speed of 8.0 m/s and stops in a time of 2.5 ms.
(a)
Calculate the momentum of the hammer head just before it hits the nail. Give the unit.
(3)
(b)
Calculate the average force exerted on the hammer head by the nail.
(3)
(c(i))
The carpenter then uses a rubber mallet with a head of the same mass to hit a tent peg. The mallet head hits the peg at 8.0 m/s and bounces back upwards at 2.0 m/s. Explain why the change in momentum of the mallet head is greater than its momentum just before it hits the peg.
(2)
(c(ii))
The mallet head is in contact with the peg for 4.0 ms. Calculate the average force exerted on the mallet head by the peg.
(3)
Show mark scheme
(a)
substitution into p = m × v;
evaluation;
kg m/s;
e.g. p = 0.60 × 8.0
(p =) 4.8
Allow
allow N s
unit mark is independent
Do not accept
kg/m/s
kg m/s²
(b)
substitution into F = Δp / t;
conversion of 2.5 ms to 0.0025 s;
evaluation;
e.g. F = 4.8 / 0.0025
(F =) 1900 (N)
Allow
allow 1920 (N)
ECF from (a)
1.92 (N) scores 2 marks (POT error)
(c(i))
idea that momentum is a vector / direction of momentum reverses;
(so) change in momentum = initial momentum + final momentum (in size) / head loses 4.8 and gains 1.2 kg m/s;
Allow
allow 'velocity changes by 10 m/s, not 8 m/s'
Do not accept
'because it bounces' alone scores 0
(c(ii))
Δp = 0.60 × (8.0 + 2.0) = 6.0 (kg m/s);
substitution into F = Δp / t;
evaluation;
e.g. F = 6.0 / 0.0040
(F =) 1500 (N)
Allow
Δp = 0.60 × (8.0 − 2.0) = 3.6 giving 900 (N) scores MP2 and MP3 only
33IGCSE-PHY-1-019Moments: tower cranePaper 2 onlyGrade 6–76 marks
A tower crane has a horizontal arm balanced on a vertical tower. A load of weight 9600 N hangs from the arm 18 m to the right of the tower. A concrete counterweight is fixed to the arm 6.0 m to the left of the tower. Ignore the weight of the arm.
(a)
State the unit of the moment of a force.
(1)
(b)
Calculate the weight of the counterweight needed to balance the arm.
(3)
(c)
The load is moved to 12 m from the tower. The counterweight is not moved. Explain what happens to the arm.
(2)
Show mark scheme
(a)
N m;
Allow
allow newton metre
Do not accept
J
N/m
Nm/s
(b)
moment of load = 9600 × 18 = 172 800 (N m);
use of principle of moments: 172 800 = W × 6.0;
evaluation;
(W =) 28 800 (N)
Allow
allow 29 000 (N)
-1 for POT error
Do not accept
9600 × 6.0 / 18 = 3200 (N) scores 0
(c)
(clockwise) moment of the load decreases;
(anticlockwise) moment of counterweight is now greater so arm rotates / tips anticlockwise / towards the counterweight;
Allow
allow 'moments are no longer balanced' for MP2 only if direction given
34IGCSE-PHY-1-020Forces on a beam supported at both endsPaper 2 onlyGrade 6–79 marks
A uniform wooden plank of length 4.0 m rests across a stream on two supports, A and B, one at each end. A person of weight 720 N stands on the plank 1.0 m from A. For parts (a), (b) and (c), assume the weight of the plank is negligible.
(a)
Calculate the upward force exerted on the plank by support B.
(3)
(b)
Determine the upward force exerted on the plank by support A.
(1)
(c)
Support B is a loose rock. It can provide an upward force of no more than 450 N. Calculate the greatest distance from A at which the person can stand without the force needed from B exceeding 450 N.
(2)
(d)
The plank actually has a weight of 200 N. The person stands 1.0 m from A again. Calculate the upward force exerted on the plank by support B.
(3)
Show mark scheme
(a)
taking moments about A: 720 × 1.0 (= 720 N m);
720 = F_B × 4.0;
evaluation;
(F_B =) 180 (N)
Allow
allow moments about any point if correct
540 (N) with A and B swapped scores 2 marks
(b)
540 (N);
Allow
ECF: 720 − answer to (a)
(c)
use of principle of moments about A: 720 × x = 450 × 4.0;
evaluation;
e.g. x = 1800 / 720
(x =) 2.5 (m)
Allow
allow 1.5 m from B
allow moments about B: 720 × (4.0 − x) = 270 × 4.0 leading to 2.5 (m)
36IGCSE-PHY-1-037Conservation of momentum: railway wagons couplingPaper 2 onlyGrade 6–710 marks
In a railway yard, two wagons move towards each other along a straight, level track. Wagon P has a mass of 30 000 kg and moves east at 2.0 m/s. Wagon Q has a mass of 20 000 kg and moves west at 1.0 m/s. The wagons collide and couple together, moving off as one object.
(a)
Calculate the total momentum of the two wagons before the collision. Give the direction of the total momentum.
(3)
(b)
Calculate the velocity of the coupled wagons just after the collision.
(3)
(c)
Calculate the change in momentum of wagon Q.
(2)
(d)
Use Newton's third law to explain why the change in momentum of wagon P has the same size as the change in momentum of wagon Q.
(2)
Show mark scheme
(a)
momentum of P = 60 000 (kg m/s) and momentum of Q = 20 000 (kg m/s);
momenta subtracted because they are in opposite directions;
40 000 (kg m/s) east;
Allow
allow −20 000 for Q's momentum for MP2
direction may be given as 'in P's direction'
80 000 (kg m/s) (momenta added) scores MP1 only
40 000 with no direction or 'west' scores 2
(b)
total momentum after = total momentum before / use of conservation of momentum;
substitution e.g. 40 000 = (30 000 + 20 000) × v;
0.80 (m/s) east;
Allow
ECF from (a)
allow 0.8 (m/s)
direction may be given as 'in P's original direction'
40 000 / 30 000 = 1.3 (m/s) scores MP1 only
(c)
change in velocity of Q = 0.80 − (−1.0) = 1.8 (m/s) / momentum after − momentum before with correct signs;
37IGCSE-PHY-1-038Moments in the human forearmPaper 2 onlyGrade 6–78 marks
A physiotherapy patient holds a 30 N weight in her hand with her forearm horizontal. The forearm pivots at the elbow. The biceps muscle pulls vertically upwards on the forearm at a point 0.040 m from the elbow. The weight in the hand is 0.35 m from the elbow. The forearm has a weight of 15 N, which acts 0.15 m from the elbow. The forearm is in equilibrium.
(a)
Explain why the weight of the forearm can be shown as a single force acting 0.15 m from the elbow.
(1)
(b)
Calculate the moment of the 30 N weight about the elbow. Give the unit.
(3)
(c)
Calculate the force F exerted by the biceps muscle.
(4)
Show mark scheme
(a)
(the weight acts through) the centre of gravity (of the forearm), which is 0.15 m from the elbow;
Allow
allow 'centre of mass'
Do not accept
'it is the middle of the forearm'
(b)
substitution into moment = force × (perpendicular) distance;
evaluation;
unit N m;
e.g. moment = 30 × 0.35
(moment =) 10.5
Allow
allow 11 (N m)
allow 1050 N cm
30 / 0.35 = 86 scores unit mark only
Do not accept
J for the unit
(c)
moment of forearm weight = 15 × 0.15 = 2.25 (N m);
total clockwise moment = 10.5 + 2.25 = 12.75 (N m);
principle of moments: F × 0.040 = total clockwise moment;
evaluation;
(F =) 320 (N)
Allow
allow 318.75 / 319 (N)
ECF from (b)
-1 for POT error e.g. 3.2 (N) from using 4.0 cm with distances in m
38IGCSE-PHY-1-018Collisions and safety: bumper carsPaper 2 onlyGrade 8–910 marks
At a fairground, bumper car X travels at 3.0 m/s towards bumper car Y, which is stationary. The mass of X (including its rider) is 220 kg. The mass of Y (including its rider) is 180 kg. After the collision, X continues in the same direction at 0.50 m/s.
(a)
State the formula linking momentum, mass and velocity.
(1)
(b)
Calculate the velocity of Y just after the collision.
(4)
(c)
A student says: 'After the collision Y moves faster than X was moving before, so momentum cannot have been conserved.' Comment on the student's statement.
(2)
(d)
Each bumper car has a thick rubber bumper around its edge. Use ideas about momentum to explain how the rubber bumper reduces the force on the riders in a collision.
(3)
Show mark scheme
(a)
momentum = mass × velocity;
Allow
p = m × v
rearrangements
Do not accept
m or M used as the symbol for momentum (unless defined)
(b)
momentum before = 220 × 3.0 = 660 (kg m/s);
momentum of X after = 220 × 0.50 = 110 (kg m/s);
use of conservation: 660 = 110 + 180 × v;
evaluation;
e.g. v = (660 − 110) / 180 = 550 / 180
(v =) 3.1 (m/s)
Allow
allow 3.06, 3.055… (m/s)
660 / 180 = 3.7 (m/s) scores MP1 only
(c)
student is incorrect / speed alone does not show momentum;
Y has a smaller mass (than X) so needs a greater velocity for the same momentum / momentum depends on mass as well as velocity;
39IGCSE-PHY-1-039Momentum and safety: airbag crash testPaper 2 onlyGrade 8–910 marks
Engineers test a new airbag using a crash-test dummy of mass 75 kg. The test car hits a barrier at 13 m/s. The airbag inflates and brings the dummy to rest. The dummy moves forward 0.65 m while it is being stopped. Assume the dummy decelerates uniformly. The engineers require the average force on the dummy to be less than 12 000 N.
(a)
Calculate the momentum of the dummy just before the collision.
(2)
(b)
Determine whether the airbag meets the engineers' requirement.
(5)
(c)
The test is repeated with the car moving at 26 m/s. The dummy is again stopped in a distance of 0.65 m. Without calculating the new force, explain how the average force on the dummy changes.
(3)
Show mark scheme
(a)
substitution into p = m × v;
evaluation;
e.g. p = 75 × 13
(p =) 975 (kg m/s)
Allow
allow 980 (kg m/s)
(b)
average speed = (13 + 0)/2 = 6.5 (m/s) / substitution into v² = u² + 2as;
stopping time = 0.65 / 6.5 = 0.10 (s) / deceleration = 130 (m/s²);
substitution into F = Δp / t (or F = m × a);
force = 9750 (N);
comparison: 9750 N is less than 12 000 N so the airbag meets the requirement;
Allow
ECF from (a)
allow 9800 (N)
MP5 can be awarded by ECF for a correct comparison with the candidate's calculated force
t = 0.65 / 13 = 0.050 (s) (initial speed used instead of average speed) giving 19 500 (N) and 'does not meet the requirement' scores 2 (MP3 and MP5 by ECF)
(c)
momentum (before the collision) doubles;
average speed doubles so the stopping time halves;
(force = change in momentum ÷ time so) the force is four times greater / 39 000 N / exceeds the limit;
Allow
allow 'deceleration is four times greater' for MP2 and MP3 if linked to F = ma
allow 'force is more than doubled' for MP3 only if linked to both changes
40IGCSE-PHY-1-040Moments: locating a patient's centre of gravity on a stretcherPaper 2 onlyGrade 8–98 marks
A hospital stretcher is used to find the position of a patient's centre of gravity. The stretcher is uniform and has a weight of 100 N. It rests horizontally on two force sensors, one under the handles at end A and one under the handles at end B. A and B are 2.0 m apart. A patient lies on the stretcher. Sensor A reads 530 N and sensor B reads 370 N.
(a)
State where the weight of the stretcher acts.
(1)
(b)
Determine the weight of the patient.
(1)
(c)
Calculate the distance of the patient's centre of gravity from end A.
(4)
(d)
The patient moves along the stretcher until both sensors give the same reading. Determine how far the patient's centre of gravity has moved, and in which direction.
(2)
Show mark scheme
(a)
at its centre (of gravity) / midpoint / 1.0 m from A;
Allow
allow 'centre of mass'
Do not accept
'at A and B'
(b)
800 (N);
Do not accept
900 (N) (stretcher weight not subtracted)
(c)
taking moments about A (or B) with all forces/distances correct;
moment of sensor B force = 370 × 2.0 = 740 (N m) / moment of stretcher = 100 × 1.0 = 100 (N m);
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