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Edexcel A Level Physics · Topic 2
Mechanics
FREE Online Physics A Level Course90minutes
78marks
10questions
How it works
- Answer each question. Type into the boxes or pick an option, and use paper for working and drawings.
- Press Check my answers at the end of the question. Tick-box and number answers are marked for you.
- For written answers, compare yours with the mark scheme and tick each point you made. We highlight the key words we spot, but you decide, just like an examiner.
You will need: paper, a pen, a ruler and a calculator. Use g = 9.81 N kg⁻¹. Your answers save automatically in this browser.
1
Question 1
Multiple choice10 marksTen quick questions from across the topic. Each part is worth 1 mark.
(a)
A constant forward force acts on a car. The car accelerates along a straight road. After a while, the speed of the car reaches a constant value. Which graph shows the variation of acceleration with time for the car?
Acceleration–time graphs
(1 mark)
Mark scheme
A, B and D never reach zero acceleration, but at constant speed the acceleration must be zero.
Answer: C
(b)
An object is falling at terminal velocity. Which of the following is not a valid conclusion from this statement?
(1 mark)
Mark scheme
Key equationΣF = materminal velocity → a = 0 → ΣF = 0A, B and D are all true. At terminal velocity the forces are balanced, so there is no resultant force.
Answer: C There is a resultant force acting on the object.
(c)
A soap bubble is being blown by the wind. P is the force of the wind, W is the weight of the bubble and U is the upthrust. Which vector diagram shows the resultant force F acting on the bubble?
Free-body diagram
Vector diagrams
(1 mark)
Mark scheme
Add the vectors tip to tail: (W − U) then P. The resultant goes from the start of the first to the end of the last. A shows the force needed for balance (it closes the loop the other way). C gives (W − U) − P; D gives P − (W − U).
Answer: B
(d)
A student determined a value for g by dropping a metal sphere from rest, measuring the distance fallen and the time taken. The student obtained a value for g of 11.2 m s⁻². Which of the following could explain the difference between the student’s value and the accepted value?
(1 mark)
Mark scheme
Key equations = ½ g t² so g = 2s ÷ t²A distance that is too big gives g too big. Air resistance (A) or a time that is too long (D) would make g smaller; the fall time doesn’t depend on horizontal motion (C).
Answer: B The student’s measured distance was greater than the actual distance.
(e)
A ball of mass m is projected vertically upwards. An air resistance F acts on the ball at all times. Which equation applies to the ball as it is travelling upwards?
(1 mark)
Mark scheme
Key equationΣF = magoing up, weight and air resistance both act downwardsThe velocity is changing, so it isn’t 0 on the right (A, B). On the way up, air resistance acts downwards, the same way as the weight (not C).
Answer: D mg + F = ma
(f)
A box moves down a ramp at constant velocity. Its free-body force diagram is shown. Which two forces acting on the box have, according to Newton’s third law, corresponding forces acting on the ramp?
Free-body diagram for the box
(1 mark)
Mark scheme
Friction and the normal contact force come from the ramp, so their pairs act on the ramp. D’s pair acts on the air; W’s pair acts on the Earth.
Answer: B F and R
(g)
A tennis ball of mass 0.06 kg moves towards a wall at 5 m s⁻¹. It hits the wall perpendicularly and rebounds at the same speed. What is the change in momentum of the ball?
The tennis ball
(1 mark)
Mark scheme
Key equationΔp = mv − mu0.06 × (−5) − 0.06 × 5 = −0.60The velocity reverses, so the change is twice the momentum, and it points back away from the wall.
Answer: D −0.60 kg m s⁻¹
(h)
A trolley of mass 0.50 kg has a speed of 2.0 m s⁻¹. A second trolley of mass 1.0 kg has a speed of 2.0 m s⁻¹. They travel in opposite directions and collide. Which could be the total momentum, in kg m s⁻¹, after the collision?
(1 mark)
Mark scheme
Key equationtotal momentum before = total momentum after1.0 × 2.0 − 0.50 × 2.0 = 1.0D is the answer if they moved the same way; C is the second trolley on its own.
Answer: B 1.0
(i)
A car of mass 1.2 × 10³ kg is travelling at 18 m s⁻¹. The driver brakes and the car comes to rest. What is the work done by the brakes?
(1 mark)
Mark scheme
Key equationEk = ½ m v²½ × 1.2 × 10³ × 18² = 194 kJA: v not squared. B: v not squared and no ½. D: the ½ left out.
Answer: C 190 kJ
(j)
A student uses a force F to push a block of weight W all the way up a frictionless ramp at a constant speed. The work done by the student can be calculated using
The ramp
(1 mark)
Mark scheme
Key equationΔEgrav = mgΔhfrictionless and constant speed: all the work becomes GPE(F × l also works, but it isn’t an option.)
Answer: C Wh
2
Question 2
3 marksThe graph shows how the velocity of an object varies with time.
Velocity–time graph
Describe how the acceleration of the object varies with time. Your answer should include calculations.
First: what is the acceleration for the first 9 seconds?
m s⁻²
(1 mark)
Mark scheme
Key equationa = (v − u) ÷ tthe gradient of a v–t graph
Acceleration from 0 to 9 s: 1.6 m s⁻² (14 ÷ 9 = 1.56; or see −4.9 to −5.2 m s⁻² for after 9 s)
💡 You multiplied. Acceleration is the gradient: change in velocity ÷ time = 14 ÷ 9.
💡 Upside down: a = Δv ÷ t = 14 ÷ 9.
💡 That’s the size of the acceleration after 9 s, which also scores. Tick the box.
Now describe how the acceleration changes.
(2 marks)
Mark scheme
Tick what you wrote (max 2):
Max 1 of these:
3
Question 3
9 marksA stunt motorcyclist drives off a ramp at the edge of a river. The ramp is at 25° to the horizontal and the end of the ramp is 3.0 m high. The river is 100 m wide. The initial velocity of the motorcyclist is 35 m s⁻¹.
The jump (not to scale)
(a)
Calculate the horizontal and vertical components of the motorcycle’s initial velocity as it leaves the ramp.
m s⁻¹
m s⁻¹
(2 marks)
Mark scheme
Key equationvx = v cos θ vy = v sin θ
Horizontal component: 32 m s⁻¹ (35 cos 25° = 31.7)
💡 That’s the vertical one. Horizontal = 35 cos 25°.
💡 Your calculator is in radians. Switch it to degrees.
Vertical component: 15 m s⁻¹ (35 sin 25° = 14.8)
💡 That’s the horizontal one. Vertical = 35 sin 25°.
💡 Your calculator is in radians. Switch it to degrees.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(b)
Deduce whether the rider lands on the other side of the river. The effects of air resistance can be ignored.
Work out how long it takes to cross 100 m, then how far the rider falls below the top of the ramp in that time.
m
(3 marks)
Mark scheme
Key equations = u t + ½ a t²horizontal: a = 0; vertical: a = −g
Distance fallen below the top of the ramp: 2.1 m (t = 100 ÷ 31.7 = 3.15 s; 14.8 × 3.15 − ½ × 9.81 × 3.15² = −2.12 m)
💡 Right size: the minus just means “below”. Tick the box.
💡 You left out the upward start. Vertically, s = uyt + ½at² with uy = 14.8 and a = −9.81.
💡 That’s just uyt. Gravity pulls it back down: add ½ × (−9.81) × t².
💡 You used a different method: finding the horizontal distance when the rider has fallen 3 m (102 m). That scores too. Tick the box.
💡 That’s a time. If you compared 3.15 s (to cross) with 3.21 s (to fall 3 m), that’s a valid method. Tick the box.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
So does the rider land on the other side?
(1 mark)
Mark scheme
2.1 m < 3.0 m, so the rider is still above the bank when they reach it. (The conclusion mark is for a conclusion consistent with your own numbers.)
Answer: A Yes: they fall less than 3.0 m while crossing
(c)
Explain how air resistance would affect the jump.
(3 marks)
Mark scheme
Key equationΣF = maa resistive force causes a deceleration
Tick each marking point your answer makes:
4
Question 4
8 marksA sledge accelerates, due to gravity, from rest down a frictionless slope. Air resistance can be ignored. The slope is at 6.9° to the horizontal and has a total length of 60 m.
The slope
(a)
On paper, draw a dot for the sledge and complete its free-body force diagram.
(2 marks)
Mark scheme
No friction (the slope is frictionless) and no air resistance. Don’t draw a “force down the slope”: that’s a component of the weight.
Compare your drawing with the answer below
(b)(i)
Show that the initial acceleration of the sledge along the slope is about 1 m s⁻².
m s⁻²
(2 marks)
Mark scheme
Key equationa = g sin θfrom ΣF = ma: mg sin θ = ma
Acceleration: 1.2 m s⁻² (9.81 × sin 6.9° = 1.18)
💡 You used cos. Along the slope it’s g sin θ.
💡 Your calculator is in radians. Switch it to degrees.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(b)(ii)
Determine the speed of the sledge at the end of the slope.
m s⁻¹
(2 marks)
Mark scheme
Key equationv² = u² + 2a s
Speed: 12 m s⁻¹ (√(2 × 1.18 × 60) = 11.9; error carried forward from (i))
💡 You used a = 1. That scores as an error carried forward if (i) gave you 1: tick the box.
💡 That’s v². Take the square root.
💡 You did a × s. Use v² = 2as, then square-root.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(b)(iii)
Determine the time taken for the sledge to travel to the end of the slope.
s
(2 marks)
Mark scheme
Key equationv = u + a t
Time: 10 s (11.9 ÷ 1.18 = 10.1; error carried forward from (i) and (ii))
💡 You used 1.18 × 11.9. v = u + at, so t = v ÷ a.
💡 That’s s ÷ v, which assumes constant speed. Use v = u + at or s = ½(u + v)t.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
5
Question 5
6 marksA sports class films a cyclist on a horizontal lawn. The cyclist starts from rest. They sketch a graph of the velocity v of the cyclist against time t.
The cyclist
Sketch graph
(a)
Explain the shape of this graph and include a consideration of force as part of your answer.
(3 marks)
Mark scheme
Key equationΣF = mathe gradient of a v–t graph is the acceleration
Tick each marking point your answer makes:
(b)
A student says:
“The work done by the cyclist is converted into the kinetic energy of the cyclist and bicycle.”
Criticise this statement.
(3 marks)
Mark scheme
Key equationΔW = FΔswork done = energy transferred
Tick each marking point your answer makes:
⚠️ Examiners prefer energy is transferred, not “converted”.
6
Question 6
7 marksA teacher uses a linear air track to provide a frictionless surface for two gliders, each of mass m. The gliders are each given a small push towards the centre. They collide and move off together.
The air track
(a)
Initial velocity of glider 1 = 0.30 m s⁻¹ to the right. Initial velocity of glider 2 = 0.70 m s⁻¹ to the left.
Calculate the velocity of the gliders after the collision, using the principle of conservation of linear momentum.
m s⁻¹
(3 marks)
Mark scheme
Key equationΣp before = Σp after0.3m − 0.7m = 2m v → v = −0.2 m s⁻¹
Magnitude: 0.2 m s⁻¹
💡 Right: the minus sign is the direction. Put 0.2 here and “left” below, then tick the box.
💡 You added the speeds. Glider 2 moves the other way, so its momentum is negative: 0.3m − 0.7m.
💡 After the collision the mass is 2m: 0.3m − 0.7m = 2m v.
Direction: to the left (accept: in glider 2’s initial direction)
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)
A student began explaining the change in velocity of glider 1 with the statement:
“During the collision there is a force on glider 2.”
Complete the explanation, making reference to Newton’s laws of motion where appropriate.
(4 marks)
Mark scheme
Key equationF1 on 2 = −F2 on 1Newton’s third lawKey equationΣF = maNewton’s second law
Tick each marking point your answer makes:
7
Question 7
10 marksA spaceship has an orbital module (O), a descent module (D) and a service module (S). Before entering the Earth’s atmosphere, the modules separate: O and S move away from D in opposite directions. The velocities are measured relative to the spaceship before the modules separated.
The spaceship
After separation
(a)
Determine the velocity v of the descent module after separation. Only consider momentum along a horizontal line through the centres of the three modules.
mass of O = 1350 kg mass of D = 2950 kg mass of S = 2100 kg
m s⁻¹
(4 marks)
Mark scheme
Key equationΣp before = Σp after = 01350 × 0.82 + 2950v − 2100 × 0.58 = 0
Magnitude of v: 0.04 m s⁻¹ (0.0376)
💡 Right size: the minus sign is the direction. Tick the box and give the direction below.
💡 O and S move in opposite directions, so one of their momenta is negative: 1350 × 0.82 − 2100 × 0.58.
💡 Check your powers of ten: 111 ÷ 2950 = 0.0376.
Direction: towards O / away from S / to the left
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)(i)
The descent module’s rocket motor ejects hot gases at high velocity. Explain why the velocity of the descent module changes when the rocket motor is used. Refer to Newton’s laws of motion.
(3 marks)
Mark scheme
Key equationF = Δp ÷ Δtthe motor changes the momentum of the gas
Tick each marking point your answer makes:
(b)(ii)
When the rocket motor is operating, the velocity of the descent module changes by 0.58 m s⁻¹ in 5.0 s. Mass of descent module = 2950 kg. Calculate the average force exerted on the hot gas during this time.
N
(3 marks)
Mark scheme
Key equationΣF = maa = Δv ÷ t
Average force: 3.4 × 10² N (a = 0.58 ÷ 5 = 0.116 m s⁻²; F = 2950 × 0.116 = 342 N)
💡 That’s the change in momentum (2950 × 0.58). Force is the rate of change: divide by 5.0 s.
💡 That’s the acceleration. Now ΣF = m a.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
8
Question 8
9 marksBeam engines helped power the Industrial Revolution. A beam rocks to and fro around a well-oiled pivot P. One rod connects the beam to a piston in a steam cylinder; the other connects it to a pump.
A beam engine
Simplified arrangement
(a)
The beam has a constant thickness and a mass of 3.05 × 10⁴ kg. It is 11.0 m long, and P is 6.0 m from the steam cylinder end. At rest, the steam cylinder rod supports the beam at 20° to the horizontal and exerts a force T on it. Neglect the force from the pump rod.
Calculate the force T.
N
(4 marks)
Mark scheme
Key equationmoment = F xx = perpendicular distance from the pivotIf cos 20 is left out of both sides, all 4 marks can still be given.
T: 25 kN (T × 6 cos 20 = 3.05 × 10⁴ × 9.81 × 0.5 × cos 20, so T = 24.9 kN)
💡 You used the mass. Turn it into a weight first: × 9.81.
💡 The beam is uniform, so its weight acts at its centre: 5.5 m from the end, which is only 0.5 m from P.
💡 That’s the whole weight of the beam. Take moments about P: T × 6.0 = W × 0.5.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 3):
(b)
The engine, running continuously, could lift 2500 kg of water through 12 m each minute. It used 1250 kg of coal a day, and 1 kg of coal releases 22.3 MJ. The engine was said to have an efficiency of 10%.
Calculate the actual efficiency.
%
(4 marks)
Mark scheme
Key equationefficiency = useful energy output ÷ total energy input
Efficiency: 1.5% (424 MJ a day out ÷ 27 900 MJ a day in)
💡 That’s a fraction. As a percentage: × 100.
💡 Out by 10. Check: useful output 2500 × 9.81 × 12 = 294 kJ per minute; × 1440 minutes = 424 MJ a day.
💡 You compared a minute of work with a day of coal. Use the same time for both.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 3):
Was the claim of 10% efficiency correct?
(1 mark)
Mark scheme
Allow the reverse argument for full credit: starting from 10% and showing the coal would last far longer than a day.
Answer: B No: it was much less efficient (about 1.5%)
9
Question 9
10 marksA portable basketball set has a base and a post arrangement (post, backboard, hoop and net). The base can be filled with water to increase stability.
The basketball set
(a)
The base has a capacity of 85.0 litres. Show that the maximum weight of the base is about 870 N.
mass of 1.00 litre of water = 1.00 kg mass of base when empty = 3.50 kg
N
(2 marks)
Mark scheme
Key equationW = m g
Weight of base: 868 N ((85.0 + 3.50) × 9.81)
💡 Don’t forget the empty base: 85.0 + 3.50 = 88.5 kg.
💡 That’s the mass. Weight = m g.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(b)
Due to the large area of the backboard, the set may topple over when the wind blows. Calculate the minimum force of the wind Fw that will cause the set to be blown over at the angle shown. Ignore the effect of the wind on the base. Weight of post arrangement = 27.0 N.
About to topple (not to scale)
N
(5 marks)
Mark scheme
Key equationΣ clockwise moments = Σ anticlockwise momentsabout the corner it tips onAccept sin 75 for cos 15, and sin 15 for cos 75.
Fw: 157 N (377.3 = 13.98 + 2.31 Fw)
💡 You left out the post arrangement. Its weight tips the set the same way as the wind: 27 × 2.0 × cos 75 = 14.0 N m.
💡 The post’s weight helps the wind, so subtract its moment from the base’s, don’t add it.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 4):
(c)
The base is filled with sand instead of water. Sand is denser than water. What would happen to the value of Fw from part (b)?
(1 mark)
Mark scheme
Answer: A It would increase
Justify your answer.
(2 marks)
Mark scheme
Key equationmoment = F x
Tick each marking point your answer makes:
10
Question 10
* extended answer6 marksThe following extract comes from a section on forces, on a website written for children.
The website extract
Forces act in pairs. The lift force on a plane pushing it up into the sky is paired with gravity, which pulls the plane back towards the centre of the Earth.
If both forces in the pair are equal, the plane will stay at rest in the same place.
*
Criticise this extract.
Marks are given for what you say and for structuring your answer logically, linking each point to the next.
(6 marks)
Mark scheme
Key equationFA on B = −FB on ANewton’s third lawKey equationΣF = 0 ⇒ a = 0Newton’s first law
Indicative content: tick each point you made
Structure (linkage): tick each point you made
| Content | 6 points → 4 marks · 4–5 → 3 · 2–3 → 2 · 1 → 1 | 0–4 |
|---|---|---|
| Structure | Coherent and logical, with linkage and sustained reasoning throughout → 2 · partly structured with some linkage → 1 · no linkage → 0 | 0–2 |
Your mark:
Show a model answer
- The two forces in a Newton’s third law pair must be the same type of force and act on different bodies. Lift and weight are different types of force and both act on the plane, so they are not a third-law pair.
- The lift on the plane (from the air) is paired with the push of the plane on the air. The Earth’s gravitational pull on the plane is paired with the plane’s gravitational pull on the Earth.
- If lift and weight are equal, the vertical resultant force is zero, so by Newton’s first law the plane does not accelerate vertically.
- That means it is either at rest vertically or moving with a constant vertical velocity: it does not have to stay still.
- A flying plane is also moving forwards horizontally, so it cannot stay in the same place.
Your results
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| Question | Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Q10 | Total |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Out of | 10 | 3 | 9 | 8 | 6 | 7 | 10 | 9 | 10 | 6 | 78 |
| Your mark | – | – | – | – | – | – | – | – | – | – | 0 |
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