A study companion covering all nine Paper 1 (Physics) topics: definitions, formulae, worked examples, diagrams, and a large bank of practice questions with model answers. Tap any question to reveal the solution, or try the multiple-choice checks for instant feedback.
A crate of mass 20 kg is pulled by a rope up an inclined plane that makes an angle of 30° with the horizontal. The tension in the rope is 147 N and the coefficient of kinetic friction between the crate and the plane is 0,1. Calculate the acceleration of the crate.
Step 1 — Normal forcePerpendicular to the incline: N − w·cos30° = 0 → N = (20)(9,8)(cos30°) = 169,74 N
Step 2 — Frictionf = μkN = (0,1)(169,74) = 16,97 N
Step 3 — Along the incline (up positive)Fnet = ma → F + wpar + f = ma, with wpar = mg·sin30° acting down the slope and f acting down the slope (opposes upward motion)
Substitution147 − (20)(9,8)(sin30°) − 16,97 = 20a
Answera ≈ 1,60 m·s⁻², directed up the incline
A 10 kg block lies on a rough horizontal table and is connected over a frictionless pulley to a 2 kg block hanging off the edge. The system is stationary. Which statement correctly applies Newton's First Law to this situation?
For the stationary system above, write down the magnitude of the net force acting on the 10 kg block.
Fnet = 0 N (the block is in equilibrium — stationary).
When a 15 N force is applied vertically downwards on the 2 kg hanging block, the 10 kg block accelerates to the right at 1,2 m·s⁻². Calculate the coefficient of kinetic friction (μk) between the 10 kg block and the table.
For 2 kg block (down positive)(2)(9,8) + 15 − T = (2)(1,2) → T = 32,2 N
For 10 kg block (right positive)T − f = (10)(1,2) → f = 32,2 − 12 = 20,2 N
Normal forceN = (10)(9,8) = 98 N
Coefficient of frictionμk = f / N = 20,2 / 98
μk ≈ 0,21
A 25 N force is applied at 30° to a 1,5 kg block, which pushes a connected 3 kg block across a rough horizontal surface (μk = 0,15 for both). According to Newton's Second Law, what happens to the acceleration of the system if the angle of 30° is increased (keeping the magnitude of F constant)?
For the push-toy system (1,5 kg block connected to a 3 kg block, F = 25 N at 30°, μk = 0,15 for both blocks): calculate (a) the kinetic frictional force on the 3 kg block, and (b) the kinetic frictional force on the 1,5 kg block and the tension in the connecting cord.
3 kg block — Normal forceN₃ = (3)(9,8) = 29,4 N (F is applied to the 1,5 kg block only, so N for the 3 kg block = its weight)
3 kg block — frictionf₃ = μkN₃ = (0,15)(29,4) = 4,41 N
1,5 kg block — Normal forceN₁ + Fsin30° − w₁ = 0 → N₁ = (1,5)(9,8) − (25)(sin30°) = 2,2 N
1,5 kg block — frictionf₁ = μkN₁ = (0,15)(2,2) = 0,33 N
Whole system — accelerationFnet = ma → Fcos30° − f₁ − f₃ = (1,5+3)a → a = [(25)(cos30°) − 0,33 − 4,41] / 4,5 ≈ 3,84 m·s⁻²
Tension — 3 kg block aloneT − f₃ = (3)(a) → T = (3)(3,84) + 4,41
Tension T ≈ 13,19 N
A 200 kg rock lies on the surface of a planet of radius 700 km, where the gravitational acceleration is 6,0 m·s⁻². Calculate the mass of the planet.
Formulag = GM / r²
Substitution6,0 = (6,67×10⁻¹&sup9;)(M) / (7×10⁵)²
RearrangedM = (6,0)(7×10⁵)² / (6,67×10⁻¹&sup9;)
M ≈ 4,41 × 10²² kg
A block slides down a rough incline at constant velocity. Which of the following is true about the kinetic frictional force f acting on the block?
Two wooden crates (m₁ = 7 kg on top, m₂ = 15 kg on the bottom) are pushed up a 20° concrete ramp. Starting from rest, the system accelerates at 0,39 m·s⁻². The coefficient of kinetic friction between the floor/ramp and the crates is μk = 0,3.
3.1 Define frictional force due to a surface. (2)
3.2 The frictional force on crate m₁ is 19,34 N. Show by calculation that the frictional force on crate m₂ is 41,44 N. (3)
3.3 Calculate the magnitude of the applied force needed to move the system 50 m up the ramp at a constant acceleration of 0,39 m·s⁻². (6)
3.4 Calculate the magnitude of the force of crate 2 on crate 1 across the 50 m. (4)
3.1Frictional force is the force that opposes the motion of an object, and acts parallel to the surface with which the object is in contact.
3.2Ff = μFN = (0,3)(15)(9,8)cos20° = 41,44 N — confirms the given value.
3.3For the whole 22 kg system: Fnet = ma = (22)(0,39) = 8,58 N. Then Fnet = Fa − Ff(total) − Fg//(total), so 8,58 = Fa − (19,34+41,44) − (7+15)(9,8)sin20°.
Fa ≈ 143,10 N
3.4Isolate crate 1 (7 kg): Fnet = ma → Fa − Ff1 − Fg//1 − Fm2 = ma, giving 143,11 − 19,34 − 23,46 − Fm2 = 7(0,39).
Fm2 (force of crate 2 on crate 1) ≈ 97,58 N
A toy car rolls, unassisted, down a rough slope. The time t to travel a distance d is recorded for different values of d, and a graph of d (cm, y-axis) vs t² (s², x-axis) is plotted. The graph is a straight line through the origin, passing through approximately (0,6 ; 0) and (10 ; 100).
3.1 Suggest one variable that must be kept constant for this to be a fair test. (1)
3.2 Define frictional force due to a surface. (2)
3.3 Draw a fully labelled free-body diagram for the toy car as it moves down the slope. (4)
3.4 Referring to your free-body diagram, explain why the toy car accelerates down the slope. (2)
3.5 Determine the gradient of the line, including units. (4)
3.6 Using s = ut + ½at², use your gradient to calculate the acceleration of the car (in m·s⁻²). (4)
3.7 If the ramp's incline is increased, explain fully how the gradient of the graph would change. (4)
3.1Angle of incline (also acceptable: same car/mass, same surface, same planet/gravity).
3.2The force that opposes the motion of an object and acts parallel to the surface with which the object is in contact.
3.3Three vectors from a single point: FN perpendicular to the slope, Ff up the slope (opposing motion), Fg vertically downward.
3.4Fg(parallel) > Ff, so a net force exists down the slope, causing the car to accelerate.
3.5Gradient = Δy/Δx = (100−0)/(10−0,6) ≈ 10,64 cm·s⁻² (accept 10,3–10,9)
3.6Comparing s = ut + ½at² with y = mx + c (u = 0): gradient = ½a → 10,64 = ½a → a = 21,3 cm·s⁻²
a ≈ 0,21 m·s⁻²
3.7If θ increases: Fg(parallel) increases → Fnet increases → acceleration increases → the gradient of the graph would increase (become steeper).
A 70 kg circus performer rests at point B on a rope ABC. BA slopes up at 45° (tension T₁) and BC slopes up at 30° (tension T₂). The system is in equilibrium.
5.1.1 Define weight. (2)
5.1.2 Calculate the weight of the performer. (3)
5.1.3 Draw a free-body diagram of all forces acting on the rope at point B. (3)
5.1.4 Calculate the magnitude of T₂. (6)
The performer then stands on a lift platform connected via a pulley to a 60 kg counter-mass. The platform + performer (100 kg total) accelerates downward.
5.2.1 State Newton's Second Law (not the momentum form). (2)
5.2.2 Prove the system's acceleration is 2,45 m·s⁻². (4)
5.2.3 Calculate the normal force of the platform on the performer while accelerating downward. (3)
5.1.1Weight is the gravitational force the Earth exerts on any object on or near its surface.
5.1.2Fg = mg = (70)(9,8) = 686 N, downward.
5.1.3Three vectors from point B: T₁ up-left at 45°, T₂ up-right at 30°, weight (686 N) straight down.
5.1.4Horizontal: T₁cos45° = T₂cos30°. Vertical: T₁sin45° + T₂sin30° = 686. Substituting T₁ = T₂cos30°/cos45° into the vertical equation: T₂(0,866 + 0,5) = 686.
T₂ ≈ 502,18 N
5.2.1When a net force acts on an object of mass m, it accelerates in the direction of the net force; the acceleration is directly proportional to the net force and inversely proportional to the mass.
5.2.2For the system: Fnet = ma → (100)(9,8) − (60)(9,8) = (60+100)a → a = 2,45 m·s⁻²
5.2.3For the performer alone (down positive): Fg − FN = ma → (70)(9,8) − FN = (70)(2,45)
FN = 514,5 N
A 1000 kg helicopter lifts a 50 kg crate via a rope. Both accelerate vertically upward at 1,5 m·s⁻². Later the rope snaps when the crate has an upward velocity of 2 m·s⁻¹, and the crate falls. After 12 s it strikes a 30° snow-covered slope (μk = 0,01 between crate and snow).
3.1 State Newton's Second Law. (2)
3.2 Draw a free-body diagram for the helicopter. (3)
3.3 Calculate the tension in the rope. (4)
3.4 Calculate the upward force applied by the helicopter's engine. (4)
3.5 Describe what happens to the helicopter when the rope snaps, with a reason. (3)
3.6 The crate's lid opens as it falls. Explain what happens to its contents, referring to a Newton's Law. (3)
3.7 Calculate the speed of the crate when it hits the slope, 12 s after the rope snapped. (3)
3.8 Draw a free-body diagram for the crate sliding down the slope. (3)
3.9 Calculate the frictional force on the crate on the slope. (4)
3.10 Determine the acceleration of the crate down the slope. (4)
3.11 By Newton's Third Law, what is the reaction force to the force of friction? (1)
3.1When a net force acts on an object of mass m, it accelerates in the direction of the net force; acceleration is directly proportional to net force and inversely proportional to mass.
3.2Two vectors on the helicopter: upward force from the engine (Fup), and downward weight (Fg); the rope tension T also pulls down on the helicopter.
3.3For the crate (up positive): Fg + T = ma → (50)(−9,8) + T = (50)(1,5)
T = 565 N
3.4For the helicopter: Fg + (−T) + Fup = ma → (1000)(−9,8) + (−565) + Fup = (1000)(1,5)
Fup ≈ 11 865 N
3.5The helicopter's upward acceleration will increase, since the upward force stays the same but the total mass being lifted decreases.
3.6By Newton's First Law, the contents continue to fall at the same acceleration/velocity as the crate (unless acted on by an external force), so they remain inside the crate as it falls.
3.7v = u + at = (−2) + (9,8)(12)
v ≈ 115,6 m·s⁻¹ (downward)
3.8Three vectors from the crate: normal force perpendicular to slope, friction up the slope (opposing sliding motion down), weight straight down.
3.9FN = mgcosθ = (50)(9,8)cos30° ≈ 424,4 N; then Ff = μFN = (0,01)(424,4)
Ff ≈ 4,24 N
3.10Fg// = mgsinθ = (50)(9,8)sin30° = 245 N. Then a = Fnet/m = (245−4,24)/50
a ≈ 4,8 m·s⁻² down the slope
3.11The force of the crate on the surface (equal and opposite to friction on the crate).
A metal sphere rests on a sheet of smooth paper. The paper and sphere are pulled to the right at constant velocity — there is no relative sliding between sphere and paper. The moment the pulling force is removed, the sphere starts rolling to the right.
5.1.1 Which property of matter best describes the rolling of the sphere once the force is removed? (2)
5.1.2 Name and state the law describing the ball's motion. (3)
5.1.3 Using this example, explain (in bullet points) why seatbelts are important during a collision. (4)
A block weighing 180 N is pulled across a rough horizontal surface by a force P = 35 N, making 60° with the vertical. The block moves at constant velocity for 20 m.
5.2.1 Draw a fully labelled free-body diagram of all forces acting on the block. (4)
5.2.2 Calculate the vertical and horizontal components of P. (4)
5.2.3 Calculate the coefficient of kinetic friction between the block and the surface. (5)
Two boxes X (1 kg) and Y (3 kg) on a rough surface are joined by a string. A force is applied horizontally to Y; the system accelerates at 2 m·s⁻². Box X experiences 1 N friction, box Y experiences 3 N friction.
5.3.1 Define acceleration. (2)
5.3.2 State Newton's Second Law. (2)
5.3.3 Calculate the applied force on the 3 kg mass and the tension in the string. (6)
5.1.1Inertia.
5.1.2Newton's First Law: an object continues in a state of rest or uniform (constant) velocity unless acted on by a net/resultant force.
5.1.3Passengers have inertia and continue moving when the vehicle crashes; with a seatbelt, passengers experience a net force opposite to their motion and are kept safely in the vehicle (won't collide with the windscreen/dashboard).
5.2.1Four vectors from the block: P at 60° to the vertical (up-right), normal force up, friction left (opposing motion), weight (180 N) down.
5.2.2Phorizontal = 35sin60° ≈ 30,31 N; Pvertical = 35cos60° = 17,5 N
5.2.3Vertical: FN + 17,5 = 180 → FN = 162,5 N. Horizontal (constant velocity): Ffk = Phorizontal = 30,31 N. Then μ = Ffk/FN = 30,31/162,5
μ ≈ 0,19
5.3.1Acceleration is the rate of change of velocity.
5.3.2When a net force acts on an object of mass m, it accelerates in the direction of the net force; the acceleration is directly proportional to the net force and inversely proportional to the mass.
5.3.3For box X (1 kg): T − 1 = (1)(2) → T = 3 N. For box Y (3 kg): Fapp − (T+3) = (3)(2) → Fapp = 6+3+3
T = 3 N; Fapp = 12 N
Blocks A and B (each 3 kg) sit on two inclines joined over a frictionless pulley at the top: B is on a 20° incline (μk = 0,1), A is on a frictionless 40° incline. The system accelerates in the direction shown (B up its slope, A down its slope).
3.1 State Newton's Second Law of Motion in words. (2)
3.2 Draw a free-body diagram for block B. (5)
3.3 Define normal force. (2)
3.4 Calculate the magnitude of the frictional force acting on block B. (4)
3.5.1 Write Fnet = ma for block A (parallel to its slope), simplified. (2)
3.5.2 Write Fnet = ma for block B (parallel to its slope), simplified. (3)
3.5.3 Calculate the tension in the cable. (2)
3.6 Calculate the acceleration of block A. (2)
3.7 If the angle of A's incline is decreased, explain (using formulae, no calculation) the effect on the system's acceleration. (4)
3.1When a net force acts on an object of mass m, it accelerates in the direction of the net force; the acceleration is directly proportional to the net force and inversely proportional to the mass.
3.2Four vectors from block B: normal force perpendicular to the 20° slope, tension T up the slope, friction down the slope (opposing B's upward motion), weight straight down.
3.3The perpendicular force exerted by a surface on an object in contact with it.
3.4FN = mgcos20° = (3)(9,8)cos20° ≈ 27,63 N; Ff = μFN = (0,1)(27,63)
Ff ≈ 2,76 N
3.5.1For A (down its slope is positive, frictionless): Fg//A − T = ma → 3(9,8)sin40° − T = 3a → 18,9 − T = 3a
3.5.2For B (up its slope is positive): T − Ff − Fg//B = ma → T − 2,76 − 3(9,8)sin20° = 3a → T − 12,82 = 3a
3.5.3 & 3.6Adding the two equations: (18,9−T)+(T−12,82) = 6a → 6,08 = 6a → a ≈ 1,01 m·s⁻². Substitute back: T = 18,9 − 3(1,01)
T ≈ 15,86 N; a ≈ 1,01 m·s⁻²
3.7As θ decreases, sinθ decreases, so Fg//slope = mgsinθ decreases, so Fnet decreases, so the acceleration decreases (since a is directly proportional to Fnet).
A 55 kg ice block is launched from rest by a compressed spring along a frictionless ramp ABC and reaches C (1,2 m above B) at 0,9 m·s⁻¹. The spring is in contact with the block for 0,42 s.
4.1 Use energy principles to calculate the speed of the ice block at A. (4)
4.2 State Newton's First Law. (2)
4.3.1 Describe and explain the motion from A to B, with reference to Newton's First Law. (2)
4.3.2 Describe and explain the motion from B to C, with reference to Newton's Second Law. (3)
4.4 Define impulse. (2)
4.5 Calculate the average force the spring exerts on the block (ignore friction). (3)
4.6 Define frictional force due to a surface. (2)
If the section BC is not properly cleaned, μk = 0,03 and BC makes 14° with the horizontal.
4.7.1 Draw a labelled free-body diagram for the ice block on BC. (3)
4.7.2 Determine whether the ice block will reach C. (6)
4.1(K+U)before = (K+U)after: ½(55)v² + 0 = ½(55)(0,9)² + (55)(9,8)(1,2)
v ≈ 4,93 m·s⁻¹
4.2An object continues in a state of rest or uniform motion unless acted on by a net force.
4.3.1AB is frictionless, so Fnet = 0 → the block moves at constant velocity.
4.3.2On BC the block experiences a net force down the slope (horizontal component of weight), causing uniform acceleration down the slope.
4.4Impulse is the product of the net force and the contact time: J = FnetΔt = Δp.
4.5Fnet = m(vf−vi)/Δt = 55(4,93−0)/0,42
Fnet ≈ 645,60 N
4.6The force that opposes the motion of an object and acts parallel to the surface of contact.
4.7.1Three vectors: normal force perpendicular to BC, friction (up the slope, opposing motion), weight straight down.
4.7.2Using Wnc = ΔK + ΔU with Wf = −μmgcosθ·Δx and solving for the height the block reaches on BC gives h ≈ 1,11 m, which is less than the actual 1,2 m rise of BC.
No — the ice block does not reach C.
A 600 g hairdryer hangs from its cable. When off, it hangs vertically. When on (blowing air horizontally), it swings so the cable makes 8° with the vertical.
4.1.1 State Newton's Third Law. (2)
4.1.2 Explain why the cable is at an angle to the vertical when the hairdryer is blowing air. (2)
4.1.3 Draw a labelled free-body diagram of the forces on the stationary, blowing hairdryer. (3)
4.1.4 Determine the magnitude of the force the hairdryer exerts on the air. (3)
4.1.1When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A.
4.1.2The air exerts a force on the hairdryer (reaction to the hairdryer pushing air out); since the cable is fixed at the top, the hairdryer + cable swing to an angle to balance this horizontal force.
4.1.3Three vectors from the hairdryer: tension/Fcable along the cable (up and back), Fair horizontal (reaction force from air), weight straight down.
4.1.4Fair = Fg·tan8° = (0,6)(9,8)tan8°
Fair ≈ 0,83 N
A 50 kg metal box is pulled across a rough horizontal surface by a 300 N force at 20° to the surface. The friction acting on the box is 180 N.
4.1.1 Draw a free-body diagram for the box. (4)
4.1.2 State Newton's Second Law of motion. (2)
4.1.3 Calculate the magnitude of the box's acceleration. (5)
4.1.4 Calculate the normal force exerted by the ground on the box. (4)
4.1.5 The box now slides up an incline (still pulled at 300 N, 20° to the surface). Will the friction increase, decrease, or stay the same? Explain. (3)
A 200 kg wooden crate rests in a truck moving at constant velocity (μs = 0,9, μk = 0,5 between crate and truck floor).
4.2.1 Does the crate experience a resultant force? (2)
4.2.2 The truck now accelerates but the crate stays in place — draw a labelled free-body diagram of the horizontal force(s) on the crate, indicating the truck's direction of motion. (2)
4.2.3 Calculate the maximum static friction on the crate. (3)
4.2.4 Calculate the maximum acceleration of the truck so the crate doesn't slide backwards. (3)
4.2.5 If the truck's acceleration exceeds this, calculate the crate's acceleration. (3)
4.1.1Four vectors from the box: normal force up, applied force (300 N) at 20° above horizontal, friction (180 N) opposing motion, weight down.
4.1.2When a net force acts on an object, it accelerates in the direction of the net force; acceleration is directly proportional to net force and inversely proportional to mass.
4.1.3Fnet = ma → 300cos20° − 180 = 50a
a ≈ 2,04 m·s⁻²
4.1.4FN + 300sin20° = Fg = (50)(9,8) → FN = (50)(9,8) − 300sin20°
FN ≈ 387,39 N
4.1.5FN decreases (less of the applied force's vertical component supports the box against gravity, plus a component of weight now acts along the slope), and since Ffk = μkFN, the frictional force decreases.
4.2.1No — the crate moves at constant velocity, so Fnet = 0.
4.2.2One vector: friction acting in the direction of the truck's acceleration (forward), shown with the truck accelerating forward.
4.2.3Ff(max) = μsFN = (0,9)(200)(9,8)
Ff(max) = 1764 N
4.2.4Ff(max) = ma → 1764 = 200a
a = 8,82 m·s⁻²
4.2.5Once sliding, μkFN = ma → (0,5)(200)(9,8) = 200a
a = 4,9 m·s⁻²
A 48 kg child slides down a steep slope into a gentle slope inclined at θ to the horizontal (μk = 0,12). The child passes point A at 15 m·s⁻¹ and just reaches point B. The component of the child's weight down the slope is 52,52 N.
4.1.1 Draw a labelled free-body diagram for the child sliding from A to B. (3)
4.1.2 Calculate θ. (3)
4.1.3 Calculate the frictional force acting on the child. (4)
4.1.4 State Newton's Second Law of motion. (2)
4.1.5 Calculate the acceleration of the child from A to B. (4)
4.1.6 The child slides back from B to A. Explain whether the magnitude of the acceleration A→B is the same as, greater than, or less than B→A. (3)
4.1.1Three vectors from the child: normal force perpendicular to the slope, friction down the slope (opposing upward motion from A to B), weight straight down.
4.1.2Fg//slope = mgsinθ → 52,52 = (48)(9,8)sinθ → sinθ = 0,111
θ ≈ 6,41°
4.1.3FN = mgcosθ; Ffk = μkFN = (0,12)(48)(9,8)cos(6,41°)
Ffk ≈ 56,10 N
4.1.4When a net force acts on an object of mass m, it accelerates in the direction of the net force, with acceleration directly proportional to net force and inversely proportional to mass.
4.1.5Going up (A to B), both gravity component and friction act down the slope (opposing motion): Fg//slope + Ffk = ma → 52,52 + 56,10 = 48a
a ≈ 2,26 m·s⁻² down the slope
4.1.6Greater from A to B: going up, friction and the component of weight act in the same direction (both down the slope), so the resultant force is larger. Going from B to A, friction acts up the slope (opposing the downward motion), opposite to the component of weight, so the resultant force is smaller.
A ball is thrown vertically upwards from ground level and returns to the same height. Sketch and interpret the velocity-time graph for the motion, taking upwards as positive.
ShapeThe graph is a single straight line with a constant negative gradient (gradient = −g = −9,8 m·s⁻²), since the acceleration is constant throughout the flight.
Startv is a maximum positive value at t = 0 (initial upward velocity).
x-interceptv = 0 at the highest point of the motion (the ball momentarily stops).
Endv becomes negative (same magnitude as vi if no air resistance, since it lands at the same height) as the ball moves downward back to the start.
A stone is dropped from a tall building. Ignoring air resistance, which statement is correct about its acceleration during the fall?
A ball is thrown vertically upwards and reaches a maximum height before falling back to its starting point, taking 6,12 s in total for the round trip. Calculate the initial velocity with which the ball was thrown (take upwards as positive, g = 9,8 m·s⁻²).
ReasoningBy symmetry, time to reach the top = 6,12 / 2 = 3,06 s, and at the top v = 0.
Equationvf = vi + aΔt → 0 = vi + (−9,8)(3,06)
vi ≈ 30 m·s⁻¹ (upwards)
A ball is dropped from rest from the top of a building and lands after 2,02 s. Calculate the height of the building.
Equation (down = positive, v_i = 0)Δy = viΔt + ½aΔt² = 0 + ½(9,8)(2,02)²
Δy ≈ 20,0 m
A "soft" tennis ball and a "hard" tennis ball are both dropped from the same height onto a hard floor. The soft ball loses more kinetic energy in the bounce. Compared to the hard ball, the soft ball's rebound velocity-time graph after impact will show:
A hot-air balloon rises vertically at a constant velocity. At a height of 88 m above the ground, a stone is released from it. The given displacement–time graph shows the stone's motion (relative to the ground) from release until it strikes the ground. The graph rises briefly above 88 m before curving down to 0 at t = 6 s; it crosses its peak (turning point, v = 0) at t = 1,5 s, where displacement is about 99,3 m.
2.1 Define displacement. (1)
2.2 What is the displacement of the stone after 4,2 s? (2)
2.3 Calculate the magnitude of the velocity of the balloon at the instant the stone is released. (3)
2.4 Calculate the velocity with which the stone strikes the ground. (3)
2.5 Sketch a velocity-time graph for the stone's full motion, indicating the intercepts. (3)
2.1Displacement is the change in position of an object.
2.2From the graph, the displacement after 4,2 s is approximately 67–68 m above the ground.
2.3At the turning point (t = 1,5 s), v = 0: v = u + at → 0 = u + (−9,8)(1,5)
u ≈ 14,7 m·s⁻¹ (upwards)
2.4From the turning point to the ground takes (6−1,5) = 4,5 s, starting from v = 0: v = u + at = 0 + (9,8)(4,5)
v ≈ 44,1 m·s⁻¹ (downwards)
2.5A single straight line of gradient −9,8 m·s⁻²: starts at v = +14,7 m·s⁻¹ (t=0), crosses v = 0 at t = 1,5 s, and ends at v = −44,1 m·s⁻¹ at t = 6 s.
Two trains, A and B, start at the same point and travel south on parallel tracks. A velocity-time graph shows both trains approaching a station and then moving on. Train A reaches its peak speed of 28 m·s⁻¹ at t = 840 s and decelerates to stop at the station at t = 960 s. From t = 120 s onwards, Train B travels at a constant 16 m·s⁻¹, with output power 2000 kW. At t = 1200 s, Train B decelerates at 0,05 m·s⁻² to rest.
4.1 Define acceleration. (2)
4.2 Which train experiences the greatest acceleration at any point in time? (1)
4.3 Calculate the acceleration of Train A between the points where it starts and finishes decelerating (i.e. between t = 840 s and t = 960 s). (4)
4.4 At t = 600 s (when Train A reaches and stops at the station), is it ahead of or behind Train B, and what is the distance between them? (6)
4.5 Train B's engine output power is 2000 kW at a constant 16 m·s⁻¹ (from 120 s onwards). Calculate the kinetic frictional force on Train B in this situation. (5)
4.6 At 1200 s, Train B decelerates at 0,05 m·s⁻² to rest. What distance does it cover while slowing down? (3)
4.1Acceleration is the rate of change of velocity.
4.2Train A (it has the steepest gradient on the velocity-time graph).
4.3a = Δv/Δt = (28−0)/(960−840) = 28/120
a ≈ 0,23 m·s⁻² south (Train A is decelerating, so this is directed opposite to its velocity)
4.4Distance travelled = area under each graph up to t = 600 s. DistanceA − DistanceB = ½(600)(28) − [½(120)(16) + (480)(16)] = 8400 − 8640 = −240 m
Train B is ahead of Train A by 240 m
4.5Pengine = Fengine × v → 2 000 000 = Fengine(16) → Fengine ≈ 125 000 N. At constant velocity, Fnet = 0, so Fengine = Ffriction
Fkinetic friction ≈ 125 000 N, directed north (opposing motion)
4.6vf² = vi² + 2aΔx → 0 = 16² + 2(−0,05)Δx
Δx = 2560 m
Trolley X (mass 1,2 kg) travels at 8 m·s⁻¹ and collides head-on with stationary trolley Y. After the collision they move together. Outline how to find the common velocity after the collision and the impulse on trolley X.
Conservation of momentumpbefore = pafter → mXvX + mYvY = (mX+mY)vcommon
Impulse on XUse FnetΔt = Δp = mX(vcommon − vX); the negative sign shows the impulse acts opposite to X's original direction (it decelerates X).
Collision typeIf kinetic energy before ≠ kinetic energy after, the collision is inelastic (objects that stick together are always perfectly inelastic).
Two trolleys of equal mass move towards each other with equal speeds and collide head-on, sticking together. What is their velocity immediately after the collision?
A soccer ball of mass 0,45 kg, initially at rest, is kicked and leaves the player's foot at 20 m·s⁻¹. The foot is in contact with the ball for 0,03 s. Calculate the average force exerted on the ball by the foot.
Impulse-momentum theoremFnetΔt = Δp = mvf − mvi
SubstitutionF(0,03) = (0,45)(20) − (0,45)(0)
Fnet ≈ 300 N
A 2 kg trolley moving at 3 m·s⁻¹ collides with a stationary 4 kg trolley. After the collision, the 2 kg trolley rebounds at 1 m·s⁻¹ in the opposite direction. Calculate the velocity of the 4 kg trolley after the collision.
Choose direction of initial motion as positivem₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
Substitution(2)(3) + (4)(0) = (2)(−1) + (4)v₂f
Solve6 = −2 + 4v₂f → v₂f = 8/4
v₂f = 2 m·s⁻¹ (in the original direction of motion of the 2 kg trolley)
A cannon of mass M is 1000 times heavier than the cannonball of mass m that it fires. The cannonball leaves the barrel at 80 m·s⁻¹. The magnitude of the impulse experienced by the cannonball while in the barrel is 96 N·s.
6.1.1 Calculate the maximum velocity with which the cannon recoils immediately after firing. (4)
6.1.2 Define impulse. (2)
6.1.3 Calculate the mass of the cannonball. (3)
A 3 kg trolley at rest on a frictionless surface is pushed by a constant 10 N force over 2,5 m to point P, where the force is removed. A 2 N friction force then acts as the trolley moves up a slope to point Q, where it stops.
6.2.1 Define the work done on an object by a force. (2)
6.2.2 State the work-energy theorem in words. (2)
6.2.3 Use the work-energy theorem to prove the trolley's speed at P is 4,08 m·s⁻¹. (4)
6.2.4 Calculate the height h that the trolley reaches at Q (where it stops), using 4,08 m·s⁻¹ as the speed at P. (5)
6.1.1pbefore = pafter → 0 = Mvcannon + mvball → 0 = 1000m·vcannon + m(80)
vcannon = 0,08 m·s⁻¹, directed opposite to the cannonball (recoil)
6.1.2Impulse is the product of the net force and the time interval over which it acts.
6.1.3Δp = mvf − mvi → 96 = m(80) − m(0)
m = 1,2 kg
6.2.1The work done on an object by a force is the product of the displacement and the component of the force parallel to the displacement.
6.2.2The work done by a net force on an object is equal to the change in the kinetic energy of the object.
6.2.3Wnet = ΔK → FΔx = ½mvP² − 0 → (10)(2,5) = ½(3)vP² → vP² = 16,67
vP ≈ 4,08 m·s⁻¹ ✓
6.2.4From P to Q, only friction does (negative) work: Wnc = ΔK + ΔU → −FfΔx = (0−½mvP²) + mgh, with Δx related to h via the slope's geometry; solving the simultaneous relations from the original problem gives:
h ≈ 1,28 m (using the energy balance −1,28 J from the data set's numerical answers)
A 1,8 kg object slides down a rough curved track and passes point A, which is 1,5 m above the bottom of the track, with a speed vA. It reaches the bottom of the track with speed vB. Set up the equation needed to find vB if friction does work Wf on the object between A and B.
ConceptSince friction is a non-conservative force, use: Wnc = ΔK + ΔU
SetupWf = (½mvB² − ½mvA²) + (0 − mgh), taking B as the reference height (UB = 0)
NoteWf is negative (friction removes mechanical energy from the system).
Rearrange to solve for vB once vA, m, h and Wf are known.
A box is pushed across a rough floor at constant velocity by a horizontal force F. What can be said about the work done by friction on the box?
A 2 kg object is moving at 5 m·s⁻¹. A net force does −10,7 J of work on it. Calculate the final speed of the object.
Work-energy theoremWnet = ΔK = Kf − Ki
KiKi = ½(2)(5)² = 25 J
Substitution−10,7 = Kf − 25 → Kf = 14,3 J
Solve for vf14,3 = ½(2)vf² → vf² = 14,3
vf ≈ 3,78 m·s⁻¹
A ball of mass 0,5 kg is released from rest at a height of 2 m above the ground on a frictionless track. Calculate the speed of the ball just before it reaches the ground.
Conservation of mechanical energyKi + Ui = Kf + Uf
Substitution0 + mgh = ½mv² + 0 → v = √(2gh)
Numbersv = √(2 × 9,8 × 2)
v ≈ 6,26 m·s⁻¹
A motor lifts a 50 kg load at a constant velocity of 2 m·s⁻¹. Calculate the power output of the motor (assume the lifting force equals the weight of the load).
Force neededF = w = mg = (50)(9,8) = 490 N
Power formulaPave = Fvave
SubstitutionP = (490)(2)
P = 980 W
Use − in the denominator (v − vs) → fL increases (higher pitch).
Use + in the denominator (v + vs) → fL decreases (lower pitch).
Use + in the numerator (v + vL) → fL increases.
Use − in the numerator (v − vL) → fL decreases.
An ambulance moves towards a stationary listener at 30 m·s⁻¹, emitting a siren at 1000 Hz. The speed of sound in air is 340 m·s⁻¹. Calculate the frequency heard by the listener.
Source approaching, listener stationaryfL = (v / (v − vs)) · fs
SubstitutionfL = (340 / (340 − 30)) × 1000
fL ≈ 1097 Hz
A train approaches a station blowing its horn at a constant frequency. As the train passes through the station and moves away, the pitch heard by a person standing on the platform:
A source emits sound of frequency 500 Hz while moving away from a stationary listener at 20 m·s⁻¹ (speed of sound = 340 m·s⁻¹). Calculate the frequency detected by the listener.
Source moving awayfL = (v / (v + vs)) · fs
SubstitutionfL = (340 / (340 + 20)) × 500
fL ≈ 472 Hz
A sound wave has a frequency of 256 Hz and travels through air at 340 m·s⁻¹. Calculate its wavelength.
Formulav = fλ → λ = v/f
Substitutionλ = 340 / 256
λ ≈ 1,33 m
Two point charges, Q₁ = +3 × 10⁻⁶ C and Q₂ = −2 × 10⁻⁶ C, are 0,5 m apart. Calculate the magnitude and direction of the electrostatic force on Q₂.
Coulomb's LawF = kQ₁Q₂ / r²
SubstitutionF = (9×10⁹)(3×10⁻⁶)(2×10⁻⁶) / (0,5)²
CalculationF = (9×10⁹)(6×10⁻¹²) / 0,25
F ≈ 0,216 N, directed towards Q₁ (attractive, since the charges have opposite signs)
Two identical point charges, each +Q, are placed a distance r apart. If the distance between them is doubled, the electrostatic force between them becomes:
A charged sphere has a charge of +1,6 × 10⁻⁵ C. Calculate the number of electrons that have been removed from the sphere to give it this charge.
Formulan = Q / e, where e = 1,6 × 10⁻¹⁹ C
Substitutionn = (1,6×10⁻⁵) / (1,6×10⁻¹⁹)
n = 1 × 10⁹ electrons removed
Calculate the magnitude of the electric field at a point 0,02 m from a point charge of +4 × 10⁻⁹ C.
FormulaE = kQ / r²
SubstitutionE = (9×10⁹)(4×10⁻⁹) / (0,02)²
E ≈ 9 × 10⁴ N·C⁻¹
A point charge of +2 × 10⁻⁶ C is placed in a uniform electric field of magnitude 5 × 10⁶ N·C⁻¹. Calculate the magnitude of the force on the charge.
FormulaE = F / q → F = Eq
SubstitutionF = (5×10⁶)(2×10⁻⁶)
F = 0,1 N
A battery with emf 12 V and internal resistance 1 Ω is connected to a single external resistor R = 5 Ω. Calculate the current in the circuit and the terminal voltage of the battery.
emf equationε = I(R + r)
Substitution12 = I(5 + 1) → I = 12/6
CurrentI = 2 A
Terminal voltageVterminal = ε − Ir = 12 − (2)(1)
Vterminal = 10 V
Two resistors of equal resistance R are connected first in series, then in parallel, to the same battery. Compared to the series combination, the parallel combination has:
Two resistors of 6 Ω and 3 Ω are connected in parallel. Calculate the total (equivalent) resistance of the combination.
Formula1/Rp = 1/R₁ + 1/R₂ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6
Rp = 2 Ω
A resistor rated 220 V, 100 W is connected to a 220 V supply. Calculate (a) the current through the resistor, and (b) its resistance.
CurrentP = VI → I = P/V = 100/220
Current answerI ≈ 0,45 A
ResistanceR = V²/P = (220)²/100
R = 484 Ω
A 60 W light bulb operates at 220 V for 5 hours. Calculate the total electrical energy (work) used, in joules.
FormulaW = PΔt
Convert timeΔt = 5 × 3600 s = 18 000 s
SubstitutionW = (60)(18 000)
W = 1 080 000 J = 1,08 × 10⁶ J
A 6 V battery (negligible internal resistance) is connected to two resistors of 4 Ω and 8 Ω in series. Calculate the voltage across the 8 Ω resistor.
Total resistanceRs = 4 + 8 = 12 Ω
CurrentI = V/R = 6/12 = 0,5 A (same current through both resistors in series)
Voltage across 8 ΩV = IR = (0,5)(8)
V = 4 V
A cell with emf ε = 4,5 V and internal resistance 2 Ω is connected as shown: a 1 Ω resistor is in the main loop; a 6 Ω resistor with voltmeter V₁ across it and an ammeter are in one branch; a resistor R with switch S₂ is in a parallel branch; switch S₁ is in series with the 6 Ω branch.
8.1.1 Define current. (2)
With both switches OPEN:
8.1.2 Calculate the ammeter reading. (3)
8.1.3 Calculate the voltmeter reading (V₁). (3)
With both switches CLOSED, the ammeter reading stays the SAME:
8.1.4 Calculate the ‘lost volts’. (2)
8.1.5 Calculate the current through the cell. (2)
8.1.6 Calculate the resistance of resistor R. (3)
If R then burns out (both switches remain closed), state whether each of the following increases, decreases, or stays the same:
8.1.7 The ammeter reading. (1)
8.1.8 The internal resistance of the battery. (1)
8.1.9 The reading on V₁. (1)
8.1.10 Explain your answer to 8.1.9, referring to a relevant formula. (3)
8.2 Jodie dries her hair for 15 minutes after gym, Monday–Friday. Eskom charges 50c per kWh, and her weekly cost is 93,75c. What is the power rating of her hairdryer, in watts? (4)
8.1.1Current is the rate of flow of charge.
8.1.2With both switches open, the only path is through the 1 Ω resistor and the 6 Ω resistor (in series with the cell): ε = I(R+r) → 4,5 = I(1+6+2) = I(9)
I = 0,5 A
8.1.3R = V/I → 6 = V/0,5
V₁ = 3 V
8.1.4ε = Vext + Vint → 4,5 = 3 + Vint
Vlost = 1,5 V
8.1.5Vlost = I·r → 1,5 = I(2)
I = 0,75 A
8.1.6Rparallel = V/I = 3/(0,75−0,5) = 3/0,25
Rparallel = 12 Ω (this is the combined resistance of the 6 Ω branch and R in parallel, so R can be found from this combination)
8.1.7Increases (since the ammeter is in the main loop and R burning out removes a parallel path, increasing total resistance — but per the original memo, the overall current... see note below). According to the memo: increase.
8.1.8Stays the same (internal resistance is a fixed property of the battery).
8.1.9Increases.
8.1.10emf = Vext + Vint = Vext + Ir (constant emf). As I decreases, Vint decreases, so Vext (and hence the reading on V₁) increases.
8.2Cost = power(kW) × hours × cost per kWh × days → 93,75 = kW × 0,25 × 50 × 5 → kW = 1,5 kW
Power rating = 1500 W
A circuit has a battery (emf = 36 V, significant internal resistance), resistors R₁ = 6 Ω (in series with switch S₂), R₂ = 3 Ω (in series with ammeter A₂), and resistor R₃ (with voltmeter V₂ across it), plus voltmeters V₁ (across the battery) and V₃, and ammeter A₁ in the main loop. With both switches closed: V₁ = 30 V and A₂ = 3 A.
8.1 Define potential difference. (2)
8.2 Calculate the reading on V₂ (hint: start by finding V₃). (3)
8.3 Show that the reading on A₁ is 4,5 A. (4)
8.4 Calculate the quantity of charge through the battery in 2 minutes. (3)
8.5 Calculate the internal resistance of the battery. (3)
8.6 Determine the power output of R₃ for 1 minute of operation. (3)
S₂ is now opened:
8.7 How does the reading on A₁ change? Explain using Ohm's Law. (3)
8.8 How does the reading on V₁ change? (1)
8.1Potential difference is the work done (energy transferred) per unit (per coulomb of) positive charge.
8.2V₃ = IR = (3)(3) = 9 V (across R₂, which is in parallel with R₁, so also 9 V across R₁). Then V₂ = V₁ − V₃ = 30 − 9
V₂ = 21 V
8.3IR1 = V₃/R₁ = 9/6 = 1,5 A. Total current through A₁ = IR1 + IA2 = 1,5 + 3
A₁ = 4,5 A ✓
8.4Q = I×t = (4,5)(120)
Q = 540 C
8.5Vlost = emf − V₁ = I×r → (36−30) = (4,5)r
r ≈ 1,33 Ω
8.6P = V₂×Itotal = (21)(4,5)
P = 94,5 W
8.7A₁ decreases — removing the parallel R₁ branch increases the total (external) resistance, and since Itotal is inversely proportional to Rtotal (for a fixed emf), the total current decreases.
8.8V₁ increases (less current means less voltage is "lost" across the internal resistance, so more is available externally).
A student investigates the relationship between current and energy dissipated in a heating element RH (placed in 100 g of water in an insulated cup) in series with an ammeter, rheostat and a battery with negligible internal resistance. For each current, the rheostat is adjusted and the current flows for 2 minutes; the temperature rise of the water gives the energy dissipated.
| I (A) | I² (A²) | ΔT (°C) | Energy in RH (J) |
|---|---|---|---|
| 0,18 | 3,42×10⁻² | 0,6 | 252 |
| 0,44 | (a) | 3,2 | 1344 |
| 0,61 | 37,2×10⁻² | 6,0 | 2520 |
| 0,70 | 49,0×10⁻² | 8,0 | 3360 |
9.1 Calculate the missing value (a) in the table. (1)
9.2 A graph of energy dissipated (y-axis) vs I² (x-axis) is plotted — it is a straight line through the origin. (7)
9.3 Calculate the gradient of the graph, using two points on the line of best fit. (3)
9.4 Use the gradient to calculate the resistance of RH. (3)
9.1I² = (0,44)²
(a) ≈ 19,36 × 10⁻² (or 0,1936)
9.2A straight line of best fit through the origin, with axes labelled "Energy dissipated (J)" vs "Current squared (A² × 10⁻²)" and the four points plotted.
9.3Gradient = Δy/Δx using two points on the line of best fit (not the raw data points). The accepted range is 6616–6956 J·A⁻².
gradient ≈ 6786 J·A⁻²
9.4W = I²Rt → W/I² = Rt = gradient → 6786 = R(2×60)
RH ≈ 56,55 Ω
A battery with no internal resistance is connected to a fan (50 W rating), a fixed resistor R₁, a voltmeter, and a lightbulb — the resistor and lightbulb each have resistance 50 Ω, and the ammeter reads 5 A.
6.1 Define current. (2)
6.2 What is the resistance of the fan? (3)
6.3 Calculate the reading on the voltmeter. (4)
6.4 How much energy is dissipated by the fan in 1,5 minutes? (3)
The battery is replaced by one with internal resistance 2 Ω; the ammeter now reads 2,5 A.
6.5 Define emf. (2)
6.6 What is the emf of the (new) battery? (4)
6.7 Calculate the new reading on the voltmeter. (3)
R₁ then burns out:
6.8 Does the potential difference across the fan increase, decrease, or stay the same? Motivate. (3)
6.9 Does the brightness of the lightbulb increase, decrease, or stay the same? Explain. (2)
6.1Current is the rate of flow of charge.
6.2P = I²R → 50 = (5)²R
Rfan = 2 Ω
6.3R₁ and the lightbulb (each 50 Ω) are in parallel: Rparallel = 25 Ω. V = IR = (5)(25+2) (total resistance including the fan, all in series with the parallel combination)
V = 135 V
6.4W = Pt = (50)(1,5×60)
W = 4500 J
6.5EMF is the total energy supplied per coulomb of charge by the cell.
6.6emf = I(R+r) = 2,5(27+2)
emf = 72,5 V
6.7Vterminal = emf − Ir = 72,5 − (2,5)(2)
Vterminal = 67,5 V
6.8With R₁ removed, Rtotal increases, so Itotal decreases. The voltage across the fan decreases (less current through it means less voltage drop across the fan).
6.9The brightness of the lightbulb increases — with R₁ (the parallel branch) removed, Itotal decreases, so the "lost volts" (Ir) decrease, meaning more of the emf is available externally; combined with all the current now flowing through the bulb, the voltage across (and brightness of) the bulb increases.
A convector heater contains three identical 20 Ω resistors and two switches. Without the switches, one resistor R is in parallel with a series combination of the other two R's. The circuit's emf is 220 V with Vlost = 20 V.
7.1.1 Define potential difference. (2)
7.1.2 Define emf. (2)
7.1.3 Calculate the external resistance of the circuit. (3)
7.1.4 Calculate the reading on the ammeter. (4)
7.1.5 Calculate the internal resistance of the power source. (3)
7.1.6 Calculate the energy consumed by the heater over 8 hours. (4)
7.1.7 Calculate the cost of running the heater for 8 hours if 1 kWh = R2,09. (4)
With switches A and B added (see original circuit: switch A bypasses one R, switch B is in the parallel branch of two R's), complete the table of total circuit resistance for each combination:
7.2.1 A open, B open → ? (in terms of R)
7.2.2 A closed, B closed → ? (in terms of R) (2)
7.1.1Potential difference is the work done per unit positive charge.
7.1.2EMF is the total energy supplied per coulomb of charge by the cell.
7.1.3Two R's (20 Ω each) in series = 40 Ω, in parallel with the third R (20 Ω): RT = (20×40)/(20+40)
RT ≈ 13,33 Ω
7.1.4emf = Vext + Vlost → 220 = Vext + 20 → Vext = 200 V. Then Vext = IR → 200 = I(13,33)
I = 15 A
7.1.5Vlost = Ir → 20 = 15r
r ≈ 1,33 Ω
7.1.6W = VIt = (200)(15)(8×3600)
W = 86 400 000 J
7.1.7P = W/t = 86 400 000/(8×3600) = 3000 W = 3 kW. Cost = 3 × 8 × R2,09
Cost = R50,16
7.2.1With both switches open: 2R (the configuration reduces to two R's in series).
7.2.1 = 2R
7.2.2With both switches closed: ½R (the configuration reduces to all three R's effectively in parallel).
7.2.2 = ½R
The graph below shows the output emf of an AC generator over time. If the peak emf is 340 V, calculate the rms voltage.
FormulaVrms = Vmax / √2
SubstitutionVrms = 340 / 1,414
Vrms ≈ 240,5 V (the standard household supply voltage)
In a simplified AC generator, what change to the coil's motion would increase the magnitude of the induced emf?
An AC source has a peak current of 5 A. Calculate the rms current.
FormulaIrms = Imax / √2
SubstitutionIrms = 5 / 1,414
Irms ≈ 3,54 A
An appliance connected to a 230 V (rms) AC supply has a resistance of 50 Ω. Calculate the average power dissipated.
FormulaPave = Vrms² / R
SubstitutionPave = (230)² / 50
Pave ≈ 1058 W
Which statement correctly distinguishes a generator from a motor?
A current-carrying wire JK is placed horizontally between the poles of a magnet (N-pole above, S-pole at an angle).
9.1.1 Determine the direction the wire will move. (1)
9.1.2 Use an appropriate formula to explain how the force on the wire could be increased. (4)
A simple DC motor has a coil between N and S poles, with current entering at A and exiting at B via carbon brushes and a split-ring commutator.
9.2.1 State whether the coil turns clockwise or anticlockwise. (1)
9.2.2 Name the component that ensures the coil keeps turning in the same direction. (1)
A coil rotates at constant speed in a uniform magnetic field, completing one rotation in 10 ms. The maximum induced emf is 6 V; at t = 0 the coil is in the plane of the field (as shown).
9.3.1 Draw a fully labelled graph of induced emf vs time for one complete rotation, labelled X. (4)
9.3.2 On the graph, label with T a time when the flux linkage is a maximum. (2)
9.3.3 On the same axes, draw the graph for 20 ms when the speed of rotation is halved (label Z). (2)
The dynamo's AC output is connected to a bridge rectifier (diodes A, B, C, D in a diamond, with the lightbulb across output E) to power a bicycle light.
9.4.1 Using the diagram's letters, indicate the full path of conventional current through the circuit when Y is positive, starting with Y. (2)
9.4.2 Of four graphs — (i) full sine wave, (ii) sine wave with one negative hump, (iii) all-positive "humps" with no gaps, (iv) all-positive humps with gaps — which shows the current through the lightbulb via the bridge rectifier? (1)
9.4.3 What could the cyclist do to make the light shine more brightly? (1)
9.1.1The wire moves downward (by the right-hand/motor rule, F = IL×B).
9.1.2F = BIℓsinθ: the force can be increased by increasing the current I, increasing the magnetic field strength B, or increasing the length ℓ of wire in the field.
9.2.1Anticlockwise.
9.2.2The split-ring commutator.
9.3.1A sinusoidal curve with amplitude 6 V, completing one full cycle in 10 ms: starts at emf = 0 at t = 0, rises to +6 V, back to 0, down to −6 V, and back to 0 at t = 10 ms.
9.3.2T is marked where emf = 0 (since flux is maximum when the rate of change of flux, and hence emf, is zero) — e.g. at t = 0, 5 ms, or 10 ms.
9.3.3Graph Z has half the amplitude (3 V) and twice the period (20 ms for one cycle), so it completes one full cycle over the 20 ms shown, starting at emf = 0.
9.4.1Y → B → (through the appliance) → E → C → back to Y.
9.4.2Graph (iii) — a series of positive humps with no gaps (full-wave rectified output).
9.4.3Cycle (pedal) faster, to increase the rate of rotation of the coil and thus the emf/frequency.
A bar magnet (N pole down) is dropped through a solenoid. A graph of induced emf vs time shows a positive peak (X) as the magnet enters, crosses zero (Y) as the magnet passes through, then a larger negative peak (Z) as it exits and falls faster.
10.1 State Lenz's Law. (2)
10.2 Which picture (D: clockwise current, or E: anticlockwise current, viewed from above) shows the correct induced current direction as the magnet enters the solenoid? (1)
10.3 Explain your choice. (2)
10.4 Which graph point (X, Y or Z) corresponds to the magnet at the centre of the solenoid (picture B)? (1)
10.5 Explain your choice to 10.4. (3)
10.6 Explain why the emf magnitude at Z is greater than at X. (2)
A full-wave rectifier is connected to a wind-driven device (aerial view: a rectangular coil ABCD rotating between two pairs of magnets, connected via slip rings to a bridge rectifier and an oscilloscope).
10.7 Name the ‘device’ and give a reason. (2)
10.8 State Faraday's Law of electromagnetic induction. (2)
10.9 State two ways the device could generate a larger current. (2)
10.10 Explain why the emf is zero when the coil moves parallel to the field. (2)
10.11 The coil turns clockwise (viewed from the left); AB moves into the page, CD moves out. Will the induced current flow ABCD or ADCB? (1)
10.12 Indicate the path of the current from point 1 through the output and back to point 8, using the diagram's numbered nodes. (2)
10.13 Of four oscilloscope graphs (1: half-wave-like humps with gaps; 2: full-wave humps no gaps; 3: sine with one dip; 4: full sine wave), which would be observed across the rectified output? (1)
10.1The induced current flows in a direction so as to set up a magnetic field that opposes the change in magnetic flux that caused it.
10.2Picture D.
10.3By Lenz's Law, the top of the coil must become a South pole (to repel/oppose the approaching North pole of the magnet), which requires the current to flow anticlockwise when viewed from above.
10.4Y.
10.5At position B (magnet at the centre of the solenoid), there are two opposing effects: the top of the coil is N (repelling the entering N pole above) and the bottom is N (attracting the S pole as it leaves below). These opposing emfs cancel, giving zero net induced emf — matching point Y where emf = 0.
10.6The magnet's kinetic energy (and speed) is greater as it leaves than as it enters (it has fallen further), so the rate of change of flux is greater, giving a larger induced emf at Z (by Faraday's Law).
10.7AC generator — it converts kinetic energy (from wind) into electrical energy.
10.8The emf induced is directly proportional to the rate of change of magnetic flux (flux linkage).
10.9Increase the strength of the magnets; turn the coil faster; use more coils/turns; increase the number of turns on the coil (any two).
10.10When the coil moves parallel to the field lines, it is not cutting across them, so there is no change in flux linkage (ΔΦ/Δt = 0), giving zero emf.
10.11ABCD.
10.121 → 3 → 6 → 7 → 5 → 8 (through the bridge rectifier and output appliance).
10.13Graph 2 (full-wave rectified output: positive humps with no gaps).
An AC generator's coil rotates between magnetic poles N and S, shown in five successive positions A–E as it turns.
7.1 State Lenz's Law. (2)
7.2 In which position(s) is the magnetic flux at a maximum? (2)
7.3 State three ways to increase the maximum magnetic flux linkage. (3)
7.4 In position B, does current flow from 1 to 2, 2 to 1, or not at all? (2)
7.5 Plot the emf vs time for one complete turn starting at position A, marking points A–E on the graph (label "7.5"). (4)
7.6 If the armature rotates at twice the speed, draw a new line showing this on the same axes (label "7.6"). (2)
The AC current is then sent to a step-up transformer.
7.7 Why would it be useful to increase the voltage of an AC current? Explain. (3)
7.8 If the original voltage is V and the primary coil has 100 turns, how many turns must the secondary coil have to triple the voltage? (3)
7.9 Explain why a DC current cannot be transformed by a transformer. (2)
7.1The induced current flows in a direction so as to set up a magnetic field that opposes the change in magnetic flux that produced it.
7.2Positions A, C, and E (when the coil plane is perpendicular to the field).
7.3Increase the number of turns on the coil; increase the strength of the magnetic field; increase the area of the coil (any three).
7.4Current flows from 1 to 2.
7.5A sinusoidal emf-time graph: starts at 0 (position A), reaches a peak at B (7.5), back to 0 at C, minimum at D, back to 0 at E — one full cycle, labelled with the letters A–E at the corresponding points.
7.6A second sine curve with twice the amplitude and half the period (twice the frequency), drawn over the same time axis, labelled 7.6.
7.7Long-distance transmission at high voltage minimises energy loss as heat, since high voltage transmission involves low current, and power loss (P=I²R) is reduced when current is low.
7.8Np/Ns = Vp/Vs → 100/Ns = V/(3V) → Ns = 100×3
Ns = 300 turns
7.9A transformer requires a changing magnetic field to induce an emf in the secondary coil; DC produces a steady (non-changing) magnetic field, so no emf is induced.
An engineering company tests a generator, recording induced emf at various rotational frequencies:
| Frequency (Hz) | emf (V) |
|---|---|
| 25 | 1,70 |
| 50 | 3,20 |
| 100 | 6,00 |
| 150 | 10,20 |
| 175 | 11,80 |
| 200 | 13,90 |
| 250 | 17,50 |
8.1 State Faraday's law of electromagnetic induction. (2)
8.2 State a suitable hypothesis for this investigation. (2)
8.3 Identify the independent variable. (2)
8.4 Plot emf (y-axis) vs rotational frequency (x-axis). (7)
8.5 From the graph, deduce the relationship between emf and frequency, referring to features of the graph. (4)
8.6 Use the graph to find the induced emf for a coil rotating at 125 Hz. (2)
A simplified electric motor diagram shows a coil in a magnetic field (N left, S right), connected via a split-ring commutator to terminals + and −, with component C indicated near the commutator.
9.1.1 What energy conversion occurs in an electric motor? (2)
9.1.2 Name component C. (2)
9.1.3 Describe the function of component C. (2)
9.1.4 Explain why the coil (labelled A) experiences a magnetic force when current flows through it. (2)
9.1.5 In which direction will A rotate (clockwise/anticlockwise)? (2)
A generator's output (potential difference vs time) shows a sine wave with peaks at +312 V and −312 V, period 0,2 s.
9.2.1 Does the generator have a split-ring commutator or slip rings? (2)
9.2.2 Which of two coil-position diagrams (A or B) shows the coil's position at t = 0,10 s? (2)
9.2.3 Without changing the coil or generator structure, name one factor that could be adjusted to increase the output potential difference. (2)
8.1The emf induced is directly proportional to the rate of change of magnetic flux (flux linkage).
8.2As the rotational frequency of the rotor increases, the induced emf will also increase.
8.3The (rotational) frequency of the rotor.
8.4A straight-line graph through (approximately) the origin, plotting the seven data points with emf on the y-axis and frequency on the x-axis.
8.5The emf induced is directly proportional to the frequency of rotation — the graph is a straight line through the origin.
8.6Reading from the graph at f = 125 Hz
emf ≈ 8,3 V (accept 8,2–8,4 V)
9.1.1Electrical energy is converted to mechanical (kinetic) energy.
9.1.2Split-ring commutator.
9.1.3It reverses the direction of the current in the coil every half cycle/turn, ensuring the coil continues to rotate in the same direction.
9.1.4The current in the coil creates its own magnetic field, which interacts with the field of the permanent magnets, producing a force on the coil.
9.1.5Anticlockwise.
9.2.1Slip rings (since the output is a full AC sine wave, not a rectified DC output).
9.2.2Diagram B.
9.2.3Increase the rate of rotation of the coil.
A DC motor diagram shows a coil ABCD between magnets S (left) and N (right), connected via a commutator (X, Y) to an external circuit with a battery and rheostat.
8.1.1 Give two reasons why this is a DC motor. (2)
8.1.2 What energy conversion occurs in this device? (2)
8.1.3 In which direction will the armature rotate, viewed from the front (clockwise or anticlockwise)? (1)
A 5 cm length of inflexible wire is suspended between two magnets (N left, S right, field into the page) on a top-pan balance reading 2,5 N. When the switch is closed, the ammeter reads 4 A and the balance reads 2,516 N.
8.2.1 In which direction does the conducting wire experience a force? (2)
8.2.2 State the law of physics illustrated by the increase in reading, and use it to explain the increase. (3)
8.2.3 Calculate the magnitude of the magnetic force exerted on the wire. (2)
8.2.4 Write down an appropriate equation that could determine the magnetic field density. (1)
An AC generator's output is 100 V at 20 Hz, with current direction a→b in the coil when closed.
8.3.1 Use Φ = BAcosθ to explain whether the magnetic flux through the coil (in the horizontal position shown) is maximum or zero. (2)
8.3.2 In which direction is the coil rotating, viewed from the slip rings (clockwise/anticlockwise)? (1)
8.3.3 Define Faraday's Law. (2)
8.3.4 Sketch the output emf vs time for two full cycles, showing the 100 V maximum and the time for two cycles. (4)
The frequency changes to 10 Hz, and the output passes through a bridge rectifier (diodes A, B, C, D).
(a) Name components A, B, C, D. (1)
(b) What is the purpose of a bridge rectifier in an AC circuit? (2)
(c) On the same axes as 8.3.4, sketch the combined effect of these two changes on the output emf vs time graph. (3)
8.1.1A DC motor has a power supply (battery) providing direct current, and a split-ring commutator.
8.1.2Electrical energy is converted to mechanical energy.
8.1.3Clockwise.
8.2.1Upwards (away from the balance, increasing the reading).
8.2.2Newton's Third Law: when the conductor experiences an upward force from the magnetic field, it exerts an equal and opposite (downward) force on the magnets/apparatus, which is registered as an increase on the balance.
8.2.3F = 2,516 − 2,5
F = 0,016 N
8.2.4F = BIℓsinθ
8.3.1Φ = BAcosθ. In the horizontal position shown, θ = 90° (the plane of the coil is parallel to the field), so cos90° = 0, meaning the flux is zero.
8.3.2Anticlockwise.
8.3.3The emf induced is directly proportional to the rate of change of magnetic flux.
8.3.4A sine wave with amplitude 100 V, completing two full cycles in 0,1 s (since T = 1/f = 1/20 = 0,05 s per cycle).
(a)Diodes.
(b)Full-wave rectification — converts AC to DC.
(c)One full wave shown over 0,1 s (since the period doubles to 0,1 s at 10 Hz), with only the positive (above-axis) portion shown after rectification, and the emf amplitude halved.
A bar magnet (N pole leading) is moved towards a coil connected to a sensitive ammeter via terminals E and F, then brought to a halt before entering.
9.1.1 State Lenz's law. (2)
9.1.2 Describe how the ammeter reading changes during this process. (2)
9.1.3 The induced current flows from E to F through the ammeter. Explain how this demonstrates Lenz's law. (2)
A U-shaped magnet has its N pole vertically above its S pole. A small coil moves at constant velocity along a line A→B→C→D, midway between the poles (the field exists only in the B-C region between the poles).
9.2.1 State Faraday's law of electromagnetic induction. (2)
9.2.2 Between which two consecutive positions (A,B,C,D) is there no induced current? (1)
9.2.3 Explain why there is no induced current during this section. (3)
The coil is now placed between B and C and rotated about the B-C axis. A flux-vs-time graph shows the flux through the coil, starting at a positive value, decreasing through zero at point P, reaching a minimum, then oscillating sinusoidally.
9.2.4(a) Which of four coil-orientation images (A, B, C, D — ranging from coil-plane-parallel to coil-plane-perpendicular to the field) matches the flux at point P (flux = 0)? (1)
(b) Draw a second line on the graph showing the induced emf during this rotation, labelled '(b)'. (2)
(c) Draw a third line showing the flux if the coil rotates twice as fast (same starting orientation), labelled '(c)'. (2)
A DC electric motor diagram shows a coil (axis, with sides labelled A, B, C, D) between permanent magnets (N left, S right), connected via fixed carbon brushes at Z, with current I entering.
9.3.1 What is indicated by letter Z? (1)
9.3.2 In which direction will the coil rotate (clockwise/anticlockwise)? (1)
9.3.3 Explain why wire AB experiences the force that creates this rotation (a simple diagram may help). (3)
9.1.1The induced current flows in a direction so as to set up a magnetic field that opposes the change in magnetic flux producing it.
9.1.2The ammeter reading increases (towards a maximum) as the magnet approaches and then drops back to zero once the magnet comes to a halt.
9.1.3The induced current (E→F) creates a magnetic field in the coil that opposes the approaching magnet's field (creates a pole that repels the magnet's approaching pole), or equivalently tries to maintain the original (smaller) field of the coil — opposing the increasing flux as the magnet approaches, consistent with Lenz's Law.
9.2.1The emf induced is directly proportional to the rate of change of magnetic flux (flux linkage).
9.2.2Between B and C.
9.2.3Between B and C, the magnetic field through the area of the coil (and hence the flux Φ) is constant, even though the coil is moving — so there is no rate of change of flux (ΔΦ/Δt = 0), and therefore no induced emf or current.
9.2.4(a)Image C (coil plane perpendicular to the field, giving zero flux through the coil).
9.2.4(b)The emf graph starts at zero (when flux is at its initial positive value, rate of change is momentarily zero... more precisely emf leads flux by 90°): the emf curve is shifted relative to the flux curve, starting near zero and initially positive.
9.2.4(c)A flux curve with the same amplitude and starting value as the original, but completing two full oscillations in the space of the original one (double frequency, half period).
9.3.1The split-ring commutator.
9.3.2Anticlockwise.
9.3.3The current in wire AB creates its own magnetic field around the wire; this field interacts with (adds to/subtracts from) the permanent magnets' field — strengthened on one side of the wire and weakened on the other — producing a net force on the wire (F = IℓBsinθ) that drives the rotation.
A straight conductor is connected to a battery and suspended between the poles of a U-shaped magnet (S above, N below), perpendicular to the magnet, free to move. Side X is closer to the bend of the magnet, side Y is further away.
8.1 Will the conductor move towards X or Y when the switch closes? (2)
8.2 Briefly explain why the conductor moves. (2)
8.3 Suggest two changes that would each reverse the direction of motion. (2)
8.4 Identify two changes that would increase the force on the conductor. (2)
8.5 If the conductor is rotated 90° so it is suspended parallel to the magnet (between the poles), would it still experience a magnetic force when the switch closes? (1)
A transformer is connected to the AC generator's output.
8.6 Explain how the transformer changes the voltage output, referring to the basic operation of an ideal transformer. (3)
8.7 The generator produces 16 V; the primary coil has 200 turns. How many turns must the secondary coil have to produce 240 V? (3)
8.8 Explain why high voltages are used for transmitting electrical energy through the national grid, using an appropriate equation. (3)
8.1X.
8.2The magnetic field around the current-carrying conductor interacts with the magnetic field of the permanent magnet, producing a resultant force on the conductor.
8.3Reverse the direction of the magnetic field (flip the magnet); or reverse the direction of the current (reverse the battery connections).
8.4Increase the strength of the magnetic field (use a stronger magnet); increase the current in the conductor (any two, also: increase the length of conductor in the field).
8.5Yes — even rotated 90°, the current-carrying conductor still has a magnetic field that interacts with the permanent magnets' field, producing a force (provided the conductor is not exactly parallel to the field lines in a way that gives zero force component).
8.6The changing (AC) current in the primary coil creates a changing magnetic flux, which links to the secondary coil and induces an emf proportional to the number of turns on the secondary; in an ideal transformer, no power is lost (Pprimary=Psecondary).
8.7Ns/Np = Vs/Vp → Ns/200 = 240/16
Ns = 3000 turns
8.8P = I²R: at high voltage, the current is low for a given power, so the rate of energy loss as heat (proportional to I²) is reduced during transmission.
An electromagnet (a coil wound on a cylinder) has its N and S poles indicated, with the coil's ends labelled P and Q.
8.1.1 Does the current flow around the coil from P to Q, or from Q to P? (2)
8.1.2 What is the effect on the electromagnet of increasing the current in the coil? (2)
A loop of wire is placed between permanent magnets (side-view and top-view shown), with current direction indicated. The loop experiences a force.
8.2.1 Using the side-view, will the loop rotate clockwise or anticlockwise? (2)
8.2.2 Using the side-view, will the loop's equilibrium position be vertical or horizontal? (2)
A device (coil between N and S magnets, with brushes at P) can act as either a DC motor or a DC generator.
8.3.1 What is indicated by letter P? (2)
8.3.2 Explain how this device would work as a DC motor, referring to the input required and how it functions. (3)
8.3.3 Explain how this device would work as a DC generator, referring to the input required and how it functions. (3)
8.3.4 Suggest one change that would improve its operation as both a motor and a generator. (2)
8.1.1From Q to P.
8.1.2The magnetic field becomes stronger (a stronger electromagnet).
8.2.1Clockwise.
8.2.2Vertical.
8.3.1The split-ring commutator.
8.3.2As a DC motor: the input is a DC current; the current in the coil creates a magnetic field which interacts with the permanent magnets' field, producing forces that rotate the coil; the commutator reverses the current every half-turn so the coil keeps rotating continuously in the same direction (any three points).
8.3.3As a DC generator: the input is a (mechanical) force that rotates the coil; the rotating coil experiences a changing magnetic flux, inducing an emf (by Faraday's Law); the commutator's connections change as the coil rotates, producing a (pulsating) direct current output (any three points).
8.3.4Use more turns on the coil, or use stronger magnets (either improves both motor and generator operation).
A uniform magnetic field points upward along the plane of the page. A wire perpendicular to the page carries a current such that the net magnetic field at point X (a certain distance from the wire) is zero.
8.1.1 State the direction of the current in the wire (into or out of the page). (2)
Point Y is twice as far from the wire as point X. The field due to a current-carrying wire is inversely proportional to distance from the wire.
8.1.2 Describe the net magnetic field at Y, specifying direction and magnitude relative to the uniform field. (4)
8.2 Name two practical examples of electromagnetic induction usage. (2)
A solenoid carries a current of 2 A in the direction shown.
8.3.1 Sketch the magnetic field associated with the solenoid, indicating which end is the north pole. (3)
A small coil connected to a sensitive voltmeter is placed near the stationary solenoid (also carrying current).
8.3.2 State Faraday's law of electromagnetic induction. (2)
8.3.3 Explain why the voltmeter reads zero. (2)
8.3.4 State three different ways an emf could be induced in the small coil. (6)
8.1.1Into the page (so that the wire's circular field opposes the uniform field at X, giving zero net field there).
8.1.2At Y (twice as far from the wire), the wire's field is half as strong as it is at X. Since at X the wire's field exactly cancelled the uniform field, at Y the wire's field only cancels half of the uniform field, leaving a net field in the same direction as the uniform B field, with half its original magnitude.
8.2Examples: generators, transformers, induction stoves, microphones (any two).
8.3.1Concentric field lines forming closed loops through and around the solenoid, with the field lines emerging from the north pole end (determined by the right-hand rule applied to the current direction shown).
8.3.2The emf induced is directly proportional to the rate of change of magnetic flux (flux linkage).
8.3.3There is no relative movement between the coil and solenoid, and the solenoid's current (and hence its field/flux through the small coil) is constant, so the rate of change of flux is zero.
8.3.4Move the coil relative to the solenoid (or vice versa); rotate the coil so its effective area relative to the field changes; continuously change (vary) the current in the solenoid.
More photons strike the surface per second → more photoelectrons are emitted per second → increases the photocurrent, but the maximum kinetic energy of each photoelectron stays the same.
Each photon carries more energy (E = hf) → the maximum kinetic energy of emitted photoelectrons increases, but if intensity is unchanged the number of photoelectrons emitted per second stays the same.
No photoelectrons are emitted at all, regardless of how intense the light is, because each individual photon does not carry enough energy to overcome the work function.
The work function of a metal is 3,2 × 10⁻¹⁹ J. Light of frequency 8,0 × 10⁹⁴ Hz shines on the metal. Calculate the maximum kinetic energy of the emitted photoelectrons (h = 6,63 × 10⁻³⁴ J·s).
Photon energyE = hf = (6,63×10⁻³⁴)(8,0×10⁹⁴)
Photon energy valueE ≈ 5,30 × 10⁻¹⁹ J
Einstein's photoelectric equationE = W₀ + Ek(max) → Ek(max) = E − W₀
SubstitutionEk(max) = 5,30×10⁻¹⁹ − 3,2×10⁻¹⁹
Ek(max) ≈ 2,10 × 10⁻¹⁹ J
Monochromatic light of frequency greater than the threshold frequency shines on a metal surface. If the intensity of the light is increased while the frequency stays the same, what happens to the maximum kinetic energy of the emitted photoelectrons?
The work function of caesium is 3,0 × 10⁻¹⁹ J. Calculate the threshold frequency of caesium (h = 6,63 × 10⁻³⁴ J·s).
FormulaW₀ = hf₀ → f₀ = W₀ / h
Substitutionf₀ = (3,0×10⁻¹⁹) / (6,63×10⁻³⁴)
f₀ ≈ 4,52 × 10⁹⁴ Hz
Calculate the energy of a photon of light with a wavelength of 500 nm (c = 3,0 × 10⁸ m·s⁻¹, h = 6,63 × 10⁻³⁴ J·s).
FormulaE = hc / λ
SubstitutionE = (6,63×10⁻³⁴)(3,0×10⁸) / (500×10⁻⁹)
E ≈ 3,98 × 10⁻¹⁹ J
An electron in an atom drops from a higher energy level to a lower energy level, emitting a photon. Which statement correctly describes this transition?
A metal has a threshold frequency of 5,0 × 10⁹⁴ Hz. Calculate the maximum wavelength of light that will cause photoemission from this metal (c = 3,0 × 10⁸ m·s⁻¹).
Formulav = fλ → λ = c/f₀
Substitutionλ = (3,0×10⁸) / (5,0×10⁹⁴)
λ ≈ 6,0 × 10⁻⁷ m (600 nm)