IGCSE Physics Paper 4 Mock 2 - Topic 1: Motion, Forces & Energy

(a) 2 marks

Define acceleration. Write the equation for acceleration, stating the meaning of each symbol.

Model Answer - Q1(a)

Acceleration is the rate of change of velocity [1]

a = (v - u) / t, where a = acceleration (m/s²), v = final velocity (m/s), u = initial velocity (m/s), t = time (s) [1]

Mark 1: Definition - rate of change of velocity

Mark 2: Correct equation with symbols defined

(b) 4 marks

Explain the difference between a scalar quantity and a vector quantity. Give two examples of each.

Model Answer - Q1(b)

A scalar quantity has magnitude only [1]

A vector quantity has both magnitude and direction [1]

Two scalar examples: e.g. speed, mass, distance, time, energy, temperature [1]

Two vector examples: e.g. velocity, force, acceleration, displacement, momentum, weight [1]

Mark 1: Scalar - magnitude only

Mark 2: Vector - magnitude AND direction

Mark 3: Two correct scalar examples

Mark 4: Two correct vector examples

A cyclist starts from rest and accelerates uniformly to 12 m/s in 8 seconds. She then travels at a constant speed of 12 m/s for 20 seconds, before decelerating uniformly to rest in 4 seconds.

Model Answer - Q1(c)(i)

a = (v - u) / t = (12 - 0) / 8 [1]

a = 1.5 m/s² [1]

Mark 1: Correct substitution into equation

Mark 2: Correct answer 1.5 m/s²

Model Answer - Q1(c)(ii)

Straight line from (0, 0) to (8, 12), horizontal line from (8, 12) to (28, 12) [1]

Straight line from (28, 12) to (32, 0) with correct labels on axes [1]

Mark 1: Correct shape - rising line, flat, falling line with correct values

Mark 2: Correct time values at each phase change and axes labelled

Model Answer - Q1(c)(iii)

Area under graph = (1/2 x 8 x 12) + (20 x 12) + (1/2 x 4 x 12) = 48 + 240 + 24 [1]

Total distance = 312 m [1]

Mark 1: Correct method (area under speed-time graph / three sections)

Mark 2: Correct answer 312 m

(a) 2 marks

Define density. Write the equation for density, including the SI units.

Model Answer - Q2(a)

Density is the mass per unit volume [1]

rho = m / V, where rho is density (kg/m³), m is mass (kg), V is volume (m³) [1]

Mark 1: Correct definition - mass per unit volume

Mark 2: Correct equation with SI units (kg/m³)

(b) 4 marks

A solid metal cylinder has a radius of 3.0 cm, a height of 8.0 cm, and a mass of 540 g.

(b)(i) Calculate the volume of the cylinder in cm³. 2 marks

[Volume of a cylinder = pi x r² x h]

Model Answer - Q2(b)(i)

V = pi x r² x h = pi x (3.0)² x 8.0 [1]

V = 226 cm³ (or 72pi = 226.2 cm³) [1]

Mark 1: Correct substitution into V = pi r² h

Mark 2: Correct answer approximately 226 cm³

(b)(ii) Calculate the density of the metal in g/cm³. 2 marks

Model Answer - Q2(b)(ii)

rho = m / V = 540 / 226 [1]

rho = 2.39 g/cm³ (accept 2.4 g/cm³) [1]

Mark 1: Correct substitution into rho = m/V

Mark 2: Correct answer 2.4 g/cm³ (accept 2.39)

Oil with a density of 800 kg/m³ is poured into a beaker of water (density 1000 kg/m³). Describe what happens and explain why.

Model Answer - Q2(c)

The oil floats on top of the water / forms a layer above the water [1]

Because the oil is less dense than the water (800 < 1000 kg/m³) / oil is lighter than the same volume of water [1]

Mark 1: Oil floats on top of water

Mark 2: Oil is less dense than water

(d) 2 marks

Steel has a density of 7800 kg/m³, which is much greater than the density of water (1000 kg/m³). Explain why a ship made of steel is able to float.

Model Answer - Q2(d)

The ship is hollow / has a large volume of air inside / the hull encloses a large volume [1]

This makes the average/overall density of the ship (steel + air) less than the density of water, so it floats / the weight of water displaced equals the weight of the ship [1]

Mark 1: Ship is hollow / contains air

Mark 2: Average density of ship is less than water / displaces enough water

(a) 2 marks

State Newton's Second Law of Motion. Write the equation relating force, mass and acceleration.

Model Answer - Q3(a)

The resultant force acting on an object is equal to the product of its mass and acceleration / the acceleration of an object is proportional to the resultant force and inversely proportional to its mass [1]

F = ma, where F = resultant force (N), m = mass (kg), a = acceleration (m/s²) [1]

Mark 1: Correct statement of Newton's Second Law (must mention resultant force)

Mark 2: F = ma with correct units/symbols

(b) 4 marks

A rocket of mass 5000 kg has an engine that produces an upward thrust of 65 000 N. The weight of the rocket is 49 000 N.

(b) (i) 2 marks

Model Answer - Q3(b)(i)

Resultant force = Thrust - Weight = 65 000 - 49 000 [1]

Resultant force = 16 000 N upwards [1]

Mark 1: Correct subtraction (thrust - weight)

Mark 2: Correct answer 16 000 N (upwards)

(b) (ii) 2 marks

Model Answer - Q3(b)(ii)

a = F / m = 16 000 / 5000 [1]

a = 3.2 m/s² [1]

Mark 1: Correct use of F = ma rearranged

Mark 2: Correct answer 3.2 m/s²

A ball is dropped from the top of a tall building. Describe and explain the motion of the ball as it falls, in terms of the forces acting, its velocity, and its acceleration. Include the concept of terminal velocity in your answer.

Model Answer - Q3(c)

Initially, the only force is weight (gravity) acting downwards, so the ball accelerates at approximately 9.8 m/s² / freely [1]

As speed increases, air resistance (drag) increases. The resultant downward force decreases, so acceleration decreases (but the ball is still speeding up) [1]

Eventually air resistance equals the weight. The resultant force is zero and the ball travels at constant velocity [1]

This constant maximum velocity is called the terminal velocity. Acceleration is now zero. [1]

Mark 1: Initially only weight acts / accelerates freely downwards

Mark 2: Air resistance increases with speed / resultant force decreases / acceleration decreases

Mark 3: Air resistance equals weight / resultant force = 0 / constant velocity

Mark 4: This is terminal velocity / acceleration = 0

(d) 2 marks

A car travels around a circular bend on a flat road. Explain what provides the centripetal force and what would happen if the car's speed increases significantly.

Model Answer - Q3(d)

The centripetal force is provided by the friction between the tyres and the road [1]

If speed increases too much, the friction is insufficient to provide the centripetal force needed and the car will skid outwards / leave the circular path / move in a straight line tangentially [1]

Mark 1: Friction between tyres and road provides centripetal force

Mark 2: Car skids / friction not sufficient if speed too great

(a) 2 marks

Three identical springs each have a spring constant of 20 N/m. A load of 6 N is hung from one of the springs. Calculate the extension of the spring.

Model Answer - Q4(a)

F = k x e, so e = F / k = 6 / 20 [1]

e = 0.3 m (or 30 cm) [1]

Mark 1: Correct rearrangement of F = ke

Mark 2: Correct answer 0.3 m or 30 cm

(b) 3 marks

Using the spring from part (a), the following data is recorded: at 0 N the extension is 0 m; at 3 N the extension is 0.15 m; at 6 N the extension is 0.30 m. The spring reaches its limit of proportionality at a load of 6 N.

(i) Plot the first three data points on the axes below.
(ii) Sketch how the graph would continue up to a load of 10 N.

Model Answer - Q4(b)

Three points plotted correctly: (0, 0), (0.15, 3), (0.30, 6) [1]

Straight line drawn through all three points (showing proportionality up to 6 N) [1]

Beyond 6 N, the line curves away from the straight line (extension increases more for each unit of force / non-linear) showing the spring has exceeded its limit of proportionality [1]

Mark 1: Three points correctly plotted

Mark 2: Straight line through points up to 6 N

Mark 3: Curve beyond 6 N showing non-proportional behaviour

A door handle is positioned 0.8 m from the hinge. A person pushes the handle with a force of 15 N perpendicular to the door. Calculate the moment of the force about the hinge.

Model Answer - Q4(c)

Moment = Force x perpendicular distance from pivot = 15 x 0.8 [1]

Moment = 12 N m [1]

Mark 1: Correct formula and substitution

Mark 2: Correct answer 12 N m

(d) 3 marks

A uniform plank of length 5.0 m and weight 200 N is supported at both ends (points A and B). A painter weighing 700 N stands on the plank at a point 2.0 m from end A. Calculate the support force at each end.

Model Answer - Q4(d)

Take moments about A: (R_B x 5.0) = (700 x 2.0) + (200 x 2.5) [1]

R_B x 5.0 = 1400 + 500 = 1900, so R_B = 380 N [1]

R_A = total weight - R_B = (700 + 200) - 380 = R_A = 520 N [1]

Mark 1: Correct moments equation about one end

Mark 2: Correct calculation of R_B = 380 N

Mark 3: Correct calculation of R_A = 520 N

(a) 2 marks

State the principle of conservation of momentum.

Model Answer - Q5(a)

The total momentum of a system of objects remains constant / is conserved [1]

...provided no external resultant force acts on the system / in a closed system [1]

Mark 1: Total momentum is conserved / stays the same

Mark 2: Condition: no external resultant force / closed system

(b) 4 marks

A truck of mass 2000 kg is moving at 15 m/s. It collides with a car of mass 1000 kg moving at 5 m/s in the same direction. After the collision, the truck and car stick together and move as one object. Calculate their common velocity after the collision.

Model Answer - Q5(b)

Total momentum before = m1 x v1 + m2 x v2 [1]

= (2000 x 15) + (1000 x 5) = 30 000 + 5 000 = 35 000 kg m/s [1]

Total momentum after = (m1 + m2) x v = (2000 + 1000) x v = 3000v [1]

35 000 = 3000v, so v = 11.7 m/s (accept 11.67 m/s or 35/3 m/s) [1]

Mark 1: Correct formula for momentum before

Mark 2: Correct total momentum before = 35 000 kg m/s

Mark 3: Correct expression for momentum after

Mark 4: Correct answer v = 11.7 m/s

A tennis ball of mass 0.060 kg is travelling at 20 m/s towards a racket. The racket hits the ball and it leaves the racket at 30 m/s in the opposite direction. The contact time between the ball and the racket is 0.005 s.

Model Answer - Q5(c)(i)

Change in momentum = m(v - u) = 0.060 x (30 - (-20)) = 0.060 x 50 [1]
(Taking direction towards racket as negative, ball approaches at -20 m/s and leaves at +30 m/s)

Change in momentum = 3.0 kg m/s [1]

Mark 1: Correct method accounting for direction change (v - u = 50)

Mark 2: Correct answer 3.0 kg m/s

Model Answer - Q5(c)(ii)

F = change in momentum / time = 3.0 / 0.005 [1]

F = 600 N [1]

Mark 1: Correct formula F = impulse / time

Mark 2: Correct answer 600 N

(d) 2 marks

Explain why seat belts are designed to stretch slightly during a crash, rather than being completely rigid.

Model Answer - Q5(d)

The stretching increases the time over which the person decelerates / increases the time for the momentum to change [1]

Since F = change in momentum / time, a longer time means a smaller force acts on the person, reducing injury [1]

Mark 1: Increases the time taken to stop / decelerate

Mark 2: Reduces the force on the person (linked to F = impulse/t)

(a) 3 marks

A high jumper of mass 70 kg clears a bar set at a height of 2.1 m above the ground. Assuming that all of the jumper's kinetic energy at take-off is converted to gravitational potential energy at the top of the jump, calculate the minimum kinetic energy needed at take-off.

[Use g = 9.8 m/s²]

Model Answer - Q6(a)

KE at take-off = GPE at the top of the jump [1]

GPE = mgh = 70 x 9.8 x 2.1 [1]

KE = 1440.6 J (accept 1441 J or 1440 J) [1]

Mark 1: Correct energy conservation principle stated

Mark 2: Correct substitution into GPE = mgh

Mark 3: Correct answer (approximately 1441 J)

(b) 3 marks

Using your answer to part (a), calculate the minimum take-off speed of the high jumper.

Model Answer - Q6(b)

KE = 1/2 mv², so v² = 2 x KE / m [1]

v² = 2 x 1440.6 / 70 = 41.16 [1]

v = 6.4 m/s (accept 6.42 m/s) [1]

Mark 1: Correct rearrangement of KE = 1/2 mv²

Mark 2: Correct substitution

Mark 3: Correct answer v = 6.4 m/s

A crane lifts a load of mass 2000 kg vertically through a height of 15 m in 30 s.

Model Answer - Q6(c)(i)

W = F x d = mgh = 2000 x 9.8 x 15 [1]

W = 294 000 J (294 kJ) [1]

Mark 1: Correct formula and substitution

Mark 2: Correct answer 294 000 J or 294 kJ

Model Answer - Q6(c)(ii)

P = W / t = 294 000 / 30 [1]

P = 9800 W (9.8 kW) [1]

Mark 1: Correct formula P = W / t

Mark 2: Correct answer 9800 W or 9.8 kW

(d) 2 marks

The crane motor uses 450 kJ of electrical energy to lift the load. Calculate the efficiency of the crane.

Model Answer - Q6(d)

Efficiency = (useful energy output / total energy input) x 100 = (294 / 450) x 100 [1]

Efficiency = 65.3% (accept 65%) [1]

Mark 1: Correct formula with correct values substituted

Mark 2: Correct answer 65.3% (accept 65%)

(a) 4 marks

Using the relationship between pressure, force and area, explain why:
(i) a tractor has wide tyres
(ii) a nail has a very sharp point

Model Answer - Q7(a)

Tractor: Wide tyres have a large surface area in contact with the ground [1]

Since pressure = force / area, a large area means the pressure on the ground is small, so the tractor does not sink into soft ground [1]

Nail: The sharp point has a very small surface area [1]

Since pressure = force / area, a small area means the pressure is very large, so the nail can penetrate the material easily [1]

Mark 1: Tractor - wide tyres = large area

Mark 2: Tractor - large area = low pressure (prevents sinking)

Mark 3: Nail - sharp point = small area

Mark 4: Nail - small area = high pressure (penetrates easily)

(b) 3 marks

A submarine dives to a depth of 200 m below the surface of the sea. The density of seawater is 1030 kg/m³. Calculate the pressure due to the water at this depth.

[Use g = 9.8 m/s²]

Model Answer - Q7(b)

p = rho x g x h [1]

p = 1030 x 9.8 x 200 [1]

p = 2 018 800 Pa (approximately 2.02 x 10⁶ Pa or 2020 kPa) [1]

Mark 1: Correct formula p = rho g h

Mark 2: Correct substitution of values

Mark 3: Correct answer approximately 2 019 000 Pa (accept 2.02 MPa)

Describe how electricity is generated in a hydroelectric dam. Include the energy transformations that take place.

Model Answer - Q7(c)

Water is stored at height behind a dam. When released, the water flows downward and its gravitational potential energy is converted to kinetic energy [1]

The moving water turns a turbine (kinetic energy of water to kinetic energy of turbine) [1]

The turbine is connected to a generator which converts kinetic energy to electrical energy [1]

Mark 1: GPE of water converts to KE as it falls

Mark 2: Water turns a turbine

Mark 3: Turbine drives generator, producing electrical energy

(d) 2 marks

State one advantage and one disadvantage of each of the following energy resources:
(i) Wind energy
(ii) Nuclear energy

Model Answer - Q7(d)

(i) Wind energy:
Advantage: Renewable / no fuel costs / no greenhouse gas emissions during operation / free energy source [any 1]
Disadvantage: Unreliable / depends on wind speed / intermittent / visual pollution / noise / large land area needed [any 1] [1 mark total for one correct advantage AND one correct disadvantage]

(ii) Nuclear energy:
Advantage: Very large energy output from small amount of fuel / reliable / does not produce greenhouse gases during operation / high power output [any 1]
Disadvantage: Produces radioactive waste / risk of nuclear accident / high cost to build and decommission / non-renewable (uranium) [any 1] [1 mark total for one correct advantage AND one correct disadvantage]

Mark 1: Wind - one valid advantage AND one valid disadvantage

Mark 2: Nuclear - one valid advantage AND one valid disadvantage

IGCSE Physics Paper 4 (Theory/Extended) - Mock 2

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