Secondary 4 Combined Science Physics Practice Paper 3

TuitionGoWhere Practice Paper - Combined Science Physics Secondary 4

Answer Key and Marking Scheme (Version 3)

Total Marks: 65

Section A: Structured Questions (45 marks)

Question 1: Kinematics and Dynamics [8 marks]

(a) Describe the motion of the car between t = 10 s and t = 30 s. [1]

Answer: The car moves with constant velocity / uniform speed of 8.0 m/s. [1]

Marking note: Accept "constant speed" or "zero acceleration". Must reference constant/uniform motion.

(b) Calculate the acceleration of the car during the first 10 seconds. [2]

Answer: a = Δv / Δt a = (8.0 - 0) / 10 [1] a = 0.80 m/s² [1]

Marking note: Award [1] for correct substitution, [1] for correct answer with units. Accept 0.8 m/s².

(c) Calculate the total distance travelled by the car in the 50 seconds. [3]

Answer: Distance = area under velocity-time graph Area = ½ × 10 × 8.0 + 20 × 8.0 + ½ × 20 × 8.0 [1] = 40 + 160 + 80 [1] = 280 m [1]

Marking note: Award [1] for correct method (sum of areas), [1] for correct calculation of areas, [1] for correct answer with units. Accept alternative methods (e.g., average velocity × time for each segment).

(d) The car has a mass of 2.5 kg. Calculate the resultant force acting on the car during the first 10 seconds. [2]

Answer: F = ma [1] F = 2.5 × 0.80 = 2.0 N [1]

Marking note: Award [1] for correct formula/substitution, [1] for correct answer with units. ECF from (b) accepted.

Question 2: Forces and Energy on an Incline [8 marks]

(a) On the diagram, draw and label the forces acting on the block just before it begins to slide. [2]

Answer:

• Weight (W or mg) acting vertically downward from centre of block [1]
• Normal reaction (N or R) perpendicular to the plank surface
• Friction (F or f) acting up the plane, parallel to the surface [1]

Marking note: Award [1] for weight correctly drawn and labelled, [1] for both normal reaction and friction correctly drawn and labelled. All three forces must be present for full marks.

(b) The block slides down the plank with constant speed after being given a gentle push. Explain, in terms of forces, why the block moves with constant speed. [2]

Answer: Constant speed means zero acceleration. [1] By Newton's First Law, the net force on the block is zero. The component of weight down the slope is balanced by the frictional force acting up the slope. [1]

Marking note: Must reference zero net force/balanced forces and link to constant speed. Accept "resultant force = 0".

(c) The vertical height of the plank's raised end is 0.45 m above the horizontal. Calculate the gravitational potential energy lost by the block as it slides from the top to the bottom of the plank. [2]

Answer: GPE lost = mgh [1] = 0.80 × 10 × 0.45 = 3.6 J [1]

Marking note: Award [1] for correct formula/substitution, [1] for correct answer with units.

(d) State the amount of work done against friction as the block slides down the plank. Explain your answer. [2]

Answer: Work done against friction = 3.6 J [1] Explanation: Since the block moves at constant speed, the loss in GPE is entirely converted to work done against friction (thermal energy). By conservation of energy, GPE lost = work done against friction. [1]

Marking note: Award [1] for correct value (ECF from (c)), [1] for explanation linking energy conservation to constant speed.

Question 3: Thermal Radiation and Cooling [7 marks]

(a) State which beaker cooled faster. Use data from the table to support your answer. [1]

Answer: Beaker B (matt black) cooled faster. After 10 minutes, Beaker B dropped by 36°C while Beaker A dropped by only 24°C. [1]

Marking note: Must state Beaker B and provide comparative data. Accept any valid comparison from the table.

(b) Explain why this beaker cooled faster, using your knowledge of thermal radiation. [2]

Answer: Matt black surfaces are better emitters of thermal radiation than shiny silver surfaces. [1] Therefore, Beaker B radiates thermal energy to the surroundings at a faster rate, causing it to cool more quickly. [1]

Marking note: Award [1] for identifying matt black as a better emitter, [1] for linking to faster rate of energy loss/cooling.

(c) Calculate the average rate of temperature decrease for Beaker B between 2 minutes and 8 minutes. Give your answer in °C/min. [2]

Answer: Temperature change = 70 - 49 = 21°C [1] Time interval = 8 - 2 = 6 min Rate = 21 / 6 = 3.5 °C/min [1]

Marking note: Award [1] for correct temperature change and time interval, [1] for correct calculation and units.

(d) The student repeats the experiment but places a lid on each beaker. Suggest and explain how this would affect the rate of cooling for both beakers. [2]

Answer: The rate of cooling would decrease for both beakers. [1] A lid reduces heat loss by convection (and evaporation), which is a significant mechanism of heat loss from hot water. This reduces the overall rate of energy transfer to the surroundings. [1]

Marking note: Award [1] for stating rate decreases, [1] for explanation referencing reduced convection/evaporation. Accept reference to trapping hot air/vapour.

Question 4: Refraction and Total Internal Reflection [6 marks]

(a) State what is meant by the term "refraction". [1]

Answer: Refraction is the bending/changing direction of light as it passes from one medium to another due to a change in speed. [1]

Marking note: Must reference change in direction and change in medium/speed.

(b) Calculate the angle of refraction in the glass. [2]

Answer: n = sin i / sin r 1.5 = sin 45° / sin r [1] sin r = sin 45° / 1.5 = 0.7071 / 1.5 = 0.4714 r = sin⁻¹(0.4714) = 28.1° (or 28°) [1]

Marking note: Award [1] for correct formula/substitution, [1] for correct answer (accept 28° to 28.1°).

(c) The ray of light then travels from the glass back into air. The angle of incidence in the glass is 42°. Explain what happens to the ray at the glass-air boundary. [2]

Answer: The critical angle for glass (n=1.5) is approximately 41.8°. Since the angle of incidence (42°) is greater than the critical angle, total internal reflection occurs. [1] The ray is completely reflected back into the glass; no light is refracted into the air. [1]

Marking note: Award [1] for identifying total internal reflection and comparing to critical angle, [1] for describing the outcome (complete reflection).

(d) State one condition necessary for the phenomenon you described in (c) to occur. [1]

Answer: Light must travel from a denser medium to a less dense medium (e.g., from glass to air). [1]

Marking note: Accept "angle of incidence must be greater than the critical angle" or "light travels from optically denser to optically less dense medium".

Question 5: Electrical Power and Efficiency [8 marks]

(a) Calculate the current flowing through the kettle when it is operating at its rated power. [2]

Answer: P = VI 2200 = 240 × I [1] I = 2200 / 240 = 9.17 A (or 9.2 A) [1]

Marking note: Award [1] for correct formula/substitution, [1] for correct answer with units.

(b) The kettle is used to heat 1.5 kg of water from 25°C to 100°C. Calculate the energy required to heat the water. [2]

Answer: Q = mcΔθ [1] Q = 1.5 × 4200 × (100 - 25) = 1.5 × 4200 × 75 = 472 500 J (or 473 kJ) [1]

Marking note: Award [1] for correct formula/substitution, [1] for correct answer with units.

(c) The kettle takes 4 minutes and 30 seconds to heat the water. Calculate the efficiency of the kettle. [3]

Answer: Time = 4 × 60 + 30 = 270 s Energy supplied by kettle = P × t = 2200 × 270 = 594 000 J [1] Efficiency = (Useful energy output / Total energy input) × 100% [1] = (472 500 / 594 000) × 100% = 79.5% (or 80%) [1]

Marking note: Award [1] for calculating energy input, [1] for correct efficiency formula, [1] for correct answer. ECF from (b) accepted.

(d) The kettle is fitted with a 13 A fuse in its plug. Explain why a 13 A fuse is suitable for this kettle. [1]

Answer: The operating current (9.17 A) is less than the fuse rating (13 A), so the fuse will not blow during normal operation. However, if a fault causes a current above 13 A, the fuse will blow and protect the appliance/user. [1]

Marking note: Must reference that normal current is below fuse rating and fuse provides protection above 13 A.

Question 6: Principle of Moments [8 marks]

(a) Calculate the weight of the 200 g mass. [1]

Answer: W = mg = 0.200 × 10 = 2.0 N [1]

Marking note: Must convert g to kg. Accept 2 N.

(b) Calculate the moment of this weight about the pivot. [2]

Answer: Distance from pivot = 50 - 20 = 30 cm = 0.30 m [1] Moment = F × d = 2.0 × 0.30 = 0.60 N m [1]

Marking note: Award [1] for correct distance, [1] for correct moment with units. Accept 60 N cm.

(c) The student hangs a second mass of 150 g on the other side of the pivot to balance the rule. Calculate the distance from the pivot where this mass should be hung. [3]

Answer: Weight of 150 g mass = 0.150 × 10 = 1.5 N [1] For equilibrium: Anticlockwise moment = Clockwise moment 2.0 × 0.30 = 1.5 × d [1] d = 0.60 / 1.5 = 0.40 m = 40 cm from pivot Position on rule = 50 + 40 = 90 cm mark [1]

Marking note: Award [1] for weight of second mass, [1] for correct equation, [1] for correct distance/position. Accept distance from pivot (40 cm) or position on rule (90 cm).

(d) The student then moves the pivot to the 40 cm mark. Explain why the rule is no longer balanced, even without any masses hanging from it. [2]

Answer: The weight of the metre rule acts at its centre of gravity (50 cm mark). [1] When the pivot is at 40 cm, the weight acts 10 cm from the pivot, producing a clockwise moment. There is no anticlockwise moment to balance this, so the rule rotates clockwise. [1]

Marking note: Award [1] for identifying weight acts at centre (50 cm), [1] for explaining the unbalanced moment.

Section B: Data-Based and Application Questions (20 marks)

Question 7: Electric Vehicles and Regenerative Braking [8 marks]

(a) State the principle of conservation of energy. [1]

Answer: Energy cannot be created or destroyed; it can only be transferred/converted from one form to another. The total energy of an isolated system remains constant. [1]

Marking note: Accept either "cannot be created or destroyed" or "total energy remains constant".

(b) Calculate the amount of energy recovered and stored in the battery when the EV decelerates from 25 m/s to rest using regenerative braking. [2]

Answer: Energy recovered = 60% of 468 750 J [1] = 0.60 × 468 750 = 281 250 J (or 281 kJ) [1]

Marking note: Award [1] for correct method, [1] for correct answer with units.

(c) Explain why friction brakes are still necessary in an EV, even with regenerative braking. [2]

Answer: Regenerative braking alone cannot bring the vehicle to a complete stop quickly enough for emergency situations. [1] Friction brakes provide additional braking force for rapid deceleration and can hold the vehicle stationary when parked. [1]

Marking note: Award [1] for identifying limitation of regenerative braking, [1] for explaining role of friction brakes.

(d) The EV accelerates from rest to 25 m/s in 8.0 seconds. Calculate the average power output of the electric motor during this acceleration, assuming no energy losses. [3]

Answer: KE gained = 468 750 J (given) [1] Power = Energy transferred / time = 468 750 / 8.0 [1] = 58 594 W ≈ 58.6 kW (or 59 kW) [1]

Marking note: Award [1] for identifying KE as energy transferred, [1] for correct formula/substitution, [1] for correct answer with units. Accept alternative method using F = ma and P = Fv.

Question 8: Pressure and Depth Investigation [8 marks]

(a) Plot a graph of pressure (y-axis) against depth (x-axis) on the grid. Draw the best-fit straight line. [3]

Answer:

• All six points plotted correctly (±½ small square) [1]
• Axes labelled with quantities and units [1]
• Best-fit straight line passing through origin and close to all points [1]

Marking note: Points: (0,0), (5.0, 0.50), (10.0, 1.00), (15.0, 1.50), (20.0, 2.00), (25.0, 2.50). Line must be straight and pass through origin.

(b) Use your graph to determine the pressure at a depth of 18.0 cm. [1]

Answer: Pressure ≈ 1.80 kPa (accept 1.78–1.82 kPa) [1]

Marking note: Must show reading from graph (interpolation). ECF from graph line.

(c) Calculate the gradient of your graph. Use the gradient to determine the density of the water. [3]

Answer: Gradient = ΔP / Δh = (2.50 - 0) / (25.0 - 0) = 0.10 kPa/cm = 10 kPa/m = 10 000 Pa/m [1] p = ρgh → gradient = ρg [1] ρ = gradient / g = 10 000 / 10 = 1000 kg/m³ [1]

Marking note: Award [1] for gradient calculation, [1] for linking gradient to ρg, [1] for correct density with units. Accept values close to 1000 kg/m³.

(d) The student repeats the experiment using a liquid of higher density. On the same axes, sketch the graph she would expect to obtain. Label this line "Liquid X". [1]

Answer: A straight line through the origin with a steeper gradient than the water line, labelled "Liquid X". [1]

Marking note: Line must be straight, through origin, and visibly steeper than the original line.

Question 9: Electromagnetic Induction [6 marks]

(a) The student pushes the north pole of the magnet into the coil. State what is observed on the galvanometer. [1]

Answer: The galvanometer needle deflects (to one side), indicating a current is induced. [1]

Marking note: Must mention deflection/movement of needle. Accept "shows a reading" or "deflects momentarily".

(b) Explain why the galvanometer shows this reading. [2]

Answer: As the magnet moves into the coil, the magnetic field lines passing through the coil change. [1] By electromagnetic induction, this changing magnetic field induces an e.m.f. (voltage) across the coil, which drives a current through the circuit, causing the galvanometer to deflect. [1]

Marking note: Award [1] for changing magnetic field/flux, [1] for linking to induced e.m.f./current.

(c) The student then pulls the magnet out of the coil at a faster speed. State and explain two differences in the galvanometer reading compared to when the magnet was pushed in slowly. [2]

Answer:

1. The galvanometer deflects in the opposite direction because the magnet is moving out (magnetic field decreasing) rather than in (magnetic field increasing). [1]
2. The deflection is larger because the faster speed causes a greater rate of change of magnetic field, inducing a larger e.m.f. and current. [1]

Marking note: Award [1] for opposite direction with explanation, [1] for larger deflection with explanation.

(d) State one way the student could increase the induced current without changing the magnet or the speed of movement. [1]

Answer: Increase the number of turns in the coil. [1]

Marking note: Accept "use a coil with more turns", "increase the number of coils", or "use a stronger magnet" (but note the constraint says "without changing the magnet"—so increasing turns is the expected answer). Also accept "use a soft iron core inside the coil".

END OF ANSWER KEY