Physics 0625 | Paper 3 Mock 01 | Theory (Core)
Question 1 Report
Fig. 21.1 shows two trolleys, A and B, on a frictionless track. Trolley A is given a push and moves toward stationary trolley B. A motion sensor records the speed of trolley A.
(a) State the initial speed of trolley B. [1]
(b) After the collision, trolley A stops and trolley B moves off at the same speed that A had before the collision. State the name of this type of collision. [1]
(c) Describe how the distance-time graph for trolley A would look before and after the collision. [2]
(d) The mass of each trolley is 0.50 kg and A was moving at 2.0 m/s. Calculate the kinetic energy of A before the collision. [2]
(e) State the kinetic energy of B after the collision. [1]
(a) Initial speed of trolley B [1]
Trolley B is stationary before the collision, so its initial speed is 0 m/s. [1]
(b) Type of collision [1]
When A stops completely and B moves off at A's original speed, both momentum and kinetic energy are conserved. This is an elastic collision. [1]
In an elastic collision between two objects of equal mass where one is initially stationary, the moving object stops and the stationary one moves off with the same speed. This is a classic result of simultaneous conservation of momentum and kinetic energy.
(c) Distance-time graph for trolley A [2]
Before the collision: Trolley A moves at constant speed, so its distance-time graph is a straight line with a positive (constant) gradient. [1]
After the collision: Trolley A is stationary, so its distance remains constant. The graph is a horizontal line. [1]
(d) Kinetic energy of A before the collision [2]
\( KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.50 \times 2.0^2 \) [1]
\( KE = \frac{1}{2} \times 0.50 \times 4.0 = 1.0 \text{ J} \) [1]
(e) Kinetic energy of B after the collision [1]
Since this is an elastic collision, kinetic energy is conserved. Trolley B moves at 2.0 m/s with the same mass as A, so its kinetic energy is 1.0 J. [1]
You can verify: \( KE_B = \frac{1}{2} \times 0.50 \times 2.0^2 = 1.0 \text{ J} \), confirming no kinetic energy was lost.
Answer Details
(a) Initial speed of trolley B [1]
Trolley B is stationary before the collision, so its initial speed is 0 m/s. [1]
(b) Type of collision [1]
When A stops completely and B moves off at A's original speed, both momentum and kinetic energy are conserved. This is an elastic collision. [1]
In an elastic collision between two objects of equal mass where one is initially stationary, the moving object stops and the stationary one moves off with the same speed. This is a classic result of simultaneous conservation of momentum and kinetic energy.
(c) Distance-time graph for trolley A [2]
Before the collision: Trolley A moves at constant speed, so its distance-time graph is a straight line with a positive (constant) gradient. [1]
After the collision: Trolley A is stationary, so its distance remains constant. The graph is a horizontal line. [1]
(d) Kinetic energy of A before the collision [2]
\( KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.50 \times 2.0^2 \) [1]
\( KE = \frac{1}{2} \times 0.50 \times 4.0 = 1.0 \text{ J} \) [1]
(e) Kinetic energy of B after the collision [1]
Since this is an elastic collision, kinetic energy is conserved. Trolley B moves at 2.0 m/s with the same mass as A, so its kinetic energy is 1.0 J. [1]
You can verify: \( KE_B = \frac{1}{2} \times 0.50 \times 2.0^2 = 1.0 \text{ J} \), confirming no kinetic energy was lost.
Question 2 Report
A student investigates the speed of sound using the apparatus shown in Fig. 15.1.
S is a starter pistol. M1 and M2 are microphones separated by a distance d of 2.40 m. Both are connected to an electronic timer. When S is fired, M1 starts the timer and M2 stops it.
(a) The timer records a time of 7.1 ms. Calculate the speed of sound. [2]
(b) Explain why the timer measures the time for sound to travel from M1 to M2, not from S to M2. [2]
(c) State one advantage of using an electronic timer rather than a stopwatch for this experiment. [1]
(d) The experiment is repeated five times. The results are: 7.1, 7.0, 7.2, 7.0, 7.1 ms. Calculate the mean time and use it to find a more accurate value for the speed of sound. [2]
(e) Suggest one source of error in this experiment, other than timing. [1]
(f) State how the accuracy of the result could be improved by changing one variable. [1]
(a) Speed of sound [2 marks]
Convert 7.1 ms to seconds: \( 7.1 \text{ ms} = 0.0071 \text{ s} \)
\( v = \frac{d}{t} = \frac{2.40}{0.0071} \) [1]
\( v = 338 \text{ m/s} \) [1]
(b) Why the timer measures M1-to-M2 time, not S-to-M2 [2 marks]
M1 is closer to the starter pistol than M2, so it receives the sound first and starts the timer. [1] The timer therefore measures only the time for the sound to travel the known distance \( d \) between M1 and M2. This eliminates the unknown and unmeasured distance from S to M1, giving a precise measurement over a known distance. [1]
(c) Advantage of an electronic timer [1 mark]
An electronic timer removes human reaction time error and provides a much more precise measurement (to the nearest millisecond or better) than a hand-operated stopwatch. [1]
(d) Mean time and improved speed of sound [2 marks]
\( \bar{t} = \frac{7.1 + 7.0 + 7.2 + 7.0 + 7.1}{5} = \frac{35.4}{5} = 7.08 \text{ ms} = 0.00708 \text{ s} \) [1]
\( v = \frac{2.40}{0.00708} \)
\( v = 339 \text{ m/s} \) [1]
Taking the mean of multiple readings reduces the impact of random errors and gives a more reliable result.
(e) Source of error other than timing [1 mark]
The distance \( d \) between the microphones may not be measured accurately (e.g. parallax when reading the ruler, or the microphones not being perfectly aligned). Alternatively, air draughts or wind could affect the speed at which sound travels. [1]
(f) How to improve accuracy [1 mark]
Increase the distance \( d \) between the two microphones. A larger distance gives a longer time interval, reducing the percentage error in the time measurement. [1]
Answer Details
(a) Speed of sound [2 marks]
Convert 7.1 ms to seconds: \( 7.1 \text{ ms} = 0.0071 \text{ s} \)
\( v = \frac{d}{t} = \frac{2.40}{0.0071} \) [1]
\( v = 338 \text{ m/s} \) [1]
(b) Why the timer measures M1-to-M2 time, not S-to-M2 [2 marks]
M1 is closer to the starter pistol than M2, so it receives the sound first and starts the timer. [1] The timer therefore measures only the time for the sound to travel the known distance \( d \) between M1 and M2. This eliminates the unknown and unmeasured distance from S to M1, giving a precise measurement over a known distance. [1]
(c) Advantage of an electronic timer [1 mark]
An electronic timer removes human reaction time error and provides a much more precise measurement (to the nearest millisecond or better) than a hand-operated stopwatch. [1]
(d) Mean time and improved speed of sound [2 marks]
\( \bar{t} = \frac{7.1 + 7.0 + 7.2 + 7.0 + 7.1}{5} = \frac{35.4}{5} = 7.08 \text{ ms} = 0.00708 \text{ s} \) [1]
\( v = \frac{2.40}{0.00708} \)
\( v = 339 \text{ m/s} \) [1]
Taking the mean of multiple readings reduces the impact of random errors and gives a more reliable result.
(e) Source of error other than timing [1 mark]
The distance \( d \) between the microphones may not be measured accurately (e.g. parallax when reading the ruler, or the microphones not being perfectly aligned). Alternatively, air draughts or wind could affect the speed at which sound travels. [1]
(f) How to improve accuracy [1 mark]
Increase the distance \( d \) between the two microphones. A larger distance gives a longer time interval, reducing the percentage error in the time measurement. [1]
Question 3 Report
Fig. 32.1 shows a circuit in which a short thick wire is accidentally connected across the terminals of a lamp.
(a) State what happens to the lamp when the short wire is connected across it. Explain why. [2]
(b) Describe what happens to the current from the battery when the short circuit occurs. [1]
(c) State the danger of a short circuit in a household mains circuit. [1]
(d) Name one safety device that would protect the circuit from damage during a short circuit. [1]
(a) When the short wire is connected across the lamp, the lamp goes out (stops working). [1]
The short wire has a very low resistance compared to the lamp. In a parallel arrangement, current takes the path of least resistance. Since nearly all the current flows through the short wire rather than through the lamp filament, the lamp receives essentially no current and produces no light. [1]
(b) The current from the battery increases greatly. [1]
The short wire effectively reduces the total resistance of the circuit to a very small value (just the wire's own tiny resistance plus the internal resistance of the battery). By \( I = \frac{V}{R} \), a much smaller R at the same voltage produces a much larger current.
(c) In a household mains circuit, the large current caused by a short circuit can make the wires overheat. This can melt the insulation surrounding the wires and potentially start a fire. [1]
Mains circuits carry much higher voltages (230 V or 240 V), so the currents during a short circuit can be extremely large and dangerous.
(d) A fuse or circuit breaker would protect the circuit from damage during a short circuit. [1]
A fuse contains a thin wire that melts and breaks the circuit when the current exceeds a safe value. A circuit breaker is an automatic switch that trips open when excessive current is detected, and can be reset after the fault is fixed.
Answer Details
(a) When the short wire is connected across the lamp, the lamp goes out (stops working). [1]
The short wire has a very low resistance compared to the lamp. In a parallel arrangement, current takes the path of least resistance. Since nearly all the current flows through the short wire rather than through the lamp filament, the lamp receives essentially no current and produces no light. [1]
(b) The current from the battery increases greatly. [1]
The short wire effectively reduces the total resistance of the circuit to a very small value (just the wire's own tiny resistance plus the internal resistance of the battery). By \( I = \frac{V}{R} \), a much smaller R at the same voltage produces a much larger current.
(c) In a household mains circuit, the large current caused by a short circuit can make the wires overheat. This can melt the insulation surrounding the wires and potentially start a fire. [1]
Mains circuits carry much higher voltages (230 V or 240 V), so the currents during a short circuit can be extremely large and dangerous.
(d) A fuse or circuit breaker would protect the circuit from damage during a short circuit. [1]
A fuse contains a thin wire that melts and breaks the circuit when the current exceeds a safe value. A circuit breaker is an automatic switch that trips open when excessive current is detected, and can be reset after the fault is fixed.
Question 4 Report
A student measures motor efficiency. The motor lifts a 2.0 N weight 0.80 m. Supply: 4.5 V, 0.30 A. Time: 6.0 s.
(a) Calculate the useful work done. [2]
(b) Calculate the electrical energy supplied. [2]
(c) Calculate the efficiency. [2]
(d) With a 5.0 N weight taking 12 s, calculate the new efficiency. [2]
(e) State what happens to the wasted energy. [1]
(a) Useful work done [2]
The useful work is the gravitational potential energy gained by the weight as it is lifted:
\(W = F \times d = 2.0 \times 0.80\) [1]
\(W = 1.6\text{ J}\) [1]
(Here \(F\) is the weight in newtons and \(d\) is the height lifted.)
(b) Electrical energy supplied [2]
\(E = VIt = 4.5 \times 0.30 \times 6.0\) [1]
\(E = 8.1\text{ J}\) [1]
(c) Efficiency [2]
\(\text{Efficiency} = \dfrac{\text{useful output}}{\text{total input}} = \dfrac{1.6}{8.1}\) [1]
\(\text{Efficiency} = 0.198 \text{ or } 19.8\%\) [1]
About 80% of the electrical energy is wasted rather than doing useful work.
(d) New efficiency with a 5.0 N weight taking 12 s [2]
Useful work: \(W = 5.0 \times 0.80 = 4.0\text{ J}\)
Electrical input: \(E = 4.5 \times 0.30 \times 12 = 16.2\text{ J}\)
\(\text{Efficiency} = \dfrac{4.0}{16.2}\) [1]
\(\text{Efficiency} = 0.247 \text{ or } 24.7\%\) [1]
The efficiency has increased from 19.8% to 24.7% with the heavier load, because the motor is doing proportionally more useful work relative to its fixed overhead losses.
(e) What happens to the wasted energy [1]
The wasted energy is transferred to thermal energy in the motor windings and bearings (the motor gets warm), with a small amount also dissipated as sound energy. [1]
Answer Details
(a) Useful work done [2]
The useful work is the gravitational potential energy gained by the weight as it is lifted:
\(W = F \times d = 2.0 \times 0.80\) [1]
\(W = 1.6\text{ J}\) [1]
(Here \(F\) is the weight in newtons and \(d\) is the height lifted.)
(b) Electrical energy supplied [2]
\(E = VIt = 4.5 \times 0.30 \times 6.0\) [1]
\(E = 8.1\text{ J}\) [1]
(c) Efficiency [2]
\(\text{Efficiency} = \dfrac{\text{useful output}}{\text{total input}} = \dfrac{1.6}{8.1}\) [1]
\(\text{Efficiency} = 0.198 \text{ or } 19.8\%\) [1]
About 80% of the electrical energy is wasted rather than doing useful work.
(d) New efficiency with a 5.0 N weight taking 12 s [2]
Useful work: \(W = 5.0 \times 0.80 = 4.0\text{ J}\)
Electrical input: \(E = 4.5 \times 0.30 \times 12 = 16.2\text{ J}\)
\(\text{Efficiency} = \dfrac{4.0}{16.2}\) [1]
\(\text{Efficiency} = 0.247 \text{ or } 24.7\%\) [1]
The efficiency has increased from 19.8% to 24.7% with the heavier load, because the motor is doing proportionally more useful work relative to its fixed overhead losses.
(e) What happens to the wasted energy [1]
The wasted energy is transferred to thermal energy in the motor windings and bearings (the motor gets warm), with a small amount also dissipated as sound energy. [1]
Question 5 Report
Fig. 6.1 shows a velocity-time graph for a bus journey.
(a) Calculate the acceleration of the bus during the first 10 s. [2]
(b) Calculate the total distance travelled by the bus. [2]
(c) Calculate the deceleration of the bus as it comes to a stop. [2]
(a) Acceleration during the first 10 s [2]
From the velocity-time graph, the bus accelerates from 0 to 15 m/s in the first 10 s:
\( a = \frac{\Delta v}{\Delta t} = \frac{15 - 0}{10} \) [1]
\( a = 1.5 \text{ m/s}^2 \) [1]
The straight line from the origin shows uniform (constant) acceleration.
(b) Total distance travelled [2]
The total distance is found from the area under the velocity-time graph. The graph forms a trapezium (or can be split into three regions):
- Region 1 (0 to 10 s): triangle, area = \( \tfrac{1}{2} \times 10 \times 15 = 75 \text{ m} \)
- Region 2 (10 to 30 s): rectangle, area = \( 20 \times 15 = 300 \text{ m} \)
- Region 3 (30 to 40 s): triangle, area = \( \tfrac{1}{2} \times 10 \times 15 = 75 \text{ m} \)
Total distance = 75 + 300 + 75 = 450 m [2]
(c) Deceleration as the bus stops [2]
From the graph, the bus decelerates from 15 m/s to 0 between t = 30 s and t = 40 s:
\( \text{deceleration} = \frac{15 - 0}{40 - 30} = \frac{15}{10} \) [1]
\( = 1.5 \text{ m/s}^2 \) [1]
The deceleration happens to equal the initial acceleration, giving the graph a symmetric shape. Deceleration is simply acceleration in the direction opposing the motion.
Answer Details
(a) Acceleration during the first 10 s [2]
From the velocity-time graph, the bus accelerates from 0 to 15 m/s in the first 10 s:
\( a = \frac{\Delta v}{\Delta t} = \frac{15 - 0}{10} \) [1]
\( a = 1.5 \text{ m/s}^2 \) [1]
The straight line from the origin shows uniform (constant) acceleration.
(b) Total distance travelled [2]
The total distance is found from the area under the velocity-time graph. The graph forms a trapezium (or can be split into three regions):
- Region 1 (0 to 10 s): triangle, area = \( \tfrac{1}{2} \times 10 \times 15 = 75 \text{ m} \)
- Region 2 (10 to 30 s): rectangle, area = \( 20 \times 15 = 300 \text{ m} \)
- Region 3 (30 to 40 s): triangle, area = \( \tfrac{1}{2} \times 10 \times 15 = 75 \text{ m} \)
Total distance = 75 + 300 + 75 = 450 m [2]
(c) Deceleration as the bus stops [2]
From the graph, the bus decelerates from 15 m/s to 0 between t = 30 s and t = 40 s:
\( \text{deceleration} = \frac{15 - 0}{40 - 30} = \frac{15}{10} \) [1]
\( = 1.5 \text{ m/s}^2 \) [1]
The deceleration happens to equal the initial acceleration, giving the graph a symmetric shape. Deceleration is simply acceleration in the direction opposing the motion.
Question 6 Report
A student connects a small coil to a sensitive galvanometer. The student moves the coil downwards between the poles of a horseshoe magnet, as shown in Fig. 11.1.
(a) State what happens to the galvanometer reading when the coil moves downwards through the field. [1]
(b) State what happens when the coil is then moved upwards through the field. [1]
(c) Name the effect being demonstrated. [1]
(d) Explain why no reading is observed when the coil is held stationary between the poles. [2]
(a) Galvanometer reading when the coil moves downwards
The galvanometer deflects (shows a reading) in one direction. [1]
Moving the coil through the magnetic field causes the coil to cut through magnetic field lines, which induces an e.m.f. and drives a current through the circuit.
(b) Galvanometer reading when the coil moves upwards
The galvanometer deflects in the opposite direction to part (a). [1]
Reversing the direction of motion reverses the direction of the induced e.m.f. and therefore the direction of the induced current.
(c) Name of the effect
Electromagnetic induction. [1]
This is the process by which a changing magnetic field through a conductor (or a conductor moving through a magnetic field) induces an e.m.f.
(d) Why no reading when the coil is stationary
When the coil is held stationary between the poles, it is not cutting through any magnetic field lines. [1]
There is no change in the magnetic field through the coil, so no e.m.f. is induced and no current flows. The galvanometer reads zero. Electromagnetic induction requires relative motion between the conductor and the magnetic field. [1]
Answer Details
(a) Galvanometer reading when the coil moves downwards
The galvanometer deflects (shows a reading) in one direction. [1]
Moving the coil through the magnetic field causes the coil to cut through magnetic field lines, which induces an e.m.f. and drives a current through the circuit.
(b) Galvanometer reading when the coil moves upwards
The galvanometer deflects in the opposite direction to part (a). [1]
Reversing the direction of motion reverses the direction of the induced e.m.f. and therefore the direction of the induced current.
(c) Name of the effect
Electromagnetic induction. [1]
This is the process by which a changing magnetic field through a conductor (or a conductor moving through a magnetic field) induces an e.m.f.
(d) Why no reading when the coil is stationary
When the coil is held stationary between the poles, it is not cutting through any magnetic field lines. [1]
There is no change in the magnetic field through the coil, so no e.m.f. is induced and no current flows. The galvanometer reads zero. Electromagnetic induction requires relative motion between the conductor and the magnetic field. [1]
Question 7 Report
Fig. 16.1 shows a coil rotating in a uniform magnetic field to produce an alternating voltage.
(a) State the position of the coil when the induced e.m.f. is at its maximum value. [1]
(b) Explain why the e.m.f. is maximum at this position. [2]
(c) State the position of the coil when the induced e.m.f. is zero. [1]
(d) Explain why the e.m.f. is zero at this position. [1]
(e) State the difference between the output of this generator if it uses slip rings compared to a split-ring commutator. [2]
(a) Position of coil at maximum e.m.f. [1]
The induced e.m.f. is maximum when the coil is parallel to the magnetic field (horizontal, with its plane aligned along the field lines).
(b) Why the e.m.f. is maximum at this position [2]
At this position, the sides of the coil are moving perpendicular to the magnetic field lines. This means they are cutting through the field lines at the greatest possible rate. Since the induced e.m.f. depends on the rate of change of magnetic flux (the rate of cutting field lines), this position gives the maximum e.m.f.
(c) Position of coil at zero e.m.f. [1]
The induced e.m.f. is zero when the coil is perpendicular to the magnetic field (vertical, as shown in position P in the diagram). The plane of the coil is at right angles to the field lines.
(d) Why the e.m.f. is zero at this position [1]
At this position, the sides of the coil are moving parallel to the magnetic field lines. They are not cutting any field lines, so there is no change in flux through the coil and no e.m.f. is induced.
(e) Difference between slip rings and split-ring commutator [2]
| Feature | Slip rings | Split-ring commutator |
|---|---|---|
| Output type | Alternating current (a.c.): the voltage alternates between positive and negative values | Direct current (d.c.): the voltage is always in the same direction, though it varies in magnitude |
The split-ring commutator reverses the connection to the external circuit every half-turn, so the output is always in the same direction (rectified). Slip rings maintain continuous connection, preserving the alternating nature of the generated e.m.f.
Answer Details
(a) Position of coil at maximum e.m.f. [1]
The induced e.m.f. is maximum when the coil is parallel to the magnetic field (horizontal, with its plane aligned along the field lines).
(b) Why the e.m.f. is maximum at this position [2]
At this position, the sides of the coil are moving perpendicular to the magnetic field lines. This means they are cutting through the field lines at the greatest possible rate. Since the induced e.m.f. depends on the rate of change of magnetic flux (the rate of cutting field lines), this position gives the maximum e.m.f.
(c) Position of coil at zero e.m.f. [1]
The induced e.m.f. is zero when the coil is perpendicular to the magnetic field (vertical, as shown in position P in the diagram). The plane of the coil is at right angles to the field lines.
(d) Why the e.m.f. is zero at this position [1]
At this position, the sides of the coil are moving parallel to the magnetic field lines. They are not cutting any field lines, so there is no change in flux through the coil and no e.m.f. is induced.
(e) Difference between slip rings and split-ring commutator [2]
| Feature | Slip rings | Split-ring commutator |
|---|---|---|
| Output type | Alternating current (a.c.): the voltage alternates between positive and negative values | Direct current (d.c.): the voltage is always in the same direction, though it varies in magnitude |
The split-ring commutator reverses the connection to the external circuit every half-turn, so the output is always in the same direction (rectified). Slip rings maintain continuous connection, preserving the alternating nature of the generated e.m.f.
Question 8 Report
Fig. 9.1 shows a simplified diagram of an optical fibre.
(a) Name the process by which light travels along the fibre. [1]
(b) Explain the role of the cladding around the core. [2]
(c) State why the core must have a higher refractive index than the cladding. [1]
(d) Give two advantages of optical fibres over copper cables for communication. [2]
(a) Name the process [1]
Total internal reflection (TIR) [1]
The light bounces off the inner surface of the core at each reflection point, staying entirely within the fibre as it travels along its length.
(b) Role of the cladding around the core [2]
The cladding has a lower refractive index than the core. [1] This ensures that when light hits the core-cladding boundary at an angle greater than the critical angle, total internal reflection occurs and the light stays within the core. [1] Without cladding, light could leak out where fibres touch or where the surface is scratched, and the signal would be lost.
(c) Why the core must have a higher refractive index than the cladding [1]
For total internal reflection to occur, light must travel from a more optically dense medium (higher refractive index) to a less optically dense medium (lower refractive index). [1] If the core had a lower refractive index than the cladding, light would refract into the cladding rather than being reflected back, and the fibre would not work.
(d) Two advantages of optical fibres over copper cables [2]
- Optical fibres can carry much more data at the same time (higher bandwidth), so they support faster communication. [1]
- Optical fibre signals do not suffer from electrical interference, making them more reliable. They are also more secure (harder to tap), lighter, and thinner than copper cables. [1]
Answer Details
(a) Name the process [1]
Total internal reflection (TIR) [1]
The light bounces off the inner surface of the core at each reflection point, staying entirely within the fibre as it travels along its length.
(b) Role of the cladding around the core [2]
The cladding has a lower refractive index than the core. [1] This ensures that when light hits the core-cladding boundary at an angle greater than the critical angle, total internal reflection occurs and the light stays within the core. [1] Without cladding, light could leak out where fibres touch or where the surface is scratched, and the signal would be lost.
(c) Why the core must have a higher refractive index than the cladding [1]
For total internal reflection to occur, light must travel from a more optically dense medium (higher refractive index) to a less optically dense medium (lower refractive index). [1] If the core had a lower refractive index than the cladding, light would refract into the cladding rather than being reflected back, and the fibre would not work.
(d) Two advantages of optical fibres over copper cables [2]
- Optical fibres can carry much more data at the same time (higher bandwidth), so they support faster communication. [1]
- Optical fibre signals do not suffer from electrical interference, making them more reliable. They are also more secure (harder to tap), lighter, and thinner than copper cables. [1]
Question 9 Report
A boat travels across a river. The boat's engine drives it at 4.0 m/s due north. The river current flows at 3.0 m/s due east.
(a) Calculate the resultant speed of the boat. [2]
(b) The river is 80 m wide. Calculate the time taken for the boat to cross the river. [2]
(c) Calculate how far downstream the boat is carried. [2]
(d) Explain why the boat does not travel in a straight line due north. [2]
(a) Resultant speed of the boat [2]
The boat's engine drives it at 4.0 m/s north and the current pushes it at 3.0 m/s east. These two velocity components are perpendicular, so the resultant is found using Pythagoras:
\( v = \sqrt{4.0^2 + 3.0^2} = \sqrt{16 + 9} = \sqrt{25} \) [1]
\( v = 5.0 \text{ m/s} \) [1]
This is a 3-4-5 right triangle. The resultant velocity is directed northeast at an angle to the north bank.
(b) Time to cross the river [2]
The river is 80 m wide (measured north-south). The boat's northward component is 4.0 m/s, which is the speed responsible for crossing:
\( t = \frac{\text{width}}{\text{speed across}} = \frac{80}{4.0} \) [1]
\( t = 20 \text{ s} \) [1]
The eastward current does not affect the time to cross, only where the boat lands.
(c) Distance carried downstream [2]
During the 20 s crossing, the current carries the boat eastward:
\( \text{downstream distance} = v_{\text{current}} \times t = 3.0 \times 20 \) [1]
\( = 60 \text{ m} \) [1]
(d) Why the boat does not travel straight north [2]
The river current exerts a sideways force on the boat, pushing it downstream (eastward). [1]
The actual path of the boat is determined by the vector sum of the boat's velocity (north) and the current's velocity (east). Since these point in different directions, the resultant velocity is at an angle, and the boat follows a diagonal path rather than going straight north. [1]
Answer Details
(a) Resultant speed of the boat [2]
The boat's engine drives it at 4.0 m/s north and the current pushes it at 3.0 m/s east. These two velocity components are perpendicular, so the resultant is found using Pythagoras:
\( v = \sqrt{4.0^2 + 3.0^2} = \sqrt{16 + 9} = \sqrt{25} \) [1]
\( v = 5.0 \text{ m/s} \) [1]
This is a 3-4-5 right triangle. The resultant velocity is directed northeast at an angle to the north bank.
(b) Time to cross the river [2]
The river is 80 m wide (measured north-south). The boat's northward component is 4.0 m/s, which is the speed responsible for crossing:
\( t = \frac{\text{width}}{\text{speed across}} = \frac{80}{4.0} \) [1]
\( t = 20 \text{ s} \) [1]
The eastward current does not affect the time to cross, only where the boat lands.
(c) Distance carried downstream [2]
During the 20 s crossing, the current carries the boat eastward:
\( \text{downstream distance} = v_{\text{current}} \times t = 3.0 \times 20 \) [1]
\( = 60 \text{ m} \) [1]
(d) Why the boat does not travel straight north [2]
The river current exerts a sideways force on the boat, pushing it downstream (eastward). [1]
The actual path of the boat is determined by the vector sum of the boat's velocity (north) and the current's velocity (east). Since these point in different directions, the resultant velocity is at an angle, and the boat follows a diagonal path rather than going straight north. [1]
Question 10 Report
Fig. 30.1 shows a circuit with a 6.0 V battery, two switches S₁ and S₂, and two lamps L₁ and L₂. S₁ is in series with L₁ and S₂ is in series with L₂. The two branches are connected in parallel.
L₁ is rated 6.0 V, 0.50 A. L₂ is rated 6.0 V, 0.30 A.
(a) Both switches are closed. State which lamps are on. [1]
(b) S₁ is opened and S₂ stays closed. State which lamp is on. [1]
(c) Calculate the power dissipated in L₁ when operating at its rated values. [2]
(d) Calculate the resistance of L₂ at its rated values. [2]
(e) State what happens to L₂ if L₁ blows. [1]
(a) With both switches closed, current flows through both parallel branches. Both L₁ and L₂ are on. [1]
(b) When S₁ is opened, it breaks the circuit through the L₁ branch, so L₁ goes off. S₂ remains closed, so current still flows through L₂. Only L₂ is on. [1]
(c) The power dissipated in L₁ at its rated values is found using \(P = IV\):
\[P = 0.50 \times 6.0\] [1]
\[P = 3.0 \text{ W}\] [1]
(d) The resistance of L₂ at its rated values is found using \(R = \frac{V}{I}\):
\[R = \frac{6.0}{0.30}\] [1]
\[R = 20 \;\Omega\] [1]
(e) If L₁ blows (its filament breaks), L₂ stays on and is unaffected. [1] Because the two lamps are on separate parallel branches, breaking one branch does not interrupt the current in the other. This is the key advantage of parallel wiring over series wiring.
Answer Details
(a) With both switches closed, current flows through both parallel branches. Both L₁ and L₂ are on. [1]
(b) When S₁ is opened, it breaks the circuit through the L₁ branch, so L₁ goes off. S₂ remains closed, so current still flows through L₂. Only L₂ is on. [1]
(c) The power dissipated in L₁ at its rated values is found using \(P = IV\):
\[P = 0.50 \times 6.0\] [1]
\[P = 3.0 \text{ W}\] [1]
(d) The resistance of L₂ at its rated values is found using \(R = \frac{V}{I}\):
\[R = \frac{6.0}{0.30}\] [1]
\[R = 20 \;\Omega\] [1]
(e) If L₁ blows (its filament breaks), L₂ stays on and is unaffected. [1] Because the two lamps are on separate parallel branches, breaking one branch does not interrupt the current in the other. This is the key advantage of parallel wiring over series wiring.
Question 11 Report
Fig. 10.1 shows the entrance to a harbour. Sea waves approach the harbour entrance from the open sea. The harbour wall has a gap of width d.
(a) The wavelength of the sea waves is much smaller than d. Describe the wave pattern inside the harbour. [2]
(b) On a different day, the wavelength of the waves is approximately equal to d. Describe how the wave pattern inside the harbour differs from part (a). [2]
(c) State the name of the wave effect observed in both cases. [1]
(d) Explain why boats moored deep inside the harbour may still experience wave motion even when the harbour entrance is narrow. [2]
(a) When the wavelength is much smaller than the gap width d:
- The waves pass through the gap and continue mostly as plane (straight) wavefronts, with a shadow zone behind each section of wall where no waves reach. [1]
- Only slight curving of the wavefronts occurs at the very edges of the gap. [1]
Diffraction is negligible when the gap is much wider than the wavelength. The waves travel through almost as if the walls were not there, and the regions directly behind the harbour walls receive little wave energy.
(b) When the wavelength is approximately equal to d:
- The waves spread out much more after passing through the gap. [1]
- The wavefronts become semicircular, reaching areas behind the harbour walls that were previously in the shadow zone. [1]
Maximum diffraction occurs when the gap width is comparable to (or smaller than) the wavelength. The gap effectively acts as a new point source, radiating circular wavefronts into the harbour.
(c) The wave effect is called diffraction. [1]
Diffraction is the spreading of waves as they pass through a gap or around an obstacle. It is a property of all waves.
(d)
- Waves diffract through the harbour entrance and spread into the enclosed area. [1]
- Even though the entrance is narrow, diffraction causes the waves to spread out and reach boats moored deep inside the harbour. [1]
In fact, a narrower entrance (closer to the wavelength of the sea waves) produces stronger diffraction, so the waves spread more effectively into the harbour interior. This is why boats far from the entrance can still bob up and down on the diffracted waves.
Answer Details
(a) When the wavelength is much smaller than the gap width d:
- The waves pass through the gap and continue mostly as plane (straight) wavefronts, with a shadow zone behind each section of wall where no waves reach. [1]
- Only slight curving of the wavefronts occurs at the very edges of the gap. [1]
Diffraction is negligible when the gap is much wider than the wavelength. The waves travel through almost as if the walls were not there, and the regions directly behind the harbour walls receive little wave energy.
(b) When the wavelength is approximately equal to d:
- The waves spread out much more after passing through the gap. [1]
- The wavefronts become semicircular, reaching areas behind the harbour walls that were previously in the shadow zone. [1]
Maximum diffraction occurs when the gap width is comparable to (or smaller than) the wavelength. The gap effectively acts as a new point source, radiating circular wavefronts into the harbour.
(c) The wave effect is called diffraction. [1]
Diffraction is the spreading of waves as they pass through a gap or around an obstacle. It is a property of all waves.
(d)
- Waves diffract through the harbour entrance and spread into the enclosed area. [1]
- Even though the entrance is narrow, diffraction causes the waves to spread out and reach boats moored deep inside the harbour. [1]
In fact, a narrower entrance (closer to the wavelength of the sea waves) produces stronger diffraction, so the waves spread more effectively into the harbour interior. This is why boats far from the entrance can still bob up and down on the diffracted waves.
