Physics 4PH1 | Paper 2 Mock 01 | Written Paper 2
Question 1 Report
Table 1 gives results from a field test on a dry lakebed. A researcher stood 90.0 m from a rock face and made a short sound using two wooden blocks. A microphone connected to an oscilloscope recorded the time until the echo returned. The sound waves travelled to the rock face and back, so their total distance was 180.0 m. Fig. 1 shows the arrangement. The temperature of the air was changed by carrying out the test at different times of day.
| air temperature / °C | echo time / s |
|---|---|
| -5 | 0.545 |
| 5 | 0.529 |
| 15 | 0.514 |
| 25 | 0.500 |
(a) Calculate the speed of sound when the air temperature was 15 °C. [3]
(b) Describe the relationship between air temperature and the speed of sound shown by Table 1. [2]
(c) Explain why increasing the temperature makes sound travel faster through air. [1]
(a) The sound travels to the rock face and back, so its distance is:
\[2\times90.0=180.0\text{ m}\]
\[v=\frac{d}{t}=\frac{180.0}{0.514}=350\text{ m s}^{-1}\]
The speed of sound at \(15^\circ\text{C}\) is \(350\text{ m s}^{-1}\). [3]
(b) As air temperature increases, the echo time decreases. Since the total distance is unchanged, this means the speed of sound increases as temperature increases. [2]
(c) At a higher temperature, air particles have more kinetic energy and move faster. They pass vibrations between particles more quickly, so sound travels faster. [1]
Answer Details
(a) The sound travels to the rock face and back, so its distance is:
\[2\times90.0=180.0\text{ m}\]
\[v=\frac{d}{t}=\frac{180.0}{0.514}=350\text{ m s}^{-1}\]
The speed of sound at \(15^\circ\text{C}\) is \(350\text{ m s}^{-1}\). [3]
(b) As air temperature increases, the echo time decreases. Since the total distance is unchanged, this means the speed of sound increases as temperature increases. [2]
(c) At a higher temperature, air particles have more kinetic energy and move faster. They pass vibrations between particles more quickly, so sound travels faster. [1]
Question 2 Report
Table 1 shows results from a student cooling 0.50 kg of melted chocolate in a mould. The student records the temperature every 2 minutes. The chocolate is stirred gently before each reading so that its temperature is uniform. The nearly constant section of the data is caused by a change of state.
| time / min | temperature / °C |
|---|---|
| 0 | 42 |
| 2 | 37 |
| 4 | 32 |
| 6 | 28 |
| 8 | 28 |
| 10 | 28 |
| 12 | 24 |
(a) State the temperature at which the chocolate changes state. [1]
(b) Complete this sentence: during the constant-temperature section, the chocolate changes from a ______ to a ______. [2]
(c) Calculate the average rate of temperature decrease between 0 min and 4 min, in °C/min. [2]
(d) Explain why energy is still transferred from the chocolate between 6 min and 10 min. [2]
(e) Describe one change to the student's method that would make the results more reliable. [2]
(a) The constant-temperature section is at 28 °C. [1 mark]
(b) During this section the chocolate changes from a liquid to a solid. [2 marks]
(c)
\[\text{temperature decrease}=42-32=10\,°\text{C}\]
\[\text{rate}=\frac{10}{4}=2.5\,°\text{C/min}\]
2.5 °C/min. [2 marks]
(d) Energy is still transferred from the warmer chocolate to the cooler surroundings. During solidification, it is released as latent heat as particles form stronger attractions, even though the temperature is constant. [2 marks]
(e) Repeat the investigation and calculate a mean temperature for each time. Repeats reduce the effect of random variation, making the results more reliable. A data logger taking readings at fixed times would also be creditworthy because it reduces reading uncertainty. [2 marks]
Answer Details
(a) The constant-temperature section is at 28 °C. [1 mark]
(b) During this section the chocolate changes from a liquid to a solid. [2 marks]
(c)
\[\text{temperature decrease}=42-32=10\,°\text{C}\]
\[\text{rate}=\frac{10}{4}=2.5\,°\text{C/min}\]
2.5 °C/min. [2 marks]
(d) Energy is still transferred from the warmer chocolate to the cooler surroundings. During solidification, it is released as latent heat as particles form stronger attractions, even though the temperature is constant. [2 marks]
(e) Repeat the investigation and calculate a mean temperature for each time. Repeats reduce the effect of random variation, making the results more reliable. A data logger taking readings at fixed times would also be creditworthy because it reduces reading uncertainty. [2 marks]
Question 3 Report
A sports technician tests an electric resistance sled on an indoor track. Fig. 1 shows the sled pulled by a student using a handle. The student pulls with a horizontal force of 165 N for 240 m. The sled travels at constant speed and takes 48 s. Its motor and control box draw an electrical power of 1.1 kW while the sled is moving. The force sensor sends data to a computer.
(a) Calculate the work done by the student on the sled. [3]
(b) Calculate the useful power transferred to the sled. [2]
(c) Calculate the efficiency of the motor and control box. [3]
(d) Explain why the horizontal resultant force on the sled is zero while its speed is constant. [2]
(e) Describe the main energy transfer as the student does work on the sled. [1]
(f) State the unit of power. [1]
(a) Work done is force multiplied by distance moved in the force direction:
\[W=Fs=165\times240=39\,600\text{ J}\]
39 600 J. [3 marks]
(b)
\[P=\frac{W}{t}=\frac{39\,600}{48}=825\text{ W}\]
825 W. [2 marks]
(c) Convert the input power first:
\[1.1\text{ kW}=1100\text{ W}\]
\[\text{efficiency}=\frac{825}{1100}=0.75=75\%\]
0.75, or 75%. [3 marks]
(d) The sled moves at constant speed, so it has no acceleration and therefore no resultant force. The pulling force must equal the resistive force. [2 marks]
(e) Chemical energy in the student's energy store is transferred mechanically to the sled. [1 mark]
(f) The unit of power is the watt, W. [1 mark]
Answer Details
(a) Work done is force multiplied by distance moved in the force direction:
\[W=Fs=165\times240=39\,600\text{ J}\]
39 600 J. [3 marks]
(b)
\[P=\frac{W}{t}=\frac{39\,600}{48}=825\text{ W}\]
825 W. [2 marks]
(c) Convert the input power first:
\[1.1\text{ kW}=1100\text{ W}\]
\[\text{efficiency}=\frac{825}{1100}=0.75=75\%\]
0.75, or 75%. [3 marks]
(d) The sled moves at constant speed, so it has no acceleration and therefore no resultant force. The pulling force must equal the resistive force. [2 marks]
(e) Chemical energy in the student's energy store is transferred mechanically to the sled. [1 mark]
(f) The unit of power is the watt, W. [1 mark]
Question 4 Report
A technician checks the support cable of a raised footbridge. Fig. 1 shows the bridge deck in balance about pivot P. A maintenance box has weight 900 N and is 3.0 m from P. The cable provides an upward force at the far end of the deck, 9.0 m from P.
(a) Calculate the clockwise moment of the box about P. [2]
(b) Calculate the upward force supplied by the cable. [2]
(a) Moment is calculated using force multiplied by perpendicular distance from the pivot:
\[\text{clockwise moment}=900\,\text{N}\times3.0\,\text{m}=2700\,\text{N m}\]
The clockwise moment is 2700 N m. [2]
(b) Because the deck is in balance, clockwise and anticlockwise moments are equal. The cable must provide an anticlockwise moment of \(2700\,\text{N m}\).
\[F\times9.0=2700\]
\[F=\frac{2700}{9.0}=300\,\text{N}\]
The upward cable force is 300 N. [2]
Answer Details
(a) Moment is calculated using force multiplied by perpendicular distance from the pivot:
\[\text{clockwise moment}=900\,\text{N}\times3.0\,\text{m}=2700\,\text{N m}\]
The clockwise moment is 2700 N m. [2]
(b) Because the deck is in balance, clockwise and anticlockwise moments are equal. The cable must provide an anticlockwise moment of \(2700\,\text{N m}\).
\[F\times9.0=2700\]
\[F=\frac{2700}{9.0}=300\,\text{N}\]
The upward cable force is 300 N. [2]
Question 5 Report
A museum conservator uses a converging lens to project an enlarged image of a small star symbol from an old glass slide onto a white screen. Fig. 1 shows the principal axis, a lens and the positions of its focal points. The object is placed between F and 2F on the left of the lens. The student needs to draw rays before deciding where to put the screen.
(a) Complete a ray diagram by drawing two suitable rays from the top of the object. [3]
(b) State the nature of the image formed on the screen. [2]
(c) Describe how the screen position changes if the object is moved closer to the lens but remains outside F. [1]
(d) Calculate the power of a lens with focal length 0.080 m. [2]
(a) A converging lens forms a real image when the object is outside its focal point. Draw one ray from the top of the object parallel to the principal axis; after the lens it passes through the focal point on the right. Draw a second ray through the centre of the lens; it continues undeviated. Where they meet is the top of the inverted image.
The rays meet on the right of the lens and the image arrow is inverted. [3]
(b) Because the rays really meet on a screen, the image is real. It is also inverted and enlarged. [2]
(c) Moving the object closer to the lens while it remains outside \(F\) makes the image distance increase. Move the screen further from the lens. [1]
(d) Lens power is the reciprocal of focal length in metres:
\[P=\frac{1}{f}=\frac{1}{0.080}=12.5\text{ D}\]
The power is 12.5 D. [2]
Answer Details
(a) A converging lens forms a real image when the object is outside its focal point. Draw one ray from the top of the object parallel to the principal axis; after the lens it passes through the focal point on the right. Draw a second ray through the centre of the lens; it continues undeviated. Where they meet is the top of the inverted image.
The rays meet on the right of the lens and the image arrow is inverted. [3]
(b) Because the rays really meet on a screen, the image is real. It is also inverted and enlarged. [2]
(c) Moving the object closer to the lens while it remains outside \(F\) makes the image distance increase. Move the screen further from the lens. [1]
(d) Lens power is the reciprocal of focal length in metres:
\[P=\frac{1}{f}=\frac{1}{0.080}=12.5\text{ D}\]
The power is 12.5 D. [2]
Question 6 Report
Fig. 1 shows a velocity-time graph for an electric delivery van travelling along a level road. The van accelerates from a depot, travels at a steady speed, then brakes at a junction. Its mass, including parcels, is 1200 kg. A student uses the graph to calculate the force needed during the first part of the journey. Ignore air resistance during this calculation.
(a) State what is represented by the gradient of a velocity-time graph. [1]
(b) Calculate the acceleration of the van during the first 8 s. [2]
(c) Calculate the distance travelled by the van during the first 20 s. [3]
(d) Calculate the resultant force on the van during the first 8 s. [2]
(e) Describe the motion of the van between 8 s and 20 s. [2]
(f) Explain why the resultant force is zero between 8 s and 20 s, even though the motor may still transfer energy. [3]
(g) State the SI unit of force. [1]
(a) The gradient of a velocity-time graph represents acceleration. [1]
(b)
\[a=\frac{12-0}{8}=1.5\text{ m/s}^2\]
[2]
(c) Distance is the area beneath the velocity-time graph.
\[\text{triangle}=\frac12\times8\times12=48\text{ m}\]
\[\text{rectangle}=12\times12=144\text{ m}\]
\[\text{total}=48+144=192\text{ m}\]
[3]
(d)
\[F=ma=1200\text{ kg}\times1.5\text{ m/s}^2=1800\text{ N}\]
[2]
(e) Between 8 s and 20 s, the van travels at constant velocity, 12 m/s. [2]
(f) Constant velocity means zero acceleration. By \(F=ma\), zero acceleration means zero resultant force. The motor can still transfer energy because its driving force balances resistive forces, such as air resistance and tyre friction. [3]
(g) The SI unit of force is the newton, N. [1]
Answer Details
(a) The gradient of a velocity-time graph represents acceleration. [1]
(b)
\[a=\frac{12-0}{8}=1.5\text{ m/s}^2\]
[2]
(c) Distance is the area beneath the velocity-time graph.
\[\text{triangle}=\frac12\times8\times12=48\text{ m}\]
\[\text{rectangle}=12\times12=144\text{ m}\]
\[\text{total}=48+144=192\text{ m}\]
[3]
(d)
\[F=ma=1200\text{ kg}\times1.5\text{ m/s}^2=1800\text{ N}\]
[2]
(e) Between 8 s and 20 s, the van travels at constant velocity, 12 m/s. [2]
(f) Constant velocity means zero acceleration. By \(F=ma\), zero acceleration means zero resultant force. The motor can still transfer energy because its driving force balances resistive forces, such as air resistance and tyre friction. [3]
(g) The SI unit of force is the newton, N. [1]
Question 7 Report
A technician seals a small amount of water in a steel chamber and warms it. Some water forms steam. Fig. 1 shows the chamber, its temperature probe and the pressure sensor.
Table 1 shows readings from the apparatus. The outside air pressure is 101 kPa. The lid has area 0.020 m2.
| temperature / degrees C | pressure in chamber / kPa |
|---|---|
| 20 | 101 |
| 40 | 107 |
| 60 | 113 |
| 80 | 119 |
(a) State the unit used for temperature and the unit used for pressure in Table 1. [2]
(b) Calculate the increase in pressure, in kPa per degrees C, between 20 degrees C and 80 degrees C. [3]
(c) Calculate the force on the lid caused by the pressure difference at 80 degrees C. [3]
(d) Explain why the pressure rises as the temperature of the water and steam rises. [2]
(e) Complete the conversions: 750 cm3 = ........ m3; 1.5 L = ........ cm3. [2]
(a) The table uses degrees Celsius, written °C, for temperature and kPa for pressure. [2 marks]
(b) Find both changes before calculating the rate:
\[\Delta p=119-101=18\text{ kPa}\]
\[\Delta T=80-20=60\,°\text{C}\]
\[\text{pressure increase per }°\text{C}=\frac{18}{60}=0.30\text{ kPa per }°\text{C}\]
The answer is 0.30 kPa per °C. [3 marks]
(c) The pressure outside is 101 kPa, so the pressure difference across the lid at 80 °C is:
\[119-101=18\text{ kPa}=18\,000\text{ Pa}\]
Use \(F=pA\), with pressure in pascals:
\[F=18\,000\times0.020=360\text{ N}\]
The force on the lid is 360 N. [3 marks]
(d) Heating gives the water and steam particles more kinetic energy, so they move faster. They collide with the chamber walls more often and/or with greater force. This increases the force per unit area, so the pressure rises. [2 marks]
(e) Since \(1\text{ cm}^3=1\times10^{-6}\text{ m}^3\):
\[750\text{ cm}^3=750\times10^{-6}=0.000750\text{ m}^3\]
Also, \(1\text{ L}=1000\text{ cm}^3\), so:
\[1.5\text{ L}=1500\text{ cm}^3\]
0.000750 m³; 1500 cm³. [2 marks]
Answer Details
(a) The table uses degrees Celsius, written °C, for temperature and kPa for pressure. [2 marks]
(b) Find both changes before calculating the rate:
\[\Delta p=119-101=18\text{ kPa}\]
\[\Delta T=80-20=60\,°\text{C}\]
\[\text{pressure increase per }°\text{C}=\frac{18}{60}=0.30\text{ kPa per }°\text{C}\]
The answer is 0.30 kPa per °C. [3 marks]
(c) The pressure outside is 101 kPa, so the pressure difference across the lid at 80 °C is:
\[119-101=18\text{ kPa}=18\,000\text{ Pa}\]
Use \(F=pA\), with pressure in pascals:
\[F=18\,000\times0.020=360\text{ N}\]
The force on the lid is 360 N. [3 marks]
(d) Heating gives the water and steam particles more kinetic energy, so they move faster. They collide with the chamber walls more often and/or with greater force. This increases the force per unit area, so the pressure rises. [2 marks]
(e) Since \(1\text{ cm}^3=1\times10^{-6}\text{ m}^3\):
\[750\text{ cm}^3=750\times10^{-6}=0.000750\text{ m}^3\]
Also, \(1\text{ L}=1000\text{ cm}^3\), so:
\[1.5\text{ L}=1500\text{ cm}^3\]
0.000750 m³; 1500 cm³. [2 marks]
Question 8 Report
Fig. 1 shows part of a photocopier. A rotating metal drum is given a positive charge. Light reflected from a printed page removes charge from selected regions of the drum. Negatively charged toner particles are then released close to the drum. The toner forms the pattern that will later be transferred to paper.
(a) State which regions of the drum attract the negatively charged toner particles most strongly. [1]
(b) Explain why those regions attract the toner particles. [2]
(c) Describe why the toner must later be heated onto the paper. [2]
(a) The negatively charged toner is attracted most strongly to regions that remain positively charged, namely the dark regions not exposed to light. [1]
(b) Toner particles are negatively charged. Opposite charges attract, so the positively charged regions exert an electrostatic force on the toner. [2]
(c) Heating melts or softens the toner. It then sticks permanently to the paper, so the copied image does not rub off easily. [2]
Answer Details
(a) The negatively charged toner is attracted most strongly to regions that remain positively charged, namely the dark regions not exposed to light. [1]
(b) Toner particles are negatively charged. Opposite charges attract, so the positively charged regions exert an electrostatic force on the toner. [2]
(c) Heating melts or softens the toner. It then sticks permanently to the paper, so the copied image does not rub off easily. [2]
