Physics 4PH1 | Paper 1 Mock 01 | Written Paper 1
Question 1 Report
Fig. 1 shows a graph from a student’s investigation of a solid wax sample. The student supplied energy at a constant rate and recorded the temperature every minute. The wax was held in a metal cup. At 4 minutes the temperature stopped rising, although energy was still transferred to the wax. The distance between the heater and the cup was kept constant.
(a) State the state of the wax before point A. [1]
(b) State the melting temperature of this wax. [1]
(c) Explain why the graph is horizontal from 4 minutes to 10 minutes. [3]
(d) Calculate the mean rate of temperature increase during the first 4 minutes. [2]
(e) Describe one difference between the particle arrangement in solid wax and in liquid wax. [1]
(a) Before point A, the wax is a solid. Its temperature is rising until it reaches its melting temperature. [1]
(b) The horizontal section is at 62 °C, so the melting temperature is 62 °C. [1]
(c) Energy continues to be transferred to the wax from 4 minutes to 10 minutes. This energy is used to overcome forces between the particles, allowing the solid to change into a liquid. It is not used to increase particle kinetic energy, so the temperature remains constant and the graph is horizontal. [3]
(d) During the first four minutes, the temperature rises from 20 °C to 62 °C:
\[\text{temperature rise}=62-20=42\text{ °C}\]
\[\text{mean rate of increase}=\frac{42\text{ °C}}{4\text{ min}}=10.5\text{ °C/min}\]
The mean rate is 10.5 °C per min. [2]
(e) In solid wax, particles are in fixed regular positions. In liquid wax, particles remain close together but can move past one another. [1]
Answer Details
(a) Before point A, the wax is a solid. Its temperature is rising until it reaches its melting temperature. [1]
(b) The horizontal section is at 62 °C, so the melting temperature is 62 °C. [1]
(c) Energy continues to be transferred to the wax from 4 minutes to 10 minutes. This energy is used to overcome forces between the particles, allowing the solid to change into a liquid. It is not used to increase particle kinetic energy, so the temperature remains constant and the graph is horizontal. [3]
(d) During the first four minutes, the temperature rises from 20 °C to 62 °C:
\[\text{temperature rise}=62-20=42\text{ °C}\]
\[\text{mean rate of increase}=\frac{42\text{ °C}}{4\text{ min}}=10.5\text{ °C/min}\]
The mean rate is 10.5 °C per min. [2]
(e) In solid wax, particles are in fixed regular positions. In liquid wax, particles remain close together but can move past one another. [1]
Question 2 Report
A marine engineer tests a radio communication system for an unmanned research boat. Fig. 1 shows a radio aerial on a cliff sending a signal to the boat. The distance from the aerial to the boat is 24 km. The electromagnetic radio waves travel through air at 3.0 × 108 m/s. The boat uses the received signal to change speed and direction.
(a) Describe how the radio signal transfers energy from the aerial to the receiver on the boat. [2]
(b) Calculate the time taken for the radio waves to travel from the aerial to the boat. [3]
(c) State one reason why radio waves are suitable for communication with the boat. [1]
(d) Explain why a very large hill between the aerial and the boat may reduce the strength of the received signal. [2]
(a) The transmitting aerial produces electromagnetic radio waves. [1] These waves carry energy through the air to the receiving aerial on the boat. [1]
(b) \[24\text{ km}=24\,000\text{ m}\] [1]
\[t=\frac{d}{v}=\frac{24\,000\text{ m}}{3.0\times10^8\text{ m/s}}\] [1]
\[t=8.0\times10^{-5}\text{ s}\]
The travel time is \(8.0\times10^{-5}\text{ s}\). [1]
(c) Radio waves are suitable because they can travel long distances through air and carry information. Either stated reason is acceptable. [1]
(d) A large hill blocks or absorbs some radio waves. [1] Less wave energy reaches the receiving aerial, so the received signal is weaker. [1]
Answer Details
(a) The transmitting aerial produces electromagnetic radio waves. [1] These waves carry energy through the air to the receiving aerial on the boat. [1]
(b) \[24\text{ km}=24\,000\text{ m}\] [1]
\[t=\frac{d}{v}=\frac{24\,000\text{ m}}{3.0\times10^8\text{ m/s}}\] [1]
\[t=8.0\times10^{-5}\text{ s}\]
The travel time is \(8.0\times10^{-5}\text{ s}\). [1]
(c) Radio waves are suitable because they can travel long distances through air and carry information. Either stated reason is acceptable. [1]
(d) A large hill blocks or absorbs some radio waves. [1] Less wave energy reaches the receiving aerial, so the received signal is weaker. [1]
Question 3 Report
Table 1 shows a quality-control record for a technetium tracer sample prepared in a hospital. Counts were measured for 60 s using the same detector geometry. The background count is 12 counts per minute. A student must plot the corrected count against time on the grid provided, then use the graph to assess whether the sample has the expected half-life.
| Time after preparation / h | Count in 60 s |
|---|---|
| 0 | 972 |
| 6 | 492 |
| 12 | 252 |
| 18 | 132 |
| 24 | 72 |
(a) Plot the corrected count against time on the grid provided and draw a smooth curve of best fit. [3]
(b) State the half-life shown by the results. [1]
(c) Calculate the activity at time 12 h in Bq. [2]
(d) Explain why background radiation is subtracted from every count. [2]
(e) Calculate when the corrected count will first be below 30 counts per minute. [2]
(f) Describe why repeated count measurements at one time may not be identical. [2]
(a) Subtract the 12 counts per minute background before plotting. The corrected points are \((0,960)\), \((6,480)\), \((12,240)\), \((18,120)\), and \((24,60)\). A smooth decreasing curve through these points is required. [3]
(b) The corrected count halves every 6 h, so the half-life is 6 h. [1]
(c)\[252-12=240\text{ counts min}^{-1}\]
\[240/60=4\text{ Bq}\]
The activity at 12 h is 4 Bq. [2]
(d) The detector records radiation from the surroundings as well as from the tracer. Background subtraction gives the count and activity due only to the tracer. [2]
(e) At 24 h the corrected count is 60 counts per minute. One more half-life gives 30 counts per minute:
\[24+6=30\text{ h}\]
It first falls below 30 counts per minute after 30 h. [2]
(f) Radioactive decay is random, so the number of particles detected in equal time intervals fluctuates. [2]
Answer Details
(a) Subtract the 12 counts per minute background before plotting. The corrected points are \((0,960)\), \((6,480)\), \((12,240)\), \((18,120)\), and \((24,60)\). A smooth decreasing curve through these points is required. [3]
(b) The corrected count halves every 6 h, so the half-life is 6 h. [1]
(c)\[252-12=240\text{ counts min}^{-1}\]
\[240/60=4\text{ Bq}\]
The activity at 12 h is 4 Bq. [2]
(d) The detector records radiation from the surroundings as well as from the tracer. Background subtraction gives the count and activity due only to the tracer. [2]
(e) At 24 h the corrected count is 60 counts per minute. One more half-life gives 30 counts per minute:
\[24+6=30\text{ h}\]
It first falls below 30 counts per minute after 30 h. [2]
(f) Radioactive decay is random, so the number of particles detected in equal time intervals fluctuates. [2]
Question 4 Report
This machine is used in a library to make copies of damaged maps without handling the original many times. Fig. 1 shows the main stages inside the copier. A rotating drum is given a positive charge. Bright light reflects from the pale areas of the map onto the drum. Negatively charged toner particles are then brought near the drum, before paper passes through the machine.
(a) State what happens to the charge on a bright area of the drum when light falls on it. [1]
(b) Explain why negatively charged toner particles stick to the dark areas of the drum. [3]
(c) Describe how the toner image is transferred from the drum to the paper. [3]
(d) State why a metal casing connected to earth is used around the charging parts. [2]
(e) Explain why the copier should not be used if its casing has a crack and a wire is exposed. [2]
(a) On a bright area of the drum, the positive charge is removed, so that area becomes uncharged or has less positive charge remaining. [1]
(b) Dark map areas receive little or no light, so the corresponding drum areas remain positively charged. Negatively charged toner particles are attracted to these oppositely charged areas and stick there. [3]
(c) The paper is given a positive charge, or another charge that attracts the toner. Toner transfers from the drum to the paper, then heating melts or fixes it permanently onto the paper. [3]
(d) The earthed metal casing provides a path for excess charge to flow to Earth. This reduces charge build-up and the risk of sparks. [2]
(e) A crack with an exposed wire could allow someone to touch a live conductor. This could cause electric shock because current could pass through the body. [2]
Answer Details
(a) On a bright area of the drum, the positive charge is removed, so that area becomes uncharged or has less positive charge remaining. [1]
(b) Dark map areas receive little or no light, so the corresponding drum areas remain positively charged. Negatively charged toner particles are attracted to these oppositely charged areas and stick there. [3]
(c) The paper is given a positive charge, or another charge that attracts the toner. Toner transfers from the drum to the paper, then heating melts or fixes it permanently onto the paper. [3]
(d) The earthed metal casing provides a path for excess charge to flow to Earth. This reduces charge build-up and the risk of sparks. [2]
(e) A crack with an exposed wire could allow someone to touch a live conductor. This could cause electric shock because current could pass through the body. [2]
Question 5 Report
A railway operator studies the motion of an unpowered maintenance cart on a straight track. Fig. 1 shows the cart before it rolls into a sand-filled stopping box. The cart has a mass of 500 kg and enters the sand at 4.0 m/s. It comes to rest after travelling 1.6 m through the sand. Assume the retarding force is constant.
(a) Calculate the initial momentum of the cart. [2]
(b) Calculate the initial kinetic energy of the cart. [3]
(c) Calculate the magnitude of the deceleration of the cart in the sand. [3]
(d) Calculate the retarding force exerted by the sand. [2]
(e) Explain why using a longer sand box reduces the force on the cart. [3]
(f) State one change to the cart that would increase its momentum at the same speed. [1]
(g) Describe the main energy transfer as the cart stops. [2]
(a) Momentum is \(p=mv\):
\[p=500\times4.0=2000\text{ kg m/s}\]
The initial momentum is 2000 kg m/s. [2]
(b) Use \(E_k=\frac{1}{2}mv^2\):
\[E_k=\frac{1}{2}\times500\times4.0^2=4000\text{ J}\]
The initial kinetic energy is 4000 J. [3]
(c) Use \(v^2=u^2+2as\):
\[0=4.0^2+2\times a\times1.6\]
\[a=-5.0\text{ m/s}^2\]
The deceleration has magnitude 5.0 m/s². [3]
(d) \[F=ma=500\times5.0=2500\text{ N}\]
The retarding force is 2500 N. [2]
(e) A longer sand box gives a greater stopping distance. [1] This increases stopping time or reduces the deceleration. [1] Since \(F=ma\), a smaller deceleration requires a smaller resultant force. [1]
(f) Increase the cart's mass. Since \(p=mv\), this increases momentum at the same speed. [1]
(g) As the cart stops, its kinetic energy is transferred mainly to thermal energy in the sand and cart. [1] Some energy may also be transferred as sound. [1]
Answer Details
(a) Momentum is \(p=mv\):
\[p=500\times4.0=2000\text{ kg m/s}\]
The initial momentum is 2000 kg m/s. [2]
(b) Use \(E_k=\frac{1}{2}mv^2\):
\[E_k=\frac{1}{2}\times500\times4.0^2=4000\text{ J}\]
The initial kinetic energy is 4000 J. [3]
(c) Use \(v^2=u^2+2as\):
\[0=4.0^2+2\times a\times1.6\]
\[a=-5.0\text{ m/s}^2\]
The deceleration has magnitude 5.0 m/s². [3]
(d) \[F=ma=500\times5.0=2500\text{ N}\]
The retarding force is 2500 N. [2]
(e) A longer sand box gives a greater stopping distance. [1] This increases stopping time or reduces the deceleration. [1] Since \(F=ma\), a smaller deceleration requires a smaller resultant force. [1]
(f) Increase the cart's mass. Since \(p=mv\), this increases momentum at the same speed. [1]
(g) As the cart stops, its kinetic energy is transferred mainly to thermal energy in the sand and cart. [1] Some energy may also be transferred as sound. [1]
Question 6 Report
Fig. 1 shows a long-period comet on an elliptical path through the Solar System. The arrow near the comet shows the direction of its motion. The tail is drawn pointing away from the Sun because particles and gas are pushed outward by solar radiation and the solar wind. At its closest point, the comet passes inside the orbit of Mars. Its orbit takes several thousand years, so it is seen from the Earth only rarely.
(a) State the name of the path followed by the comet. [1]
(b) State whether the tail points towards or away from the Sun. [1]
(c) Explain why the tail does not always trail behind the comet's direction of motion. [2]
(d) Describe how the speed of the comet changes as it travels from the farthest part of its orbit towards the Sun. [2]
(e) Explain why the gravitational force on the comet changes during its orbit. [2]
(f) Calculate the time in seconds for an orbit period of 3200 years. Use 365 days per year. [3]
(g) Calculate the mean orbital speed if the comet travels 1.20 × 1013 km in this time. [2]
(h) State why a comet can return after thousands of years instead of escaping permanently from the Solar System. [1]
(a) The comet follows an elliptical orbit, or ellipse. [1]
(b) Its tail points away from the Sun. [1]
(c) The tail direction is set by radiation pressure and the solar wind from the Sun. This is independent of the comet's direction of motion, so the tail does not necessarily trail behind it. [2]
(d) The comet's speed increases as it moves towards the Sun, and it is fastest near the Sun. [2]
(e) The separation between the comet and Sun changes during the elliptical orbit. A smaller separation gives a stronger gravitational force. [2]
(f) \[t=3200\times365\times24\times3600=1.01\times10^{11}\text{ s}\]
[3]
(g) \[v=\frac{1.20\times10^{13}\text{ km}}{1.01\times10^{11}\text{ s}}=119\text{ km/s}\approx1.2\times10^2\text{ km/s}\]
[2]
(h) The Sun gravitationally attracts the comet and keeps it in a bound orbit, allowing it to return. [1]
Answer Details
(a) The comet follows an elliptical orbit, or ellipse. [1]
(b) Its tail points away from the Sun. [1]
(c) The tail direction is set by radiation pressure and the solar wind from the Sun. This is independent of the comet's direction of motion, so the tail does not necessarily trail behind it. [2]
(d) The comet's speed increases as it moves towards the Sun, and it is fastest near the Sun. [2]
(e) The separation between the comet and Sun changes during the elliptical orbit. A smaller separation gives a stronger gravitational force. [2]
(f) \[t=3200\times365\times24\times3600=1.01\times10^{11}\text{ s}\]
[3]
(g) \[v=\frac{1.20\times10^{13}\text{ km}}{1.01\times10^{11}\text{ s}}=119\text{ km/s}\approx1.2\times10^2\text{ km/s}\]
[2]
(h) The Sun gravitationally attracts the comet and keeps it in a bound orbit, allowing it to return. [1]
Question 7 Report
Fig. 1 shows part of the circuit inside a plug fitted to a 2.3 kW kettle. The kettle is connected to a 230 V mains supply. The wires in the flex are insulated and the plug has a fuse in the live wire.
(a) State the function of the fuse in this electric circuit. [2]
(b) Explain why the fuse is placed in the live wire rather than the neutral wire. [2]
(c) Calculate the current taken by the kettle when it is operating normally. [3]
(d) Which fuse rating is most suitable for this kettle: 3 A, 5 A or 13 A? [2]
(a) A fuse melts when the current becomes too large, breaking the circuit and stopping current to the kettle. [2]
(b) If a fuse were in the neutral wire, a fault could leave the appliance connected to the live supply. A fuse in the live wire disconnects the live supply when it melts, making the appliance safer. [2]
(c) Convert \(2.3\text{ kW}\) to \(2300\text{ W}\), then use \(I=P/V\):
\[I=\frac{2300}{230}=10\text{ A}\]
The normal current is \(10\text{ A}\). [3]
(d) A 13 A fuse is most suitable. It is the smallest available rating above the normal operating current of 10 A, so it should not melt in normal use but protects against excessive current. [2]
Answer Details
(a) A fuse melts when the current becomes too large, breaking the circuit and stopping current to the kettle. [2]
(b) If a fuse were in the neutral wire, a fault could leave the appliance connected to the live supply. A fuse in the live wire disconnects the live supply when it melts, making the appliance safer. [2]
(c) Convert \(2.3\text{ kW}\) to \(2300\text{ W}\), then use \(I=P/V\):
\[I=\frac{2300}{230}=10\text{ A}\]
The normal current is \(10\text{ A}\). [3]
(d) A 13 A fuse is most suitable. It is the smallest available rating above the normal operating current of 10 A, so it should not melt in normal use but protects against excessive current. [2]
Question 8 Report
Fig. 1 shows water waves in a ripple tank moving towards a narrow gap in a barrier. A student uses a stroboscope to see the wave pattern clearly. The waves have frequency 5.0 Hz and wavelength 3.0 cm before reaching the gap.
(a) State the type of wave made on the water surface. [1]
(b) Describe how the student could measure the wavelength from the diagram or ripple tank pattern. [2]
(c) Explain why the waves spread out more if the width of the gap is reduced. [2]
(d) Calculate the speed of the water waves. [1]
(a) Water-surface waves are transverse waves. [1]
(b) Measure the distance across several complete gaps between equivalent wavefronts, then divide this total distance by the number of wavelengths measured. Measuring several wavelengths reduces percentage uncertainty. [2]
(c) A narrower gap has a width closer to the wavelength. Diffraction is greater when the gap size is comparable with the wavelength, so the waves spread out more. [2]
(d) Convert \(3.0\text{ cm}\) to \(0.030\text{ m}\): \[v=f\lambda=5.0\times0.030=0.15\text{ m/s}\] [1]
Answer Details
(a) Water-surface waves are transverse waves. [1]
(b) Measure the distance across several complete gaps between equivalent wavefronts, then divide this total distance by the number of wavelengths measured. Measuring several wavelengths reduces percentage uncertainty. [2]
(c) A narrower gap has a width closer to the wavelength. Diffraction is greater when the gap size is comparable with the wavelength, so the waves spread out more. [2]
(d) Convert \(3.0\text{ cm}\) to \(0.030\text{ m}\): \[v=f\lambda=5.0\times0.030=0.15\text{ m/s}\] [1]
Question 9 Report
The diagram shows an iceberg floating in seawater beside a research boat. The ice has density 920 kg/m3. Seawater has density 1030 kg/m3. Most of the iceberg is below the water surface.
(a) State why the iceberg floats. [2]
(b) Explain why only part of the iceberg is above the water surface. [3]
(c) Calculate the mass of an iceberg with volume 250 m3. [2]
(d) State whether the iceberg would float higher or lower in fresh water. [1]
(e) Explain your answer to part (d). [3]
(f) Give one danger caused by the part of the iceberg below water. [2]
(a) Ice has lower density than seawater. [1] It floats when the upthrust from the seawater balances its weight. [1]
(b) The iceberg displaces seawater. [1] As more of it is submerged, the volume of displaced water, and therefore the upthrust, increases. [1] It settles when upthrust equals its weight, so only enough ice is submerged to provide that upthrust. [1]
(c) \[m=\rho V=920\text{ kg/m}^3\times250\text{ m}^3=230\,000\text{ kg}\]
Correct formula [1]; mass = 230 000 kg. [1]
(d) The iceberg would float lower, with more of its volume submerged, in fresh water. [1]
(e) Fresh water has lower density than seawater. [1] Each cubic metre of fresh water displaced provides a smaller upthrust. [1] Therefore more of the iceberg must be submerged for the upthrust to balance its unchanged weight. [1]
(f) The underwater part can be difficult for boats to see [1] and may damage or sink a vessel in a collision. [1]
Answer Details
(a) Ice has lower density than seawater. [1] It floats when the upthrust from the seawater balances its weight. [1]
(b) The iceberg displaces seawater. [1] As more of it is submerged, the volume of displaced water, and therefore the upthrust, increases. [1] It settles when upthrust equals its weight, so only enough ice is submerged to provide that upthrust. [1]
(c) \[m=\rho V=920\text{ kg/m}^3\times250\text{ m}^3=230\,000\text{ kg}\]
Correct formula [1]; mass = 230 000 kg. [1]
(d) The iceberg would float lower, with more of its volume submerged, in fresh water. [1]
(e) Fresh water has lower density than seawater. [1] Each cubic metre of fresh water displaced provides a smaller upthrust. [1] Therefore more of the iceberg must be submerged for the upthrust to balance its unchanged weight. [1]
(f) The underwater part can be difficult for boats to see [1] and may damage or sink a vessel in a collision. [1]
Question 10 Report
Fig. 1 shows a student releasing a steel ball from a platform above a tray of sand. The ball has a mass of 0.50 kg and is released from rest. As it falls, its gravitational potential energy changes. The student notices that the sand becomes slightly warmer after several drops.
(a) State the main energy store of the ball when it is held on the platform. [1]
(b) Describe the energy transfer that occurs while the ball is falling. [2]
(c) State one reason why not all of the energy transferred becomes kinetic energy of the ball. [1]
(a) The ball has a gravitational potential energy store when held above the ground. Its energy is due to its position in Earth’s gravitational field. [1]
(b) As the ball falls, energy is transferred from its gravitational potential energy store to its kinetic energy store. The ball speeds up, so its kinetic energy increases. [2]
(c) Not all of this transfer becomes kinetic energy because air resistance transfers some energy by heating to the thermal energy store of the air. [1]
Exam point: Energy is conserved, but it may be dissipated into thermal stores rather than all becoming useful kinetic energy.
Answer Details
(a) The ball has a gravitational potential energy store when held above the ground. Its energy is due to its position in Earth’s gravitational field. [1]
(b) As the ball falls, energy is transferred from its gravitational potential energy store to its kinetic energy store. The ball speeds up, so its kinetic energy increases. [2]
(c) Not all of this transfer becomes kinetic energy because air resistance transfers some energy by heating to the thermal energy store of the air. [1]
Exam point: Energy is conserved, but it may be dissipated into thermal stores rather than all becoming useful kinetic energy.
Question 11 Report
A regional water service checks the pressure in pipes supplied by an elevated storage tank. Fig. 1 is a simplified diagram of the system. Gauge L is beside a workshop near the valley floor. Gauge H is at a clinic on higher ground. The water surface in the tank is 36 m above gauge L and 14 m above gauge H. Take the density of water as 1000 kg/m3 and take g as 10 N/kg. Ignore atmospheric pressure.
(a) State which gauge, L or H, has the greater water pressure. [1]
(b) Calculate the pressure caused by the water at gauge L. [3]
(c) Explain why the pressure at gauge H is less than the pressure at gauge L. [2]
(d) Calculate the difference in water pressure between the two gauges. [3]
(a) Gauge L has the greater water pressure because it has a greater depth of water above it. [1]
(b) \[p=h\rho g=36\text{ m}\times1000\text{ kg/m}^3\times10\text{ N/kg}\] [1]
\[p=360\,000\text{ Pa}=360\text{ kPa}\]
The pressure at L is 360 000 Pa or 360 kPa. [2]
(c) Gauge H has a smaller depth, or shorter vertical column, of water above it. [1] The water above a unit area therefore has less weight and exerts less force per unit area, so H has lower pressure. [1]
(d) The depth difference is:
\[36-14=22\text{ m}\] [1]
\[\Delta p=22\text{ m}\times1000\text{ kg/m}^3\times10\text{ N/kg}\] [1]
\[\Delta p=220\,000\text{ Pa}=220\text{ kPa}\]
The pressure difference is 220 000 Pa or 220 kPa. [1]
Answer Details
(a) Gauge L has the greater water pressure because it has a greater depth of water above it. [1]
(b) \[p=h\rho g=36\text{ m}\times1000\text{ kg/m}^3\times10\text{ N/kg}\] [1]
\[p=360\,000\text{ Pa}=360\text{ kPa}\]
The pressure at L is 360 000 Pa or 360 kPa. [2]
(c) Gauge H has a smaller depth, or shorter vertical column, of water above it. [1] The water above a unit area therefore has less weight and exerts less force per unit area, so H has lower pressure. [1]
(d) The depth difference is:
\[36-14=22\text{ m}\] [1]
\[\Delta p=22\text{ m}\times1000\text{ kg/m}^3\times10\text{ N/kg}\] [1]
\[\Delta p=220\,000\text{ Pa}=220\text{ kPa}\]
The pressure difference is 220 000 Pa or 220 kPa. [1]
