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Question 1 Report
Fig. 1 shows a simplified sequence from a computer simulation of a red giant. The outer gas becomes less tightly held as the star expands. Eventually a glowing shell is left around the small central remnant. This shell is called a planetary nebula, although it has no connection with a planet. Students use the simulation to identify the path followed by a Sun-like star.
(a) Label the small object at the far right as a white dwarf. [1]
(b) What happens to the outer layers of the red giant? [1]
(c) Describe why the central object remains visible after the gas shell has formed. [2]
A telescope team photographs the remains of a star in a nearby galaxy. Fig. 1 shows a shell of hot gas expanding away from a compact object at its centre. The team detects radiation from the shell and uses the image in a news report. The original star had a much greater mass than the Sun.
(a) What event produced the expanding shell? [1]
(b) Label X by writing the likely compact object for a very massive original star. [1]
(c) Describe what happens to the outer layers during the event in part (a). [2]
(d) Give one reason why this event can briefly be seen from a very large distance. [1]
Planetary nebula sequence
Remains of a very massive star
Planetary nebula sequence
Remains of a very massive star
Question 2 Report
Table 1 gives results from a workshop test of two magnetic catches used to hold cupboard doors closed. A force meter pulls each catch directly away from a steel plate. For each gap, students record the greatest force before the catch separates. The same steel plate, force meter and direction of pull are used each time. A small cardboard spacer controls the gap. The students also draw the apparatus in Fig. 1 so another group can repeat the method safely.
| Gap between magnet and plate / mm | Separation force / N |
|---|---|
| 0 | 18 |
| 1 | 12 |
| 2 | 7 |
| 4 | 2 |
(a) Describe the relationship between gap and separation force. [2]
(b) Calculate the decrease in separation force when the gap changes from 0 mm to 2 mm. [2]
(c) Give one reason for repeating each measurement. [1]
(d) What variable is changed in this test? [2]
(a) As the gap between the magnet and steel plate increases, the separation force decreases. [2]
(b)
\[18\text{ N}-7\text{ N}=11\text{ N}\]
The separation force decreases by 11 N. [2]
(c) Repeating measurements helps identify anomalous results, calculate a mean, and improve reliability. [1]
(d) The variable changed is the distance, or gap, between the magnet and the steel plate. [2]
(a) As the gap between the magnet and steel plate increases, the separation force decreases. [2]
(b)
\[18\text{ N}-7\text{ N}=11\text{ N}\]
The separation force decreases by 11 N. [2]
(c) Repeating measurements helps identify anomalous results, calculate a mean, and improve reliability. [1]
(d) The variable changed is the distance, or gap, between the magnet and the steel plate. [2]
Question 3 Report
A science museum displays Fig. 1 to compare the apparent movement of stars during one night. The camera was fixed on a tripod and its shutter was left open. The curved trails are centred on a point close to the north celestial pole. Visitors ask why the stars appear to move although their positions change very little over one night.
(a) What motion of Earth causes the trails in Fig. 1? [1]
(b) When does Earth complete one rotation? [1]
(c) Describe the direction of Earth's rotation as seen from above the north pole. [1]
(d) Give two effects of Earth's rotation on people at one location. [2]
A group in the school observatory records the appearance of the Moon at sunset for several evenings. Fig. 1 shows three positions of the Moon in its orbit, viewed from above the north pole of Earth. Half of the Moon is always lit by the Sun. The students use the diagram to plan a public viewing session.
(a) Which position, A, B or C, gives a full Moon as seen from Earth? [1]
(b) What is the name of the phase at position A? [1]
(c) Describe how the Moon produces the light seen by the students. [1]
(d) Give two reasons why a Moon phase is not caused by Earth's shadow. [2]
Star trails
(a) The trails are caused by Earth’s rotation on its axis [1].
(b) Earth completes one rotation in 24 hours, or one day [1].
(c) Seen from above the north pole, Earth rotates anticlockwise [1].
(d) Two effects at one location are day and night [1] and the apparent daily movement of the Sun or stars [1]. Changing shadow direction is also accepted.
Moon phases
(a) A full Moon is at C [1]. At this position, Earth is between the Sun and Moon, so the illuminated side faces Earth.
(b) At A the phase is a new Moon [1].
(c) The Moon reflects light from the Sun [1]. It does not produce its own visible light.
(d) Earth’s shadow reaches the Moon only during a lunar eclipse [1], whereas phases occur every month. The changing angle from which Earth sees the Moon’s sunlit half causes the phases [1]. Half of the Moon is illuminated by the Sun at all times.
Star trails
(a) The trails are caused by Earth’s rotation on its axis [1].
(b) Earth completes one rotation in 24 hours, or one day [1].
(c) Seen from above the north pole, Earth rotates anticlockwise [1].
(d) Two effects at one location are day and night [1] and the apparent daily movement of the Sun or stars [1]. Changing shadow direction is also accepted.
Moon phases
(a) A full Moon is at C [1]. At this position, Earth is between the Sun and Moon, so the illuminated side faces Earth.
(b) At A the phase is a new Moon [1].
(c) The Moon reflects light from the Sun [1]. It does not produce its own visible light.
(d) Earth’s shadow reaches the Moon only during a lunar eclipse [1], whereas phases occur every month. The changing angle from which Earth sees the Moon’s sunlit half causes the phases [1]. Half of the Moon is illuminated by the Sun at all times.
Question 4 Report
The diagram in Fig. 1 is from a navigation lesson for a vessel travelling near Jupiter. The planet has a strong gravitational field. The arrows show two possible paths for an unmanned probe. Path A passes close to Jupiter and path B remains far away. The navigation computer uses a gravity assist to change the probe's velocity without using much fuel.
(a) Which path experiences the stronger gravitational force from Jupiter? [1]
(b) What happens to the direction of the probe's velocity on path A? [1]
(c) Describe why the gravitational field is stronger closer to Jupiter. [2]
(d) Give two advantages of changing a probe's path using gravity rather than rocket fuel. [2]
(a) Path A experiences the stronger gravitational force [1], because it passes nearer to Jupiter.
(b) The probe’s velocity changes direction: it turns towards Jupiter [1]. Gravity is a force, so it can change velocity even if the probe’s speed is not the focus of the question.
(c) Gravitational force decreases as distance increases [1]. On the closer path, the probe is less distant from Jupiter’s mass [1], so Jupiter’s gravitational pull is stronger.
(d) Using gravity can save fuel [1] and therefore reduce launch mass [1]. Other accepted advantages include allowing a higher speed or extending the mission duration.
(a) Path A experiences the stronger gravitational force [1], because it passes nearer to Jupiter.
(b) The probe’s velocity changes direction: it turns towards Jupiter [1]. Gravity is a force, so it can change velocity even if the probe’s speed is not the focus of the question.
(c) Gravitational force decreases as distance increases [1]. On the closer path, the probe is less distant from Jupiter’s mass [1], so Jupiter’s gravitational pull is stronger.
(d) Using gravity can save fuel [1] and therefore reduce launch mass [1]. Other accepted advantages include allowing a higher speed or extending the mission duration.
Question 5 Report
Fig. 1 shows a small electric cart travelling along a level warehouse floor. Its motor produces a driving force of 500 N. At one speed, friction and air resistance total 500 N. The cart carries boxes of water and moves past a cooling fan. The operator then adds more boxes but keeps the cart at the same speed.
(a) What is the resultant force at the stated speed? [1]
(b) Describe the motion of the cart at this speed. [1]
(c) When extra boxes are added, which force increases: the weight or the air resistance? [1]
(d) Calculate the new resultant force if resistance becomes 560 N and the motor force stays at 500 N. [2]
(e) What happens to the cart's velocity in this case? [1]
This question tests the link between resultant force and motion. Equal forces in opposite directions give a zero resultant force.
Exam reminder: A force opposite to motion does not necessarily make an object move backwards. It first causes deceleration.
This question tests the link between resultant force and motion. Equal forces in opposite directions give a zero resultant force.
Exam reminder: A force opposite to motion does not necessarily make an object move backwards. It first causes deceleration.
Question 6 Report
This bicycle is fitted with two lamps so a rider can be seen at night. Fig. 1 shows the circuit planned by a technician. Each lamp is connected directly across the 6.0 V battery. The technician chooses this arrangement because a rear lamp may fail during rain, but the front lamp should still receive energy from the battery. A fuse is also placed close to the positive terminal in the cable that carries current to both lamps.
(a) Which type of circuit connection is shown for the lamps? [1]
(b) What happens to the front lamp if the rear lamp breaks? [1]
(c) Calculate the current from the battery when each lamp has a resistance of 12 ohms. [2]
(d) Give one reason for fitting the fuse near the battery. [1]
This circuit is used by a repair shop to compare two metal wires for a model railway. Fig. 1 shows wire X and wire Y connected one at a time to the same 6.0 V supply. Both wires have equal length and diameter. The ammeter reading for X is 0.50 A, while the reading for Y is 0.20 A. The shop needs the wire that gives a larger resistance for a speed-control circuit.
(a) Calculate the resistance of wire X. [2]
(b) Calculate the resistance of wire Y. [1]
(c) Which wire should the shop choose for the speed-control circuit? [1]
(d) Give one variable that must be controlled for this comparison to be fair. [1]
Bicycle lamps
(a) The lamps are connected in parallel. [1]
(b) If the rear lamp breaks, the front lamp stays on because it remains on its own complete branch and continues to have 6.0 V across it. [1]
(c) For each lamp:
\[I=\frac{V}{R}=\frac{6.0\ \mathrm{V}}{12\ \Omega}=0.50\ \mathrm{A}\]
There are two parallel branches, so:
\[I_\text{total}=0.50\ \mathrm{A}+0.50\ \mathrm{A}=1.0\ \mathrm{A}\]
The battery current is 1.0 A. [2]
(d) The fuse protects the cable and circuit if an excessive current flows, for example during a fault. [1]
Comparing metal wires
(a) \[R_X=\frac{V}{I}=\frac{6.0\ \mathrm{V}}{0.50\ \mathrm{A}}=12\ \Omega\]
Wire X has resistance 12 \(\Omega\). [2]
(b) \[R_Y=\frac{6.0\ \mathrm{V}}{0.20\ \mathrm{A}}=30\ \Omega\]
Wire Y has resistance 30 \(\Omega\). [1]
(c) Choose wire Y, because 30 \(\Omega\) is greater than 12 \(\Omega\). [1]
(d) Control the temperature of the wires. Length and diameter are also acceptable, because each can affect resistance. [1]
Bicycle lamps
(a) The lamps are connected in parallel. [1]
(b) If the rear lamp breaks, the front lamp stays on because it remains on its own complete branch and continues to have 6.0 V across it. [1]
(c) For each lamp:
\[I=\frac{V}{R}=\frac{6.0\ \mathrm{V}}{12\ \Omega}=0.50\ \mathrm{A}\]
There are two parallel branches, so:
\[I_\text{total}=0.50\ \mathrm{A}+0.50\ \mathrm{A}=1.0\ \mathrm{A}\]
The battery current is 1.0 A. [2]
(d) The fuse protects the cable and circuit if an excessive current flows, for example during a fault. [1]
Comparing metal wires
(a) \[R_X=\frac{V}{I}=\frac{6.0\ \mathrm{V}}{0.50\ \mathrm{A}}=12\ \Omega\]
Wire X has resistance 12 \(\Omega\). [2]
(b) \[R_Y=\frac{6.0\ \mathrm{V}}{0.20\ \mathrm{A}}=30\ \Omega\]
Wire Y has resistance 30 \(\Omega\). [1]
(c) Choose wire Y, because 30 \(\Omega\) is greater than 12 \(\Omega\). [1]
(d) Control the temperature of the wires. Length and diameter are also acceptable, because each can affect resistance. [1]
Question 7 Report
Fig. 1 shows a safety check being carried out in a theatre. A technician places a flat mirror on the floor so that a camera can view a cable tray above the stage. The cable tray is 1.6 m in front of the mirror. The camera is beside the tray, on the same side of the mirror. Light from the tray reflects from the mirror into the camera. The mirror does not produce sound, but the reflected light allows the technician to check the tray before electrical current is switched on.
(a) What is meant by the term virtual image? [2]
(b) Describe two properties of the image of the cable tray, other than that it is virtual. [2]
(c) Calculate the distance between the cable tray and its image. Give your answer in m. [2]
(d) Draw on Fig. 1 one further ray from the cable tray that reflects from the mirror and enters the camera. [2]
(e) Explain why the image cannot be formed on a screen placed behind the mirror. [2]
(a) A virtual image is formed at a position where light rays appear to come from. [1] The rays do not actually pass through or meet at that image position behind the mirror. [1]
(b) Two properties are that the image is upright [1] and the same size as the cable tray. [1] It is also laterally inverted and is the same distance behind the mirror as the object is in front.
(c) The cable tray is 1.6 m in front of the mirror, so its image is 1.6 m behind the mirror. [1]
\[1.6\text{ m}+1.6\text{ m}=3.2\text{ m}\]
The distance between the cable tray and its image is 3.2 m. [1]
(d) A further ray must travel from the cable tray to the mirror and then reflect into the camera. [1] It must obey \(i=r\). [1]
(e) The reflected rays diverge and do not actually meet behind the mirror. [1] Therefore no light falls on a screen at the apparent image position, so the image cannot be formed on that screen. [1]
(a) A virtual image is formed at a position where light rays appear to come from. [1] The rays do not actually pass through or meet at that image position behind the mirror. [1]
(b) Two properties are that the image is upright [1] and the same size as the cable tray. [1] It is also laterally inverted and is the same distance behind the mirror as the object is in front.
(c) The cable tray is 1.6 m in front of the mirror, so its image is 1.6 m behind the mirror. [1]
\[1.6\text{ m}+1.6\text{ m}=3.2\text{ m}\]
The distance between the cable tray and its image is 3.2 m. [1]
(d) A further ray must travel from the cable tray to the mirror and then reflect into the camera. [1] It must obey \(i=r\). [1]
(e) The reflected rays diverge and do not actually meet behind the mirror. [1] Therefore no light falls on a screen at the apparent image position, so the image cannot be formed on that screen. [1]
Question 8 Report
A technician calibrates a current balance used in a satellite laboratory. Fig. 1 shows a straight wire of length 0.080 m in a uniform magnetic field. The current is 3.0 A. The balance measures a magnetic force of 0.12 N on the wire.
(a) Calculate the magnetic flux density of the field.
Use the equation: force = magnetic flux density × current × length. [3]
(b) Which direction is the magnetic field in Fig. 1? [1]
(c) Give one change that would double the force on the wire. [1]
(d) Describe why there is no force if the wire is parallel to the magnetic field. [1]
The motor-effect equation applies when the wire is perpendicular to the magnetic field.
The motor-effect equation applies when the wire is perpendicular to the magnetic field.
Question 9 Report
Fig. 1 shows a solar garden light installed beside a footpath. In daylight, a solar cell transfers energy from radiation to electrical energy and charges a rechargeable battery. At night, a light sensor closes the circuit to an LED. The battery has a potential difference of 3.6 V. When the LED is operating, the current is 0.080 A. A resident wants the light to operate for as long as possible after a cloudy day.
(a) Which energy store increases when the battery is charged? [1]
(b) Describe the useful energy transfer made by the solar cell. [2]
(c) Calculate the power transferred to the LED. [2]
(d) Calculate the charge supplied by the battery in 3.0 hours. [3]
(e) Give two ways the resident could increase the operating time of the light. [2]
(a) Charging the battery increases its chemical energy store. [1]
(b) The solar cell transfers energy from solar radiation to electrical energy. This electrical energy charges the chemical energy store of the battery. [2]
(c)
\[P=VI=3.6\text{ V}\times0.080\text{ A}=0.288\text{ W}\]
The power transferred to the LED is 0.288 W, or 0.29 W. [2]
(d)
\[3.0\text{ hours}=3.0\times3600=10800\text{ s}\]
\[Q=It=0.080\text{ A}\times10800\text{ s}=864\text{ C}\]
The battery supplies 864 C of charge in 3.0 hours. [3]
(e) Two valid ways are to use a larger battery and use a lower-current LED. A more efficient LED or positioning the solar cell where it receives more radiation would also increase operating time. [2]
(a) Charging the battery increases its chemical energy store. [1]
(b) The solar cell transfers energy from solar radiation to electrical energy. This electrical energy charges the chemical energy store of the battery. [2]
(c)
\[P=VI=3.6\text{ V}\times0.080\text{ A}=0.288\text{ W}\]
The power transferred to the LED is 0.288 W, or 0.29 W. [2]
(d)
\[3.0\text{ hours}=3.0\times3600=10800\text{ s}\]
\[Q=It=0.080\text{ A}\times10800\text{ s}=864\text{ C}\]
The battery supplies 864 C of charge in 3.0 hours. [3]
(e) Two valid ways are to use a larger battery and use a lower-current LED. A more efficient LED or positioning the solar cell where it receives more radiation would also increase operating time. [2]
Question 10 Report
A research balloon samples high-altitude air where atoms can be ionised by radiation from space. Fig. 1 shows an oxygen atom before and after it gains two electrons. The charged oxygen ion can affect instruments that measure electrical current in the thin air. Each diagram gives the number of protons, and the shells show the electron arrangement. The balloon team needs to calculate the charge correctly before calibrating its detector.
(a) Calculate the charge on the oxygen ion. [1]
(b) What is the formula of this ion? [1]
(c) Which particles have moved into the oxygen atom? [1]
(d) Describe why the ion has a different charge from the atom. [2]
(e) Give the number of protons in the oxygen ion. [1]
(a) The oxygen ion has 8 protons but, after gaining two electrons, 10 electrons. It has two extra negative charges, so its charge is \(-2\). [1]
(b) The formula is \(\mathrm{O^{2-}}\). [1]
(c) The particles that moved into the atom are electrons. [1]
(d) The ion has gained two electrons. It therefore has two more electrons than protons, giving an overall negative charge. [2]
(e) Ion formation changes electrons, not the nucleus. The oxygen ion still has 8 protons. [1]
(a) The oxygen ion has 8 protons but, after gaining two electrons, 10 electrons. It has two extra negative charges, so its charge is \(-2\). [1]
(b) The formula is \(\mathrm{O^{2-}}\). [1]
(c) The particles that moved into the atom are electrons. [1]
(d) The ion has gained two electrons. It therefore has two more electrons than protons, giving an overall negative charge. [2]
(e) Ion formation changes electrons, not the nucleus. The oxygen ion still has 8 protons. [1]
Question 11 Report
Fig. 1 shows a simplified magnetic field surrounding a rapidly rotating star. A science team uses this model to predict the path of charged particles in the thin air-free region above the star. The arrows show the direction of the magnetic field. A proton enters the region at right angles to the field lines with a velocity of 2.4 × 106 m/s. At this point, the magnetic flux density is 0.30 T. The charge on a proton is 1.6 × 10−19 C.
(a) What is the name of the region around the star in which a magnetic force can act on the proton? [1]
(b) Describe the direction of the magnetic field at the position of the proton in Fig. 1. [2]
(c) Calculate the magnetic force on the proton. Use the equation:
force = magnetic flux density × charge × velocity
Give your answer in N. [3]
(d) Use ideas about forces to describe the proton’s path as it travels through a uniform magnetic field. [2]
A moving charged particle in a magnetic field experiences a force when it moves at right angles to the field. That force changes its direction, rather than simply increasing its speed.
A moving charged particle in a magnetic field experiences a force when it moves at right angles to the field. That force changes its direction, rather than simply increasing its speed.
Question 12 Report
During a theatre refit, a scenery platform is raised from below the stage by two vertical steel cables. Fig. 1 shows the platform at one point in its journey. The load, including lamps and a sound system, has a mass of 780 kg. It rises through 12 m at a constant velocity. The stage manager records the time as 15 s. Take gravitational field strength, g, as 10 N/kg. The cables are identical and share the load equally. Students are asked to use the figure to consider forces and energy in the lifting system.
(a) Calculate the weight of the platform and its load. [2]
(b) Calculate the tension in one cable. [2]
(c) Calculate the gain in gravitational potential energy of the load. [3]
(d) Calculate the useful power supplied in raising the load. [2]
(e) Explain why the upward force from both cables equals the weight while the platform travels at constant velocity. [2]
(a)
\[W=mg=780\text{ kg}\times10\text{ N/kg}=7800\text{ N}\]
The weight is 7800 N. [2]
(b) At constant velocity, total upward force equals weight, so the two cables together provide 7800 N. They share this equally:
\[\frac{7800\text{ N}}{2}=3900\text{ N}\]
The tension in one cable is 3900 N. [2]
(c)
\[\Delta E=mgh=780\text{ kg}\times10\text{ N/kg}\times12\text{ m}=93600\text{ J}\]
The gain in gravitational potential energy is 93 600 J. [3]
(d)
\[P=\frac{E}{t}=\frac{93600\text{ J}}{15\text{ s}}=6240\text{ W}\]
The useful power is 6240 W. [2]
(e) Constant velocity means zero acceleration. Therefore the resultant force is zero, so the total upward tension from both cables equals the downward weight. [2]
(a)
\[W=mg=780\text{ kg}\times10\text{ N/kg}=7800\text{ N}\]
The weight is 7800 N. [2]
(b) At constant velocity, total upward force equals weight, so the two cables together provide 7800 N. They share this equally:
\[\frac{7800\text{ N}}{2}=3900\text{ N}\]
The tension in one cable is 3900 N. [2]
(c)
\[\Delta E=mgh=780\text{ kg}\times10\text{ N/kg}\times12\text{ m}=93600\text{ J}\]
The gain in gravitational potential energy is 93 600 J. [3]
(d)
\[P=\frac{E}{t}=\frac{93600\text{ J}}{15\text{ s}}=6240\text{ W}\]
The useful power is 6240 W. [2]
(e) Constant velocity means zero acceleration. Therefore the resultant force is zero, so the total upward tension from both cables equals the downward weight. [2]
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