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Question 1 Report
Fig. 1 shows a safety test of a bicycle helmet on a dummy head. The head is dropped onto a pad and comes to rest in a short time. A high-speed camera provides the velocity before impact and the stopping time. The helmet increases the time taken for the head to stop. The mass of the dummy head is kept the same in every test.
(a) What happens to the velocity of the head during the impact? [1]
(b) Which quantity is reduced when the helmet makes the stopping time longer: force on the head or mass of the head? [1]
(c) What happens to the acceleration magnitude if the same change in velocity occurs in a longer time? [1]
(d) Sketch an arrow to show the direction of the contact force from the pad on the head. [1]
The motion of a marble is filmed as it rolls down a shallow ramp and onto a horizontal track. Fig. 1 shows positions of the marble in photographs taken every 0.20 s. The spaces between the first positions increase. The spaces on the horizontal section are nearly equal because friction is small. A ruler beside the track is used to measure distance.
(a) What does the increasing separation of the first three marble positions show? [1]
(b) Which section shows approximately constant velocity? [1]
(c) Calculate the average speed on the horizontal track if the marble moves 0.50 m in 1.0 s. [2]
Helmet test
(a) The head's velocity decreases to zero during impact. [1]
(b) A longer stopping time reduces the force on the head; the mass is unchanged. [1]
(c) For the same change in velocity, \(a=\Delta v/\Delta t\), so acceleration magnitude decreases when time increases. [1]
(d) The pad exerts an upward contact force on the head.
[1]
Marble motion
(a) Increasing separation in equal time intervals shows that the marble is accelerating, so its speed is increasing. [1]
(b) The horizontal section shows approximately constant velocity because successive positions are nearly equally spaced. [1]
(c) \(v=d/t=0.50/1.0=0.50\text{ m/s}\). [2]
Helmet test
(a) The head's velocity decreases to zero during impact. [1]
(b) A longer stopping time reduces the force on the head; the mass is unchanged. [1]
(c) For the same change in velocity, \(a=\Delta v/\Delta t\), so acceleration magnitude decreases when time increases. [1]
(d) The pad exerts an upward contact force on the head.
[1]
Marble motion
(a) Increasing separation in equal time intervals shows that the marble is accelerating, so its speed is increasing. [1]
(b) The horizontal section shows approximately constant velocity because successive positions are nearly equally spaced. [1]
(c) \(v=d/t=0.50/1.0=0.50\text{ m/s}\). [2]
Question 2 Report
Fig. 1 shows a simplified d.c. motor in an electric toothbrush. The toothbrush battery is connected to brushes, a split-ring commutator and a coil. The labelled arrows show the force on each side of the coil at one instant. These forces turn the axle, which moves the brush head. The casing is sealed to keep water away from the electrical parts. A designer is considering using a stronger permanent magnet, without changing the battery or the resistance of the coil.
(a) What produces the turning effect on the coil? [1]
(b) When does the split-ring commutator change the connections to the coil? [2]
(c) What effect would a stronger magnet have on the force and the motor speed? [2]
Fig. 1 shows the coil and brushes inside a battery-powered hand fan. Two carbon brushes press against the split-ring commutator. This maintains an electrical connection while the coil rotates. The permanent magnets provide a fixed magnetic field. The fan blades are attached to the axle, so a steady turning effect is needed. During a repair, one brush becomes worn and no longer touches the commutator reliably. The fan then runs in short bursts instead of at a constant speed.
(a) What is the function of the carbon brushes? [1]
(b) Which part reverses the current supplied to the rotating coil? [2]
(c) When the fan is operating normally, what energy transfer makes the blades move? [2]
Electric toothbrush motor
(a) Opposite motor-effect forces on opposite sides of the coil produce the turning effect [1].
(b) The split-ring commutator changes the connections every half-turn [1], reversing current so the turning effect remains in the same direction [1]. [2 marks]
(c) A stronger magnet increases the force [1], giving a greater turning effect so the motor can rotate faster [1]. [2 marks]
Hand fan
(a) Carbon brushes provide electrical contact with the rotating commutator [1].
(b) The split-ring commutator reverses the current [1] every half-turn [1]. [2 marks]
(c) Electrical energy is transferred [1] to kinetic or mechanical energy of the rotating blades [1]. [2 marks]
Electric toothbrush motor
(a) Opposite motor-effect forces on opposite sides of the coil produce the turning effect [1].
(b) The split-ring commutator changes the connections every half-turn [1], reversing current so the turning effect remains in the same direction [1]. [2 marks]
(c) A stronger magnet increases the force [1], giving a greater turning effect so the motor can rotate faster [1]. [2 marks]
Hand fan
(a) Carbon brushes provide electrical contact with the rotating commutator [1].
(b) The split-ring commutator reverses the current [1] every half-turn [1]. [2 marks]
(c) Electrical energy is transferred [1] to kinetic or mechanical energy of the rotating blades [1]. [2 marks]
Question 3 Report
Fig. 1 shows a classroom card-sort activity. Each card names one stage in the life cycle of a star with a mass like the Sun. A student has put the cards in an incorrect order. The cards describe the object after a cloud of gas and dust has collapsed. The teacher asks students to sketch a corrected sequence in their answer box. The final object is very small and hot at first, but no longer has fusion reactions taking place. It will cool over an extremely long time.
(a) Sketch the correct order of the four cards, starting with protostar. [2]
(b) What happens to a white dwarf over a very long time? [2]
Fig. 1 shows a simplified graph from an outreach article. It compares the mass of a star at birth with its possible final remnant. The boundaries are approximate because other factors can affect a real star. A student uses the graph to predict the future of a star with mass 12 solar masses. The star is already a red supergiant. It will not form a planetary nebula because its mass is too great. The graph shows that it must first release a very large amount of energy in an explosion.
(a) Which remnant is most likely for the 12 solar-mass star? [2]
(b) When does the supernova occur: before or after the red supergiant stage? [2]
Sun-like star sequence
(a) The correct order is:
protostar → main sequence → red giant → white dwarf. [2]
(b) A white dwarf cools over a very long time and eventually becomes a black dwarf. [2]
Massive-star remnant
(a) A star with a birth mass of 12 solar masses is most likely to form a neutron star. [2]
(b) The supernova occurs after the red supergiant stage. [2]
Sun-like star sequence
(a) The correct order is:
protostar → main sequence → red giant → white dwarf. [2]
(b) A white dwarf cools over a very long time and eventually becomes a black dwarf. [2]
Massive-star remnant
(a) A star with a birth mass of 12 solar masses is most likely to form a neutron star. [2]
(b) The supernova occurs after the red supergiant stage. [2]
Question 4 Report
The diagram shows a student using a screwdriver to lift the lid from a paint tin. The rim of the tin is the pivot. The student pushes down on the handle with 60 N. The distance from the pivot to the hand is 0.18 m. The screwdriver blade touches the lid 0.015 m from the pivot. The lid is stuck because dried paint forms a seal around its edge.
The forces are perpendicular to the screwdriver.
(a) What is the name of this simple machine? [1]
(b) Calculate the moment from the student's hand. [2]
(c) Calculate the upward force on the lid. [2]
Fig. 1 shows a laboratory test of a metre rule used as a lever. A 2.0 N load hangs from the 20 cm mark. The rule is supported at the 50 cm mark. A spring balance pulls down at the 80 cm mark until the rule is horizontal. The student records the force shown by the spring balance. The weight of the rule is ignored in this simple model.
The distances from the pivot are measured along the rule.
(a) What is the distance of the load from the pivot? [1]
(b) Calculate the moment of the load about the pivot. [1]
(c) Calculate the spring-balance force needed for balance. [2]
First lever system
Metre-rule lever system
Exam reminder: use perpendicular distance in metres, and equate clockwise and anticlockwise moments when a lever is balanced.
First lever system
Metre-rule lever system
Exam reminder: use perpendicular distance in metres, and equate clockwise and anticlockwise moments when a lever is balanced.
Question 5 Report
Fig. 1 shows a diagram from a science centre display about the future of the Sun. The boxes are not drawn to scale. The Sun is now a main-sequence star and fuses hydrogen in its core. In about 5000 million years, the supply of hydrogen in the core will decrease. The display explains that the outer layers will then grow much larger. Earth may be inside these outer layers, so the energy received at Earth would change greatly. The final stages shown apply only to a star that is not very massive.
(a) When the Sun leaves the main sequence, which named stage follows? [2]
(b) What is the name of the gas shell stage in the diagram? [2]
Fig. 1 shows the data path of a space telescope measuring radiation from a newly discovered pulsar. A pulsar is a neutron star rotating very quickly. The telescope records a pulse of radio radiation each time the narrow beam sweeps across Earth. The graph has evenly spaced peaks, showing that the source rotates at a regular rate. The original star was much more massive than the Sun and ended its life in a supernova. The graph is used to calculate the time for one rotation.
(a) What is the time between two successive pulses? [2]
(b) Which stellar remnant is the pulsar? [2]
Future of the Sun
(a) After the main-sequence stage, the Sun becomes a red giant. [2]
(b) The gas shell stage is a planetary nebula. [2]
Pulsar data
(a) The time between successive peaks is 0.8 s. This is the pulsar’s rotation period, because one pulse is detected each time its beam sweeps past Earth. [2]
(b) A pulsar is a neutron star. [2]
Future of the Sun
(a) After the main-sequence stage, the Sun becomes a red giant. [2]
(b) The gas shell stage is a planetary nebula. [2]
Pulsar data
(a) The time between successive peaks is 0.8 s. This is the pulsar’s rotation period, because one pulse is detected each time its beam sweeps past Earth. [2]
(b) A pulsar is a neutron star. [2]
Question 6 Report
Fig. 1 shows a uniform metal beam being lifted by two cables at a theatre. The beam has a weight of 800 N acting at its centre. Cable A is attached 1.0 m to the left of the centre and cable B is attached 2.0 m to the right of the centre. The stage manager needs the beam to remain horizontal while a light is fitted. The upward forces in the cables produce moments about the beam's centre.
The beam is in equilibrium.
(a) What is the total upward force supplied by cables A and B? [1]
(b) Calculate the force in cable A. [3]
(c) Calculate the force in cable B. [2]
For equilibrium, upward force equals downward force and clockwise moments equal anticlockwise moments.
(a) The total upward force is 800 N, balancing the beam's weight. [1]
(b) Take moments about cable B. Cable A is \(3.0\text{ m}\) from B and the weight is \(2.0\text{ m}\) from B:
\[F_A\times3.0=800\times2.0\]
\[F_A=\frac{1600}{3.0}=533\text{ N}\]
Force in A is 533 N (about 530 N). [3]
(c) \[F_B=800-533=267\text{ N}\] Force in B is 267 N (about 270 N). [2]
For equilibrium, upward force equals downward force and clockwise moments equal anticlockwise moments.
(a) The total upward force is 800 N, balancing the beam's weight. [1]
(b) Take moments about cable B. Cable A is \(3.0\text{ m}\) from B and the weight is \(2.0\text{ m}\) from B:
\[F_A\times3.0=800\times2.0\]
\[F_A=\frac{1600}{3.0}=533\text{ N}\]
Force in A is 533 N (about 530 N). [3]
(c) \[F_B=800-533=267\text{ N}\] Force in B is 267 N (about 270 N). [2]
Question 7 Report
The table below gives data from a laboratory check of a radon monitor in the basement of a library. Radon is a radioactive gas that can enter buildings from some rocks and soil. The monitor records alpha particles from radon decay products. Each result is the mean of several readings made for the same time. The facilities manager compares the values before and after a ventilation fan is installed. Windows are kept shut during both sets of measurements so that the test is controlled.
| Condition | Mean count rate / counts per minute |
|---|---|
| Fan off | 94 |
| Fan on | 39 |
| Outdoor background | 22 |
(a) What is the corrected count rate with the fan on? [1]
(b) Which condition gives the greatest radon level? [1]
(c) When the fan is used, explain why the count rate decreases but does not become zero. [2]
A geologist takes a rock sample from a volcanic ash layer beneath an old settlement. Fig. 1 is a simplified graph from the laboratory report. It shows the percentage of potassium-40 remaining in minerals in the sample. The geologist compares the graph with the age of pottery found above the ash. The graph uses percentage rather than mass because different samples began with different masses of potassium-40. The radioactive decay process is random for one nucleus, although the trend for a large number of nuclei is predictable.
(a) What percentage of potassium-40 remains after 1.25 billion years? [1]
(b) What is the half-life shown by Fig. 1? [1]
(c) Sketch the shape of a graph of count rate against time for this sample. Include labelled axes. [3]
Radon monitor
(a) Corrected count rate \(=39-22=\mathbf{17}\) counts per minute. Background must be removed because it is not caused by radon in the basement. [1]
(b) Fan off gives the greatest radon level because it has the greatest measured count rate. [1]
(c) Ventilation removes or dilutes some radon, so fewer alpha particles are detected. The count does not become zero because background radiation remains and/or some radon remains in the building. [2]
Potassium-40 dating
(a) At 1.25 billion years, 50% of the potassium-40 remains. [1]
(b) A half-life is the time for the amount remaining to fall to 50%. The graph gives a half-life of 1.25 billion years. [1]
(c) Count rate is proportional to activity, so it has the same exponential decay shape: high initially, decreasing, becoming less steep, and approaching zero. [3]
Radon monitor
(a) Corrected count rate \(=39-22=\mathbf{17}\) counts per minute. Background must be removed because it is not caused by radon in the basement. [1]
(b) Fan off gives the greatest radon level because it has the greatest measured count rate. [1]
(c) Ventilation removes or dilutes some radon, so fewer alpha particles are detected. The count does not become zero because background radiation remains and/or some radon remains in the building. [2]
Potassium-40 dating
(a) At 1.25 billion years, 50% of the potassium-40 remains. [1]
(b) A half-life is the time for the amount remaining to fall to 50%. The graph gives a half-life of 1.25 billion years. [1]
(c) Count rate is proportional to activity, so it has the same exponential decay shape: high initially, decreasing, becoming less steep, and approaching zero. [3]
Question 8 Report
During a science-club demonstration, students send a narrow light beam through three transparent blocks and towards air. They rotate each block until no ray can be seen leaving its upper face. At this point the light is reflected inside the block, so the angle inside the material is its critical angle. Table 1 records the results. The students use the same laser, a clean dry surface and the same angle scale for each block. A fibre-optic cable uses repeated internal reflection to keep light energy inside its glass core. The table can help the students decide which material would be suitable for a cable that needs light to remain trapped easily.
| Material | Critical angle / degrees |
|---|---|
| Acrylic | 42 |
| Dense glass | 38 |
| Flint glass | 35 |
Table 1
(a) Which material has the greatest refractive index? [1]
(b) Calculate the refractive index of flint glass. Use n = 1 ÷ sin c, where c is the critical angle. [2]
(c) Explain why a ray in flint glass at an angle of incidence of 36 degrees can undergo total internal reflection at a flint glass-air boundary. [1]
Fig. 1 shows a laboratory setup used to compare the apparent depth of a coin in a beaker of water. The coin is fixed to the base. A pin is moved vertically beside the beaker until its tip appears to be at the same position as the image of the coin. The observer keeps one eye at the marked position and does not move the beaker. The glass wall is thin and is ignored in this model. The real depth is the distance from the water surface to the coin.
(a) What is meant by the apparent depth of the coin? [1]
(b) Which is greater, the real depth or the apparent depth? [1]
(c) What happens to the light ray as it leaves the water? [1]
(d) Sketch a dashed line backwards from the ray in air to show where the observer sees the coin. [1]
Critical angle investigation
(a) Flint glass has the greatest refractive index. For light entering air, a smaller critical angle corresponds to a greater refractive index; flint glass has the smallest critical angle, \(35^\circ\). [1]
(b) \[n=\frac{1}{\sin c}=\frac{1}{\sin 35^\circ}=1.74\]
The refractive index is 1.74 (allow about 1.7). [2]
(c) \(36^\circ\) is greater than flint glass’s critical angle of \(35^\circ\), so the ray can undergo total internal reflection at a flint glass-air boundary. [1]
Apparent depth
(a) The apparent depth is the depth at which the coin, or its image, appears to be. [1]
(b) The real depth is greater than the apparent depth. [1]
(c) As the ray leaves water and enters air, it refracts away from the normal. [1]
(d) The dashed line is the backward extension of the emergent ray. It locates a virtual image above the real coin. [1]
Critical angle investigation
(a) Flint glass has the greatest refractive index. For light entering air, a smaller critical angle corresponds to a greater refractive index; flint glass has the smallest critical angle, \(35^\circ\). [1]
(b) \[n=\frac{1}{\sin c}=\frac{1}{\sin 35^\circ}=1.74\]
The refractive index is 1.74 (allow about 1.7). [2]
(c) \(36^\circ\) is greater than flint glass’s critical angle of \(35^\circ\), so the ray can undergo total internal reflection at a flint glass-air boundary. [1]
Apparent depth
(a) The apparent depth is the depth at which the coin, or its image, appears to be. [1]
(b) The real depth is greater than the apparent depth. [1]
(c) As the ray leaves water and enters air, it refracts away from the normal. [1]
(d) The dashed line is the backward extension of the emergent ray. It locates a virtual image above the real coin. [1]
Question 9 Report
Fig. 1 shows a cool box used by a field biologist to carry ice packs and small water samples. The box has a shiny inner lining, a thick foam wall and a tight lid. At the start, an ice pack is at 0 °C. Several hours later, some ice has melted but the temperature of the ice-water mixture is still 0 °C. The biologist wants to keep the samples cool without using an electrical current from a car battery. Energy enters the box from warmer surroundings.
(a) What energy transfer occurs while the ice melts? [2]
(b) Which feature of the box reduces transfer by radiation? [2]
(c) When the lid is opened often, what energy transfer becomes greater? [2]
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