Physics 4PH1 | Paper 2 Mock 01 | Written…

Question 1 Report

A research group uses type Ia supernovae as standard candles. Each explosion is assumed to have a power output of 1.0 × 10³⁶ W at maximum brightness. Table 1 gives measurements from four host galaxies. The redshift has been found from spectral lines, and intensity is the power received per square metre at Earth.

supernova	redshift, z	peak intensity / W/m²
A	0.025	7.2 × 10^-15
B	0.040	2.8 × 10^-15
C	0.055	1.5 × 10^-15
D	0.070	9.2 × 10^-16

For these small redshifts, use v = zc, c = 3.0 × 10⁵ km/s and H₀ = 70 km s^-1 Mpc^-1. Also use I = P/(4πd²) and 1 Mpc = 3.09 × 10²² m.

(a) Calculate the recession speed of the host galaxy of supernova C. [2]
(b) Calculate the distance to the host galaxy of supernova C using Hubble’s law. [2]
(c) Calculate the distance to supernova B from its peak intensity. Give your answer in Mpc. [4]
(d) Describe the trend in Table 1 between redshift and the measured intensity. [2]
(e) Explain why a standard candle can be used to test the expansion of the universe. [3]

Answer

Answer Details

(a)

\[v=zc=0.055\times3.0\times10^5=16\,500\text{ km/s}\]

16 500 km/s. [2 marks]

(b) Use Hubble's law, \(d=v/H_0\):

\[d=\frac{16\,500}{70}=236\text{ Mpc}\]

236 Mpc, with 240 Mpc also acceptable. [2 marks]

(c) Rearrange the intensity equation:

\[d=\sqrt{\frac{P}{4\pi I}}\]

\[d=\sqrt{\frac{1.0\times10^{36}}{4\pi\times2.8\times10^{-15}}}=5.3\times10^{24}\text{ m}\]

Convert to megaparsecs:

\[d=\frac{5.3\times10^{24}}{3.09\times10^{22}}=172\text{ Mpc}\]

172 Mpc, with 170 Mpc also acceptable. [4 marks]

(d) As redshift increases from 0.025 to 0.070, the measured intensity decreases. Therefore more red-shifted supernovae appear dimmer. [2 marks]

(e) A standard candle has a known power output, or luminosity. Its measured intensity can therefore be used to calculate its distance. Comparing this distance with redshift or recession speed tests Hubble's relationship and hence the expansion of the universe. [3 marks]

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Cosmology

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Question 2 Report

A technician uses a rubber suction cup to hold a warning sign against a smooth glass panel. Fig. 1 shows a cross-section through the cup. Most of the air is pushed out when the cup is pressed onto the glass. The area inside the circular rim is 0.0060 m². Atmospheric pressure outside the cup is 100 000 Pa and the pressure of the trapped air is 82 000 Pa.

(a) Calculate the pressure difference across the suction cup. [2]
(b) State which side of the cup has the greater pressure. [1]
(c) Explain why the suction cup is held against the glass. [3]

Answer

Answer Details

(a) The pressure difference is:

\[100000\text{ Pa}-82000\text{ Pa}=18000\text{ Pa}\]

[2]

(b) The outside, atmospheric-air side of the cup has the greater pressure. [1]

(c) Atmospheric air exerts a greater pressure on the outside of the cup than the trapped air exerts inside it. This pressure difference creates a resultant force towards the glass, holding the cup in place. The smooth rim forms a seal that prevents air entering under the cup and removing the pressure difference. [3]

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Density And Pressure

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Question 3 Report

The graph in Fig. 1 shows measurements from a tracer used to find a small leak in a district-heating water pipe. A student measures the activity of a sample of the tracer at regular intervals. The measuring system records gamma radiation, which can pass through the pipe wall.

(a) State the activity of the sample at 12 h. [1]
(b) Calculate the half-life of this radioactive isotope. [2]
(c) Calculate the activity after 18 h as a fraction of its activity at 0 h. [2]
(d) Describe the change in the number of undecayed nuclei in the sample during the first 18 h. [2]
(e) Explain why the activity cannot be predicted exactly for one particular nucleus. [2]

Answer

Answer Details

(a) At 12 h, the graph gives an activity of 20 kBq. [1]

(b) The initial activity is 80 kBq. It falls to 40 kBq, half its initial value, at 6 h. Therefore the half-life is 6 h. [2]

(c) At 18 h the activity is 10 kBq:

\[\frac{10}{80}=\frac18\]

The activity is one eighth of its activity at 0 h. [2]

(d) The number of undecayed nuclei decreases. It halves every 6 h, so after 18 h, which is three half-lives, one eighth remains. [2]

(e) Decay of an individual nucleus is random. It is only possible to state the probability that it will decay during a particular time interval, not exactly when it will decay. [2]

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Radioactivity

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Question 4 Report

Fig. 1 shows a physics club demonstration using a thin aluminium bead suspended by an insulating thread. A negatively charged acrylic sheet is moved slowly towards the bead. At first the bead swings towards the sheet. It is then allowed to touch the sheet and swings away. The bead has a very small mass, so the electrostatic force causes a clear movement.

Fig. 1

(a) State the overall charge on the aluminium bead before the sheet is brought close to it. [1]
(b) Describe the movement of electrons in the bead while the negative sheet is close but has not touched it. [2]
(c) Explain why the initially neutral bead moves towards the sheet. [2]
(d) Explain why the bead moves away after touching the sheet. [2]

Answer

Answer Details

(a) Before the sheet approaches, the aluminium bead is neutral, with no overall charge. [1]

(b) Electrons in the bead are repelled by the negatively charged sheet. They move to the side of the bead furthest from the sheet. [2]

(c) The near side of the bead is left with a shortage of electrons, so it becomes positive. Opposite charges attract, producing a force that moves the bead towards the negative sheet. [2]

(d) When the bead touches the sheet, electrons transfer from the sheet onto the bead. The bead and sheet are then both negatively charged, so they repel and the bead moves away. [2]

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Electric Charge

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Question 5 Report

Fig. 1 shows a student making ice cream during a science festival. The student turns a handle about its pivot axis so that a metal container of ice cream mixture is surrounded by crushed ice and salt. The mixture changes from a liquid to a solid. The temperature of the mixture falls while some of the ice outside the container melts.

(a) State the change of state taking place in the ice surrounding the container. [1]
(b) Explain why the temperature of the ice cream mixture falls when the surrounding ice melts. [2]
(c) State the direction of energy transfer between the ice cream mixture and the melting ice. [1]

Answer

Answer Details

(a) The ice around the container changes state by melting. [1 mark]

(b) Melting ice needs energy, called latent heat. It takes this energy from the warmer ice cream mixture, so the mixture loses thermal energy and its temperature falls. [2 marks]

(c) Thermal energy is transferred from the ice cream mixture to the melting ice. [1 mark]

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Change Of State

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Question 6 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]

Answer

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]

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Change Of State

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Question 7 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]

Answer

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]

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Work And Power

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Question 8 Report

The diagram shows a velocity-time graph for a cyclist riding a short stage of a charity event. Positive velocity is in the direction of the cyclist's motion. The cyclist increases speed, rides steadily, then uses the brakes near a water station.

(a) State the maximum velocity of the cyclist. [1]
(b) Calculate the acceleration during the first 4 s. [2]
(c) Calculate the distance travelled during the 18 s shown on the graph. [3]
(d) Explain why the cyclist has a deceleration from 14 s to 18 s. [2]
(e) Which section of the graph shows constant velocity? [2]

Answer

Answer Details

(a) The maximum velocity is 6.0 m/s. [1]

(b) The gradient of a velocity-time graph is acceleration:

\[a=\frac{6.0-0}{4.0}=1.5\text{ m/s}^2\]

[2]

(c) Distance is the area under a velocity-time graph.

\[\text{first triangle}=\frac12\times4\times6=12\text{ m}\]

\[\text{rectangle}=10\times6=60\text{ m}\]

\[\text{final triangle}=\frac12\times4\times6=12\text{ m}\]

\[\text{total distance}=12+60+12=84\text{ m}\]

[3]

(d) Braking, and possibly other resistive forces, acts opposite to the cyclist's motion. This gives a resultant force opposite to the velocity, so the velocity decreases. [2]

(e) From 4 s to 14 s the cyclist has constant velocity. The line is horizontal, so the velocity is unchanged. [2]

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Movement And Position

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Physics 4PH1 | Paper 2 Mock 01 | Written Paper 2