Question 1 (Sub-topic – 1.2)
(a) Determine the distance between:
(i) the engineer’s home and her office
▶️Answer/Explanation
60 km
To determine the distance between the engineer’s home and her office, we look at the distance-time graph (Fig. 1.1). The graph shows that at time = 0 h, the distance from home is 0 km, and at time = 2 h, the distance from home is 60 km. Therefore, the distance between the engineer’s home and her office is 60 km.
(ii) the engineer’s office and her friend’s house
▶️Answer/Explanation
40 km
To determine the distance between the engineer’s office and her friend’s house, we look at the distance-time graph (Fig. 1.1). The graph shows that at time = 2 h, the distance from home is 60 km, and at time = 5 h, the distance from home is 100 km. Therefore, the distance between the engineer’s office and her friend’s house is 100 km – 60 km = 40 km.
(b) Determine the time taken to travel between:
(i) the engineer’s home and her office
▶️Answer/Explanation
2 h
To determine the time taken to travel between the engineer’s home and her office, we look at the distance-time graph (Fig. 1.1). The graph shows that the journey starts at time = 0 h and ends at time = 2 h. Therefore, the time taken is 2 h.
(ii) the engineer’s office and her friend’s house
▶️Answer/Explanation
3 h
To determine the time taken to travel between the engineer’s office and her friend’s house, we look at the distance-time graph (Fig. 1.1). The graph shows that the journey starts at time = 2 h and ends at time = 5 h. Therefore, the time taken is 5 h – 2 h = 3 h.
(c) Calculate the speed of the car between time = 7 h and time = 10 h.
▶️Answer/Explanation
13 km/h
To calculate the speed of the car between time = 7 h and time = 10 h, we use the formula:
\[ \text{speed} = \frac{\text{distance}}{\text{time}} \]
From the graph, the distance at time = 7 h is 60 km, and at time = 10 h, the distance is 100 km. Therefore, the distance traveled is 100 km – 60 km = 40 km. The time taken is 10 h – 7 h = 3 h. Thus, the speed is:
\[ \text{speed} = \frac{40 \text{ km}}{3 \text{ h}} = 13.\overline{3} \text{ km/h} \]
Rounding to the nearest whole number, the speed is approximately 13 km/h.
Question 2 (Sub-topic – 1.7.3)
Fig. 2.1 shows an engineer working with wind turbines.
(a)(i) Complete the sentence describing how electrical power is generated by energy in the wind.
The source of the wind energy is
(ii) When the blades turn, electrical power is generated in the
(b) Describe two advantages, apart from cost, of generating electrical power by using wind turbines compared with using a coal-fired power station.
▶️Answer/Explanation
2(a)(i) The source of the wind energy is the Sun.
2(a)(ii) When the blades turn, electrical power is generated in the generator.
2(b) Two advantages of using wind turbines compared to coal-fired power stations are:
- No gaseous or SO₂ emissions, which means less air pollution.
- Wind turbines do not contribute to global warming as they do not emit CO₂.
- Wind turbines do not use fossil fuels, making them a renewable energy source.
Question 3
(a) A student balances a beam on a pivot. They then balance block A and block B on the beam, as shown in Fig. 3.1. (Sub-topic – 1.5.2)
(i) The weight of block A is 0.14 N. Show that the moment of block A about the pivot is approximately 0.8 N cm.
(ii) The weight of block B is 0.19 N. Calculate the distance d between the pivot and the centre of block B.
▶️Answer/Explanation
(i) Solution:
The moment of a force is calculated using the formula:
\[ \text{Moment} = \text{Force} \times \text{Perpendicular distance from the pivot} \]
Given:
Force (F) = 0.14 N
Distance (d) = 5.5 cm
\[ \text{Moment} = 0.14 \, \text{N} \times 5.5 \, \text{cm} = 0.77 \, \text{N cm} \]
The moment is approximately 0.8 N cm.
(ii) Solution:
For the beam to be balanced, the sum of the moments about the pivot must be zero. Therefore:
\[ \text{Moment of block A} = \text{Moment of block B} \]
\[ 0.14 \, \text{N} \times 5.5 \, \text{cm} = 0.19 \, \text{N} \times d \]
\[ 0.77 \, \text{N cm} = 0.19 \, \text{N} \times d \]
\[ d = \frac{0.77 \, \text{N cm}}{0.19 \, \text{N}} = 4.05 \, \text{cm} \]
The distance \( d \) is approximately 4.1 cm.
(b) The weight of block B is 0.19 N. Calculate the mass of block B. (Sub-topic – 1.3)
▶️Answer/Explanation
Solution:
The weight of an object is related to its mass by the formula:
\[ \text{Weight} = \text{Mass} \times \text{Acceleration due to gravity} \]
Given:
Weight (W) = 0.19 N
Acceleration due to gravity (g) = 9.8 m/s²
\[ \text{Mass} = \frac{\text{Weight}}{g} = \frac{0.19 \, \text{N}}{9.8 \, \text{m/s}^2} = 0.0194 \, \text{kg} \]
The mass of block B is approximately 0.019 kg.
Question 4
A tight-fitting lid keeps air inside a metal can. An airtight rubber bung holds a liquid-in-glass thermometer that is inserted through a hole in the lid, as shown in Fig. 4.1. (Sub-topic – 2.2.2)
(a) (i) State what happens to the liquid in the thermometer when the air temperature rises.
▶️Answer/Explanation
The liquid in the thermometer expands when the air temperature rises. This is because the liquid gains thermal energy, causing its particles to move more vigorously and occupy more space.
(ii) The temperature of the air in the can is 18°C. Calculate the temperature of the air in kelvin.
▶️Answer/Explanation
To convert Celsius to Kelvin, use the formula: \( T(K) = T(°C) + 273 \).
Given \( T(°C) = 18 \),
\( T(K) = 18 + 273 = 291 \, K \).
So, the temperature of the air in kelvin is 291 K.
(b) The can is placed in a refrigerator. The temperature of the air inside the can decreases. State and explain what happens to the pressure exerted by the air in the can. Use your ideas about gas particles.
(Sub-topic – 2.1.3)
▶️Answer/Explanation
When the temperature of the air inside the can decreases, the pressure exerted by the air also decreases. This is because the gas particles lose kinetic energy, move more slowly, and collide with the walls of the can less frequently and with less force. As a result, the pressure inside the can decreases.
(c) The air in another can exerts a pressure of 102,000 N/m² on the lid. The area of the can lid is 0.0082 m². Calculate the force on the lid due to the air in the can. (Sub-topic – 1.8)
▶️Answer/Explanation
To calculate the force on the lid, use the formula: \( F = P \times A \), where \( F \) is the force, \( P \) is the pressure, and \( A \) is the area.
Given \( P = 102,000 \, N/m² \) and \( A = 0.0082 \, m² \),
\( F = 102,000 \times 0.0082 = 836.4 \, N \).
So, the force on the lid due to the air in the can is 836.4 N.
Question 5
A teacher demonstrates the behaviour of waves by using water waves in a ripple tank.
Fig. 5.1 shows a cross-section through part of the water waves.
(a) Calculate the wavelength of the water waves. Use the information in Fig. 5.1. (Sub-topic – 3.1)
▶️Answer/Explanation
Answer: 2.6 cm
Explanation: The wavelength is calculated by dividing the total distance by the number of waves. In Fig. 5.1, the total distance is 18.2 cm, and there are 7 waves. Therefore, the wavelength is \( \frac{18.2}{7} = 2.6 \) cm.
(b) The teacher places a pointer above the water waves as shown in Fig. 5.2.
Three students use stop-watches to measure the time taken for 50 peaks to pass the pointer. Fig. 5.3 shows the measurements.
(Sub-topic – 3.1)
(i) On the line below each stop-watch, state the time measurement, in seconds.
(ii) Calculate the average of the three time measurements in (b)(i).
(iii) Calculate the frequency of the water waves using your result in (b)(ii).
▶️Answer/Explanation
Answer (i): 17.24 s, 17.14 s, 17.16 s
Answer (ii): 17.18 s
Explanation (ii): The average time is calculated by summing the three time measurements and dividing by 3: \( \frac{17.24 + 17.14 + 17.16}{3} = 17.18 \) s.
Answer (iii): 2.9 Hz
Explanation (iii): The frequency is calculated by dividing the number of peaks (50) by the average time (17.18 s): \( \frac{50}{17.18} = 2.9 \) Hz.
(c) The teacher repeats the demonstration using a different ripple tank and obtains these results for the waves. (Sub-topic – 3.1)
wavelength = 0.025 m
frequency = 2.4 Hz
Calculate the speed of the wave.
▶️Answer/Explanation
Answer: 0.06 m/s
Explanation: The speed of the wave is calculated using the formula \( \text{speed} = \text{frequency} \times \text{wavelength} \). Given the frequency is 2.4 Hz and the wavelength is 0.025 m, the speed is \( 2.4 \times 0.025 = 0.06 \) m/s.
Question 6
Table 6.1 shows regions of the electromagnetic (e.m.) spectrum.
Two of the regions are not labelled.
(a)(i) Complete Table 6.1 by writing the name of each region that is not labelled. (Sub-topic – 3.3)
Table 6.1 shows regions of the electromagnetic (e.m.) spectrum. Two of the regions are not labelled. Complete the table by writing the name of each region that is not labelled.
▶️Answer/Explanation
The two missing regions in Table 6.1 are:
- Ultraviolet (UV)
- Microwaves
(ii) State two properties that are the same for waves in all regions of the e.m. spectrum. (Sub-topic – 3.3)
State two properties that are common to all waves in the electromagnetic spectrum.
▶️Answer/Explanation
Two properties that are the same for waves in all regions of the electromagnetic spectrum are:
- They are all transverse waves.
- They all travel at the same speed in a vacuum (approximately \(3 \times 10^8 \, \text{m/s}\)).
(b) (i) State one other use for X-rays. (Sub-topic – 3.3)
X-rays are used in hospitals to check for broken bones. State one other use for X-rays.
▶️Answer/Explanation
One other use for X-rays is:
- Security scanners at airports.
(ii) State one precaution taken by people who work with X-rays. (Sub-topic – 3.3)
State one precaution that people who work with X-rays take to protect themselves.
▶️Answer/Explanation
One precaution taken by people who work with X-rays is:
- Wearing lead aprons or using lead shields to protect against radiation exposure.
Question 7
(a) Students are investigating the refraction of light as it travels from air into glass. Their task is to measure the angle of incidence and the angle of refraction at the surface of the glass block. The students have the equipment shown in Fig. 7.1. (Sub-topic – 3.2.1)
Describe the method for the task. You may draw a diagram as part of your answer.
▶️Answer/Explanation
Method:
- Place the glass block on a sheet of paper and draw around it to mark its position.
- Use a ray box to shine a ray of light at the glass block at a certain angle (angle of incidence).
- Mark the path of the incident ray and the refracted ray on the paper.
- Draw a normal (perpendicular line) at the point where the light enters the glass block.
- Measure the angle of incidence (between the incident ray and the normal) using a protractor.
- Measure the angle of refraction (between the refracted ray and the normal) using a protractor.
- Repeat the experiment for different angles of incidence and record the results.
Diagram:
(A diagram showing the glass block, incident ray, refracted ray, and normal line should be drawn here.)
(b) Fig. 7.2 and Fig. 7.3 show two identical lenses, each forming an image. The images \( I_1 \) and \( I_2 \) have different characteristics. (Sub-topic – 3.2.3)
One difference in the characteristics of the two images is:
Image \( I_1 \) is diminished but image \( I_2 \) is enlarged.
State two more differences in the characteristics of the images:
Image \( I_1 \) is _____ but image \( I_2 \) is _____.
Image \( I_1 \) is _____ but image \( I_2 \) is _____.
▶️Answer/Explanation
Differences:
- Image \( I_1 \) is real but image \( I_2 \) is virtual.
- Image \( I_1 \) is inverted but image \( I_2 \) is upright.
Explanation:
Image \( I_1 \) is formed by a converging lens when the object is placed beyond the focal point, resulting in a real, inverted, and diminished image. Image \( I_2 \) is formed when the object is placed between the focal point and the lens, resulting in a virtual, upright, and enlarged image.
Question 8
(a) Fig. 8.1 shows the electrical symbols for some circuit components.
Draw a line from each electrical symbol to the name of the circuit component it represents. (Sub-topic – 4.3.1)
▶️Answer/Explanation
The correct matching of electrical symbols to circuit components is as follows:
- Fuse
- Lamp
- Heater
- Thermistor
(b) Fig. 8.2 shows a circuit including a battery, a fixed resistor R and an ammeter.
The reading on the ammeter is 0.38 A.
The potential difference across the fixed resistor R is 12 V.
(i) Calculate the resistance of the fixed resistor R. (Sub-topic – 4.2.4)
▶️Answer/Explanation
To calculate the resistance \( R \), we use Ohm’s Law: \[ R = \frac{V}{I} \] Given: \[ V = 12 \, \text{V}, \quad I = 0.38 \, \text{A} \] Substituting the values: \[ R = \frac{12}{0.38} \approx 31.58 \, \Omega \] Therefore, the resistance of the fixed resistor \( R \) is approximately \( 31.58 \, \Omega \).
(ii) Calculate the electrical power transferred in the fixed resistor R. Include the unit. (Sub-topic – 4.2.5)
▶️Answer/Explanation
To calculate the electrical power \( P \), we use the formula: \[ P = IV \] Given: \[ I = 0.38 \, \text{A}, \quad V = 12 \, \text{V} \] Substituting the values: \[ P = 0.38 \times 12 = 4.56 \, \text{W} \] Therefore, the electrical power transferred in the fixed resistor \( R \) is \( 4.56 \, \text{W} \).
Question 9
Fig. 9.1 represents an atom of beryllium. The labels A, B and C indicate three types of particle. (Sub-topic – 5.1.1)
(a) (i) Complete Table 9.1. Name each type of particle and state the sign of its charge. One row is done for you.
| Type of particle | Name | Sign of charge |
|---|---|---|
| A | ||
| B | ||
| C | proton | positive (+) |
(ii) There are several different isotopes of beryllium. State what is meant by the term isotope.
▶️Answer/Explanation
(i)
| Type of particle | Name | Sign of charge |
|---|---|---|
| A | electron | negative (-) |
| B | neutron | neutral (0) |
| C | proton | positive (+) |
(ii)
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons. This means they have the same atomic number but different mass numbers.
(b) Fig. 9.2 shows sources of background radiation that affect people. (Sub-topic – 5.2.1)
Suggest the source of background radiation in region D.
▶️Answer/Explanation
The source of background radiation in region D is cosmic radiation. Cosmic radiation comes from outer space and includes high-energy particles that reach the Earth’s surface.
(c) The nuclide notation for an atom of radon is: (Sub-topic – 5.1.2)
\(\frac{222}{86} \text{Rn}\)
(i) State the number of protons in this atom of radon.
(ii) State the number of particles in the nucleus of this atom of radon.
▶️Answer/Explanation
(i) The number of protons in this atom of radon is 86. This is given by the atomic number (bottom number) in the nuclide notation.
(ii) The number of particles in the nucleus of this atom of radon is 222. This is given by the mass number (top number) in the nuclide notation, which represents the total number of protons and neutrons in the nucleus.
Question 10
Fig. 10.1 represents part of the Solar System. (Sub-topic – 6.1.2)
(a) (i) State the name of planet A and the name of planet B.
(ii) On Fig. 10.1, draw an X to represent a moon of Jupiter. Draw a line to show how this moon moves.
(iii) State two ways in which the four planets nearest to the Sun are different from the four planets furthest away from the Sun.
(iv) Complete the following sentences:
The galaxy that includes the Solar System is called the …………………………………………. .
The …………………………………. includes billions of galaxies.
▶️Answer/Explanation
(i) Planet A is Mercury, and planet B is Mars.
(ii) The moon of Jupiter should be represented by an X, and a circular or elliptical path should be drawn around Jupiter to show its orbit.
(iii)
1. The four planets nearest to the Sun are rocky and small, while the four planets furthest from the Sun are gaseous and large.
2. The four planets nearest to the Sun have shorter orbital periods compared to the four planets furthest from the Sun.
(iv) The galaxy that includes the Solar System is called the Milky Way. The Universe includes billions of galaxies.
(b) The distance between the Sun and the Earth is \(1.5 \times 10^{11} \, \text{m}\). (Sub-topic – 6.1.2)
The speed of an electromagnetic wave is \(3.0 \times 10^8 \, \text{m/s}\).
Calculate the time taken for an electromagnetic wave to travel from the Sun to the Earth.
▶️Answer/Explanation
To calculate the time taken for an electromagnetic wave to travel from the Sun to the Earth, we use the formula: \[ \text{time} = \frac{\text{distance}}{\text{speed}} \] Given: \[ \text{distance} = 1.5 \times 10^{11} \, \text{m}, \quad \text{speed} = 3.0 \times 10^8 \, \text{m/s} \] Substitute the values into the formula: \[ \text{time} = \frac{1.5 \times 10^{11}}{3.0 \times 10^8} = 500 \, \text{s} \] Therefore, the time taken is 500 seconds.