Question 1
A car accelerates uniformly in a straight line from rest at time t = 0. At t = 3.2 s, the speed of the car is 13.0 m / s.
(a) (i) Calculate the acceleration of the car. (Sub-topic – 1.2)
▶️Answer/Explanation
\[ \boxed{4.1 \, \text{m/s}^2} \]
Given:
- Initial speed, \( u = 0 \, \text{m/s} \)
- Final speed, \( v = 13.0 \, \text{m/s} \)
- Time, \( t = 3.2 \, \text{s} \)
Using the formula for acceleration:
\[ a = \frac{v – u}{t} \]
Substitute the given values:
\[ a = \frac{13.0 – 0}{3.2} = 4.0625 \, \text{m/s}^2 \]
Therefore, the acceleration of the car is approximately \( 4.1 \, \text{m/s}^2 \).
(ii) Explain in words what is meant by the term acceleration. (Sub-topic – 1.2)
▶️Answer/Explanation
Acceleration is the rate at which an object’s velocity changes over time. It can be an increase or decrease in speed, or a change in direction. In this case, the car’s speed increases uniformly from rest to 13.0 m/s in 3.2 seconds.
(b) The car travels at 13.0 m / s from t = 3.2 s to t = 12.0 s.
(i) Plot the speed–time graph for the car from \( t = 0 \) to \( t = 12.0 \, \text{s} \). (Sub-topic – 1.2)
▶️Answer/Explanation
The graph should show a straight line from (0, 0) to (3.2, 13.0) and a horizontal line from (3.2, 13.0) to (12.0, 13.0).
The speed–time graph will consist of two parts:
- From \( t = 0 \) to \( t = 3.2 \, \text{s} \), the car accelerates uniformly from 0 m/s to 13.0 m/s. This will be a straight line with a positive slope.
- From \( t = 3.2 \, \text{s} \) to \( t = 12.0 \, \text{s} \), the car travels at a constant speed of 13.0 m/s. This will be a horizontal line.
(ii) Determine the distance travelled by the car between \( t = 0 \) and \( t = 3.2 \, \text{s} \). (Sub-topic – 1.2)
▶️Answer/Explanation
\[ \boxed{20.8 \, \text{m}} \]
(c) The car decelerates from 13.0 m / s to 0 m / s at a constant deceleration. The mass of the car is 1350 kg. The car travels 13 m in 2.0 s as it decelerates.
Show that the work done by the car as it decelerates is approximately \( 1.1 \times 10^5 \, \text{J} \). (Sub-topic – 1.7)
▶️Answer/Explanation
\[ \boxed{1.1 \times 10^5 \, \text{J}} \]
Given:
- Mass of the car, \( m = 1350 \, \text{kg} \)
- Initial speed, \( u = 13.0 \, \text{m/s} \)
- Final speed, \( v = 0 \, \text{m/s} \)
- Distance travelled during deceleration, \( d = 13 \, \text{m} \)
First, calculate the deceleration using the equation:
\[ v^2 = u^2 + 2ad \]
Rearrange to solve for \( a \):
\[ a = \frac{v^2 – u^2}{2d} = \frac{0 – (13.0)^2}{2 \times 13} = \frac{-169}{26} = -6.5 \, \text{m/s}^2 \]
The negative sign indicates deceleration.
Next, calculate the force using Newton’s second law:
\[ F = ma = 1350 \times 6.5 = 8775 \, \text{N} \]
Finally, calculate the work done:
\[ W = F \times d = 8775 \times 13 = 114075 \, \text{J} \approx 1.1 \times 10^5 \, \text{J} \]
Therefore, the work done by the car as it decelerates is approximately \( 1.1 \times 10^5 \, \text{J} \).
(d) On another day, the car in (c) travels a longer distance while it decelerates from 13.0 m / s to 0 m / s. The deceleration is constant.
Suggest and explain what causes the stopping distance to increase. (Sub-topic – 1.2)
▶️Answer/Explanation
The stopping distance increases if the deceleration decreases, which could be due to factors like reduced friction or less effective brakes.
Suggestion: The stopping distance increases if the deceleration decreases.
Explanation: If the deceleration is less, the car takes longer to come to a stop, thus covering a greater distance before stopping. This could be due to factors such as wet or icy road conditions, which reduce the friction between the tires and the road, or a less effective braking system.
Question 2
Fig. 2.1 shows an electric tumble dryer used to dry wet clothes. (Sub-topic – 2.2.3)
(a) Hot air blows into the drum. The air gains water vapour from the clothes and then leaves the drum. The moist air enters the condenser. Cool air leaves the condenser, passes through the heating element and enters the drum again.
(i) State the process by which the hot air removes water from the wet clothes.
▶️Answer/Explanation
Answer: Evaporation
Explanation: The hot air removes water from the wet clothes through the process of evaporation. When hot air blows into the drum, it increases the temperature of the water in the clothes, causing the water molecules to gain enough energy to change from the liquid phase to the gas phase, thus removing moisture from the clothes.
(ii) The air is cooled as it passes through the condenser. Describe and explain one other way in which the air leaving the condenser is different from the air entering the condenser.
▶️Answer/Explanation
Answer: The air leaving the condenser is drier.
Explanation: As the moist air passes through the condenser, the water vapour in the air condenses into liquid water. This process removes moisture from the air, making the air leaving the condenser drier than the air entering it.
(b) The drum of the tumble dryer rotates, lifting up the wet clothes which then fall down through the hot air. (Sub-topic – 1.5.1)
(i) Name the force that causes the clothes to fall down.
▶️Answer/Explanation
Answer: Gravitational force (or weight)
Explanation: The force that causes the clothes to fall down is the gravitational force, also known as weight. This force acts downward, pulling the clothes toward the bottom of the drum.
(ii) When the drum rotates too fast, the clothes remain in contact with the wall of the drum. State the direction of the resultant force on the clothes during the circular motion.
▶️Answer/Explanation
Answer: The resultant force is perpendicular to the motion of the clothes.
Explanation: During circular motion, the resultant force (centripetal force) acts perpendicular to the direction of motion, toward the center of the circular path. This force keeps the clothes moving in a circular path along the wall of the drum.
(c) Suggest why using a clothesline to dry clothes in the open air is better for the environment than using an electric tumble dryer. (Sub-topic – 1.7.3)
▶️Answer/Explanation
Answer: Using a clothesline to dry clothes in the open air is better for the environment because it uses renewable energy (solar and wind) and does not produce greenhouse gases.
Explanation: Electric tumble dryers consume electricity, which is often generated from non-renewable sources like fossil fuels, leading to greenhouse gas emissions. In contrast, drying clothes on a clothesline uses natural energy from the sun and wind, which are renewable and do not contribute to environmental pollution.
Question 3
(a) A balloon of mass 15 g is glued to a straw. The straw is threaded onto a horizontal string, as shown in Fig. 3.1.
The balloon is filled with air and then the air is released. (Sub-topic – 1.6)
As the air leaves the balloon, the balloon experiences a force. The balloon accelerates from rest until it reaches a constant speed. It then travels 0.67 m in 0.18 s at this constant speed.
(i) Explain in words what is meant by the term impulse.
(ii) Calculate the resultant impulse on the balloon while it is accelerating.
(iii) Explain how momentum is conserved as the balloon accelerates.
▶️Answer/Explanation
(i) Impulse is defined as the product of the force acting on an object and the time for which the force acts. It is equal to the change in momentum of the object.
(ii) The resultant impulse on the balloon can be calculated using the formula: \[ \text{Impulse} = \Delta p = m \Delta v \] First, calculate the final velocity (\(v\)) of the balloon: \[ v = \frac{\text{distance}}{\text{time}} = \frac{0.67 \, \text{m}}{0.18 \, \text{s}} = 3.72 \, \text{m/s} \] The change in velocity (\(\Delta v\)) is \(3.72 \, \text{m/s}\) (since the balloon starts from rest). The mass (\(m\)) of the balloon is \(15 \, \text{g} = 0.015 \, \text{kg}\). \[ \text{Impulse} = 0.015 \, \text{kg} \times 3.72 \, \text{m/s} = 0.0558 \, \text{Ns} \]
(iii) Momentum is conserved as the balloon accelerates because the momentum gained by the balloon is equal to the momentum lost by the air being expelled from the balloon. The total momentum of the system (balloon + air) remains constant.
(b) Fig. 3.2 shows the directions of two forces acting on a different balloon as it moves. (Sub-topic – 1.5)
Determine the magnitude and direction of the resultant force on the balloon.
▶️Answer/Explanation
To determine the magnitude and direction of the resultant force, we can use vector addition. The two forces acting on the balloon are \(0.40 \, \text{N}\) and \(0.74 \, \text{N}\). The resultant force (\(F_R\)) can be calculated using the Pythagorean theorem: \[ F_R = \sqrt{(0.40 \, \text{N})^2 + (0.74 \, \text{N})^2} = \sqrt{0.16 + 0.5476} = \sqrt{0.7076} = 0.84 \, \text{N} \] The direction of the resultant force relative to the horizontal force can be found using trigonometry: \[ \theta = \tan^{-1}\left(\frac{0.74}{0.40}\right) = \tan^{-1}(1.85) \approx 62^\circ \] Therefore, the resultant force is \(0.84 \, \text{N}\) at an angle of \(62^\circ\) below the horizontal.
Question 4
Fig. 4.1 shows a bottle part-filled with water. The air inside the bottle is at the same pressure as the air outside the bottle. The bottle and its contents are at room temperature. (Sub-topic – 2.1.2)
(a) The temperature of the bottle and its contents are increased.
(i) Explain, in terms of particles, how the air pressure inside the bottle changes as the temperature increases.
(ii) The lid is removed from the bottle. State and explain how the air pressure inside the bottle changes.
▶️Answer/Explanation
(i) As the temperature increases, the average kinetic energy of the air particles inside the bottle increases. This causes the particles to move faster and collide more frequently with the walls of the bottle. The increased frequency and force of these collisions result in an increase in air pressure inside the bottle.
(ii) When the lid is removed, the air particles inside the bottle can escape into the surrounding environment. As particles escape, the number of particles inside the bottle decreases, leading to a reduction in the frequency of collisions with the walls of the bottle. This results in a decrease in air pressure inside the bottle until it equalizes with the external atmospheric pressure.
(b) The mass of water in the bottle is 0.18 kg. The specific heat capacity of water is 4200 J/(kg °C). Calculate the thermal energy needed to increase the temperature of the water by 20 °C. (Sub-topic – 2.2.2)
▶️Answer/Explanation
The thermal energy required can be calculated using the formula: \[ Q = mc\Delta\theta \] where:
- \( Q \) = thermal energy (J)
- \( m \) = mass of water (kg)
- \( c \) = specific heat capacity of water (J/(kg °C))
- \( \Delta\theta \) = change in temperature (°C)
Substituting the given values: \[ Q = 0.18 \times 4200 \times 20 \] \[ Q = 15120 \, \text{J} \] Therefore, the thermal energy needed is 15120 J.
(c) Another plastic bottle is filled to the top with water. The height of the bottle is 40.0 cm. The density of water is \(1.0 \times 10^3 \, \text{kg/m}^3\). Calculate the pressure difference between the top and bottom of the water. (Sub-topic – 1.8)
▶️Answer/Explanation
The pressure difference due to a column of liquid is given by: \[ \Delta p = \rho g h \] where:
- \( \Delta p \) = pressure difference (Pa)
- \( \rho \) = density of water (kg/m³)
- \( g \) = acceleration due to gravity (9.8 m/s²)
- \( h \) = height of the water column (m)
First, convert the height from cm to m: \[ h = 40.0 \, \text{cm} = 0.40 \, \text{m} \] Now, substitute the values into the formula: \[ \Delta p = 1.0 \times 10^3 \times 9.8 \times 0.40 \] \[ \Delta p = 3920 \, \text{Pa} \] Therefore, the pressure difference between the top and bottom of the water is 3920 Pa.
Question 5
Fig. 5.1 shows a road junction, a moving car and a stationary truck. The road has high walls on each side. (Sub-topic – 3.2.1)
(a) The driver of the truck is at position X. The car moves around the corner.
On Fig. 5.1, label a point Y on the road where the truck driver first sees the car.
▶️Answer/Explanation
Answer: Point Y should be labeled at the position where the car becomes visible to the truck driver as it moves around the corner. This point is where the line of sight from the truck driver’s position X intersects the road after the corner.
(b) A plane mirror is placed at the road junction as shown in Fig. 5.2. (Sub-topic – 3.2.1)
Show how this mirror allows the driver of the truck to see the car when it is at the position shown in Fig. 5.2.
▶️Answer/Explanation
Answer: The mirror reflects the light from the car to the truck driver’s eyes. To show this, draw an incident ray from the car to the mirror and a reflected ray from the mirror to the truck driver’s position X. The angle of incidence should be equal to the angle of reflection.
(c) The truck driver wears spectacles to correct long-sightedness. Fig. 5.3 shows how a blurred image of an object O forms on the retina. Any effect of the cornea on the rays of light can be ignored.
(Sub-topic – 3.2.3)
On Fig. 5.4, show how long-sightedness is corrected by:
- adding a suitable lens in front of the eye
- continuing the path of the three rays of light until they meet to form an image.
▶️Answer/Explanation
Answer: To correct long-sightedness, a converging lens is placed in front of the eye. The lens refracts the light rays so that they converge on the retina. On Fig. 5.4, draw the converging lens and show the three rays of light refracting through the lens and meeting at a point on the retina to form a clear image.
Question 6
Fig. 6.1 shows the circuit diagram for a flashlight (torch).
The electromotive force (e.m.f.) of the battery is 4.5 V. The circuit contains a 60 Ω fixed resistor. The current in the light-emitting diode (LED) is 0.020 A.
(a) Calculate the potential difference (p.d.) across the LED. (Sub-topic – 4.3.2)
▶️Answer/Explanation
Answer: 3.3 V
Explanation:
The potential difference (p.d.) across the LED can be calculated using Ohm’s Law, which states that \( V = IR \), where \( V \) is the potential difference, \( I \) is the current, and \( R \) is the resistance.
Given:
- Current (\( I \)) = 0.020 A
- Resistance (\( R \)) = 60 Ω
The p.d. across the resistor is: \[ V = IR = 0.020 \times 60 = 1.2 \, \text{V} \] Since the total e.m.f. of the battery is 4.5 V, the p.d. across the LED is: \[ \text{p.d. across LED} = 4.5 – 1.2 = 3.3 \, \text{V} \]
(b) Explain why the LED does not light up if the battery is reversed. (Sub-topic – 4.3.3)
▶️Answer/Explanation
Answer: The LED is a diode, which only allows current to flow in one direction. When the battery is reversed, the LED is reverse-biased and has a very high resistance, preventing current from flowing and causing the LED not to light up.
Explanation:
An LED (Light Emitting Diode) is a semiconductor device that allows current to flow in only one direction. When the battery is connected correctly, the LED is forward-biased, allowing current to flow and the LED to light up. However, when the battery is reversed, the LED becomes reverse-biased, which means it has a very high resistance and does not allow current to flow. As a result, the LED does not light up.
(c) The chemical energy stored in the battery is 1050 J.
Show that the flashlight operates for approximately 3 hours. (Sub-topic – 4.2.5)
▶️Answer/Explanation
Answer: The flashlight operates for approximately 3 hours.
Explanation:
The total energy stored in the battery is 1050 J. The power consumed by the circuit can be calculated using the formula \( P = IV \), where \( P \) is power, \( I \) is current, and \( V \) is voltage.
Given:
- Current (\( I \)) = 0.020 A
- Voltage (\( V \)) = 4.5 V
The power consumed by the circuit is: \[ P = IV = 0.020 \times 4.5 = 0.09 \, \text{W} \] The total operating time (\( t \)) can be calculated using the formula \( E = Pt \), where \( E \) is energy and \( P \) is power. \[ t = \frac{E}{P} = \frac{1050}{0.09} = 11666.67 \, \text{s} \] Converting seconds to hours: \[ t = \frac{11666.67}{3600} \approx 3.24 \, \text{hours} \] Therefore, the flashlight operates for approximately 3 hours.
(d) Calculate the total charge that flows through the LED in 3600 s. (Sub-topic – 4.2.2)
▶️Answer/Explanation
Answer: 72 C
Explanation:
The total charge (\( Q \)) that flows through the LED can be calculated using the formula \( Q = It \), where \( I \) is current and \( t \) is time.
Given:
- Current (\( I \)) = 0.020 A
- Time (\( t \)) = 3600 s
The total charge is: \[ Q = It = 0.020 \times 3600 = 72 \, \text{C} \]
Question 7
Fig. 7.1 shows some uses of electromagnetic radiation and different regions of the electromagnetic spectrum. (Sub-topic – 3.3)
(a) Draw a line from each use to the correct region of the spectrum. Each region of the spectrum is used once. One line has been completed for you.
▶️Answer/Explanation
- Bluetooth headset – Radio waves
- Thermal imaging – Infrared
- Photography of people’s faces – Visible light
- Sterilising medical equipment – Gamma rays
(b) State the speed of electromagnetic waves in a vacuum. (Sub-topic – 3.3)
▶️Answer/Explanation
The speed of electromagnetic waves in a vacuum is \(3.0 \times 10^8 \, \text{m/s}\).
(c) (i) A Bluetooth headset can be used to listen to music on a mobile (cell) phone without the need for wires to connect the headset to the phone. The headset uses frequencies in the range 2.40–2.48 GHz. Calculate the wavelength of the radio waves when the frequency is in the middle of the frequency range. (Sub-topic – 3.3)
▶️Answer/Explanation
To calculate the wavelength (\(\lambda\)), we use the formula: \[ \lambda = \frac{v}{f} \] where \(v\) is the speed of light (\(3.0 \times 10^8 \, \text{m/s}\)) and \(f\) is the frequency.
The middle of the frequency range is: \[ f = \frac{2.40 + 2.48}{2} = 2.44 \, \text{GHz} = 2.44 \times 10^9 \, \text{Hz} \] Now, substitute the values into the formula: \[ \lambda = \frac{3.0 \times 10^8}{2.44 \times 10^9} = 0.123 \, \text{m} = 12.3 \, \text{cm} \] Therefore, the wavelength of the radio waves is approximately 12.3 cm.
(ii) Suggest why a Bluetooth headset only works well over short distances. (Sub-topic – 3.3)
▶️Answer/Explanation
A Bluetooth headset only works well over short distances because radio waves lose energy or get weaker as they pass through walls and other obstacles. The signal strength decreases with distance, making it difficult to maintain a stable connection over longer distances.
Question 8
The isotope uranium-235 is represented by. (Sub-topic – 5.1.2)
\( _{92}^{235}\textrm{U}\)
(a) State what the numbers 92 and 235 represent in this symbol.
▶️Answer/Explanation
Answer:
92 is the proton number (atomic number), which represents the number of protons in the nucleus.
235 is the nucleon number (mass number), which represents the total number of protons and neutrons in the nucleus.
(b) Uranium-235 is a fuel used in nuclear reactors. (Sub-topic – 5.1.2)
(i) State the process by which energy is released from uranium-235 in a nuclear reactor.
▶️Answer/Explanation
Answer:
The process is nuclear fission, where the nucleus of uranium-235 splits into smaller nuclei, releasing energy.
(ii) A nuclide equation for this process is:
\( _{92}^{235}\textrm{U}\) + \( _{0}^{1}\textrm{n}\) → \( _{54}^{140}\textrm{Xe}\) + \( _{38}^{94}\textrm{Sr}\) + 2 \( _{0}^{1}\textrm{n}\)
Describe the mass and energy changes that take place during this process in a nuclear reactor.
▶️Answer/Explanation
Answer:
During nuclear fission, the total mass of the products is slightly less than the mass of the original uranium-235 nucleus and the neutron. This mass difference is converted into energy according to Einstein’s equation \( E = mc^2 \). The energy is released as kinetic energy of the fission fragments and neutrons, and as thermal energy.
(c) (i) Describe how thermal energy from nuclear reactions is used to generate electricity in a power station. (Sub-topic – 5.2.3)
▶️Answer/Explanation
Answer:
Thermal energy from nuclear reactions is used to heat water, producing steam. The steam drives a turbine, which is connected to a generator. The generator converts the mechanical energy from the turbine into electrical energy.
(ii) State one advantage and one disadvantage of using nuclear fuels in a power station instead of using fossil fuels. (Sub-topic – 5.2.3)
▶️Answer/Explanation
Answer:
Advantage: Nuclear fuels produce a much larger amount of energy per unit mass compared to fossil fuels, and they do not produce greenhouse gases during operation.
Disadvantage: Nuclear power plants produce hazardous radioactive waste that requires long-term storage and poses environmental and health risks.
Question 9
Table 9.1 gives information about three planets in the Solar System. (Sub-topic – 6.1.2)
(a) State the name of planet X.
▶️Answer/Explanation
Venus
(b) Describe the relationship shown in Table 9.1 between the mass of a planet and the gravitational field strength at its surface. (Sub-topic – 6.1.2)
▶️Answer/Explanation
The larger the mass of the planet, the larger the gravitational field strength at the surface.
(c) Explain why ‘distance from Sun’ in Table 9.1 is an average value. (Sub-topic – 6.1.2)
▶️Answer/Explanation
The orbit of planets is elliptical (not circular), so the distance from the Sun varies over time. Therefore, the value given is an average distance.
(d) Show that the average orbital speed of the Earth is approximately 30 km/s. (Sub-topic – 6.1.1)
▶️Answer/Explanation
To calculate the average orbital speed of the Earth, we use the formula: \[ v = \frac{2\pi r}{T} \] where:
- \( r \) is the average radius of the orbit (149.6 million km or \( 149.6 \times 10^6 \) km)
- \( T \) is the orbital period (365.2 days or \( 365.2 \times 24 \times 60 \times 60 \) seconds)
First, convert the orbital period \( T \) into seconds: \[ T = 365.2 \times 24 \times 60 \times 60 = 3.16 \times 10^7 \, \text{s} \] Now, calculate the orbital speed: \[ v = \frac{2\pi \times 149.6 \times 10^6 \, \text{km}}{3.16 \times 10^7 \, \text{s}} \approx 29.8 \, \text{km/s} \] This is approximately 30 km/s.
Question 10
(a) Protostars are formed from? (Sub-topic – 6.2.2)
▶️Answer/Explanation
Protostars are formed from interstellar clouds of gas and dust or stellar nebula.
(b) A protostar becomes a stable star when …… is balanced by …… (Sub-topic – 6.2.2)
▶️Answer/Explanation
A protostar becomes a stable star when the inward force of gravitational attraction is balanced by the outward force due to the high temperature in the centre of the star.
(c) The initial fuel used to power nuclear reactions in stars is …… (Sub-topic – 6.2.2)
▶️Answer/Explanation
The initial fuel used to power nuclear reactions in stars is hydrogen.
(d) Stars that are approximately the same size as the Sun become red giant stars which then form a …… with a white dwarf star at its centre. (Sub-topic – 6.2.2)
▶️Answer/Explanation
Stars that are approximately the same size as the Sun become red giant stars which then form a planetary nebula with a white dwarf star at its centre.