Question 1
(a) Fig. 1.1 shows a helicopter which is stationary at a height of 1500 m above the ground.
(i) State the two conditions necessary for the helicopter to remain in equilibrium. (Sub-topic – 1.1)
▶️Answer/Explanation
Condition 1: No resultant/net force acting on the helicopter.
Condition 2: No resultant/net moment acting on the helicopter.
(ii) The mass of the helicopter is 3200 kg. Calculate the change in the gravitational potential energy of the helicopter as it rises from the ground to 1500 m. (Sub-topic – 1.2)
▶️Answer/Explanation
The change in gravitational potential energy (ΔEp) is calculated using the formula:
ΔEp = mgh
where:
m = mass of the helicopter = 3200 kg
g = acceleration due to gravity = 9.8 m/s²
h = height = 1500 m
ΔEp = 3200 × 9.8 × 1500 = 4.7 × 107 J or 47 MJ
(b) Fig. 1.2 shows a vertical speed–time graph for a parachutist who jumps from a stationary hot-air balloon.
The parachutist jumps from the balloon at time = 0 and reaches the ground at B. The point A indicates when the parachute opens.
(i) On Fig. 1.2, label a point on the graph where the acceleration is: (Sub-topic – 1.3)
- zero with ‘1’
- negative with ‘2’
- decreasing with ‘3’.
▶️Answer/Explanation
Point labelled ‘1’ should be on either of the horizontal sections of the graph (where speed is constant, indicating zero acceleration).
Point labelled ‘2’ should be on the graph between point A (when the parachute opens) and the start of the horizontal section near the ground (where the parachutist is decelerating, indicating negative acceleration).
Point labelled ‘3’ should be on the graph between the start of the curved section (just after the jump) and the start of the horizontal section before point A (where the acceleration is decreasing as air resistance increases).
(ii) Explain, in terms of forces, the changes in motion which occur from when the parachutist leaves the hot-air balloon until point A. (Sub-topic – 1.7)
▶️Answer/Explanation
Initially, the parachutist accelerates due to the force of gravity (weight) acting downward.
As the parachutist gains speed, air resistance (drag force) increases, opposing the motion.
The resultant force downward decreases as air resistance increases, causing the acceleration to decrease.
Eventually, the air resistance equals the weight, and the parachutist reaches terminal velocity, moving at a constant speed until the parachute opens at point A.
Question 2
A student catches a cricket ball. The speed of the ball immediately before it is caught is 18 m/s. The mass of the cricket ball is 160 g.
(a) Calculate the kinetic energy stored in the cricket ball immediately before it is caught. (Sub-topic – 1.7.1)
▶️Answer/Explanation
Solution:
The kinetic energy (Ek) of an object is given by the formula:
\[ E_k = \frac{1}{2}mv^2 \]
where:
– \( m \) is the mass of the object (in kg),
– \( v \) is the velocity of the object (in m/s).
Given:
– Mass \( m = 160 \, \text{g} = 0.16 \, \text{kg} \),
– Velocity \( v = 18 \, \text{m/s} \).
Substituting the values into the formula:
\[ E_k = \frac{1}{2} \times 0.16 \times (18)^2 \]
\[ E_k = \frac{1}{2} \times 0.16 \times 324 \]
\[ E_k = 0.08 \times 324 \]
\[ E_k = 25.92 \, \text{J} \]
Therefore, the kinetic energy stored in the cricket ball is 25.92 J.
(b) It takes 0.12 s to catch the ball and bring it to rest. Calculate the average force exerted on the ball. (Sub-topic – 1.5.1)
▶️Answer/Explanation
Solution:
The average force (F) exerted on the ball can be calculated using the impulse-momentum theorem:
\[ F = \frac{\Delta p}{\Delta t} \]
where:
– \( \Delta p \) is the change in momentum,
– \( \Delta t \) is the time taken to bring the ball to rest.
The change in momentum (\( \Delta p \)) is given by:
\[ \Delta p = m \Delta v \]
where:
– \( m \) is the mass of the ball,
– \( \Delta v \) is the change in velocity.
Given:
– Mass \( m = 0.16 \, \text{kg} \),
– Initial velocity \( v = 18 \, \text{m/s} \),
– Final velocity \( v = 0 \, \text{m/s} \),
– Time \( \Delta t = 0.12 \, \text{s} \).
The change in velocity (\( \Delta v \)) is:
\[ \Delta v = 18 – 0 = 18 \, \text{m/s} \]
The change in momentum (\( \Delta p \)) is:
\[ \Delta p = 0.16 \times 18 = 2.88 \, \text{kg m/s} \]
The average force (F) is:
\[ F = \frac{2.88}{0.12} = 24 \, \text{N} \]
Therefore, the average force exerted on the ball is 24 N.
(c) As the student catches the ball, she moves her hands backwards. Explain the effect of this action on the student’s hands. (Sub-topic – 1.5.1)
▶️Answer/Explanation
Explanation:
When the student moves her hands backward while catching the ball, she increases the time over which the ball comes to rest. According to the impulse-momentum theorem:
\[ F = \frac{\Delta p}{\Delta t} \]
By increasing the time (\( \Delta t \)), the force (F) exerted on her hands decreases. This reduces the impact force on her hands, making it less likely to cause injury or discomfort. Therefore, moving her hands backward helps to reduce the force experienced by her hands during the catch.
Question 3
(a) Fig. 3.1 shows a person moving across an ice-covered pond to reach a ball on the ice. (Subtopic – 1.5.1)
Explain why this way of moving across the ice is safer than walking. Use your understanding of pressure in your answer.
▶️Answer/Explanation
When a person moves across the ice by spreading their body weight over a larger area (e.g., crawling or sliding), the force of their weight is distributed over a greater surface area. According to the formula for pressure \( P = \frac{F}{A} \), where \( P \) is pressure, \( F \) is force, and \( A \) is area, increasing the area reduces the pressure exerted on the ice. This reduces the risk of the ice cracking or breaking, making it safer than walking, where the force is concentrated on a smaller area (the feet), resulting in higher pressure.
(b) Fig. 3.2 shows a side view of the pond with a layer of ice floating freely on the water. (Subtopic – 1.8)
The surface area of the pond is \(5.0 \, \text{m}^2\).
The mass of the ice is \(690 \, \text{kg}\).
The density of water is \(1000 \, \text{kg/m}^3\).
Point X is \(0.45 \, \text{m}\) below the ice.
Calculate the pressure at point X due to the ice and the water.
▶️Answer/Explanation
To calculate the pressure at point X, we need to consider the pressure due to the ice and the pressure due to the water above point X.
Step 1: Calculate the pressure due to the ice
The weight of the ice is given by \( W = mg \), where \( m = 690 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
\( W = 690 \times 9.8 = 6762 \, \text{N} \).
The pressure due to the ice is \( P_{\text{ice}} = \frac{F}{A} = \frac{6762}{5.0} = 1352.4 \, \text{Pa} \).
Step 2: Calculate the pressure due to the water
The pressure due to the water is given by \( P_{\text{water}} = \rho gh \), where \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), and \( h = 0.45 \, \text{m} \).
\( P_{\text{water}} = 1000 \times 9.8 \times 0.45 = 4410 \, \text{Pa} \).
Step 3: Total pressure at point X
The total pressure at point X is the sum of the pressure due to the ice and the pressure due to the water:
\( P_{\text{total}} = P_{\text{ice}} + P_{\text{water}} = 1352.4 + 4410 = 5762.4 \, \text{Pa} \).
Therefore, the pressure at point X is approximately \( 5.8 \times 10^3 \, \text{Pa} \).
Question 4
(a) The temperature of a fixed mass of gas at constant volume is decreased. State and explain, in terms of particles, how the pressure of the gas changes. (Sub-topic – Gas Laws)
▶️Answer/Explanation
Answer: The pressure of the gas decreases. When the temperature is decreased, the particles of the gas move more slowly, resulting in a lower kinetic energy. This means that the particles collide with the walls of the container less frequently and with less force, leading to a decrease in pressure.
(b)(i) State the value of absolute zero in °C. (Sub-topic – Absolute Zero)
▶️Answer/Explanation
Answer: The value of absolute zero is -273 °C.
(ii) Explain what is meant by the term absolute zero. Refer to particles in your answer.
▶️Answer/Explanation
Answer: Absolute zero is the lowest possible temperature where the particles of a substance have the minimum possible kinetic energy. At this temperature, the particles stop moving entirely, and no further cooling is possible.
(c) Cylinder 1 contains 350 cm³ of gas at a pressure of 9.0 × 10⁴ Pa. The gas is transferred to cylinder 2 and the pressure increases to 1.6 × 10⁵ Pa. The temperature remains constant. Calculate the volume of cylinder 2. (Sub-topic – Boyle’s Law)
▶️Answer/Explanation
Answer: Using Boyle’s Law, \( P_1V_1 = P_2V_2 \), where:
- \( P_1 = 9.0 \times 10^4 \, \text{Pa} \)
- \( V_1 = 350 \, \text{cm}^3 \)
- \( P_2 = 1.6 \times 10^5 \, \text{Pa} \)
- \( V_2 \) is the volume of cylinder 2.
Rearranging the formula to solve for \( V_2 \): \[ V_2 = \frac{P_1V_1}{P_2} = \frac{9.0 \times 10^4 \times 350}{1.6 \times 10^5} \] \[ V_2 = \frac{3.15 \times 10^7}{1.6 \times 10^5} = 196.875 \, \text{cm}^3 \] Therefore, the volume of cylinder 2 is approximately 197 cm³.
Question 5
(a) Fig. 5.1 shows an electric heater used to heat a room.
The dimensions of the room are \(4.5 \, \text{m} \times 6.1 \, \text{m} \times 2.4 \, \text{m}\). The density of air is \(1.2 \, \text{kg/m}^3\).
(i) Show that the mass of air in the room is \(79 \, \text{kg}\). (Sub-topic – 1.7.1)
(ii) The power of the heater is \(1100 \, \text{W}\). The specific heat capacity of air is \(1000 \, \text{J/(kg}^\circ \text{C)}\). Calculate the time taken to increase the temperature of the air in the room from \(16.0^\circ \text{C}\) to \(20.0^\circ \text{C}\). (Sub-topic – 1.7.4)
(iii) Suggest one reason why the time calculated in (a)(ii) is the minimum time needed to increase the temperature of the air in the room from \(16.0^\circ \text{C}\) to \(20.0^\circ \text{C}\). (Sub-topic – 2.3.1 Conduction)
(b) Fig. 5.2 shows a cross-section of a double-glazed window in the room. (Sub-topic – 2.3.2)
State the main methods of thermal energy transfer from the room to outside which are reduced by this type of window.
▶️Answer/Explanation
(a)(i) To find the mass of air in the room, we first calculate the volume of the room: \[ \text{Volume} = 4.5 \, \text{m} \times 6.1 \, \text{m} \times 2.4 \, \text{m} = 65.88 \, \text{m}^3 \] Next, we use the density of air to find the mass: \[ \text{Mass} = \text{Density} \times \text{Volume} = 1.2 \, \text{kg/m}^3 \times 65.88 \, \text{m}^3 = 79.056 \, \text{kg} \approx 79 \, \text{kg} \] Thus, the mass of air in the room is \(79 \, \text{kg}\).
(a)(ii) The energy required to increase the temperature of the air is given by: \[ \Delta E = mc\Delta\theta \] where \(m\) is the mass of air, \(c\) is the specific heat capacity, and \(\Delta\theta\) is the change in temperature. Substituting the values: \[ \Delta E = 79 \, \text{kg} \times 1000 \, \text{J/(kg}^\circ \text{C)} \times (20.0^\circ \text{C} – 16.0^\circ \text{C}) = 316,000 \, \text{J} \] The time taken to supply this energy at a power of \(1100 \, \text{W}\) is: \[ t = \frac{\Delta E}{P} = \frac{316,000 \, \text{J}}{1100 \, \text{W}} \approx 287.27 \, \text{s} \approx 290 \, \text{s} \] Thus, the time taken is approximately \(290 \, \text{s}\).
(a)(iii) One reason why the calculated time is the minimum is that some of the energy from the heater is lost to the surroundings, such as through the walls, windows, and furniture, rather than being used solely to heat the air.
(b) The main methods of thermal energy transfer reduced by a double-glazed window are conduction and convection. The narrow air gap between the two panes of glass reduces heat transfer by conduction, and the sealed gap minimizes convection currents, thereby reducing heat loss from the room to the outside.
Question 6
Two types of seismic waves are P-waves and S-waves.
(a) State the types of wave that P-waves and S-waves can be modelled as. (Sub-topic – 3.1)
▶️Answer/Explanation
P-waves: longitudinal
S-waves: transverse
(b) The velocity of a P-wave in the Earth’s solid crust is 7.2 km/s and its frequency is 4.5 Hz. Calculate the wavelength of this P-wave. (Sub-topic – 3.1)
▶️Answer/Explanation
Solution:
The wave speed \( v \) is related to the frequency \( f \) and wavelength \( \lambda \) by the equation:
\[ v = f \lambda \]
Rearranging for \( \lambda \):
\[ \lambda = \frac{v}{f} \]
Substituting the given values:
\[ \lambda = \frac{7.2 \, \text{km/s}}{4.5 \, \text{Hz}} = 1.6 \, \text{km} \]
Therefore, the wavelength of the P-wave is 1.6 km.
Question 7 (Subtopics – 3.2.2, 3.3)
Fig. 7.1 shows a container of oil.
A ray of light shines on the surface of the oil. The refractive index of the oil is 1.47.
(a) On Fig. 7.1, draw the normal at the point where the ray enters the oil.
▶️Answer/Explanation
The normal should be drawn perpendicular to the surface of the oil at the point where the ray enters the oil.
(b) The angle \( x \) is \( 56^\circ \). Calculate the value of the angle of refraction.
▶️Answer/Explanation
Using Snell’s Law: \( n_1 \sin i = n_2 \sin r \), where \( n_1 = 1 \) (air), \( n_2 = 1.47 \) (oil), and \( i = 56^\circ \).
\( \sin r = \frac{\sin 56^\circ}{1.47} \)
\( \sin r = \frac{0.829}{1.47} \approx 0.564 \)
\( r = \sin^{-1}(0.564) \approx 34.3^\circ \)
The angle of refraction is approximately \( 34.3^\circ \).
(c) State the approximate speed of light in air.
▶️Answer/Explanation
The speed of light in air is approximately \( 3.0 \times 10^8 \, \text{m/s} \).
(d) Calculate the speed of light in the oil. Give your answer to three significant figures.
▶️Answer/Explanation
The speed of light in a medium is given by \( v = \frac{c}{n} \), where \( c \) is the speed of light in air and \( n \) is the refractive index of the medium.
\( v = \frac{3.0 \times 10^8 \, \text{m/s}}{1.47} \approx 2.04 \times 10^8 \, \text{m/s} \)
The speed of light in the oil is approximately \( 2.04 \times 10^8 \, \text{m/s} \).
Question 8
(a) (i) State what is meant by a magnetic field. (Subtopic – 4.1)
▶️Answer/Explanation
A magnetic field is a region in which a magnetic pole experiences a force.
(ii) Define the direction of a magnetic field.
▶️Answer/Explanation
The direction of a magnetic field at a point is the direction of the force on the north pole (N pole) of a magnet placed at that point.
(b) Fig. 8.1 shows a negatively charged metal sphere. (Subtopic – 4.2.1)
On Fig. 8.1, draw four lines to show the electric field and its direction.
▶️Answer/Explanation
The electric field lines should be drawn as radial lines pointing towards the negatively charged sphere. The lines should be equally spaced and touch the surface of the sphere. The direction of the electric field lines is towards the sphere because the sphere is negatively charged, and electric field lines point towards negative charges.
(c) Fig. 8.2 shows a circuit.
The three cells are identical and have zero resistance.
The resistors R1, R2 and R3 are identical.
The reading on the voltmeter is 6.0 V.
When the diode is conducting, it has zero resistance and zero potential difference (p.d.) across it
(i) Determine the e.m.f. of one cell. (Subtopic – 4.3.3)
▶️Answer/Explanation
The e.m.f. of one cell is 2.0 V. This is because the total e.m.f. of the three identical cells is 6.0 V (as given by the voltmeter reading), and since the cells are identical, the e.m.f. of one cell is 6.0 V divided by 3, which equals 2.0 V.
(ii) Determine the ratio of the p.d. across \( R_2 \) to the p.d. across \( R_3 \).
▶️Answer/Explanation
The ratio of the p.d. across \( R_2 \) to the p.d. across \( R_3 \) is 1:2. This is because \( R_2 \) and \( R_3 \) are identical resistors, and the potential difference across resistors in series is proportional to their resistance. Since \( R_3 \) has twice the resistance of \( R_2 \), the p.d. across \( R_3 \) is twice that across \( R_2 \).
(iii) 1. State and explain the change in current in \( R_1 \) when all the cells are reversed.
▶️Answer/Explanation
The current in \( R_1 \) becomes zero when all the cells are reversed. This is because the diode is now in the wrong direction (reverse-biased), and it will not allow current to flow through the circuit. Therefore, no current will pass through \( R_1 \).
2. Determine the new value of the ratio of the p.d. across \( R_2 \) to the p.d. across \( R_3 \) when all the cells are reversed.
▶️Answer/Explanation
The new ratio of the p.d. across \( R_2 \) to the p.d. across \( R_3 \) is 1:1. This is because, with the cells reversed, the diode blocks the current, and no current flows through the circuit. As a result, there is no potential difference across either \( R_2 \) or \( R_3 \), making the ratio 1:1.
Question 9 (Subtopics: 5.2.1, 5.2.2, 5.2.3, 5.2.4, 5.2.5)
(a) Table 9.1 shows some properties and values for α-particles, β-particles, and γ-radiation. Complete Table 9.1.
| Type of Radiation | Number of Protons | Number of Neutrons | Charge (C) | Stopped By |
|---|---|---|---|---|
| α | 2 | +3.2 × 10-19 | Thin sheet of paper | |
| β | 0 | Thin sheet of aluminium | ||
| γ | 0 |
▶️Answer/Explanation
Explanation:
– α-particles: Consist of 2 protons and 2 neutrons, giving them a charge of +3.2 × 10-19 C. They are stopped by a thin sheet of paper.
– β-particles: High-energy electrons with no protons or neutrons, giving them a charge of -1.6 × 10-19 C. They are stopped by a thin sheet of aluminium.
– γ-radiation: Electromagnetic waves with no charge or mass. They are stopped by very thick concrete or thick lead.
(b) State how β-decay changes the nucleus of an atom.
▶️Answer/Explanation
Answer: β-decay changes the nucleus by converting a neutron into a proton and an electron. The electron (β-particle) is emitted, and the proton remains in the nucleus.
Explanation: This process increases the atomic number by 1 while the mass number remains the same.
(c) A radiation detector used in a laboratory detects a background count rate of 30 counts/min. A radioactive source is placed in front of the radiation detector. The initial reading on the detector is 550 counts/min. The half-life of the source is 25 minutes. Calculate the expected reading on the detector after 75 minutes.
▶️Answer/Explanation
Answer: 95 counts/min
Explanation:
1. First, calculate the initial count rate due to the source: 550 counts/min (total) – 30 counts/min (background) = 520 counts/min.
2. The half-life is 25 minutes, so after 75 minutes, 3 half-lives have passed (75 ÷ 25 = 3).
3. After each half-life, the count rate is halved:
– After 1st half-life: 520 ÷ 2 = 260 counts/min
– After 2nd half-life: 260 ÷ 2 = 130 counts/min
– After 3rd half-life: 130 ÷ 2 = 65 counts/min
4. Add the background count rate: 65 + 30 = 95 counts/min.
(d) State two safety precautions taken when moving, using, or storing radioactive sources in a laboratory.
▶️Answer/Explanation
Answer:
1. Limit the time of exposure to radioactive sources.
2. Store radioactive sources in lead boxes to shield radiation.
Explanation: These precautions minimize exposure to harmful radiation and ensure safe handling of radioactive materials.
Question 10
(a) State the equation that defines the average orbital speed \( v \) of a planet. State the meaning of any symbols you use. (Sub-topic – 6.1.1)
▶️Answer/Explanation
The equation that defines the average orbital speed \( v \) of a planet is: \[ v = \frac{2\pi r}{T} \] Where:
- \( v \) is the average orbital speed of the planet.
- \( r \) is the average radius of the planet’s orbit.
- \( T \) is the orbital period of the planet.
(b) Suggest why countries that are a significant distance from the Equator experience significant temperature variation throughout the year. (Sub-topic – 6.1.1)
▶️Answer/Explanation
Countries that are a significant distance from the Equator experience significant temperature variation throughout the year because the angle at which the Sun’s rays strike the Earth’s surface changes with the seasons. During summer, the Sun’s rays are more direct, leading to higher temperatures, while during winter, the Sun’s rays are more oblique, leading to lower temperatures. This variation is more pronounced in regions farther from the Equator due to the tilt of the Earth’s axis.
(c) Fill in the gaps in the paragraph about a star much more massive than the Sun. (Sub-topic – 6.2.2)
The stage that follows the stable state in the life cycle of the star is the red supergiant stage. It then explodes as a supernova to form a nebula, this leaves behind a neutron star or a black hole.
▶️Answer/Explanation
The stage that follows the stable state in the life cycle of the star is the red supergiant stage. It then explodes as a supernova to form a nebula, this leaves behind a neutron star or a black hole.
(d) A galaxy is moving away from the Earth with a speed of 33 000 km/s. The value of the Hubble constant is \(2.2 \times 10^{-18}\) per second. Calculate the distance from the galaxy to the Earth. Give your answer in light-years. (Sub-topic – 6.2.3)
▶️Answer/Explanation
To calculate the distance from the galaxy to the Earth, we use the Hubble’s Law equation: \[ v = H_0 \times d \] Where:
- \( v \) is the speed of the galaxy (33,000 km/s).
- \( H_0 \) is the Hubble constant (\(2.2 \times 10^{-18}\) per second).
- \( d \) is the distance to the galaxy.
Rearranging the equation to solve for \( d \): \[ d = \frac{v}{H_0} \] Substituting the given values: \[ d = \frac{33,000 \, \text{km/s}}{2.2 \times 10^{-18} \, \text{s}^{-1}} = 1.5 \times 10^{22} \, \text{km} \] To convert this distance into light-years, we use the conversion factor \(1 \, \text{light-year} = 9.46 \times 10^{12} \, \text{km}\): \[ d = \frac{1.5 \times 10^{22} \, \text{km}}{9.46 \times 10^{12} \, \text{km/light-year}} \approx 1.6 \times 10^9 \, \text{light-years} \] Therefore, the distance from the galaxy to the Earth is approximately \(1.6 \times 10^9\) light-years.