An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Paper 22 選擇題 (Extended)
Answer all forty questions. Choose the correct option from A, B, C, or D.
40 题目 · 40 分
题目 1 · 選擇題
1 分
A block of mass \(1.5\text{ kg}\) moving at a speed of \(4.0\text{ m/s}\) to the right collides with a vertical wall. It rebounds with a speed of \(2.0\text{ m/s}\) to the left. The collision lasts for a time interval of \(0.15\text{ s}\). What is the average force exerted by the wall on the block?
A.20\text{ N}
B.30\text{ N}
C.40\text{ N}
D.60\text{ N}
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解题
To find the average force, we use the impulse-momentum theorem: \(F = \frac{\Delta p}{\Delta t}\). Taking the right direction as positive: \(p_{\text{initial}} = m \cdot v_i = 1.5\text{ kg} \times 4.0\text{ m/s} = 6.0\text{ kg m/s}\). \(p_{\text{final}} = m \cdot v_f = 1.5\text{ kg} \times (-2.0\text{ m/s}) = -3.0\text{ kg m/s}\). The change in momentum is \|\Delta p| = |-3.0 - 6.0| = 9.0\text{ kg m/s}\). Therefore, the average force is \(F = \frac{9.0\text{ N s}}{0.15\text{ s}} = 60\text{ N}\).
评分标准
D is correct because \(F = \Delta p / \Delta t = (6.0 - (-3.0)) / 0.15 = 60\text{ N}\). A is incorrect because it uses only initial momentum: \(6.0 / 0.15 = 40\text{ N}\) but then subtracts final momentum incorrectly. B is incorrect because it subtracts the magnitudes instead of vector values: \((6.0 - 3.0) / 0.15 = 20\text{ N}\). C is incorrect because of an arithmetic error.
题目 2 · 選擇題
1 分
An electric heater rated at \(800\text{ W}\) is used to heat a \(2.0\text{ kg}\) block of metal. The temperature of the block rises from \(20^\circ\text{C}\) to \(50^\circ\text{C}\) in \(120\text{ s}\). Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the metal?
A.400\text{ J / (kg }^\circ\text{C)}
B.800\text{ J / (kg }^\circ\text{C)}
C.1600\text{ J / (kg }^\circ\text{C)}
D.3200\text{ J / (kg }^\circ\text{C)}
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解题
The electrical energy supplied by the heater is \(E = P \cdot t = 800\text{ W} \times 120\text{ s} = 96\text{ }000\text{ J}\). The temperature change of the block is \(\Delta T = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}\). Using the formula \(E = m \cdot c \cdot \Delta T\), we get \(96\text{ }000 = 2.0\text{ kg} \times c \times 30^\circ\text{C}\). Solving for \(c\): \(c = \frac{96\text{ }000}{60} = 1600\text{ J / (kg }^\circ\text{C)}\).
评分标准
C is correct as calculated from \(c = \frac{P \cdot t}{m \cdot \Delta T}\). A is incorrect because it uses the wrong temperature value (using \(50 + 20 = 70\) or a similar combination). B is incorrect because it misses the factor of 2.0 kg. D is incorrect due to dividing by a factor of 2 instead of multiplying.
题目 3 · 選擇題
1 分
A potential divider circuit consists of an NTC thermistor and a fixed resistor of \(10\text{ k}\Omega\) connected in series across a constant \(12\text{ V}\) d.c. power supply. A voltmeter is connected across the thermistor. The temperature of the surroundings increases. What happens to the resistance of the thermistor and the reading on the voltmeter?
A.The resistance decreases and the voltmeter reading decreases.
B.The resistance decreases and the voltmeter reading increases.
C.The resistance increases and the voltmeter reading decreases.
D.The resistance increases and the voltmeter reading increases.
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解题
For an negative temperature coefficient (NTC) thermistor, its resistance decreases as the temperature increases. In a series potential divider circuit, as the resistance of the thermistor decreases, it takes a smaller fraction of the total supply voltage. Consequently, the reading on the voltmeter connected across the thermistor decreases.
评分标准
A is correct because the resistance of the NTC thermistor decreases, which reduces the potential difference across it. B is incorrect because a lower resistance cannot increase the potential difference across that component in series. C is incorrect because the resistance of an NTC thermistor decreases with temperature, not increases. D is incorrect for the same reason.
题目 4 · 選擇題
1 分
A specific spectral line in the light from a distant galaxy is analyzed. In the laboratory, the wavelength of this line is \(600\text{ nm}\). When observed from the galaxy, the wavelength of this line is \(612\text{ nm}\). What is the recession speed of this galaxy? (Take the speed of light to be \(3.0 \times 10^8\text{ m/s}\)).
A.2.0 \times 10^6\text{ m/s}
B.6.0 \times 10^6\text{ m/s}
C.1.2 \times 10^7\text{ m/s}
D.3.0 \times 10^8\text{ m/s}
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解题
Using the redshift formula: \(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\). First calculate the change in wavelength: \(\Delta \lambda = 612\text{ nm} - 600\text{ nm} = 12\text{ nm}\). Now substitute the values into the formula: \(\frac{12\text{ nm}}{600\text{ nm}} = \frac{v}{3.0 \times 10^8\text{ m/s}}\). This gives \(v = 0.02 \times 3.0 \times 10^8\text{ m/s} = 6.0 \times 10^6\text{ m/s}\).
评分标准
B is correct as \(v = (12/600) \times 3.0 \times 10^8 = 6.0 \times 10^6\text{ m/s}\). A is incorrect due to a calculation error using 612 directly or wrong subtraction. C is incorrect because it doubles the calculated speed. D is incorrect because it is the speed of light itself.
题目 5 · 選擇題
1 分
An ideal step-down transformer is connected to a \(240\text{ V}\) a.c. mains supply. The primary coil has \(1200\text{ turns}\) and the secondary coil has \(60\text{ turns}\). The secondary coil is connected to a resistor, producing a current of \(4.0\text{ A}\) in the secondary circuit. What is the current in the primary coil?
A.0.10\text{ A}
B.0.20\text{ A}
C.8.0\text{ A}
D.80\text{ A}
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解题
For an ideal transformer, the power input equals the power output: \(V_p \cdot I_p = V_s \cdot I_s\). Also, we know \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). This can be rearranged to give \(I_p = I_s \cdot \frac{N_s}{N_p}\). Substituting the values: \(I_p = 4.0\text{ A} \times \frac{60}{1200} = 4.0\text{ A} \times \frac{1}{20} = 0.20\text{ A}\).
评分标准
B is correct because \(I_p = I_s \cdot (N_s / N_p) = 4.0 \times (60 / 1200) = 0.20\text{ A}\). A is incorrect due to a scale error. C is incorrect because it multiplies by the turns ratio instead of dividing. D is incorrect because it represents the secondary current multiplied by the ratio reversed.
题目 6 · 選擇題
1 分
A ray of light passes from air into a rectangular glass block. The angle of incidence at the air-glass boundary is \(48.0^\circ\) and the angle of refraction is \(30.0^\circ\). What is the critical angle for light travelling inside this glass block towards a boundary with air?
A.30.0^\circ
B.42.3^\circ
C.45.0^\circ
D.48.0^\circ
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解题
First, calculate the refractive index \(n\) of the glass using Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(48.0^\circ)}{\sin(30.0^\circ)} = \frac{0.7431}{0.5000} \approx 1.486\). Next, find the critical angle \(c\) using \(\sin(c) = \frac{1}{n}\): \(\sin(c) = \frac{1}{1.486} \approx 0.6729\). Therefore, \(c = \sin^{-1}(0.6729) \approx 42.3^\circ\).
评分标准
B is correct as \(c = \sin^{-1}(\sin(30.0^\circ) / \sin(48.0^\circ)) = 42.3^\circ\). A is incorrect because it is just the angle of refraction. C is incorrect because it is a common guess close to critical angles of typical glass. D is incorrect because it is the angle of incidence.
题目 7 · 選擇題
1 分
A radioactive isotope has a half-life of \(6.0\text{ hours}\). A sample initially contains \(8.0 \times 10^{20}\) active nuclei of this isotope. How many of these nuclei decay during a time period of \(24\text{ hours}\)?
A.5.0 \times 10^{19}
B.2.0 \times 10^{20}
C.7.5 \times 10^{20}
D.8.0 \times 10^{20}
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解题
The number of half-lives that pass in \(24\text{ hours}\) is \(n = \frac{24\text{ hours}}{6.0\text{ hours}} = 4\). After 4 half-lives, the fraction of nuclei remaining is \(\left(\frac{1}{2}\right)^4 = \frac{1}{16}\). The number of active nuclei remaining is \(N_{\text{remaining}} = \frac{8.0 \times 10^{20}}{16} = 5.0 \times 10^{19}\). Therefore, the number of nuclei that have decayed is \(N_{\text{decayed}} = 8.0 \times 10^{20} - 5.0 \times 10^{19} = 7.5 \times 10^{20}\).
评分标准
C is correct because \(7.5 \times 10^{20}\) nuclei decayed. A is incorrect because it is the number of remaining nuclei. B is incorrect because it assumes a linear decay over time. D is incorrect because it is the initial amount of active nuclei.
题目 8 · 選擇題
1 分
A container is filled with water of density \(1.0 \times 10^3\text{ kg/m}^3\). At a certain depth, the pressure due to the water only is \(1.5 \times 10^5\text{ Pa}\). The container is then emptied and refilled with oil of density \(8.0 \times 10^2\text{ kg/m}^3\) to the same depth. What is the pressure due to the oil only at this depth?
A.1.2 \times 10^5\text{ Pa}
B.1.5 \times 10^5\text{ Pa}
C.1.9 \times 10^5\text{ Pa}
D.2.4 \times 10^5\text{ Pa}
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解题
The pressure due to a liquid column is given by \(p = \rho \cdot g \cdot h\). Since the depth \(h\) and the gravitational field strength \(g\) remain constant, the pressure is directly proportional to the density \(\rho\). Therefore, \(p_{\text{oil}} = p_{\text{water}} \times \frac{\rho_{\text{oil}}}{\rho_{\text{water}}} = 1.5 \times 10^5\text{ Pa} \times \frac{8.0 \times 10^2\text{ kg/m}^3}{1.0 \times 10^3\text{ kg/m}^3} = 1.2 \times 10^5\text{ Pa}\).
评分标准
A is correct because pressure scales proportionally with density: \(1.5 \times 10^5 \times 0.8 = 1.2 \times 10^5\text{ Pa}\). B is incorrect because it is the pressure of the water. C is incorrect because it incorrectly increases the pressure. D is incorrect because it is a result of a multiplier inversion: \(1.5 \times 10^5 / 0.8 = 1.875 \times 10^5\text{ Pa}\) rounded.
题目 9 · 選擇題
1 分
A cyclist travels along a straight road. She accelerates uniformly from rest for 4.0 s to a speed of 10 m/s. She then travels at this constant speed of 10 m/s for 12 s, and finally decelerates uniformly to rest in 4.0 s.
What is the average speed of the cyclist for the entire journey?
A.5.0 m/s
B.8.0 m/s
C.10 m/s
D.16 m/s
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解题
First, calculate the total distance travelled by finding the area under the speed-time graph: - During acceleration (0 to 4.0 s): distance = 0.5 * 4.0 s * 10 m/s = 20 m. - During constant speed (4.0 to 16 s): distance = 12 s * 10 m/s = 120 m. - During deceleration (16 to 20 s): distance = 0.5 * 4.0 s * 10 m/s = 20 m.
Total distance = 20 + 120 + 20 = 160 m. Total time = 4.0 + 12 + 4.0 = 20 s.
Average speed = (total distance) / (total time) = 160 m / 20 s = 8.0 m/s.
评分标准
B is correct. Finding total distance (160 m) and dividing by total time (20 s) yields 8.0 m/s.
题目 10 · 選擇題
1 分
A toy car of mass 0.40 kg is travelling at a velocity of 3.0 m/s to the right. It collides with a wall and rebounds with a velocity of 2.0 m/s to the left. The collision lasts for a time of 0.050 s.
What is the average force exerted by the wall on the toy car?
A.8.0 N
B.24 N
C.40 N
D.100 N
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解题
Taking the direction to the right as positive: - Initial velocity, u = +3.0 m/s - Final velocity, v = -2.0 m/s
Change in momentum, Δp = m(v - u) = 0.40 kg * (-2.0 - 3.0) m/s = -2.0 kg m/s (or N s in magnitude).
Average force, F = Δp / Δt = -2.0 N s / 0.050 s = -40 N.
The magnitude of the force is 40 N.
评分标准
C is correct. Calculates impulse as 2.0 N s by taking direction change into account, then divides by 0.050 s to get 40 N.
题目 11 · 選擇題
1 分
An electric heater rated at 60 W is used to heat a block of metal of mass 1.2 kg. The heater is switched on for 5.0 minutes. The temperature of the block increases from 20°C to 50°C.
Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the metal?
A.8.3 J / (kg °C)
B.300 J / (kg °C)
C.500 J / (kg °C)
D.15 000 J / (kg °C)
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解题
Thermal energy supplied, E = P * t = 60 W * (5.0 * 60) s = 18 000 J.
Temperature rise, Δθ = 50°C - 20°C = 30°C.
Using E = mcΔθ: c = E / (m * Δθ) = 18 000 / (1.2 * 30) = 500 J / (kg °C).
评分标准
C is correct. Converting time to 300 s, calculating thermal energy as 18 000 J, and dividing by mass (1.2 kg) and temperature change (30 °C) gives 500 J / (kg °C).
题目 12 · 選擇題
1 分
A detector is used to measure the count rate from a radioactive source. The background count rate is constant at 15 counts / second.
At time t = 0, the measured count rate is 135 counts / second. At time t = 6.0 hours, the measured count rate is 30 counts / second.
What is the half-life of the source?
A.1.5 hours
B.2.0 hours
C.3.0 hours
D.4.5 hours
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解题
Subtract background radiation from the measured count rates to find the count rates due to the source alone: - At t = 0, source count rate = 135 - 15 = 120 counts / second. - At t = 6.0 hours, source count rate = 30 - 15 = 15 counts / second.
The fraction of the source remaining is 15 / 120 = 1/8. Since 1/8 = (1/2)^3, exactly 3 half-lives have passed in 6.0 hours.
One half-life = 6.0 hours / 3 = 2.0 hours.
评分标准
B is correct. Subtracting background (15) to get source activity (120 to 15), recognizing this as 3 half-lives, and dividing 6.0 hours by 3 gives 2.0 hours.
题目 13 · 選擇題
1 分
A wire of length L and circular cross-section of diameter d has a resistance R.
Another wire is made of the same metal. It has a length of 3L and a diameter of 2d.
What is the resistance of the second wire in terms of R?
A.0.375 R
B.0.75 R
C.1.5 R
D.6.0 R
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解题
Resistance R of a wire is given by R = ρ * (L / A). Since the cross-sectional area A is proportional to d^2, the resistance R is proportional to L / d^2.
For the second wire: R' is proportional to 3L / (2d)^2 = 3L / 4d^2 = 0.75 * (L / d^2).
Therefore, the resistance of the second wire is 0.75 R.
评分标准
B is correct. Using the relation that resistance is proportional to length and inversely proportional to the square of diameter gives a factor of 3/4 = 0.75.
题目 14 · 選擇題
1 分
Light travels from a glass block into air. The refractive index of the glass is 1.5. The angle of incidence of the light inside the glass is 35°.
What is the angle of refraction of the light in the air?
A.23°
B.35°
C.59°
D.No light emerges because total internal reflection occurs.
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解题
Using Snell's law for refraction from glass into air: n_glass * sin(θ_glass) = n_air * sin(θ_air)
(Note: The critical angle θ_c is arcsin(1/1.5) ≈ 41.8°. Since 35° < 41.8°, refraction occurs and light emerges into the air).
评分标准
C is correct. Applying Snell's law correctly to find the angle of refraction as 59°.
题目 15 · 選擇題
1 分
Light from a distant galaxy is observed on Earth. A specific spectral line, which has a wavelength of 400 nm when measured in a laboratory on Earth, is detected at a wavelength of 412 nm in the light from the galaxy.
The speed of light is 3.0 * 10^8 m/s.
What is the recession speed of the galaxy?
A.9.0 * 10^5 m/s
B.9.0 * 10^6 m/s
C.1.2 * 10^7 m/s
D.3.1 * 10^8 m/s
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解题
First, calculate the change in wavelength: Δλ = 412 nm - 400 nm = 12 nm.
Using the redshift formula: Δλ / λ_0 = v / c
12 / 400 = v / (3.0 * 10^8)
v = 0.030 * 3.0 * 10^8 = 9.0 * 10^6 m/s.
评分标准
B is correct. Calculating redshift as 0.030 and multiplying by the speed of light yields 9.0 * 10^6 m/s.
题目 16 · 選擇題
1 分
A submarine is submerged in seawater of density 1020 kg/m^3. The atmospheric pressure at the sea surface is 1.0 * 10^5 Pa.
The total pressure at the depth of the submarine is 6.0 * 10^5 Pa. The acceleration of free fall g is 9.8 m/s^2.
What is the depth of the submarine below the surface of the sea?
A.10 m
B.50 m
C.60 m
D.510 m
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解题
The total pressure P_total is the sum of atmospheric pressure P_atm and the hydrostatic pressure of the seawater: P_total = P_atm + ρ * g * h
6.0 * 10^5 Pa = 1.0 * 10^5 Pa + (1020 kg/m^3 * 9.8 m/s^2 * h)
5.0 * 10^5 Pa = 9996 * h
h = 500 000 / 9996 ≈ 50 m.
评分标准
B is correct. Subtracts atmospheric pressure to find hydrostatic pressure (5.0 * 10^5 Pa), then divides by density and g to obtain approximately 50 m.
题目 17 · 選擇題
1 分
A student determines the period of a simple pendulum. She starts a digital stopwatch when the pendulum bob passes the center of its swing. She stops the stopwatch after 25 complete oscillations. The stopwatch reading is \(35.5\text{ s}\).
What is the period of the pendulum, and why does she time 25 oscillations rather than just one?
A.\(0.70\text{ s}\); to eliminate systematic errors from the stopwatch
B.\(0.70\text{ s}\); to reduce the effect of random errors due to reaction time
C.\(1.42\text{ s}\); to eliminate systematic errors from the stopwatch
D.\(1.42\text{ s}\); to reduce the effect of random errors due to reaction time
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解题
1. Calculate the period \(T\): \[T = \frac{\text{total time}}{\text{number of oscillations}} = \frac{35.5\text{ s}}{25} = 1.42\text{ s}\]
2. The reason for timing multiple oscillations (like 25) instead of just one is to reduce the percentage uncertainty (or random error) introduced by human reaction time when starting and stopping the stopwatch.
评分标准
1 mark: Correct calculation of period (1.42 s) and correct justification (reduces the effect of random error due to reaction time).
题目 18 · 選擇題
1 分
A car travels along a straight road at a constant speed of \(15\text{ m/s}\) for \(6.0\text{ s}\). The brakes are then applied and it decelerates uniformly to rest in a further \(4.0\text{ s}\).
What is the magnitude of the deceleration of the car and the total distance traveled during the \(10.0\text{ s}\)?
A.deceleration \(= 2.7\text{ m/s}^2\), total distance \(= 90\text{ m}\)
B.deceleration \(= 3.8\text{ m/s}^2\), total distance \(= 120\text{ m}\)
C.deceleration \(= 3.8\text{ m/s}^2\), total distance \(= 150\text{ m}\)
D.deceleration \(= 15\text{ m/s}^2\), total distance \(= 120\text{ m}\)
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解题
1. **Deceleration** during the last \(4.0\text{ s}\): \[a = \frac{v - u}{t} = \frac{0 - 15}{4.0} = -3.75\text{ m/s}^2\] Rounding to two significant figures gives a magnitude of \(3.8\text{ m/s}^2\).
2. **Total Distance**: The motion consists of two parts: - Constant speed phase: \(d_1 = v \times t = 15 \times 6.0 = 90\text{ m}\). - Deceleration phase (area of a triangle): \(d_2 = \frac{1}{2} \times u \times t = \frac{1}{2} \times 15 \times 4.0 = 30\text{ m}\). - Total distance \(d_{\text{total}} = 90 + 30 = 120\text{ m}\).
评分标准
1 mark: Correct deceleration magnitude (3.8 m/s²) and correct total distance (120 m).
题目 19 · 選擇題
1 分
An object moves in a circular path at a constant speed.
Which statement about the resultant force acting on the object is correct?
A.The force is zero because the speed is constant.
B.The force is directed along the tangent to the circular path, in the direction of motion.
C.The force is directed along the tangent to the circular path, opposite to the direction of motion.
D.The force is directed perpendicular to the direction of motion, towards the center of the circular path.
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解题
For an object in uniform circular motion, the speed is constant but the direction of velocity is continuously changing, which means the object is accelerating. This acceleration (and the resultant force causing it) is centripetal, meaning it is directed towards the center of the circle, perpendicular to the instantaneous velocity (direction of motion) at any point.
评分标准
1 mark: Identify that the centripetal force is perpendicular to motion, pointing towards the center of the circular path.
题目 20 · 選擇題
1 分
An electric motor is used to lift a metal block of mass \(15\text{ kg}\) vertically upwards through a height of \(8.0\text{ m}\) in \(5.0\text{ s}\).
The electrical power input to the motor is \(360\text{ W}\).
Take the gravitational field strength \(g\) to be \(9.8\text{ N/kg}\).
What is the efficiency of the motor?
A.\(6.7\%\)
B.\(33\%\)
C.\(65\%\)
D.\(83\%\)
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解题
1. **Work done against gravity (useful energy output)**: \[E = mgh = 15 \times 9.8 \times 8.0 = 1176\text{ J}\]
3. **Efficiency**: \[\text{Efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}} \times 100\% = \frac{235.2}{360} \times 100\% \approx 65.3\%\] Rounding to two significant figures gives \(65\%\).
评分标准
1 mark: Correct calculation of useful power output (235.2 W) and overall efficiency (~65%).
题目 21 · 選擇題
1 分
A fixed mass of gas is held in a cylinder by a piston. The initial volume of the gas is \(120\text{ cm}^3\) and its pressure is \(1.5 \times 10^5\text{ Pa}\).
The piston is slowly pushed in, so that the temperature of the gas remains constant, until the volume is reduced to \(45\text{ cm}^3\).
What is the new pressure of the gas?
A.\(5.6 \times 10^4\text{ Pa}\)
B.\(1.5 \times 10^5\text{ Pa}\)
C.\(4.0 \times 10^5\text{ Pa}\)
D.\(5.4 \times 10^6\text{ Pa}\)
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解题
According to Boyle's law for a constant mass of gas at a constant temperature: \[p_1 V_1 = p_2 V_2\] Substitute the given values: \[(1.5 \times 10^5\text{ Pa}) \times 120\text{ cm}^3 = p_2 \times 45\text{ cm}^3\] \[p_2 = \frac{1.8 \times 10^7}{45} = 4.0 \times 10^5\text{ Pa}\]
评分标准
1 mark: Identify and correctly apply Boyle's law equation to obtain \(4.0 \times 10^5\text{ Pa}\).
题目 22 · 選擇題
1 分
Water waves in a ripple tank approach a gap in a barrier.
Which change increases the amount of diffraction (spreading) of the waves as they pass through the gap?
A.decreasing the wavelength of the waves
B.decreasing the width of the gap in the barrier
C.increasing the amplitude of the waves
D.increasing the frequency of the waves
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解题
Diffraction is the spreading of waves when they pass through a gap or around an obstacle. It is most significant when the wavelength of the waves is comparable to or larger than the size of the gap. Decreasing the gap width makes the gap closer in size to the wavelength, thereby increasing the amount of diffraction (spreading).
评分标准
1 mark: Correctly identify that a narrower gap relative to wavelength leads to greater wave diffraction.
题目 23 · 選擇題
1 分
A \(12\text{ V}\) battery of negligible internal resistance is connected in series with a resistor of resistance \(4.0\ \Omega\) and a parallel combination of two resistors of resistances \(3.0\ \Omega\) and \(6.0\ \Omega\).
What is the current in the \(4.0\ \Omega\) resistor?
A.\(1.0\text{ A}\)
B.\(2.0\text{ A}\)
C.\(3.0\text{ A}\)
D.\(4.0\text{ A}\)
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解题
1. **Find the equivalent resistance of the parallel combination (\(R_p\))**: \[R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{3.0 \times 6.0}{3.0 + 6.0} = 2.0\ \Omega\]
2. **Find the total resistance of the circuit (\(R_{\text{total}}\))**: Since the \(4.0\ \Omega\) resistor is in series with the parallel combination: \[R_{\text{total}} = 4.0\ \Omega + R_p = 4.0 + 2.0 = 6.0\ \Omega\]
3. **Find the total current from the battery**: \[I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\] Since the \(4.0\ \Omega\) resistor is connected in series with the main loop, the full total current of \(2.0\text{ A}\) flows through it.
评分标准
1 mark: Correct parallel resistance calculation (2.0 Ω), total circuit resistance (6.0 Ω), and total current (2.0 A).
题目 24 · 選擇題
1 分
Which sequence correctly describes the life cycle of a star with a mass much larger than the mass of the Sun?
A.protostar \(\rightarrow\) stable star (main sequence) \(\rightarrow\) red giant \(\rightarrow\) white dwarf
B.protostar \(\rightarrow\) stable star (main sequence) \(\rightarrow\) red supergiant \(\rightarrow\) supernova \(\rightarrow\) neutron star or black hole
C.stable star (main sequence) \(\rightarrow\) protostar \(\rightarrow\) planetary nebula \(\rightarrow\) white dwarf
D.stable star (main sequence) \(\rightarrow\) red supergiant \(\rightarrow\) supernova \(\rightarrow\) planetary nebula \(\rightarrow\) black hole
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解题
High-mass stars begin as protostars, become stable main sequence stars, then expand into red supergiants when their core hydrogen is depleted. The core then collapses in a massive explosion called a supernova, leaving behind either a neutron star or a black hole depending on the remaining core mass.
评分标准
1 mark: Identify the correct sequence of stages for a high-mass star.
题目 25 · 選擇題
1 分
A toy car travels along a straight track. It has an initial velocity of 5.0 m / s and accelerates uniformly at 4.0 m / s^{2} over a distance of 18 m. What is the final velocity of the toy car?
A.12 m / s
B.13 m / s
C.17 m / s
D.77 m / s
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解题
Using the equations of uniform acceleration: v^{2} = u^{2} + 2as. Substituting the given values: v^{2} = 5.0^{2} + 2 \times 4.0 \times 18 = 25 + 144 = 169. Therefore, v = \sqrt{169} = 13 m / s.
评分标准
1 mark for the correct option B.
题目 26 · 選擇題
1 分
An open-topped cylindrical beaker contains a layer of oil floating on water. The depth of the oil layer is 0.12 m and its density is 800 kg / m^{3}. The depth of the water layer is 0.25 m and its density is 1000 kg / m^{3}. The gravitational field strength g is 9.8 N / kg. What is the pressure exerted by the two liquids on the bottom of the beaker?
A.940 Pa
B.2500 Pa
C.3400 Pa
D.3600 Pa
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解题
The pressure due to a liquid column is given by p = \rho g h. The total pressure at the bottom is the sum of the pressures due to each liquid layer: p_{\text{total}} = (\rho_{\text{oil}} \times g \times h_{\text{oil}}) + (\rho_{\text{water}} \times g \times h_{\text{water}}). p_{\text{total}} = (800 \times 9.8 \times 0.12) + (1000 \times 9.8 \times 0.25) = 940.8 + 2450 = 3390.8 Pa, which is approximately 3400 Pa to two significant figures.
评分标准
1 mark for the correct option C.
题目 27 · 選擇題
1 分
An atom of carbon-14 has a proton number of 6 and a nucleon number of 14. This nucleus absorbs a single neutron to form a new nucleus. Which row correctly identifies the nucleon number, proton number, and neutron number of the new nucleus?
| | Nucleon number | Proton number | Neutron number | |---|---|---|---| | A | 15 | 6 | 9 | | B | 15 | 7 | 8 | | C | 14 | 5 | 9 | | D | 14 | 6 | 8 |
A.Row A
B.Row B
C.Row C
D.Row D
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解题
The original carbon-14 nucleus contains 6 protons and 8 neutrons (since 14 - 6 = 8). When it absorbs a neutron, the number of protons remains 6, so it is still carbon. The number of neutrons increases by 1 to become 9. The nucleon number (total protons + neutrons) becomes 15 (since 14 + 1 = 15).
评分标准
1 mark for the correct option A.
题目 28 · 選擇題
1 分
A radioactive sample has an initial activity of 1200 Bq. After a time of 18 hours, the activity of the sample has decreased to 150 Bq. What is the half-life of the radioactive isotope in this sample?
A.3.0 hours
B.4.5 hours
C.6.0 hours
D.9.0 hours
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解题
The activity decreases from 1200 Bq to 150 Bq. Let's find the number of half-lives: 1200 -> 600 (1 half-life) -> 300 (2 half-lives) -> 150 (3 half-lives). Thus, 18 hours is equal to 3 half-lives: 3 \times T_{1/2} = 18 \implies T_{1/2} = 6.0 hours.
评分标准
1 mark for the correct option C.
题目 29 · 選擇題
1 分
A ray of light in air is incident on the surface of a glass block at an angle of incidence of 40^{\circ}. The refractive index of the glass is 1.50. What is the angle of refraction inside the glass block?
A.25^{\circ}
B.27^{\circ}
C.40^{\circ}
D.74^{\circ}
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解题
Snell's Law states n = \sin(i) / \sin(r). Rearranging for \sin(r): \sin(r) = \sin(i) / n = \sin(40^{\circ}) / 1.50 = 0.6428 / 1.50 = 0.4285. Thus, r = \sin^{-1}(0.4285) \approx 25.37^{\circ}, which is approximately 25^{\circ}.
评分标准
1 mark for the correct option A.
题目 30 · 選擇題
1 分
An electric motor is used to lift a load of mass 50 kg vertically upwards through a height of 12 m in a time of 8.0 s. The efficiency of the motor is 60%. The gravitational field strength g is 9.8 N / kg. What is the electrical power input to the motor?
A.440 W
B.740 W
C.1200 W
D.3500 W
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解题
The useful work output is the gain in gravitational potential energy: W_{\text{out}} = m g h = 50 \times 9.8 \times 12 = 5880 J. The useful power output is: P_{\text{out}} = W_{\text{out}} / t = 5880 / 8.0 = 735 W. Since the efficiency is 60%, the electrical power input is: P_{\text{in}} = P_{\text{out}} / 0.60 = 735 / 0.60 = 1225 W. To two significant figures, this is 1200 W.
评分标准
1 mark for the correct option C.
题目 31 · 選擇題
1 分
A metal block of mass 2.5 kg is heated by an electric heater of power 150 W for a time of 4.0 minutes. The temperature of the block rises by 16^{\circ}C. Assume that all the thermal energy from the heater is transferred to the block. What is the specific heat capacity of the metal?
A.230 J / (kg^{\circ}C)
B.900 J / (kg^{\circ}C)
C.1400 J / (kg^{\circ}C)
D.36000 J / (kg^{\circ}C)
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解题
The energy supplied by the heater is: E = P \times t = 150 \times (4.0 \times 60) = 150 \times 240 = 36000 J. Using the thermal energy equation: E = m c \Delta T \implies 36000 = 2.5 \times c \times 16 \implies 36000 = 40 c \implies c = 36000 / 40 = 900 J / (kg^{\circ}C).
评分标准
1 mark for the correct option B.
题目 32 · 選擇題
1 分
A battery of electromotive force (e.m.f.) 12 V is connected to a circuit. Over a period of 5.0 minutes, a charge of 150 C flows through the battery. How much chemical energy is converted to electrical energy in the battery during this time?
A.6.0 J
B.60 J
C.1800 J
D.9000 J
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解题
By definition, the e.m.f. of a battery is the energy transferred per unit charge: E = W / Q. Therefore, the energy converted is: W = e.m.f. \times Q = 12 V \times 150 C = 1800 J. Note that the time of 5.0 minutes is extra information and is not required for this calculation.
评分标准
1 mark for the correct option C.
题目 33 · multiple_choice
1 分
A cyclist accelerates uniformly from a speed of 2.0 m / s to a speed of 8.0 m / s with a constant acceleration of 1.5 m / s^{2}. What is the distance travelled by the cyclist while accelerating?
A.4.0 m
B.20 m
C.22 m
D.30 m
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解题
The relationship between initial speed u, final speed v, acceleration a, and distance s is given by: v^{2} = u^{2} + 2as. Substituting the given values: 8.0^{2} = 2.0^{2} + 2(1.5)s which simplifies to 64 = 4.0 + 3.0s, so 3.0s = 60, resulting in s = 20 m.
评分标准
B is correct. 1 mark for correct selection of 20 m using v^2 = u^2 + 2as.
题目 34 · multiple_choice
1 分
A toy car of mass 0.50 kg travelling at 6.0 m / s collides with a stationary toy truck of mass 1.5 kg. After the collision, the toy car rebounds with a speed of 1.5 m / s in the opposite direction. What is the speed of the toy truck after the collision?
A.1.5 m / s
B.2.0 m / s
C.2.5 m / s
D.3.0 m / s
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解题
Using the principle of conservation of momentum: Total initial momentum = Total final momentum. m1*u1 + m2*u2 = m1*v1 + m2*v2. Taking the initial direction of the car as positive: (0.50 kg * 6.0 m / s) + 0 = (0.50 kg * (-1.5 m / s)) + (1.5 kg * v2) which gives 3.0 = -0.75 + 1.5*v2, so 3.75 = 1.5*v2, leading to v2 = 2.5 m / s.
评分标准
C is correct. 1 mark for calculating final speed of truck as 2.5 m/s using conservation of momentum.
题目 35 · multiple_choice
1 分
A fixed mass of gas is cooled in a container of constant volume. Which statement describes the behaviour of the gas particles?
A.They hit the walls of the container harder and more frequently.
B.They hit the walls of the container harder but less frequently.
C.They hit the walls of the container less hard and less frequently.
D.They hit the walls of the container less hard but more frequently.
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解题
When a gas is cooled, its temperature decreases, meaning the average kinetic energy and speed of the gas particles decrease. As a result, the particles collide with the walls of the container with less force (less hard) and, since they are moving slower, they collide less frequently.
评分标准
C is correct. 1 mark for identifying that cooler particles collide less hard and less frequently.
题目 36 · multiple_choice
1 分
A battery transfers 45 J of electrical energy when a charge of 15 C passes through it. What is the electromotive force (e.m.f.) of the battery?
A.0.33 V
B.3.0 V
C.30 V
D.675 V
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解题
The electromotive force (e.m.f.) is defined as the work done per unit charge: E = W / Q. Substituting the values: E = 45 J / 15 C = 3.0 V.
评分标准
B is correct. 1 mark for calculating e.m.f. as 3.0 V.
题目 37 · multiple_choice
1 分
A ray of light in air is incident on the surface of a glass block. The angle of incidence is 45 degrees. The refractive index of the glass is 1.52. What is the angle of refraction in the glass?
A.28 degrees
B.30 degrees
C.45 degrees
D.68 degrees
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解题
According to Snell's Law: n = sin(i) / sin(r), where n is the refractive index, i is the angle of incidence, and r is the angle of refraction. Rearranging for r: sin(r) = sin(i) / n = sin(45 degrees) / 1.52 = 0.7071 / 1.52 = 0.4652. Thus, r = sin^-1(0.4652) = 27.7 degrees, which rounds to 28 degrees.
评分标准
A is correct. 1 mark for applying Snell's law to find the angle of refraction as 28 degrees.
题目 38 · multiple_choice
1 分
The count rate from a radioactive source is measured. The background count rate is constant at 20 counts / minute. The measured count rates (including background) are 340 counts / minute at 0 hours, 180 counts / minute at 2 hours, 100 counts / minute at 4 hours, and 60 counts / minute at 6 hours. What is the half-life of the radioactive source?
A.1.5 hours
B.2.0 hours
C.2.7 hours
D.3.0 hours
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解题
First, subtract the background count rate of 20 counts / minute to find the corrected count rates of the source: At 0 hours: 340 - 20 = 320 counts / minute; At 2 hours: 180 - 20 = 160 counts / minute; At 4 hours: 100 - 20 = 80 counts / minute; At 6 hours: 60 - 20 = 40 counts / minute. The corrected count rate halves every 2.0 hours (320 to 160, then 80, then 40). Therefore, the half-life is 2.0 hours.
评分标准
B is correct. 1 mark for subtracting background to find corrected counts and determining the half-life is 2.0 hours.
题目 39 · multiple_choice
1 分
A satellite orbits a planet at an altitude of 600 km above the planet's surface. The radius of the planet is 3400 km. The orbital period of the satellite is 2.0 hours. What is the average orbital speed of the satellite?
A.0.52 km / s
B.2.9 km / s
C.3.5 km / s
D.21 km / s
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解题
The radius of the orbit, r, is the sum of the planet's radius and the altitude: r = 3400 km + 600 km = 4000 km = 4.0 * 10^6 m. The orbital period, T, in seconds is: T = 2.0 hours = 2.0 * 3600 s = 7200 s. The average orbital speed, v, is: v = 2*pi*r / T = 2 * pi * 4.0 * 10^6 m / 7200 s = 3491 m / s = 3.5 km / s.
评分标准
C is correct. 1 mark for finding orbital radius as 4000 km and calculating speed as 3.5 km/s.
题目 40 · multiple_choice
1 分
The gravitational field strength on the Earth is 9.8 N / kg and on Mars is 3.7 N / kg. An astronaut has a mass of 75 kg on the Earth. What are the astronaut's mass and weight on Mars?
A.Mass = 75 kg, Weight = 280 N
B.Mass = 75 kg, Weight = 740 N
C.Mass = 28 kg, Weight = 280 N
D.Mass = 28 kg, Weight = 740 N
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解题
The mass of an object is constant and does not change with location, so the mass on Mars is still 75 kg. The weight on Mars is calculated using W = m * g: W = 75 kg * 3.7 N / kg = 277.5 N, which rounds to 280 N.
评分标准
A is correct. 1 mark for stating mass is unchanged at 75 kg and weight is 280 N.
Answer all questions. Show all working and write answers in the spaces provided.
10 题目 · 80 分
题目 1 · Structured Theory
8 分
A vehicle of mass \( 1200\text{ kg} \) is travelling along a straight horizontal road. (a) The vehicle accelerates from rest with a constant acceleration of \( 2.5\text{ m/s}^2 \) for a time of \( 6.0\text{ s} \). (i) Calculate the speed of the vehicle at \( t = 6.0\text{ s} \). (ii) Determine the distance travelled by the vehicle during these \( 6.0\text{ s} \). (b) At \( t = 6.0\text{ s} \), the driver sees an obstacle ahead and applies the brakes. The vehicle decelerates to a stop over a distance of \( 45\text{ m} \) with a constant deceleration. (i) Calculate the time taken for the vehicle to decelerate to rest. (ii) State and explain how a wet road surface affects the stopping distance of the vehicle.
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解题
(a)(i) using \( v = u + at \): \( v = 0 + (2.5 \times 6.0) = 15\text{ m/s} \). (a)(ii) using \( s = ut + \frac{1}{2}at^2 \): \( s = 0 + (0.5 \times 2.5 \times 6.0^2) = 45\text{ m} \). (b)(i) Average speed during deceleration \( = \frac{15 + 0}{2} = 7.5\text{ m/s} \). Since \( s = \text{average speed} \times t \), we have \( 45 = 7.5 \times t \implies t = 6.0\text{ s} \). (b)(ii) A wet road reduces the friction/grip between the tyres and the road surface, which reduces the decelerating force and deceleration rate, thereby increasing the stopping distance.
A block of ice of mass \( 0.25\text{ kg} \) at a temperature of \( -10^\circ\text{C} \) is heated until it becomes liquid water at \( 20^\circ\text{C} \). The specific heat capacity of ice is \( 2100\text{ J/(kg}^\circ\text{C)} \). The specific latent heat of fusion of ice is \( 3.3 \times 10^5\text{ J/kg} \). The specific heat capacity of water is \( 4200\text{ J/(kg}^\circ\text{C)} \). (a) Calculate: (i) the thermal energy required to raise the temperature of the ice to \( 0^\circ\text{C} \). (ii) the thermal energy required to melt the ice at \( 0^\circ\text{C} \). (iii) the thermal energy required to raise the temperature of the liquid water to \( 20^\circ\text{C} \). (b) State and explain one difference in the molecular structure of ice compared to liquid water.
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解题
(a)(i) \( E_1 = mc_{\text{ice}}\Delta\theta = 0.25 \times 2100 \times (0 - (-10)) = 5250\text{ J} \). (a)(ii) \( E_2 = mL = 0.25 \times (3.3 \times 10^5) = 82500\text{ J} \). (a)(iii) \( E_3 = mc_{\text{water}}\Delta\theta = 0.25 \times 4200 \times (20 - 0) = 21000\text{ J} \). (b) In ice, the molecules are held in fixed positions inside a regular lattice structure by strong intermolecular bonds. In liquid water, the molecules are in a random arrangement and are free to move and slide past one another because the bonds are weaker.
A ray of monochromatic light passes from air into a semi-circular glass block. (a) The angle of incidence in air is \( 52^\circ \) and the angle of refraction in the glass is \( 31^\circ \). (i) Calculate the refractive index of the glass. (ii) Calculate the speed of light in the glass. The speed of light in air is \( 3.0 \times 10^8\text{ m/s} \). (b) The ray of light inside the glass block travels towards the flat boundary from inside the block. (i) Describe what is meant by the term critical angle. (ii) Calculate the critical angle for this glass-to-air boundary.
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解题
(a)(i) using Snell's Law: \( n = \frac{\sin i}{\sin r} = \frac{\sin 52^\circ}{\sin 31^\circ} = \frac{0.7880}{0.5150} \approx 1.53 \). (a)(ii) using \( v = \frac{c}{n} \): \( v = \frac{3.0 \times 10^8}{1.53} \approx 1.96 \times 10^8\text{ m/s} \). (b)(i) The critical angle is the angle of incidence in the optically denser medium at which the angle of refraction in the less dense medium is \( 90^\circ \). (b)(ii) using \( \sin c = \frac{1}{n} \): \( \sin c = \frac{1}{1.53} = 0.6536 \implies c = \sin^{-1}(0.6536) \approx 41^\circ \).
评分标准
(a)(i) Correct formula or substitution: \( \sin 52 / \sin 31 \) [1 mark], Correct calculation: \( 1.53 \) [1 mark]. (a)(ii) Correct formula or substitution: \( 3.0 \times 10^8 / 1.53 \) [1 mark], Correct calculation: \( 1.96 \times 10^8\text{ m/s} \) [1 mark]. (b)(i) Mention of angle of incidence in denser medium [1 mark], Refracted ray travels along the boundary / angle of refraction is 90 degrees [1 mark]. (b)(ii) Correct formula or substitution: \( \sin c = 1 / 1.53 \) [1 mark], Correct calculation: \( 41^\circ \) (allow 40.8 to 41.2) [1 mark].
题目 4 · Structured Theory
8 分
A total charge of \( 450\text{ C} \) flows through a resistor in a circuit during a time interval of \( 5.0\text{ minutes} \). The potential difference across the resistor is \( 12\text{ V} \). (a) Calculate: (i) the current in the resistor. (ii) the resistance of the resistor. (iii) the electrical energy transferred to the resistor. (b) The resistor consists of a uniform metallic wire. State the effect, if any, on the resistance if it is replaced with: (i) a wire of the same material but twice the length. (ii) a wire of the same material and length but twice the cross-sectional area.
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解题
(a)(i) Convert time to seconds: \( t = 5.0 \times 60 = 300\text{ s} \). Current \( I = \frac{Q}{t} = \frac{450}{300} = 1.5\text{ A} \). (a)(ii) Resistance \( R = \frac{V}{I} = \frac{12}{1.5} = 8.0\ \Omega \). (a)(iii) Energy \( E = QV = 450 \times 12 = 5400\text{ J} \). (b)(i) Since \( R \propto L \), the resistance doubles. (b)(ii) Since \( R \propto 1/A \), the resistance is halved.
评分标准
(a)(i) Conversion of minutes to seconds (300 s) [1 mark], Correct current: \( 1.5\text{ A} \) [1 mark]. (a)(ii) Substitution: \( 12 / 1.5 \) [1 mark], Correct resistance: \( 8.0\ \Omega \) [1 mark]. (a)(iii) Substitution: \( 450 \times 12 \) or equivalent [1 mark], Correct energy: \( 5400\text{ J} \) [1 mark]. (b)(i) Doubles / increases by a factor of 2 [1 mark]. (b)(ii) Halves / decreases by a factor of 2 [1 mark].
题目 5 · Structured Theory
8 分
A sample of a radioactive isotope decays. A detector is used to measure the count rate from the sample. At \( t = 0 \), the count rate is \( 820\text{ counts/minute} \). After \( 12\text{ hours} \), the count rate has fallen to \( 120\text{ counts/minute} \). The background count rate is constant at \( 20\text{ counts/minute} \). (a) Calculate: (i) the initial count rate due only to the radioactive isotope. (ii) the count rate due only to the radioactive isotope after \( 12\text{ hours} \). (iii) the half-life of the isotope. (b) The isotope emits \(\beta^-\) particles. (i) Describe the nature of a \(\beta^-\) particle. (ii) State the effect of \(\beta^-\) emission on the proton number and the nucleon number of a nucleus.
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解题
(a)(i) Corrected initial count rate \( = 820 - 20 = 800\text{ counts/minute} \). (a)(ii) Corrected final count rate \( = 120 - 20 = 100\text{ counts/minute} \). (a)(iii) Since the corrected activity decreases from \( 800 \rightarrow 400 \rightarrow 200 \rightarrow 100 \), there are exactly 3 half-lives during the 12 hours. Therefore, the half-life is \( 12 / 3 = 4.0\text{ hours} \). (b)(i) A \(\beta^-\) particle is a high-speed electron emitted from the nucleus. (b)(ii) The proton number increases by 1, and the nucleon number remains unchanged.
评分标准
(a)(i) Correct calculation: \( 800\text{ counts/minute} \) [1 mark]. (a)(ii) Correct calculation: \( 100\text{ counts/minute} \) [1 mark]. (a)(iii) Identify that activity decreases by a factor of 8 (or 3 half-lives) [1 mark], Set up equation \( 3 \times T_{1/2} = 12\text{ hours} \) [1 mark], Correct calculation: \( 4.0\text{ hours} \) [1 mark]. (b)(i) High-speed electron [1 mark]. (b)(ii) Proton number increases by 1 [1 mark], Nucleon number remains unchanged [1 mark].
题目 6 · Structured Theory
8 分
A planetary space probe is orbiting a planet in the Solar System. (a) The planet has an orbital period around the Sun of \( 687\text{ Earth days} \). (i) Identify this planet. (ii) State what is meant by one light-year. (b) The average radius of this planet's orbit around the Sun is \( 2.3 \times 10^8\text{ km} \). (i) Calculate the average orbital speed of this planet in \(\text{km/s}\). Assume the orbit is circular. (ii) State two reasons why the distance of the planet from the Sun is described as an 'average' value. (c) Define the term gravitational field strength.
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解题
(a)(i) Mars has an orbital period of 687 Earth days. (a)(ii) One light-year is the distance travelled by light in a vacuum in one Earth year. (b)(i) Orbit period in seconds \( T = 687 \times 24 \times 3600 = 5.936 \times 10^7\text{ s} \). Radius \( r = 2.3 \times 10^8\text{ km} \). Speed \( v = \frac{2\pi r}{T} = \frac{2 \times \pi \times 2.3 \times 10^8}{5.936 \times 10^7} \approx 24.3\text{ km/s} \) (or \( 24\text{ km/s} \)). (b)(ii) The planet's orbit is elliptical rather than circular, so the distance to the Sun varies continuously. The Sun is not at the exact geometric centre of the orbit. (c) Gravitational field strength is the gravitational force acting per unit mass on an object.
评分标准
(a)(i) Mars [1 mark]. (a)(ii) Distance light travels in one year in a vacuum [1 mark]. (b)(i) Calculate orbital period in seconds (\( 5.94 \times 10^7\text{ s} \)) [1 mark], Correct formula \( v = 2\pi r / T \) [1 mark], Correct calculation: \( 24\text{ km/s} \) (accept 24.3) [1 mark]. (b)(ii) Elliptical orbit / not a perfect circle [1 mark], Distance varies / Sun is not at the centre [1 mark]. (c) Force per unit mass [1 mark].
题目 7 · Structured Theory
8 分
A sample of gas is sealed inside a rigid container of fixed volume. (a) Describe, in terms of the kinetic particle model, how the gas particles exert a pressure on the inner walls of the container. (b) The temperature of the gas is increased. (i) State and explain, in terms of the particles, how this temperature increase affects the pressure inside the container. (ii) State the value of absolute zero in degrees Celsius (\(^\circ\text{C}\)). (c) State what happens to the average kinetic energy of the gas particles at absolute zero.
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解题
(a) Gas particles are in rapid, continuous, random motion. They collide with the inner walls of the container. During these collisions, their momentum changes, resulting in a force on the walls. The total force per unit area on the walls is the gas pressure. (b)(i) The pressure increases. As temperature rises, particles gain kinetic energy and move faster. Consequently, they collide with the container walls more frequently and with greater force per collision. (b)(ii) Absolute zero is \( -273^\circ\text{C} \). (c) The average kinetic energy becomes zero because the particles stop moving.
评分标准
(a) Gas particles are in random motion [1 mark], Particles collide with container walls [1 mark], Collisions cause momentum change, exerting a force per unit area [1 mark]. (b)(i) Pressure increases [1 mark], Particles move faster / have more kinetic energy [1 mark], Collide more frequently / with greater force per collision [1 mark]. (b)(ii) \( -273^\circ\text{C} \) [1 mark]. (c) Average kinetic energy becomes zero [1 mark].
题目 8 · Structured Theory
8 分
A small electric motor is used to lift a load of mass \( 12\text{ kg} \) vertically upwards through a height of \( 3.0\text{ m} \) in a time of \( 8.0\text{ s} \). The electrical power input to the motor is \( 60\text{ W} \). (a) Calculate: (i) the weight of the load. Take the gravitational field strength \( g = 9.8\text{ m/s}^2 \). (ii) the useful work done in lifting the load. (iii) the useful power output of the motor. (iv) the efficiency of the motor system. (b) Suggest one reason why the efficiency of the motor system is less than \( 100\% \).
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解题
(a)(i) Weight \( W = mg = 12 \times 9.8 = 117.6\text{ N} \). (a)(ii) Work done \( W = Fd = 117.6 \times 3.0 = 352.8\text{ J} \). (a)(iii) Useful power output \( P_{\text{out}} = \frac{\text{Work}}{t} = \frac{352.8}{8.0} = 44.1\text{ W} \). (a)(iv) Efficiency \( = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{44.1}{60} \times 100\% = 73.5\% \). (b) Energy is lost as thermal energy due to friction in the bearings or resistance in the motor coils.
A tennis ball of mass 0.060 kg is moving horizontally with a speed of 22 m/s. It is struck by a racket. Immediately after leaving the racket, the ball travels horizontally in the opposite direction with a speed of 28 m/s.
(a) (i) Calculate the magnitude of the change in momentum of the tennis ball. [2] (ii) The racket is in contact with the ball for 4.5 ms. Calculate the average force exerted on the ball by the racket. [2]
(b) (i) Explain what is meant by the term elastic deformation. [1] (ii) State the main energy transfer that occurs in the racket strings during the first half of the collision, as the ball is being compressed. [1]
(c) After extensive use, the racket strings undergo plastic deformation. Explain, in terms of work done and energy, how this affects the performance of the racket. [2]
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解题
(a) (i) Initial momentum \(p_i = m u = 0.060 \text{ kg} \times 22 \text{ m/s} = +1.32 \text{ kg m/s}\). Final momentum \(p_f = m v = 0.060 \text{ kg} \times (-28 \text{ m/s}) = -1.68 \text{ kg m/s}\). Change in momentum \(\Delta p = p_f - p_i = -1.68 - 1.32 = -3.00 \text{ kg m/s}\). Magnitude of change in momentum = \(3.0 \text{ kg m/s}\) (or \(\text{N s}\)).
(ii) Average force \(F = \frac{\Delta p}{\Delta t} = \frac{3.0 \text{ N s}}{4.5 \times 10^{-3} \text{ s}} = 667 \text{ N} \approx 670 \text{ N}\) (or \(6.7 \times 10^2 \text{ N}\)).
(b) (i) Elastic deformation is a temporary change in shape or size where the object returns to its original dimensions when the deforming force is removed. (ii) Kinetic energy (of the ball) is transferred to elastic potential energy (or strain energy) in the racket strings.
(c) Work is done in permanently stretching/deforming the strings (plastic deformation), which dissipates some of the energy as thermal energy. Consequently, less elastic potential energy is stored in the strings, meaning less kinetic energy is transferred back to the ball, resulting in a lower exit speed of the ball.
评分标准
**(a) (i)** - \(p = m \Delta v\) or \(0.060 \times (28 - (-22))\) or equivalent substitution [1] - \(3.0 \text{ kg m/s}\) or \(\text{N s}\) [1]
**(c)** - work is done in permanently deforming strings / energy is wasted as thermal energy [1] - less elastic potential energy stored, so less kinetic energy transferred back to ball / lower speed [1]
题目 10 · Structured Theory
8 分
An electric heater of power 120 W is used to heat a 0.40 kg block of a metal. The block is initially at a temperature of \(-10\ ^\circ\text{C}\).
(a) The specific heat capacity of the metal is \(380\text{ J}/(\text{kg}\ ^\circ\text{C})\). Calculate the thermal energy required to raise the temperature of the block from \(-10\ ^\circ\text{C}\) to its melting point of \(230\ ^\circ\text{C}\). [2]
(b) Calculate the time taken by the heater to supply this thermal energy, assuming no energy is lost to the surroundings. [2]
(c) When the metal block reaches its melting point, it begins to melt. The heater remains switched on. (i) Describe the change, if any, in the temperature of the metal block as it melts. Explain your answer in terms of the arrangement and forces between the particles. [2] (ii) State the term used to describe the thermal energy required to melt a unit mass of a substance at constant temperature. [1] (iii) Suggest why, in practice, the actual time taken to heat the block to its melting point is longer than the time calculated in (b). [1]
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解题
(a) The temperature rise is \(\Delta \theta = 230\ ^\circ\text{C} - (-10\ ^\circ\text{C}) = 240\ ^\circ\text{C}\). \(E = m c \Delta \theta = 0.40\text{ kg} \times 380\text{ J}/(\text{kg}\ ^\circ\text{C}) \times 240\ ^\circ\text{C} = 36480\text{ J} \approx 3.6 \times 10^4\text{ J}\) (or \(3.65 \times 10^4\text{ J}\)).
(c) (i) The temperature of the block remains constant as it melts. The thermal energy supplied is used to break or weaken the forces of attraction between the particles to change the regular solid lattice arrangement into a liquid, rather than increasing the kinetic energy of the particles. (ii) Specific latent heat of fusion. (iii) Some thermal energy is lost to the surrounding environment (the process is not 100% efficient).
评分标准
**(a)** - \(\Delta \theta = 240\ ^\circ\text{C}\) seen or implied [1] - \(3.6 \times 10^4\text{ J}\) (allow \(3.65 \times 10^4\text{ J}\) or \(36480\text{ J}\)) [1]
**(c) (i)** - temperature remains constant [1] - energy is used to break/weaken forces between particles / bonds [1]
**(c) (ii)** - specific latent heat of fusion [1]
**(c) (iii)** - thermal energy is lost to the surroundings / heater is not 100% efficient [1]
Paper 62 Alternative to Practical
Answer all questions. Use a ruler, sharp pencil, and show experimental calculations.
4 题目 · 40 分
题目 1 · structured
10 分
A student investigates the balancing of a metre rule to determine its mass.
(a) The student first balances the metre rule, on its own, on a pivot to locate its centre of gravity \(G\).
State why the centre of gravity of a standard metre rule may not be exactly at the 50.0 cm mark. [1]
(b) The student places the pivot at the 40.0 cm mark.
A load of mass \(m = 100\text{ g}\) is placed on the rule. The rule is balanced when the load is at the 25.4 cm mark. The centre of gravity \(G\) of the rule is at the 50.3 cm mark.
(i) Calculate the distance \(d\) between the pivot and the centre of the load. [1]
(ii) Calculate the distance \(x\) between the pivot and the centre of gravity \(G\). [1]
(c) Calculate the mass \(M\) of the metre rule using the equation:
\[M = \frac{m \cdot d}{x}\]
Show your working and give your answer to an appropriate number of significant figures. [2]
(d) The student moves the pivot to the 35.0 cm mark and balances the rule again using the same load \(m = 100\text{ g}\). The balanced position of the load is at the 13.3 cm mark.
(i) Calculate the new distance \(d_2\) from the load to the pivot. [1]
(ii) Calculate the new distance \(x_2\) from the pivot to the centre of gravity \(G\). [1]
(iii) Calculate a second value for the mass \(M_2\) of the rule. [1]
(e) State whether the two values for the mass of the metre rule, \(M\) and \(M_2\), are equal within the limits of experimental accuracy. Explain your answer with reference to the values. [1]
(f) Suggest one precaution the student should take when performing this experiment to ensure the readings are as accurate as possible. [1]
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解题
(a) Wood is non-uniform in density or the ruler is worn/chipped at one end. (b)(i) \(d = 40.0\text{ cm} - 25.4\text{ cm} = 14.6\text{ cm}\) (b)(ii) \(x = 50.3\text{ cm} - 40.0\text{ cm} = 10.3\text{ cm}\) (c) \(M = \frac{100 \times 14.6}{10.3} = 141.75\text{ g} \approx 142\text{ g}\) (correctly rounded to 3 significant figures). (d)(i) \(d_2 = 35.0\text{ cm} - 13.3\text{ cm} = 21.7\text{ cm}\) (d)(ii) \(x_2 = 50.3\text{ cm} - 35.0\text{ cm} = 15.3\text{ cm}\) (d)(iii) \(M_2 = \frac{100 \times 21.7}{15.3} = 141.83\text{ g} \approx 142\text{ g}\). (e) Yes, they are equal because the difference is negligible / much less than 10%. (f) Avoid parallax error by looking perpendicularly at the markings of the ruler, or align the load's centre of mass carefully with the ruler division.
评分标准
(a) Metre rule has non-uniform density OR is chipped/worn at one end. [1] (b)(i) \(14.6\text{ cm}\). [1] (b)(ii) \(10.3\text{ cm}\). [1] (c) Substitution of values: \(M = \frac{100 \times 14.6}{10.3}\) [1]; mass = \(142\text{ g}\) (correctly rounded to 3 significant figures). [1] (d)(i) \(21.7\text{ cm}\). [1] (d)(ii) \(15.3\text{ cm}\). [1] (d)(iii) \(142\text{ g}\) (or \(141.8\text{ g}\)). [1] (e) Correct statement ('Yes') and comparison explanation showing the values are within 10% / very close. [1] (f) View ruler perpendicularly to avoid parallax error / align the centre of the mass carefully with the mark. [1]
题目 2 · structured
10 分
A student investigates the path of light through a semi-circular glass block.
(a) The student uses two pins, \(P_1\) and \(P_2\), to mark the incident ray.
State the minimum distance that should be left between the two pins, and explain why this distance is necessary. [2]
(b) The student records the angles of refraction \(r\) for different angles of incidence \(i\) in a table:
(i) Calculate the missing ratio value for \(i = 35.0^{\circ}\). Give your answer to 3 significant figures. [1]
(ii) Suggest an appropriate unit for the ratio column in the table, if any. [1]
(iii) Calculate the average value of the ratio \(\frac{\sin i}{\sin r}\) using the four values. [1]
(c) The student plots a graph of \(\sin i\) (y-axis) against \(\sin r\) (x-axis) and draws a best-fit straight line through the origin.
Two points on the line are selected to find the gradient \(G\): - Point 1: \((x_1, y_1) = (0.12, 0.18)\) - Point 2: \((x_2, y_2) = (0.48, 0.72)\)
(i) Calculate the gradient \(G\) of the line. Show your working. [2]
(ii) State what physical quantity the gradient \(G\) represents. [1]
(d) The student observes that at very large angles of incidence, the ray of light does not emerge from the curved side of the block, but is reflected back inside.
State the name of this wave phenomenon. [1]
(e) Suggest one safety precaution the student should take when using the ray box. [1]
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解题
(a) The pins should be placed at least 5.0 cm apart. This reduces the effect of any small alignment error when drawing the line. (b)(i) Ratio = \(\frac{\sin(35.0^{\circ})}{\sin(22.0^{\circ})} = \frac{0.5736}{0.3746} = 1.531 \approx 1.53\). (b)(ii) Since both sine values are dimensionless, the ratio has no unit. (b)(iii) Average = \(\frac{1.52 + 1.53 + 1.49 + 1.51}{4} = 1.5125 \approx 1.51\). (c)(i) \(G = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0.72 - 0.18}{0.48 - 0.12} = \frac{0.54}{0.36} = 1.50\). (c)(ii) The gradient of the graph of \(\sin i\) against \(\sin r\) is the refractive index of the glass block. (d) Total internal reflection. (e) Ray box lamps get very hot, so avoid touching the bulb/casing directly, or switch it off when not in use.
评分标准
(a) Distance \(\ge 5.0\text{ cm}\) (or \(50\text{ mm}\)) [1]; to increase accuracy of the drawn ray/line. [1] (b)(i) \(1.53\) (allow \(1.531\)). [1] (b)(ii) None / no unit / blank. [1] (b)(iii) \(1.51\) (consistent with calculation). [1] (c)(i) Substitution: \(\frac{0.72 - 0.18}{0.48 - 0.12}\) [1]; \(1.5\) (or \(1.50\)). [1] (c)(ii) Refractive index (of the glass block). [1] (d) Total internal reflection. [1] (e) Avoid touching hot lamp / do not stare directly into the light source. [1]
题目 3 · structured
10 分
A student investigates the resistance of three identical resistors connected in different combinations.
(a) Draw a circuit diagram of a single resistor \(R_1\) connected to a cell, an ammeter, and a voltmeter connected to measure the potential difference across \(R_1\). [2]
(b) The student closes the switch and records: - Current \(I_1 = 0.24\text{ A}\) - Potential difference \(V_1 = 1.45\text{ V}\)
Calculate the resistance \(R_1\). Include the unit in your final answer. [2]
(c) The student then connects all three resistors in series.
They measure the total series potential difference \(V_S = 1.35\text{ V}\) and the current \(I_S = 0.075\text{ A}\).
(i) Calculate the total series resistance \(R_S\). [1]
(ii) Assuming the resistors are identical, state the relationship between \(R_1\) and \(R_S\). State if your calculated values support this relationship. [2]
(d) The student connects the three resistors in parallel.
They measure the total parallel resistance \(R_P = 2.02\text{ }\Omega\).
A theory suggests that the resistance of three identical resistors in parallel is given by:
\[R_P = \frac{R_1}{3}\]
State whether the value of \(R_P\) supports this theory. Explain your answer with reference to the values. [2]
(e) Suggest one reason why a student might include a variable resistor (rheostat) in this circuit. [1]
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解题
(a) A series circuit containing a cell/battery, switch, ammeter, and the resistor \(R_1\). A voltmeter must be connected in parallel across the resistor \(R_1\). (b) \(R_1 = \frac{V_1}{I_1} = \frac{1.45\text{ V}}{0.24\text{ A}} = 6.04\text{ }\Omega\) (accept \(6.0\text{ }\Omega\)). (c)(i) \(R_S = \frac{V_S}{I_S} = \frac{1.35\text{ V}}{0.075\text{ A}} = 18.0\text{ }\Omega\). (c)(ii) The relationship is \(R_S = 3 R_1\). Yes, \(3 \times 6.04\text{ }\Omega = 18.12\text{ }\Omega\), which is extremely close to the measured value of \(18.0\text{ }\Omega\). (d) Yes, the calculated theoretical value \(\frac{R_1}{3} = \frac{6.04\text{ }\Omega}{3} = 2.01\text{ }\Omega\) is almost identical to the experimental value of \(2.02\text{ }\Omega\). This is well within \(10\%\) experimental limits. (e) It allows the current to be varied so multiple sets of current and potential difference values can be recorded, or to limit the current to prevent the resistors from heating up and changing resistance.
评分标准
(a) Correct symbols for cell, ammeter, resistor in series [1]; voltmeter in parallel across resistor \(R_1\). [1] (b) Formula or substitution: \(R = V/I\) [1]; \(6.0\text{ }\Omega\) or \(6.04\text{ }\Omega\) with the correct unit symbol \(\Omega\). [1] (c)(i) \(18.0\text{ }\Omega\) (or \(18\text{ }\Omega\)). [1] (c)(ii) State relationship: \(R_S = 3 R_1\) [1]; comparison showing \(18.0\text{ }\Omega\) and \(18.12\text{ }\Omega\) are close enough to support it. [1] (d) State 'Yes' [1]; explanation showing calculation \(\frac{6.04}{3} = 2.01\text{ }\Omega\) and stating it is within the limits of experimental accuracy / close to \(2.02\text{ }\Omega\). [1] (e) To vary the current/voltage OR to prevent overheating of resistors. [1]
题目 4 · structured
10 分
A student wants to investigate how the rate of thermal energy transfer (cooling rate) from a beaker of hot water depends on the type of insulating material wrapped around the beaker.
The following apparatus is available: - identical glass beakers - thermometer - stopwatch - kettle to heat water - a selection of different sheet materials (wool, cotton, paper, bubble wrap) - rubber bands (to secure the wrapping)
Plan an experiment to investigate this.
Your plan should describe: - the arrangement of the apparatus (you may draw a diagram if needed) - the key variables that must be kept constant - the detailed procedure, including the measurements to take - how the results should be analyzed to reach a conclusion. [10]
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解题
### Plan: 1. **Apparatus & Set-up:** Wrap a beaker in one of the insulating materials (e.g., wool) and secure it with rubber bands. Place a thermometer inside the beaker.
2. **Variables to keep constant:** - Volume of water in each beaker (e.g., 150 ml) - Initial temperature of the hot water at the start of each trial (e.g., 80 °C) - Thickness of wrapping / number of layers of wrapping - Ambient room temperature (avoid draughts) - Material of the beaker
3. **Procedure:** - Boil water in a kettle. - Pour a fixed volume of hot water into the first wrapped beaker. - Insert the thermometer and wait for it to reach a high temperature, then record this initial temperature (at \(t = 0\)). - Start the stopwatch immediately. - Record the temperature of the water at regular intervals (e.g., every 60 seconds) for a total of 10 minutes. - Repeat the entire process for identical beakers wrapped with different materials (cotton, paper, bubble wrap, and one unwrapped control beaker).
4. **Analysis & Conclusion:** - Calculate the total temperature drop \(\Delta \theta = \theta_{\text{initial}} - \theta_{\text{final}}\) for each beaker. - Draw a table with columns: 'Type of insulation', 'Initial temperature / °C', 'Final temperature / °C', and 'Temperature drop / °C'. - Alternatively, plot a graph of temperature against time for each material on the same axes. - The material that shows the smallest temperature drop (or the shallowest curve) is the best insulator, as it has the lowest rate of thermal energy transfer.
评分标准
Award marks as follows (up to 10 marks total): - **Set-up & Diagram:** Labelled diagram or description showing beaker wrapped in insulating material with a thermometer inside. [1] - **Procedure (First Run):** Pour hot water and record the starting temperature. [1] - **Time measurement:** Use stopwatch to record temperature at regular time intervals (e.g., every 1 min). [1] - **Duration:** Continue recording for a specified sensible time limit (e.g., 5 to 15 minutes). [1] - **Repetition:** Repeat with different insulating materials. [1] - **Control Variable 1:** Keep the volume/mass of hot water constant. [1] - **Control Variable 2:** Keep the starting temperature of the hot water constant. [1] - **Control Variable 3:** Keep the thickness of wrapping / number of layers constant (or room temperature / beaker type). [1] - **Data Presentation:** Suggest a suitable table with columns for Insulation Type, Time, and Temperature. [1] - **Analysis/Conclusion:** Determine the temperature change (or plot temperature-time graph) and state that the material with the smallest temperature drop is the best insulator. [1]
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