An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
卷二 (Extended MCQ)
Answer all 40 multiple-choice questions. For each question, choose the correct answer from the four options A, B, C or D.
40 题目 · 40 分
题目 1 · MCQ
1 分
A space probe of mass \(1200\text{ kg}\) is coasting at \(15\text{ m/s}\) in deep space. An internal explosion splits the probe into two parts. A larger part of mass \(800\text{ kg}\) is projected forward in the same direction of motion at \(22\text{ m/s}\). What is the velocity of the remaining \(400\text{ kg}\) part?
A.\(1.0\text{ m/s}\) in the opposite direction to the original motion
B.\(1.0\text{ m/s}\) in the same direction as the original motion
C.\(4.0\text{ m/s}\) in the opposite direction to the original motion
D.\(4.0\text{ m/s}\) in the same direction as the original motion
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解题
Using the law of conservation of momentum: \(m_{\text{total}} u = m_1 v_1 + m_2 v_2\). Substituting the given values: \(1200 \times 15 = 800 \times 22 + 400 \times v_2\). This simplifies to \(18000 = 17600 + 400 v_2\), which gives \(400 = 400 v_2\). Thus, \(v_2 = +1.0\text{ m/s}\). The positive sign indicates that the velocity is in the same direction as the initial motion.
评分标准
1 mark for the correct option B.
题目 2 · MCQ
1 分
A cylindrical container with a flat base of area \(0.050\text{ m}^2\) is filled with a liquid of density \(1200\text{ kg/m}^3\) to a depth of \(0.80\text{ m}\). What is the total force exerted by the liquid on the base of the container? (Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\).)
A.\(47\text{ N}\)
B.\(470\text{ N}\)
C.\(940\text{ N}\)
D.\(9400\text{ N}\)
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解题
The hydrostatic pressure at the base is given by \(p = \rho g h = 1200 \times 9.8 \times 0.80 = 9408\text{ Pa}\). The force exerted on the base is \(F = p \times A = 9408 \times 0.050 = 470.4\text{ N}\), which rounds to \(470\text{ N}\).
评分标准
1 mark for the correct option B.
题目 3 · MCQ
1 分
Light from a distant galaxy is observed to have its hydrogen emission spectral lines shifted towards longer wavelengths (redshifted). Which statement correctly describes what this redshift indicates about the galaxy's motion and its relation to distance from Earth?
A.The galaxy is moving towards Earth, and its speed of recession is inversely proportional to its distance.
B.The galaxy is moving away from Earth, and its speed of recession is inversely proportional to its distance.
C.The galaxy is moving towards Earth, and its speed of recession is directly proportional to its distance.
D.The galaxy is moving away from Earth, and its speed of recession is directly proportional to its distance.
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解题
A redshift indicates that the galaxy is moving away from Earth (receding). According to Hubble's Law, the speed of recession of a distant galaxy is directly proportional to its distance from Earth.
评分标准
1 mark for the correct option D.
题目 4 · MCQ
1 分
A metal block of mass \(0.50\text{ kg}\) is heated by an electric heater. When \(6000\text{ J}\) of thermal energy is supplied to the block, its temperature rises from \(20\text{ }^\circ\text{C}\) to \(52\text{ }^\circ\text{C}\). What is the specific heat capacity of the metal?
A.\(120\text{ J / (kg }^\circ\text{C)}\)
B.\(190\text{ J / (kg }^\circ\text{C)}\)
C.\(375\text{ J / (kg }^\circ\text{C)}\)
D.\(750\text{ J / (kg }^\circ\text{C)}\)
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解题
The formula for thermal energy is \(Q = m c \Delta \theta\), where \(\Delta \theta = 52\text{ }^\circ\text{C} - 20\text{ }^\circ\text{C} = 32\text{ }^\circ\text{C}\). Rearranging the formula for specific heat capacity gives \(c = \frac{Q}{m \Delta \theta} = \frac{6000}{0.50 \times 32} = 375\text{ J / (kg }^\circ\text{C)}\).
评分标准
1 mark for the correct option C.
题目 5 · MCQ
1 分
A step-down transformer has a primary voltage of \(240\text{ V}\) a.c. and a secondary voltage of \(12\text{ V}\) a.c. The primary coil has \(4000\text{ turns}\). Assuming the transformer is \(100\%\) efficient and the secondary current is \(2.0\text{ A}\), what are the secondary turns and primary current?
A.secondary turns = \(200\), primary current = \(0.10\text{ A}\)
B.secondary turns = \(200\), primary current = \(40\text{ A}\)
C.secondary turns = \(80000\), primary current = \(0.10\text{ A}\)
D.secondary turns = \(80000\), primary current = \(40\text{ A}\)
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解题
Using the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\), we get \(\frac{240}{12} = \frac{4000}{N_s} \implies N_s = 200\text{ turns}\). For a \(100\%\) efficient transformer, \(I_p V_p = I_s V_s \implies I_p \times 240 = 2.0 \times 12 \implies I_p = 0.10\text{ A}\).
评分标准
1 mark for the correct option A.
题目 6 · MCQ
1 分
A freshly prepared radioactive sample initially contains \(8.0 \times 10^{10}\) nuclei of a certain radioisotope. After \(12\text{ hours}\) have elapsed, only \(1.0 \times 10^{10}\) of these radioactive nuclei remain. What is the half-life of this radioisotope?
A.\(1.5\text{ hours}\)
B.\(3.0\text{ hours}\)
C.\(4.0\text{ hours}\)
D.\(6.0\text{ hours}\)
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解题
The fraction of remaining nuclei is \(\frac{1.0 \times 10^{10}}{8.0 \times 10^{10}} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly three half-lives have passed in \(12\text{ hours}\). Therefore, the half-life is \(T_{1/2} = \frac{12\text{ hours}}{3} = 4.0\text{ hours}\).
评分标准
1 mark for the correct option C.
题目 7 · MCQ
1 分
A uniform metallic wire of length \(L\) and cross-sectional area \(A\) has an electrical resistance of \(16.0\text{ }\Omega\). A second wire of the exact same metal has a length of \(2L\) and a diameter that is twice the diameter of the first wire. What is the resistance of the second wire?
A.\(4.0\text{ }\Omega\)
B.\(8.0\text{ }\Omega\)
C.\(16.0\text{ }\Omega\)
D.\(32.0\text{ }\Omega\)
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解题
The resistance of a wire is given by \(R = \rho \frac{L}{A}\). The area is proportional to the square of the diameter, so doubling the diameter increases the cross-sectional area by a factor of 4 (i.e., \(A_2 = 4A\)). Thus, the new resistance is \(R_2 = \rho \frac{2L}{4A} = 0.5 \times R_1 = 0.5 \times 16.0\text{ }\Omega = 8.0\text{ }\Omega\).
评分标准
1 mark for the correct option B.
题目 8 · MCQ
1 分
A narrow ray of monochromatic light travels through a glass prism and meets the boundary with air. The refractive index of the glass is \(1.50\). What is the critical angle for light at this boundary?
A.\(30^\circ\)
B.\(42^\circ\)
C.\(48^\circ\)
D.\(90^\circ\)
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解题
The critical angle \(c\) is given by the relation \(\sin c = \frac{1}{n}\). Substituting \(n = 1.50\) gives \(\sin c = \frac{1}{1.50} \approx 0.6667\). Calculating the inverse sine gives \(c \approx 41.8^\circ\), which is closest to \(42^\circ\).
评分标准
1 mark for the correct option B.
题目 9 · 選擇題
1 分
Light from a distant galaxy is analysed. A spectral line of wavelength \(656.3 \text{ nm}\) is observed to be shifted to a wavelength of \(689.1 \text{ nm}\). The speed of light is \(3.00 \times 10^8 \text{ m/s}\). Using the Hubble constant \(H_0 = 2.2 \times 10^{-18} \text{ s}^{-1}\), what is the estimated distance of this galaxy from Earth?
A.\(1.5 \times 10^7 \text{ m}\)
B.\(1.4 \times 10^{17} \text{ m}\)
C.\(6.8 \times 10^{24} \text{ m}\)
D.\(3.1 \times 10^{26} \text{ m}\)
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解题
First, calculate the redshift \(z\): \(z = \frac{\Delta \lambda}{\lambda} = \frac{689.1 - 656.3}{656.3} \approx 0.0500\). Next, calculate the recession velocity \(v\): \(v = z \times c = 0.0500 \times 3.00 \times 10^8 \text{ m/s} = 1.50 \times 10^7 \text{ m/s}\). Finally, use Hubble's Law to find the distance \(d\): \(d = \frac{v}{H_0} = \frac{1.50 \times 10^7}{2.2 \times 10^{-18}} \approx 6.8 \times 10^{24} \text{ m}\).
评分标准
1 mark for the correct option C.
题目 10 · 選擇題
1 分
A sky-diver of mass \(60 \text{ kg}\) falls vertically. At time \(t_1\), her speed is \(40 \text{ m/s}\). She opens her parachute, and by time \(t_2\), she reaches a new lower terminal velocity of \(5.0 \text{ m/s}\). Which statement describes her motion and the forces acting on her between \(t_1\) and \(t_2\)?
A.Her deceleration is constant because the resultant force remains constant.
B.Her deceleration decreases because the upward drag force decreases as her speed decreases.
C.Her deceleration increases because the downward gravitational force increases.
D.Her deceleration decreases because the upward drag force increases as her speed decreases.
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解题
When the parachute opens, the upward drag force is much larger than her weight, causing a significant upward resultant force (and thus deceleration). As her speed decreases, the air resistance (drag) also decreases, which reduces the resultant upward force, hence her deceleration decreases until it reaches zero at the new terminal velocity.
评分标准
1 mark for the correct option B.
题目 11 · 選擇題
1 分
A rubber ball of mass \(0.15 \text{ kg}\) is moving horizontally to the right at \(12 \text{ m/s}\). It hits a wall and bounces back horizontally to the left at \(8.0 \text{ m/s}\). The ball is in contact with the wall for \(0.050 \text{ s}\). What is the average force exerted by the wall on the ball?
A.\(12 \text{ N}\) to the left
B.\(12 \text{ N}\) to the right
C.\(60 \text{ N}\) to the left
D.\(60 \text{ N}\) to the right
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解题
Taking the direction to the right as positive: initial velocity \(u = +12 \text{ m/s}\) and final velocity \(v = -8.0 \text{ m/s}\). The change in momentum is \(\Delta p = m(v - u) = 0.15 \times (-8.0 - 12) = -3.0 \text{ kg m/s}\). The average force is \(F = \frac{\Delta p}{\Delta t} = \frac{-3.0}{0.050} = -60 \text{ N}\), which means \(60 \text{ N}\) to the left.
评分标准
1 mark for the correct option C.
题目 12 · 選擇題
1 分
A rectangular hatch of dimensions \(0.40 \text{ m} \times 0.60 \text{ m}\) is situated on the horizontal deck of a submarine. The submarine is submerged in seawater of density \(1025 \text{ kg/m}^3\) at a depth of \(80 \text{ m}\). The atmospheric pressure inside the submarine is \(1.0 \times 10^5 \text{ Pa}\), and the gravitational field strength \(g\) is \(9.8 \text{ N/kg}\). What is the magnitude of the resultant force acting on the hatch?
A.\(1.9 \times 10^4 \text{ N}\)
B.\(1.9 \times 10^5 \text{ N}\)
C.\(2.2 \times 10^5 \text{ N}\)
D.\(2.4 \times 10^5 \text{ N}\)
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解题
The pressure outside the submarine is \(p_{\text{ext}} = p_{\text{atm}} + \rho g h\). The pressure inside is \(p_{\text{atm}}\). Thus, the net pressure difference across the hatch is \(\Delta p = \rho g h = 1025 \times 9.8 \times 80 = 803\,600 \text{ Pa}\). The area of the hatch is \(A = 0.40 \times 0.60 = 0.24 \text{ m}^2\). The magnitude of the resultant force is \(F = \Delta p \times A = 803\,600 \times 0.24 \approx 1.9 \times 10^5 \text{ N}\).
评分标准
1 mark for the correct option B.
题目 13 · 選擇題
1 分
Two solid blocks, X and Y, are made of different metals and are heated by identical heaters supplying thermal energy at the same rate. Block X has mass \(m\) and specific heat capacity \(c\). Its temperature increases by \(20 \ ^\circ\text{C}\) in time \(t\). Block Y has twice the mass of X, but its specific heat capacity is only \(\frac{1}{3}\) of the specific heat capacity of X. What is the time taken for the temperature of block Y to increase by \(20 \ ^\circ\text{C}\)?
A.\(0.33t\)
B.\(0.67t\)
C.\(1.5t\)
D.\(6.0t\)
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解题
The energy supplied to X is \(E_X = P \cdot t = m \cdot c \cdot \Delta T\). For block Y, the required energy is \(E_Y = P \cdot t_Y = m_Y \cdot c_Y \cdot \Delta T = (2m) \cdot (\frac{1}{3}c) \cdot \Delta T = \frac{2}{3} (m \cdot c \cdot \Delta T)\). Therefore, \(P \cdot t_Y = \frac{2}{3} (P \cdot t)\), which simplifies to \(t_Y = \frac{2}{3}t \approx 0.67t\).
评分标准
1 mark for the correct option B.
题目 14 · 選擇題
1 分
A potential divider circuit consists of a fixed resistor of resistance \(2.0 \text{ k}\Omega\) in series with a negative temperature coefficient (NTC) thermistor, powered by a stable \(12 \text{ V}\) d.c. supply. Initially, at room temperature, the resistance of the thermistor is \(4.0 \text{ k}\Omega\). The temperature of the thermistor is then significantly increased. What happens to the potential difference (p.d.) across the fixed resistor?
A.It decreases from \(8.0 \text{ V}\) towards \(0 \text{ V}\).
B.It decreases from \(4.0 \text{ V}\) towards \(0 \text{ V}\).
C.It increases from \(4.0 \text{ V}\) towards \(12 \text{ V}\).
D.It increases from \(8.0 \text{ V}\) towards \(12 \text{ V}\).
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解题
Initially, total resistance is \(2.0 \text{ k}\Omega + 4.0 \text{ k}\Omega = 6.0 \text{ k}\Omega\). The current is \(I = \frac{12 \text{ V}}{6.0 \text{ k}\Omega} = 2.0 \text{ mA}\). The initial p.d. across the fixed resistor is \(V = 2.0 \text{ mA} \times 2.0 \text{ k}\Omega = 4.0 \text{ V}\). As the temperature increases, the resistance of the NTC thermistor decreases towards \(0\). This causes the total resistance to decrease, increasing the current. Consequently, the p.d. across the fixed resistor increases from \(4.0 \text{ V}\) towards the supply voltage of \(12 \text{ V}\).
评分标准
1 mark for the correct option C.
题目 15 · 選擇題
1 分
An ideal step-down transformer has a primary coil with \(1500\) turns and a secondary coil with \(60\) turns. The primary coil is connected to a \(230 \text{ V}\) a.c. mains supply. An electrical appliance of resistance \(1.2 \ \Omega\) is connected across the secondary coil. What is the current in the primary coil?
A.\(0.31 \text{ A}\)
B.\(7.7 \text{ A}\)
C.\(9.2 \text{ A}\)
D.\(19 \text{ A}\)
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解题
First, find the secondary voltage \(V_s = V_p \left(\frac{N_s}{N_p}\right) = 230 \times \frac{60}{1500} = 9.2 \text{ V}\). Next, find the secondary current \(I_s = \frac{V_s}{R} = \frac{9.2 \text{ V}}{1.2 \ \Omega} \approx 7.67 \text{ A}\). For an ideal transformer, primary power equals secondary power: \(V_p I_p = V_s I_s \implies I_p = I_s \left(\frac{V_s}{V_p}\right) = 7.67 \times \frac{9.2}{230} \approx 0.31 \text{ A}\).
评分标准
1 mark for the correct option A.
题目 16 · 選擇題
1 分
A student measures the count rate from a radioactive source. The background count rate is constant at \(15 \text{ counts/s}\). At time \(t = 0\), the measured count rate is \(255 \text{ counts/s}\). After a time of \(4.0 \text{ hours}\), the measured count rate is \(45 \text{ counts/s}\). What is the half-life of the radioactive source?
A.\(40 \text{ minutes}\)
B.\(80 \text{ minutes}\)
C.\(120 \text{ minutes}\)
D.\(160 \text{ minutes}\)
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解题
Subtract background to get the corrected count rates: initial corrected rate \(R_0 = 255 - 15 = 240 \text{ counts/s}\); final corrected rate \(R_t = 45 - 15 = 30 \text{ counts/s}\). The fraction remaining is \(\frac{30}{240} = \frac{1}{8}\), which represents exactly 3 half-lives. Therefore, \(3 \times t_{1/2} = 4.0 \text{ hours} = 240 \text{ minutes} \implies t_{1/2} = 80 \text{ minutes}\).
评分标准
1 mark for the correct option B.
题目 17 · 選擇題
1 分
An object is projected vertically upwards through the air. Air resistance opposes its motion. Which statement describes the acceleration of the object as it rises?
A.It remains constant at \(9.8\text{ m/s}^2\).
B.It decreases as the object rises.
C.It increases as the object rises.
D.It is zero at the maximum height of its flight.
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解题
As the object rises, there are two downward forces acting on it: the constant weight \(W\) and the speed-dependent air resistance force \(F_d\). The net downward force is \(W + F_d\), which gives a downward acceleration of \(a = g + \frac{F_d}{m}\). As the object rises, its speed decreases, which in turn reduces the air resistance force \(F_d\). Consequently, the net force decreases, and the acceleration of the object decreases as it rises, reaching its minimum value of \(g\) at the highest point where the speed is zero.
评分标准
Award 1 mark for identifying that air resistance decreases as speed decreases, leading to a decrease in the net downward force and thus a decrease in acceleration.
题目 18 · 選擇題
1 分
An object of mass \(M\) moving at speed \(v\) collides head-on with a stationary object of mass \(2M\). The two objects stick together after the collision. What fraction of the initial kinetic energy is lost as thermal energy during the collision?
A.\(\frac{1}{3}\)
B.\(\frac{1}{2}\)
C.\(\frac{2}{3}\)
D.\(\frac{8}{9}\)
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解题
From the conservation of linear momentum: \(M v = (M + 2M) v_f \implies v_f = \frac{v}{3}\). The initial kinetic energy is \(E_{ki} = \frac{1}{2} M v^2\). The final kinetic energy is \(E_{kf} = \frac{1}{2} (3M) \left(\frac{v}{3}\right)^2 = \frac{1}{6} M v^2 = \frac{1}{3} E_{ki}\). Therefore, the fraction of kinetic energy lost is \(1 - \frac{1}{3} = \frac{2}{3}\).
评分标准
Award 1 mark for finding the final velocity using momentum conservation, calculating the final kinetic energy, and determining the fraction lost.
题目 19 · 選擇題
1 分
A uniform beam of mass \(3.0\text{ kg}\) and length \(2.0\text{ m}\) is pivoted at one end. It is held horizontally by a vertical wire attached to the beam at a distance of \(1.5\text{ m}\) from the pivot. Taking the acceleration of free fall \(g = 9.8\text{ m/s}^2\), what is the tension in the wire?
A.\(9.8\text{ N}\)
B.\(14.7\text{ N}\)
C.\(19.6\text{ N}\)
D.\(29.4\text{ N}\)
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解题
The weight of the beam is \(W = mg = 3.0 \times 9.8 = 29.4\text{ N}\). Since the beam is uniform, this weight acts at its center of gravity, which is at the midpoint \(1.0\text{ m}\) from the pivot. Taking moments about the pivot for rotational equilibrium: Clockwise moment = Anticlockwise moment \implies W \times 1.0\text{ m} = T \times 1.5\text{ m} \implies 29.4 \times 1.0 = T \times 1.5 \implies T = \frac{29.4}{1.5} = 19.6\text{ N}\).
评分标准
Award 1 mark for identifying the correct center of gravity, calculating the weight, and applying the principle of moments to solve for tension.
题目 20 · 選擇題
1 分
A \(500\text{ W}\) electric heater is used to heat a \(2.0\text{ kg}\) block of metal. In \(4.0\text{ minutes}\), the temperature of the block rises from \(20\text{ }^\circ\text{C}\) to \(50\text{ }^\circ\text{C}\). Assuming no heat is lost to the surroundings, what is the specific heat capacity of the metal?
A.\(1200\text{ J / (kg }^\circ\text{C)}" (the denominator is \)50\text{ }^\circ\text{C}\) instead of the temperature difference)
C.\(4000\text{ J / (kg }^\circ\text{C)}" (using a factor incorrect by 2)
D.\(6000\text{ J / (kg }^\circ\text{C)}" (not converting minutes to seconds)
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解题
Energy supplied by the heater: \(E = P \times t = 500\text{ W} \times (4.0 \times 60\text{ s}) = 120,000\text{ J}\). The temperature change is \(\Delta T = 50 - 20 = 30\text{ }^\circ\text{C}\). Using the formula \(E = m c \Delta T\): \(120,000\text{ J} = 2.0\text{ kg} \times c \times 30\text{ }^\circ\text{C} \implies 120,000 = 60 c \implies c = 2000\text{ J / (kg }^\circ\text{C)}\).
评分标准
Award 1 mark for converting time to seconds, calculating total energy, and using the heat capacity formula to find the correct value.
题目 21 · 選擇題
1 分
A wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). A second wire made of the same metal has twice the length and half the diameter of the first wire. What is the resistance of the second wire?
A.\(2R\)
B.\(4R\)
C.\(8R\)
D.\(16R\)
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解题
Resistance is given by \(R = \rho \frac{L}{A}\). The cross-sectional area is \(A = \pi \frac{d^2}{4}\). Halving the diameter reduces the cross-sectional area to \(\frac{A}{4}\). For the second wire: \(L' = 2L\) and \(A' = \frac{A}{4}\). Therefore, the new resistance is \(R' = \rho \frac{2L}{A/4} = 8 \rho \frac{L}{A} = 8R\).
评分标准
Award 1 mark for identifying that halving the diameter quarter-sizes the area, and combining this with the doubled length to show the resistance increases by a factor of 8.
题目 22 · 選擇題
1 分
An ideal step-up transformer has 200 turns on its primary coil and 1000 turns on its secondary coil. The primary coil is connected to a \(12\text{ V}\) alternating current (a.c.) supply, and the primary current is \(5.0\text{ A}\). What are the voltage and current in the secondary coil?
A.Voltage = \(2.4\text{ V}\), Current = \(1.0\text{ A}\)
B.Voltage = \(60\text{ V}\), Current = \(1.0\text{ A}\)
C.Voltage = \(60\text{ V}\), Current = \(25\text{ A}\)
D.Voltage = \(240\text{ V}\), Current = \(5.0\text{ A}\)
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解题
For a transformer: \(\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 12 \times \frac{1000}{200} = 60\text{ V}\). For an ideal transformer, input power equals output power: \(V_p I_p = V_s I_s \implies 12 \times 5.0 = 60 \times I_s \implies I_s = 1.0\text{ A}\).
评分标准
Award 1 mark for correctly calculating the secondary voltage using turn ratio and secondary current using power conservation.
题目 23 · 選擇題
1 分
A GM tube is used to measure the count rate from a radioactive sample in a laboratory where the background radiation is constant at \(20\text{ counts / minute}\). At the start, the measured total count rate is \(340\text{ counts / minute}\). After \(3.0\text{ hours}\), the measured total count rate is \(60\text{ counts / minute}\). What is the half-life of the radioactive source?
A.\(45\text{ minutes}\)
B.\(60\text{ minutes}\)
C.\(90\text{ minutes}\)
D.\(120\text{ minutes}\)
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解题
First, subtract background radiation to find the corrected count rates: Initial corrected count rate = \(340 - 20 = 320\text{ counts / minute}\). Final corrected count rate = \(60 - 20 = 40\text{ counts / minute}\). The ratio of final to initial activity is \(\frac{40}{320} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\), which represents 3 half-lives. Since 3 half-lives equal \(3.0\text{ hours}\) (or 180 minutes), one half-life is \(\frac{180\text{ minutes}}{3} = 60\text{ minutes}\).
评分标准
Award 1 mark for subtracting the background rate, identifying that 3 half-lives have elapsed, and calculating the single half-life duration in minutes.
题目 24 · 選擇題
1 分
A distant galaxy is located at a distance of \(1.5 \times 10^{24}\text{ m}\) from Earth. Assuming the Hubble constant is \(2.2 \times 10^{-18}\text{ s}^{-1}\), what is the recessional speed of this galaxy?
A.\(1.5 \times 10^6\text{ m/s}\)
B.\(3.3 \times 10^6\text{ m/s}\)
C.\(6.8 \times 10^6\text{ m/s}\)
D.\(3.3 \times 10^7\text{ m/s}\)
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解题
According to Hubble's Law: \(v = H_0 d\). Substituting the given values: \(v = (2.2 \times 10^{-18}\text{ s}^{-1}) \times (1.5 \times 10^{24}\text{ m}) = 3.3 \times 10^6\text{ m/s}\).
评分标准
Award 1 mark for applying Hubble's Law formula and completing the multiplication correctly to find the recessional speed.
题目 25 · multiple_choice
1 分
A car accelerates from rest along a straight road with a constant acceleration of \(1.5\text{ m/s}^2\) for \(10\text{ s}\). It then travels at a constant speed for \(20\text{ s}\), and finally decelerates uniformly to rest in \(5.0\text{ s}\).
What is the total distance travelled by the car?
A.\(300\text{ m}\)
B.\(375\text{ m}\)
C.\(412.5\text{ m}\)
D.\(450\text{ m}\)
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解题
The motion is divided into three stages:
1. **Acceleration stage:** Using \(v = u + at\): \(v = 0 + (1.5 \times 10) = 15\text{ m/s}\).
The distance travelled during this stage (area of the triangle under the speed-time graph): \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10\text{ s} \times 15\text{ m/s} = 75\text{ m}\).
Award 1 mark for the correct option C. - 1 mark for calculating correct maximum velocity of \(15\text{ m/s}\) and individual stage distances. - 1 mark for summing the distances correctly to obtain \(412.5\text{ m}\).
题目 26 · multiple_choice
1 分
An electric motor with an efficiency of \(60\%\) is used to lift a load of mass \(80\text{ kg}\) vertically upwards through a height of \(12\text{ m}\) in \(4.0\text{ s}\).
What is the electrical power input to the motor? (Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\).)
A.\(1.4\text{ kW}\)
B.\(2.4\text{ kW}\)
C.\(3.9\text{ kW}\)
D.\(6.5\text{ kW}\)
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解题
1. First, calculate the useful work output (gravitational potential energy gained by the load): \(W_{\text{out}} = mgh = 80\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 9408\text{ J}\).
2. Next, calculate the useful power output: \(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{9408\text{ J}}{4.0\text{ s}} = 2352\text{ W}\).
3. Use the efficiency formula to find the electrical power input: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\) \(0.60 = \frac{2352\text{ W}}{P_{\text{in}}}\) \(P_{\text{in}} = \frac{2352}{0.60} = 3920\text{ W} \approx 3.9\text{ kW}\).
评分标准
Award 1 mark for the correct option C. - 1 mark for calculating useful work output as \(9408\text{ J}\). - 1 mark for using efficiency to correctly determine input power as \(3.9\text{ kW}\).
题目 27 · multiple_choice
1 分
A spring obeys Hooke's law up to its limit of proportionality. A load of \(6.0\text{ N}\) causes an extension of \(3.0\text{ cm}\). Two of these identical springs are connected in parallel, and a load of \(15.0\text{ N}\) is suspended from the combination.
What is the total extension of this parallel spring combination?
A.\(1.5\text{ cm}\)
B.\(3.75\text{ cm}\)
C.\(7.5\text{ cm}\)
D.\(15.0\text{ cm}\)
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解题
1. Find the spring constant \(k\) of a single spring: \(k = \frac{F}{x} = \frac{6.0\text{ N}}{3.0\text{ cm}} = 2.0\text{ N/cm}\).
2. When two identical springs are connected in parallel, they share the load equally. The effective spring constant \(k_{\text{parallel}}\) of the system is the sum of their individual spring constants: \(k_{\text{parallel}} = 2k = 2 \times 2.0\text{ N/cm} = 4.0\text{ N/cm}\).
3. Find the total extension under a load of \(15.0\text{ N}\): \(x_{\text{total}} = \frac{F_{\text{load}}}{k_{\text{parallel}}} = \frac{15.0\text{ N}}{4.0\text{ N/cm}} = 3.75\text{ cm}\).
评分标准
Award 1 mark for the correct option B. - 1 mark for determining individual spring constant as \(2.0\text{ N/cm}\). - 1 mark for parallel combination calculation yielding \(3.75\text{ cm}\).
题目 28 · multiple_choice
1 分
A ball of mass \(0.40\text{ kg}\) travelling horizontally at a velocity of \(8.0\text{ m/s}\) strikes a vertical wall and bounces back horizontally with a speed of \(5.0\text{ m/s}\).
What is the magnitude of the impulse exerted by the wall on the ball?
A.\(1.2\text{ N s}\)
B.\(2.0\text{ N s}\)
C.\(3.2\text{ N s}\)
D.\(5.2\text{ N s}\)
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解题
1. Impulse is equal to the change in momentum: \(I = \Delta p = m(v - u)\). 2. Defining the initial direction of velocity as positive: Initial velocity, \(u = +8.0\text{ m/s}\). Final velocity (rebounded in the opposite direction), \(v = -5.0\text{ m/s}\). 3. Calculate the change in momentum: \(\Delta p = 0.40\text{ kg} \times (-5.0\text{ m/s} - 8.0\text{ m/s}) = 0.40 \times (-13.0) = -5.2\text{ N s}\). 4. The magnitude of the impulse is therefore \(5.2\text{ N s}\).
评分标准
Award 1 mark for the correct option D. - 1 mark for showing change in momentum with correct vector signs (summing the speeds to \(13.0\text{ m/s}\)).
题目 29 · multiple_choice
1 分
A cylindrical tank is filled with a liquid of density \(1200\text{ kg/m}^3\) to a depth of \(0.80\text{ m}\). The atmospheric pressure acting on the surface of the liquid is \(1.0 \times 10^5\text{ Pa}\).
What is the total pressure acting at the bottom of the cylinder? (Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\).)
A.\(9.4 \times 10^3\text{ Pa}\)
B.\(9.4 \times 10^4\text{ Pa}\)
C.\(1.09 \times 10^5\text{ Pa}\)
D.\(1.94 \times 10^5\text{ Pa}\)
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解题
1. Calculate the pressure exerted by the liquid column using \(p_{\text{liquid}} = \rho g h\): \(p_{\text{liquid}} = 1200\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 0.80\text{ m} = 9408\text{ Pa}\).
2. Add the atmospheric pressure to find the total pressure at the bottom: \(p_{\text{total}} = p_{\text{atmosphere}} + p_{\text{liquid}}\) \(p_{\text{total}} = 1.0 \times 10^5\text{ Pa} + 9408\text{ Pa} = 109\,408\text{ Pa} \approx 1.09 \times 10^5\text{ Pa}\).
评分标准
Award 1 mark for the correct option C. - 1 mark for calculating the hydrostatic pressure as \(9408\text{ Pa}\). - 1 mark for correctly adding atmospheric pressure to find total pressure.
题目 30 · multiple_choice
1 分
A ray of light is travelling inside a solid glass block of refractive index \(1.52\) towards the boundary with air.
What is the critical angle for light at this boundary?
A.\(34.1^\circ\)
B.\(41.1^\circ\)
C.\(48.2^\circ\)
D.\(56.8^\circ\)
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解题
1. The relationship between critical angle \(c\) and refractive index \(n\) is given by: \(\sin c = \frac{1}{n}\).
3. Calculate the angle: \(c = \sin^{-1}(0.6579) \approx 41.14^\circ\).
Therefore, the critical angle is approximately \(41.1^\circ\).
评分标准
Award 1 mark for the correct option B. - 1 mark for stating and correctly applying the critical angle formula \(\sin c = 1/n\).
题目 31 · multiple_choice
1 分
A uniform metal wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). Another wire of the same material is drawn out so that its length becomes \(3L\), while its total volume remains unchanged.
What is the resistance of the new wire in terms of \(R\)?
A.\(\frac{1}{3}R\)
B.\(R\)
C.\(3R\)
D.\(9R\)
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解题
1. Volume \(V = L \times A\) must remain constant. If the length becomes \(3L\), the cross-sectional area must decrease to \(\frac{A}{3}\) to keep volume unchanged.
2. The formula for resistance is: \(R = \rho \frac{L}{A}\).
3. For the new wire, substituting the new length and area: \(R_{\text{new}} = \rho \frac{3L}{(A/3)} = 9 \left(\rho \frac{L}{A}\right) = 9R\).
评分标准
Award 1 mark for the correct option D. - 1 mark for identifying that area decreases by a factor of 3. - 1 mark for obtaining the final resistance value of \(9R\).
题目 32 · multiple_choice
1 分
Light from a distant galaxy is observed to have a fractional increase in wavelength (redshift) of \(0.040\). The Hubble constant is given as \(H_0 = 2.2 \times 10^{-18}\text{ s}^{-1}\), and the speed of light is \(c = 3.0 \times 10^8\text{ m/s}\).
What is the estimated distance of this galaxy from Earth?
A.\(5.5 \times 10^{24}\text{ m}\)
B.\(1.2 \times 10^{25}\text{ m}\)
C.\(5.5 \times 10^{25}\text{ m}\)
D.\(1.4 \times 10^{26}\text{ m}\)
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解题
1. Calculate the recession velocity \(v\) using the redshift formula: \(v = z \times c\) \(v = 0.040 \times 3.0 \times 10^8\text{ m/s} = 1.2 \times 10^7\text{ m/s}\).
2. Use Hubble's Law to estimate the distance \(d\): \(v = H_0 \times d\) \(d = \frac{v}{H_0} = \frac{1.2 \times 10^7\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 5.45 \times 10^{24}\text{ m} \approx 5.5 \times 10^{24}\text{ m}\).
评分标准
Award 1 mark for the correct option A. - 1 mark for finding the recession velocity as \(1.2 \times 10^7\text{ m/s}\). - 1 mark for correctly applying Hubble's Law to calculate the distance.
题目 33 · mcq
1 分
A skydiver jumps from an airplane and falls through the air. Before opening her parachute, she reaches terminal velocity.
How do her acceleration and the air resistance acting on her change during the first few seconds of her fall?
A.Acceleration increases, and air resistance increases.
B.Acceleration decreases, and air resistance increases.
C.Acceleration decreases, and air resistance decreases.
D.Acceleration remains constant, and air resistance increases.
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解题
When the skydiver first jumps, her speed is relatively low, so the air resistance is small, and her acceleration is close to the acceleration of free fall, \(g\).
As she falls and her speed increases: 1. The air resistance increases because air resistance is directly proportional to speed (or a power of speed). 2. The resultant downward force decreases because the upward air resistance force increases while her downward weight remains constant. 3. Since the resultant force decreases, her acceleration decreases (according to \(F = ma\)).
Therefore, her acceleration decreases while the air resistance increases.
评分标准
Award 1 mark for the correct option (B). - Reject other options that state acceleration increases, remains constant, or air resistance decreases.
题目 34 · mcq
1 分
An electric water pump has an efficiency of \(75\%\). It is used to raise \(200\text{ kg}\) of water through a vertical height of \(12\text{ m}\) in a time of \(10\text{ s}\).
What is the electrical power input to the pump? Take the weight of \(1.0\text{ kg}\) to be \(9.8\text{ N}\).
A.\(0.32\text{ kW}\)
B.\(1.8\text{ kW}\)
C.\(2.4\text{ kW}\)
D.\(3.1\text{ kW}\)
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解题
First, calculate the useful work done by the pump to raise the water: \[W = mgh = 200\text{ kg} \times 9.8\text{ N/kg} \times 12\text{ m} = 23\,520\text{ J}\]
Next, calculate the useful power output of the pump: \[P_{\text{out}} = \frac{W}{t} = \frac{23\,520\text{ J}}{10\text{ s}} = 2352\text{ W}\]
Using the efficiency formula: \[\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\] \[0.75 = \frac{2352\text{ W}}{P_{\text{in}}}\] \[P_{\text{in}} = \frac{2352}{0.75} = 3136\text{ W} \approx 3.1\text{ kW}\]
评分标准
Award 1 mark for the correct option (D). - Distractor A: calculation without using gravitational field strength \(g\). - Distractor B: multiplying by \(0.75\) instead of dividing. - Distractor C: calculating only the power output \(P_{\text{out}}\) without dividing by efficiency.
题目 35 · mcq
1 分
A student hangs a load of weight \(F\) from a single spring and measures its extension. She then connects two of these identical springs in parallel and hangs the same load of weight \(F\) from the combination.
The spring constant of a single spring is \(k\).
What is the effective spring constant of the parallel combination, and how does its extension compare to the extension of the single spring?
A.Effective spring constant is \(k/2\), and the extension is half as much.
B.Effective spring constant is \(2k\), and the extension is half as much.
C.Effective spring constant is \(2k\), and the extension is twice as much.
D.Effective spring constant is \(k/2\), and the extension is twice as much.
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解题
When two identical springs are in parallel, they share the applied load equally, meaning each spring experiences a force of \(F/2\).
The extension of each spring (and therefore the combination) is: \[x_{\text{parallel}} = \frac{F/2}{k} = \frac{F}{2k}\]
Compared to a single spring under load \(F\) (which extends by \(x_{\text{single}} = \frac{F}{k}\)), the parallel combination extends by half as much.
The effective spring constant \(k_{\text{eff}}\) of the combination is: \[k_{\text{eff}} = \frac{F}{x_{\text{parallel}}} = 2k\]
评分标准
Award 1 mark for the correct option (B). - Reject options stating that the spring constant is halved or that the extension is doubled.
题目 36 · mcq
1 分
A tennis ball of mass \(0.060\text{ kg}\) is moving horizontally at a speed of \(25\text{ m/s}\) when it is struck by a racket. It recoils in the opposite direction at a speed of \(35\text{ m/s}\).
The racket is in contact with the ball for a time of \(4.0\text{ ms}\).
What is the average force exerted on the ball by the racket during the collision?
A.\(0.90\text{ N}\)
B.\(150\text{ N}\)
C.\(525\text{ N}\)
D.\(900\text{ N}\)
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解题
First, find the change in momentum (impulse) of the tennis ball. Taking the initial direction of motion as positive: - Initial velocity \(u = +25\text{ m/s}\) - Final velocity \(v = -35\text{ m/s}\)
The magnitude of the change in momentum (impulse) is \(3.6\text{ N s}\).
Now, calculate the average force exerted on the ball: \[F = \frac{\Delta p}{\Delta t} = \frac{3.6\text{ N s}}{4.0 \times 10^{-3}\text{ s}} = 900\text{ N}\]
评分标准
Award 1 mark for the correct option (D). - Distractor A: forgetting to convert \(4.0\text{ ms}\) into seconds. - Distractor B: subtracting the speeds instead of adding (neglecting vector direction, giving \(\Delta v = 10\text{ m/s}\)). - Distractor C: using only the final speed \(35\text{ m/s}\) to find momentum change.
题目 37 · mcq
1 分
A metal block of mass \(2.0\text{ kg}\) is heated by an electrical heater with a power rating of \(50\text{ W}\). The heater is switched on for a duration of \(8.0\text{ minutes}\).
The specific heat capacity of the metal is \(400\text{ J / (kg }^{\circ}\text{C)}\).
Assuming there is no heat loss to the surroundings, what is the temperature rise of the block?
A.\(0.50\text{ }^{\circ}\text{C}\)
B.\(15\text{ }^{\circ}\text{C}\)
C.\(30\text{ }^{\circ}\text{C}\)
D.\(60\text{ }^{\circ}\text{C}\)
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解题
First, calculate the thermal energy supplied by the heater: \[E = P \times t = 50\text{ W} \times (8.0 \times 60\text{ s}) = 50 \times 480 = 24\,000\text{ J}\]
Using the specific heat capacity formula: \[E = m c \Delta\theta\] \[24\,000\text{ J} = 2.0\text{ kg} \times 400\text{ J/(kg }^{\circ}\text{C)} \times \Delta\theta\] \[24\,000 = 800 \times \Delta\theta\] \[\Delta\theta = \frac{24\,000}{800} = 30\text{ }^{\circ}\text{C}\]
评分标准
Award 1 mark for the correct option (C). - Distractor A: forgetting to convert minutes to seconds in the time calculation. - Distractor B: arithmetic mistake (e.g., dividing by \(2.0\) twice). - Distractor D: multiplying rather than dividing, or neglecting the mass \(m = 2.0\text{ kg}\).
题目 38 · mcq
1 分
A ray of light in air is incident on a transparent plastic block at an angle of incidence of \(50^{\circ}\). The refractive index of the plastic is \(1.45\).
What is the angle of refraction inside the plastic, and what is the critical angle for light travelling from the plastic into air?
A.Angle of refraction = \(32^{\circ}\), critical angle = \(44^{\circ}\)
B.Angle of refraction = \(32^{\circ}\), critical angle = \(50^{\circ}\)
C.Angle of refraction = \(34^{\circ}\), critical angle = \(44^{\circ}\)
D.Angle of refraction = \(34^{\circ}\), critical angle = \(50^{\circ}\)
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解题
1. Use Snell's Law to find the angle of refraction \(r\): \[n = \frac{\sin i}{\sin r} \implies \sin r = \frac{\sin i}{n} = \frac{\sin 50^{\circ}}{1.45} = \frac{0.7660}{1.45} \approx 0.5283\] \[r = \arcsin(0.5283) \approx 31.9^{\circ} \approx 32^{\circ}\]
2. Use the critical angle formula to find \(c\): \[\sin c = \frac{1}{n} = \frac{1}{1.45} \approx 0.6897\] \[c = \arcsin(0.6897) \approx 43.6^{\circ} \approx 44^{\circ}\]
评分标准
Award 1 mark for the correct option (A). - Distractors C and D use a common student error of simply dividing the angles without taking the sines (i.e., \(50^{\circ} / 1.45 \approx 34^{\circ}\)).
题目 39 · mcq
1 分
A cylindrical metal wire of length \(L\) and diameter \(d\) has a resistance of \(8.0\ \Omega\).
A second wire of the same metal has a length of \(3L\) and a diameter of \(0.50d\).
What is the resistance of the second wire?
A.\(12\ \Omega\)
B.\(24\ \Omega\)
C.\(48\ \Omega\)
D.\(96\ \Omega\)
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解题
The resistance of a wire is given by: \[R = \rho \frac{L}{A}\] Since \(A = \pi \left(\frac{d}{2}\right)^2\), the resistance is proportional to length and inversely proportional to the square of the diameter: \[R \propto \frac{L}{d^2}\]
Comparing the second wire to the first wire: - Length is multiplied by \(3\). - Diameter is multiplied by \(0.50\).
Therefore, the new resistance \(R_2\) is: \[R_2 = R_1 \times \frac{3}{(0.50)^2} = 8.0\ \Omega \times \frac{3}{0.25} = 8.0 \times 12 = 96\ \Omega\]
评分标准
Award 1 mark for the correct option (D). - Distractor A: dividing the length multiplier by \(2\). - Distractor B: multiplying only by length factor \(3\). - Distractor C: not squaring the diameter factor (dividing by \(0.50\) instead of \(0.25\)).
题目 40 · mcq
1 分
Which row correctly describes the final evolutionary stages in the life cycles of a star similar in mass to the Sun, and a star much more massive than the Sun?
A.Sun-like star: planetary nebula \(\rightarrow\) white dwarf \(\rightarrow\) black dwarf | Massive star: red supergiant \(\rightarrow\) supernova \(\rightarrow\) neutron star or black hole
B.Sun-like star: planetary nebula \(\rightarrow\) white dwarf \(\rightarrow\) black dwarf | Massive star: red giant \(\rightarrow\) planetary nebula \(\rightarrow\) white dwarf
C.Sun-like star: red supergiant \(\rightarrow\) supernova \(\rightarrow\) neutron star or black hole | Massive star: planetary nebula \(\rightarrow\) white dwarf \(\rightarrow\) black dwarf
D.Sun-like star: red giant \(\rightarrow\) supernova \(\rightarrow\) white dwarf | Massive star: red supergiant \(\rightarrow\) planetary nebula \(\rightarrow\) black hole
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解题
According to the Cambridge IGCSE syllabus (Section 6.1): 1. A star of mass up to about 8 solar masses (similar to the Sun) evolves from a stable star to a red giant, then to a planetary nebula, white dwarf, and finally a black dwarf. 2. A star of mass much greater than 8 solar masses (massive star) evolves into a red supergiant, then explodes as a supernova, leaving behind either a neutron star or a black hole.
Option A correctly identifies these two paths.
评分标准
Award 1 mark for the correct option (A). - Reject options that switch the two life cycles or use incorrect sequences of stages.
Answer all structured questions in the spaces provided on the question paper. Show all your working and use appropriate units.
10 题目 · 80 分
题目 1 · structured
8 分
In a cosmology study, astronomers analyze the light emitted from distant galaxies.
(a) State two astronomical observations that support the Big Bang theory. [2]
(b) A galaxy is observed at a distance of \( 4.5 \times 10^{24} \text{ m} \) from Earth. The Hubble constant \( H_0 \) is taken to be \( 2.2 \times 10^{-18} \text{ s}^{-1} \).
(i) Calculate the recessional speed of this galaxy. [3]
(ii) Estimate the age of the Universe in years, using the Hubble constant. [3]
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解题
(a) Observations supporting the Big Bang theory: 1. Redshift of light from distant galaxies (showing that the Universe is expanding). 2. Cosmic Microwave Background Radiation (CMBR) (remnant thermal energy from the early Universe).
(b)(i) Using Hubble's Law: \( v = H_0 \times d \) \( v = (2.2 \times 10^{-18} \text{ s}^{-1}) \times (4.5 \times 10^{24} \text{ m}) \) \( v = 9.9 \times 10^6 \text{ m/s} \)
(b)(ii) The age of the Universe \( T \) can be estimated as: \( T \approx \frac{1}{H_0} \) \( T \approx \frac{1}{2.2 \times 10^{-18} \text{ s}^{-1}} \approx 4.545 \times 10^{17} \text{ s} \)
Convert this time to years: \( T \text{ (years)} = \frac{4.545 \times 10^{17} \text{ s}}{365 \times 24 \times 3600 \text{ s/year}} \) \( T \text{ (years)} \approx \frac{4.545 \times 10^{17}}{3.154 \times 10^7} \approx 1.44 \times 10^{10} \text{ years} \) To 2 significant figures, the age is \( 1.4 \times 10^{10} \text{ years} \).
评分标准
(a) Award 1 mark for each correct observation: - Redshift of distant galaxies [1] - Cosmic Microwave Background Radiation (CMBR) [1]
(b)(i) Award marks as follows: - Recall of \( v = H_0 d \) [1] - Correct substitution [1] - Correct final answer with unit \( (9.9 \times 10^6 \text{ m/s}) \) [1]
(b)(ii) Award marks as follows: - Recall of \( T \approx 1 / H_0 \) [1] - Calculation of age in seconds \( (4.5 \times 10^{17} \text{ s}) \) [1] - Conversion to years with correct final answer \( (1.4 \times 10^{10} \text{ years}) \) [1] (Accept range 1.4-1.5 x 10^10 years)
题目 2 · structured
8 分
A model rocket of mass \( 0.85 \text{ kg} \) is launched vertically upwards. At \( t = 0 \), the rocket engine is ignited and provides a constant upward thrust of \( 24 \text{ N} \). The acceleration of free fall \( g \) is \( 9.8 \text{ m/s}^2 \).
(a) (i) Show that the weight of the model rocket is approximately \( 8.3 \text{ N} \). [1]
(ii) Calculate the initial vertical acceleration of the rocket. [3]
(b) After some time, the fuel is completely used up and the engine stops. The rocket continues to move upwards before falling back to the ground. Air resistance cannot be ignored.
(i) Describe the motion of the rocket from the moment the engine stops until it reaches its maximum height. [2]
(ii) State and explain how the acceleration of the rocket at its maximum height compares with its acceleration just after the engine stops. [2]
Using Newton's Second Law: \( a = \frac{F_{\text{res}}}{m} \) \( a = \frac{15.67 \text{ N}}{0.85 \text{ kg}} \approx 18.4 \text{ m/s}^2 \) (or \( 18 \text{ m/s}^2 \) to 2 s.f.)
(b)(i) The rocket is moving upwards but decelerating. The deceleration is not constant; it decreases as the rocket slows down because air resistance decreases as speed decreases. Finally, the speed reaches zero at maximum height.
(b)(ii) Just after the engine stops, the rocket is moving fast, so there is a large downward air resistance force in addition to its downward weight. This produces a large deceleration (acceleration much greater than \( 9.8 \text{ m/s}^2 \) downwards). At maximum height, the speed is zero, so air resistance is zero. The only force acting on the rocket is its weight, so its acceleration is exactly \( 9.8 \text{ m/s}^2 \) downwards. Therefore, the acceleration at maximum height is less than the acceleration just after the engine stops.
(a)(ii) - Calculation of resultant force \( (15.7 \text{ N}) \) [1] - Use of \( a = F/m \) [1] - Correct final acceleration with unit \( (18 \text{ m/s}^2) \) [1] (Accept 18.4)
(b)(i) - Rocket decelerates / slows down [1] - Deceleration decreases / is non-uniform because air resistance decreases [1]
(b)(ii) - At maximum height, acceleration is less (exactly \( 9.8 \text{ m/s}^2 \)) [1] - Because velocity is zero, so air resistance is zero (whereas just after engine stops, air resistance is acting downwards, increasing deceleration) [1]
题目 3 · structured
8 分
Sodium-24 (\( {}_{11}^{24}\text{Na} \)) is a radioactive isotope that decays to stable Magnesium-24 (\( {}_{12}^{24}\text{Mg} \)) by emitting a beta-minus (\( \beta^{-} \)) particle and a gamma (\( \gamma \)) ray.
(a) (i) Complete the decay equation for this process:
(ii) State one similarity and one difference between the structure of a sodium-24 nucleus and a magnesium-24 nucleus. [2]
(b) A hospital receives a sample of sodium-24 with an initial activity of \( 3.2 \times 10^6 \text{ Bq} \). The half-life of sodium-24 is \( 15 \text{ hours} \).
(i) Calculate the activity of the sample after \( 45 \text{ hours} \). [2]
(ii) Explain why sodium-24, with its half-life of 15 hours, is suitable for use as a medical tracer inside the human body, whereas an isotope with a half-life of 15 years would not be suitable. [2]
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解题
(a)(i) In beta-minus decay, a neutron converts to a proton, emitting an electron. For the \( \beta^{-} \) particle: Nucleon number \( A = 0 \) Proton number \( Z = -1 \)
(a)(ii) Similarity: Both nuclei have the same nucleon number (or both contain 24 nucleons in total). Difference: They have different proton/neutron numbers. Sodium-24 has 11 protons and 13 neutrons, while Magnesium-24 has 12 protons and 12 neutrons.
(b)(i) Number of half-lives in 45 hours: \( n = \frac{45 \text{ hours}}{15 \text{ hours}} = 3 \) Activity after 3 half-lives: \( \text{Activity} = \frac{3.2 \times 10^6 \text{ Bq}}{2^3} = \frac{3.2 \times 10^6}{8} = 4.0 \times 10^5 \text{ Bq} \)
(b)(ii) A half-life of 15 hours is long enough to allow the medical tracer to distribute and the scan to be completed, but short enough to decay quickly, minimizing radiation exposure and tissue damage. An isotope with a half-life of 15 years remains active for way too long, continuing to emit radiation and causing long-term harm to the patient.
评分标准
(a)(i) - Correct value of \( A = 0 \) [1] - Correct value of \( Z = -1 \) [1]
(a)(ii) - Similarity: Same mass number / number of nucleons / 24 nucleons [1] - Difference: Different proton numbers (11 vs 12) / different neutron numbers (13 vs 12) [1]
(b)(i) - Determine that 45 hours is 3 half-lives [1] - Correct calculation of final activity: \( 4.0 \times 10^5 \text{ Bq} \) (with unit) [1]
(b)(ii) - 15 hours is long enough to complete medical measurements but short enough that active material decays quickly to safe levels [1] - 15 years remains in the body for far too long, delivering a dangerously high/unnecessary radiation dose [1]
题目 4 · structured
8 分
In a mountain hydroelectric power station, water falls from an upper reservoir through a vertical height of \( 120 \text{ m} \) to drive a turbine located below.
Every second, \( 1500 \text{ kg} \) of water passes through the turbines. The acceleration of free fall \( g \) is \( 9.8 \text{ m/s}^2 \).
(a) (i) Calculate the rate at which gravitational potential energy is lost by the falling water. [2]
(ii) The electrical power output from the generators is \( 1.4 \times 10^6 \text{ W} \). Calculate the efficiency of the power station. [3]
(b) State and explain two environmental disadvantages of generating electricity using large-scale hydroelectric power schemes. [3]
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解题
(a)(i) Rate of loss of GPE (Power input): \( P_{\text{in}} = \frac{\Delta E_p}{t} = \frac{mgh}{t} \) Since mass per second \( \frac{m}{t} = 1500 \text{ kg/s} \): \( P_{\text{in}} = 1500 \text{ kg/s} \times 9.8 \text{ m/s}^2 \times 120 \text{ m} \) \( P_{\text{in}} = 1.764 \times 10^6 \text{ W} \approx 1.8 \times 10^6 \text{ W} \) (or \( 1.76 \times 10^6 \text{ W} \))
(a)(ii) Efficiency \( \eta \) is given by: \( \eta = \frac{\text{Useful power output}}{\text{Total power input}} \times 100\% \) \( \eta = \frac{1.4 \times 10^6 \text{ W}}{1.764 \times 10^6 \text{ W}} \times 100\% \approx 79.4\% \) (or \( 79\% \) to 2 s.f.) *(Self-correction on the calculation during development: 1.4 / 1.764 = 0.7936 = 79% efficiency)*
(b) Two environmental disadvantages of large-scale hydroelectric schemes: 1. Flooding of valleys upstream: Large areas of land are submerged to create the reservoir, which destroys terrestrial ecosystems and habitats. 2. Disruption to aquatic migration: The dam blocks river pathways, preventing fish species (like salmon) from migrating upstream to spawn.
评分标准
(a)(i) - Use of \( E_p = mgh \) or \( P = \frac{m}{t}gh \) [1] - Correct calculation: \( 1.76 \times 10^6 \text{ W} \) or \( 1.8 \times 10^6 \text{ W} \) (or J/s) [1]
(a)(ii) - Recall of efficiency formula [1] - Correct substitution of values [1] - Correct final answer: \( 79\% \) or \( 79.4\% \) [1]
(b) - Submerging of land / flooding destroys forest and wildlife habitats [1] - Disruption of migration / blocks fish pathways up or down the river [1] - Explanatory detail of at least one of these points (e.g., rotting flooded vegetation releases methane, a greenhouse gas; or loss of agricultural land) [1]
题目 5 · structured
8 分
A ray of monochromatic light is incident on the flat face of a glass prism at an angle of incidence of \( 42^\circ \). The refractive index of the glass is \( 1.5 \).
(a) (i) Calculate the angle of refraction inside the glass. [3]
(ii) Describe what happens to the speed, frequency, and wavelength of the light as it enters the glass prism from air. [3]
(b) State what is meant by the critical angle for light travelling from glass to air. [2]
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解题
(a)(i) Using Snell's Law: \( n = \frac{\sin i}{\sin r} \) \( 1.5 = \frac{\sin(42^\circ)}{\sin r} \) \( \sin r = \frac{\sin(42^\circ)}{1.5} \) \( \sin r = \frac{0.6691}{1.5} = 0.4461 \) \( r = \sin^{-1}(0.4461) \approx 26.5^\circ \) (or \( 26^\circ \) to 2 s.f.)
(a)(ii) As the light enters the optically denser medium (glass): 1. Speed: Decreases (as \( v = c/n \)). 2. Frequency: Remains constant / unchanged. 3. Wavelength: Decreases (as \( \lambda = v/f \)).
(b) The critical angle is the angle of incidence in the optically denser medium (glass) [1] for which the angle of refraction in the less dense medium (air) is \( 90^\circ \) / the refracted ray travels along the boundary [1].
评分标准
(a)(i) - Recall of \( n = \sin i / \sin r \) [1] - Correct substitution of values [1] - Correct calculation of angle of refraction: \( 26^\circ \) to \( 27^\circ \) [1]
(b) - Angle of incidence in the denser medium [1] - Results in an angle of refraction of 90 degrees (or ray traveling along the boundary) [1]
题目 6 · structured
8 分
An alternating current (a.c.) generator is used to produce electrical energy.
(a) (i) State the name of the component that connects the rotating coil to the external stationary circuit while allowing continuous rotation. [1]
(ii) Explain, in terms of magnetic field lines, how an electromotive force (e.m.f.) is induced in the rotating coil. [3]
(b) In a particular test, the coil is rotated. The maximum induced e.m.f. is \( 6.0 \text{ V} \) and the period of rotation is \( 0.040 \text{ s} \).
(i) State one change to the design of the generator, other than rotating the coil faster, that would increase the maximum induced e.m.f. [1]
(ii) The speed of rotation of the coil is now doubled.
State and explain the effect of this change on the new maximum e.m.f. and the new period of the induced voltage. [3]
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解题
(a)(i) Slip rings (with carbon brushes).
(a)(ii) As the coil of wire rotates in the magnetic field, the sides of the coil cut through the magnetic field lines [1]. This causes a change in the magnetic flux linkage through the coil [1], which induces an electromotive force (e.m.f.) across the ends of the coil [1] according to Faraday's law.
(b)(i) To increase the maximum e.m.f. without rotating faster: - Use a stronger magnet / increase magnetic field strength. - Increase the number of turns on the coil. - Increase the cross-sectional area of the coil.
(b)(ii) Doubling the speed of rotation: - The maximum e.m.f. is doubled to \( 12 \text{ V} \) [1] because the coil cuts magnetic field lines at twice the rate [1]. - The period of the induced a.c. is halved to \( 0.020 \text{ s} \) [1] because the time for one complete revolution is halved.
评分标准
(a)(i) - Slip rings [1]
(a)(ii) - Coil cuts magnetic field lines / changes flux linkage [1] - Induces an e.m.f. / voltage [1] - Explanation of alternating nature or link to rate of cutting [1]
(b)(i) - Stronger magnets / more turns on the coil / larger area [1]
(b)(ii) - Maximum e.m.f. doubles to \( 12 \text{ V} \) [1] - Rate of cutting of field lines is doubled [1] - Period is halved to \( 0.020 \text{ s} \) [1]
题目 7 · structured
8 分
Trolley A of mass \( 1.2 \text{ kg} \) is moving to the right along a horizontal friction-free track at a constant velocity of \( 3.0 \text{ m/s} \). It collides head-on with a stationary trolley B of mass \( 0.80 \text{ kg} \).
After the collision, trolley A continues to move to the right with a velocity of \( 1.0 \text{ m/s} \).
(a) (i) Show that the momentum of trolley A before the collision is \( 3.6 \text{ kg m/s} \). [1]
(ii) Calculate the velocity of trolley B after the collision. [3]
(b) During the collision, trolley A is in contact with trolley B for a short time and exerts an average force of \( 12 \text{ N} \) on trolley B.
(i) Calculate the duration of the contact during the collision. [2]
(ii) State the magnitude and direction of the average force exerted by trolley B on trolley A during the collision. [2]
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解题
(a)(i) \( p = mv \) \( p = 1.2 \text{ kg} \times 3.0 \text{ m/s} = 3.6 \text{ kg m/s} \)
(a)(ii) Using the Principle of Conservation of Momentum: \( \text{Total momentum before} = \text{Total momentum after} \) \( m_A u_A + m_B v_B(\text{initial}) = m_A v_A + m_B v_B(\text{final}) \) \( 3.6 \text{ kg m/s} + 0 = (1.2 \text{ kg} \times 1.0 \text{ m/s}) + (0.80 \text{ kg} \times v_B) \) \( 3.6 = 1.2 + 0.80 v_B \) \( 0.80 v_B = 2.4 \) \( v_B = 3.0 \text{ m/s} \) to the right
(b)(i) Using the relationship between impulse, force, and momentum for trolley B: \( \text{Impulse} = F \Delta t = \Delta p \) \( \Delta p_B = m_B v_B - 0 = 0.80 \text{ kg} \times 3.0 \text{ m/s} = 2.4 \text{ N s} \) \( 12 \text{ N} \times \Delta t = 2.4 \text{ N s} \) \( \Delta t = \frac{2.4}{12} = 0.20 \text{ s} \)
(b)(ii) According to Newton's Third Law (equal and opposite reaction): - Magnitude of force = \( 12 \text{ N} \) - Direction = to the left (opposite to the motion of trolley A)
(a)(ii) - Statement of conservation of momentum [1] - Substitution of values: \( 3.6 = 1.2 + 0.80 v_B \) [1] - Correct final velocity with unit \( (3.0 \text{ m/s}) \) [1]
(b)(i) - Use of Impulse \( F \Delta t = \Delta p \) (or equivalent) [1] - Correct calculation: \( 0.20 \text{ s} \) [1]
(b)(ii) - Magnitude: \( 12 \text{ N} \) [1] - Direction: to the left / opposite direction to trolley A's velocity [1]
题目 8 · structured
8 分
A copper block of mass \( 2.0 \text{ kg} \) is heated using an electrical heater rated at \( 150 \text{ W} \) for \( 5.0 \text{ minutes} \). The specific heat capacity of copper is \( 385 \text{ J / (kg }^\circ\text{C)} \).
(a) (i) Calculate the thermal energy supplied by the heater in \( 5.0 \text{ minutes} \). [2]
(ii) Calculate the temperature rise of the copper block, assuming no thermal energy is lost to the surroundings. [3]
(b) In a real-world scenario, some thermal energy is lost to the surroundings during heating.
(i) State how the actual temperature rise of the copper block compares with your calculated value in (a)(ii). [1]
(ii) Explain, in terms of conduction and convection, how wrapping the copper block in cotton wool reduces the thermal energy lost to the surroundings. [2]
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解题
(a)(i) \( E = P \times t \) Convert minutes to seconds: \( t = 5.0 \text{ minutes} \times 60 \text{ s/minute} = 300 \text{ s} \) \( E = 150 \text{ W} \times 300 \text{ s} = 4.5 \times 10^4 \text{ J} \) (or \( 45000 \text{ J} \))
(b)(i) The actual temperature rise is lower / less than calculated.
(b)(ii) Cotton wool traps air in small pockets. Air is a very poor conductor of thermal energy, which reduces thermal energy loss by conduction [1]. Additionally, trapping the air prevents air currents from moving, which stops convection currents from forming and reduces convection loss [1].
评分标准
(a)(i) - Conversion of time to seconds (300 s) [1] - Correct thermal energy calculation with unit: \( 4.5 \times 10^4 \text{ J} \) or \( 45000 \text{ J} \) [1]
(a)(ii) - Recall of \( E = mc\Delta\theta \) [1] - Correct substitution of values [1] - Correct calculation of temperature rise: \( 58^\circ\text{C} \) [1] (Accept 58.4)
(b)(i) - Actual temperature rise is less / lower [1]
(b)(ii) - Trapped air is a poor conductor, reducing conduction [1] - Trapped air cannot move / circulate, preventing convection currents [1]
题目 9 · Structured
8 分
A cylindrical heating element is made from a high-resistance alloy wire.
The element has a length \(L = 1.5\text{ m}\) and a cross-sectional area \(A = 2.4 \times 10^{-7}\text{ m}^2\). The resistance of this heating element is \(12\\ \Omega\).
(a) The heating element is connected to a \(24\text{ V}\) d.c. power supply.
(i) Calculate the current in the heating element.
current = ..................................................... [2]
(ii) Calculate the charge that flows through the heating element in \(5.0\text{ minutes}\).
(ii) Convert the time into seconds: \[t = 5.0 \times 60 = 300\text{ s}\] Using the formula for charge: \[Q = I \times t = 2.0 \times 300 = 600\text{ C}\]
(b) The resistance \(R\) of a wire is given by the relation: \[R = \rho \frac{L}{A}\] Since the wire is cylindrical, its cross-sectional area \(A\) in terms of diameter \(d\) is: \[A = \frac{\pi d^2}{4}\] This shows that: \[R \propto \frac{L}{d^2}\] Let the first heating element have length \(L_1\), diameter \(d_1\), and resistance \(R_1 = 12\\ \Omega\). For the second heating element: \[L_2 = 2L_1\] \[d_2 = 2d_1\] Substituting these into the proportional relationship gives: \[R_2 = R_1 \times \frac{L_2}{L_1} \times \left(\frac{d_1}{d_2}\right)^2 = 12 \times 2 \times \left(\frac{1}{2}\right)^2 = 12 \times 2 \times \frac{1}{4} = 6.0\\ \Omega\]
评分标准
(a) (i) - \(I = V / R\) or substitution \(24 / 12\) [C1] - \(2.0\text{ A}\) (with unit) [A1]
(ii) - \(t = 300\text{ s}\) seen or used [C1] - \(600\text{ C}\) (with unit; allow ecf from (a)(i)) [A1]
(b) - \(R \propto L\) (or resistance doubles when length doubles) [B1] - \(A \propto d^2\) (or area is four times larger when diameter doubles) [B1] - \(R \propto 1/A\) (or resistance is quartered when area is quadrupled) [B1] - \(6.0\\ \Omega\) (with unit) [A1]
题目 10 · Structured
8 分
The light emitted from a distant galaxy is observed on Earth to have a longer wavelength than when it was emitted.
(ii) The Hubble constant \(H_0\) is estimated to be \(2.2 \times 10^{-18}\text{ s}^{-1}\). Use this value to estimate the age of the Universe in years.
estimated age = ..................................................... [2]
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解题
(a) (i) The name given to this effect is redshift (or cosmological redshift).
(ii) Redshift indicates that galaxies are moving away from us and from each other. Observations show that galaxies further away are moving faster (showing greater redshift), which means the Universe is expanding. If the Universe is expanding, extrapolating backward in time implies that all matter in the Universe was once concentrated at a single, extremely dense and hot point before it began expanding.
(b) (i) Cosmic Microwave Background Radiation (CMBR) is microwave radiation received from all directions in space. It is the remnant electromagnetic radiation left over from the very early, hot, dense stages of the Universe following the Big Bang.
(ii) The age of the Universe \(T\) can be estimated using the reciprocal of the Hubble constant: \[T = \frac{1}{H_0} = \frac{1}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 4.55 \times 10^{17}\text{ s}\] Converting this time into years: \[T = \frac{4.55 \times 10^{17}\text{ s}}{365.25 \times 24 \times 3600\text{ s/year}} \approx 1.44 \times 10^{10}\text{ years}\] (Accept values in the range \(1.4 \times 10^{10}\text{ years}\) to \(1.44 \times 10^{10}\text{ years}\)).
评分标准
(a) (i) - (cosmological) redshift [B1]
(ii) - Redshift shows galaxies are moving away from Earth / from each other [B1] - Galaxies further away are moving faster / have greater redshift [B1] - This shows Universe is expanding, meaning in the past it started from a single point / was much closer together [B1]
(b) (i) - Electromagnetic radiation / microwave radiation that is detected coming from all directions in space [B1] - Remnant / leftover radiation from the Big Bang / early stages of the Universe [B1]
(ii) - Correct formula \(T = 1 / H_0\) or calculation of age in seconds as \(4.5\text{ to }4.6 \times 10^{17}\text{ s}\) [C1] - Correct value in years as \(1.4 \times 10^{10}\text{ years}\) (allow range \(1.4 \times 10^{10}\text{ years}\) to \(1.44 \times 10^{10}\text{ years}\)) [A1]
Paper 6 (Alternative to Practical)
Answer all questions. You will need to show understanding of experimental techniques, graph plotting, gradient calculations, and practical design planning.
4 题目 · 40 分
题目 1 · practical structured
10 分
Fig. 1.1 shows a student balancing a uniform metre rule on a pivot at the 50.0 cm mark.
They place a known load \(P\) of mass \(150\text{ g}\) on one side of the pivot, and an unknown block \(X\) on the other side.
(a) The rule is balanced when load \(P\) is centered at the 12.0 cm mark. Calculate the distance \(d_1\) from the 50.0 cm pivot to the centre of load \(P\).
(b) The rule is balanced when block \(X\) is centered at the 72.5 cm mark. Calculate the distance \(d_2\) from the 50.0 cm pivot to the centre of block \(X\).
(c) Suggest why it might be difficult to position the block \(X\) precisely at its intended mark on the metre rule.
(d) The student repeats the procedure for several positions of the load \(P\) to obtain further pairs of \(d_1\) and \(d_2\) values. The results are shown in Table 1.1.
Plot a graph of \(d_1\text{ / cm}\) (y-axis) against \(d_2\text{ / cm}\) (x-axis). Start both axes from the origin \((0,0)\). Draw the best-fit straight line.
(e) Determine the gradient \(G\) of the graph. Show clearly on your graph how you obtained the necessary information.
(f) The mass \(M\) of block \(X\) is given by the equation: \(M = \frac{150}{G}\) Using your value of \(G\) from (e), calculate the mass \(M\) of block \(X\).
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解题
(a) \(d_1 = 50.0\text{ cm} - 12.0\text{ cm} = 38.0\text{ cm}\). (b) \(d_2 = 72.5\text{ cm} - 50.0\text{ cm} = 22.5\text{ cm}\). (c) The block has a non-negligible width, so its center of mass must be estimated visually. (d) Plotting the points: - (25.3, 38.0) - (22.1, 33.0) - (18.6, 28.0) - (15.3, 23.0) - (12.0, 18.0) A straight line of best fit starting from (0,0) with gradient approximately 1.5. (e) Gradient \(G = \frac{\Delta y}{\Delta x}\). Using extreme points or a large triangle: \(G \approx \frac{38.0 - 0}{25.3 - 0} \approx 1.50\). (f) \(M = \frac{150}{1.50} = 100\text{ g}\).
评分标准
- (a) [1] \(d_1 = 38.0\text{ cm}\). - (b) [1] \(d_2 = 22.5\text{ cm}\). - (c) [1] Any sensible practical reason: e.g., block has a wide base / center of mass is difficult to judge / ruler may slide. - (d) [4] Graph criteria: - Axes correctly labelled with quantity and unit [1] - Suitable scales, points occupy at least half the grid [1] - All 5 points plotted accurately within half a small square [1] - Good best-fit straight line, thin, continuous [1] - (e) [2] Gradient determination: - Large triangle used (at least half the line) [1] - Correct calculation with correct value of \(G \approx 1.5 \pm 0.05\) [1] - (f) [1] Correct calculation of \(M\) using candidate's \(G\) value, unit (g) included [1]
题目 2 · practical structured
10 分
A student investigates the rate of cooling of hot water in two different beakers. Beaker A is wrapped in black paper. Beaker B is wrapped in shiny aluminum foil.
Both beakers are filled with hot water and their temperatures are recorded at 30-second intervals.
(a) The room temperature \(\theta_R\) during the experiment is \(21.5^\circ\text{C}\). Write down this value.
(b) (i) Describe one precaution that the student should take to ensure that the temperature readings are as accurate as possible. (ii) State one variable that must be kept constant to ensure a fair comparison of the cooling rates between the two beakers.
(c) (i) Calculate the total temperature drop \(\Delta\theta_A\) of the water in Beaker A over the 180 s. (ii) Calculate the total temperature drop \(\Delta\theta_B\) of the water in Beaker B over the 180 s.
(d) Explain, with reference to the mechanisms of thermal energy transfer, why Beaker A cools faster than Beaker B.
(e) The student suggests that the rate of cooling depends on the temperature difference between the water and the room. Using the data for Beaker A: - Calculate the temperature difference \(\theta_A - \theta_R\) at \(t = 0\text{ s}\). - Calculate the temperature difference \(\theta_A - \theta_R\) at \(t = 180\text{ s}\).
(f) Suggest one way the experiment could be modified to reduce heat loss specifically from the top surface of the water.
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解题
(a) \(\theta_R = 21.5^\circ\text{C}\). (b) (i) Stirring ensures uniform temperature throughout the liquid. Parallax error is avoided by looking at eye level. (ii) Volume of water is crucial as larger volumes have greater thermal capacity and cool slower. (c) (i) \(\Delta\theta_A = 85.5 - 61.5 = 24.0^\circ\text{C}\). (ii) \(\Delta\theta_B = 85.5 - 68.5 = 17.0^\circ\text{C}\). (d) Radiation is the main heat loss mechanism affected by surface finish. Matte black (Beaker A) is a better emitter than polished silver/shiny foil (Beaker B). (e) At \(t = 0\text{ s}\): \(85.5 - 21.5 = 64.0^\circ\text{C}\). At \(t = 180\text{ s}\): \(61.5 - 21.5 = 40.0^\circ\text{C}\). (f) Evaporation and convection from the top surface can be minimised by using a lid.
评分标准
- (a) [1] Correct room temperature \(21.5^\circ\text{C}\) with unit. - (b) (i) [1] Any one valid precaution: e.g., stir before reading / keep thermometer off bottom or sides of beaker / view perpendicular to scale. - (b) (ii) [1] Any one valid control variable: e.g., volume of water / surface area of water / initial temperature / thickness of wrap. - (c) (i) [1] \(\Delta\theta_A = 24.0^\circ\text{C}\). - (c) (ii) [1] \(\Delta\theta_B = 17.0^\circ\text{C}\). - (d) [2] Detailed explanation: - Mention of radiation/infrared [1] - Black paper is a better emitter of thermal radiation than shiny foil (or shiny foil is a poor emitter) [1] - (e) [2] Calculations: - At \(t=0\text{ s}\): \(64.0^\circ\text{C}\) [1] - At \(t=180\text{ s}\): \(40.0^\circ\text{C}\) [1] - (f) [1] Use a lid / cover.
题目 3 · practical structured
10 分
A student investigates the refraction of a ray of light passing through a semi-circular glass block.
The ray enters the flat face at the center point \(O\) at an angle of incidence \(i\), and refracts into the glass at an angle of refraction \(r\).
(a) Complete the column headings in Table 3.1 by adding the appropriate units or symbols (if any).
(b) Plot a graph of \(\sin i\) (y-axis) against \(\sin r\) (x-axis). Start both axes from the origin \((0,0)\). Draw the best-fit straight line.
(c) Determine the gradient of the graph. Show clearly on your graph how you obtained the necessary information.
(d) The refractive index \(n\) of the glass is equal to the gradient of this graph. State the value of \(n\) obtained from your gradient.
(e) State one precaution the student must take when using a ray box to ensure the rays of light are clearly visible and easy to trace.
(f) The student reverses the direction of the ray so that it enters the curved surface radially and meets the flat face at the center point \(O\) from inside the glass.
Explain what happens to the refracted ray when the angle of incidence inside the glass is increased beyond the critical angle of the glass.
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解题
(a) The angles \(i\) and \(r\) are in degrees (\(^\circ\)). The sines have no units. (b) Points to plot: - (0.17, 0.26) - (0.33, 0.50) - (0.47, 0.71) - (0.57, 0.87) - (0.64, 0.97) A straight line passing through origin with gradient \(\approx 1.5\). (c) Gradient = \(\frac{0.97 - 0}{0.64 - 0} \approx 1.52\). (d) \(n = \text{gradient} \approx 1.52\). (e) Use a thin slit / perform in a dark room to see the ray clearly. (f) Total internal reflection occurs, meaning all light is reflected back into the glass block.
评分标准
- (a) [1] Column headings: degrees or symbol (\(^\circ\)) for angles; no unit or '/' for sines. - (b) [4] Graph criteria: - Axes labelled correctly with quantities [1] - Scales are sensible (starting from 0,0) and fill at least half of the grid [1] - All 5 points plotted accurately to within half a small square [1] - Best-fit straight line drawn, passing through origin, thin and single line [1] - (c) [2] Gradient: - Triangle used to find gradient is large (covering more than half the line) [1] - Correct calculation to give gradient \(1.5 \pm 0.05\) [1] - (d) [1] Refractive index \(n\) matching the candidate's gradient value, given to 2 or 3 sig figs with no unit [1]. - (e) [1] Precautions: e.g., perform in dark room / use a thin slit / make sure ray box is flat on paper. - (f) [1] Total internal reflection occurs / no light escapes flat face.
题目 4 · practical structured
10 分
A student investigates how the resistance of a wire depends on its diameter. They are provided with three wires made of the same alloy, each \(100.0\text{ cm}\) long, with the following diameters: - Wire A: diameter \(D = 0.25\text{ mm}\) - Wire B: diameter \(D = 0.40\text{ mm}\) - Wire C: diameter \(D = 0.60\text{ mm}\)
(a) Draw a circuit diagram showing how the student should connect a cell, a switch, an ammeter, a voltmeter, and one of the test wires to measure the current in the wire and the potential difference across it.
(b) The student tests Wire A first. - The ammeter reading is \(I_A = 0.35\text{ A}\). - The voltmeter reading is \(V_A = 2.10\text{ V}\).
State the reading of the current and potential difference, including their units.
(c) Calculate the resistance \(R_A\) of Wire A using the equation: \(R_A = \frac{V_A}{I_A}\)
(d) The resistance values for Wire B and Wire C are determined as: - \(R_B = 4.1\text{ }\Omega\) - \(R_C = 1.8\text{ }\Omega\)
With reference to these values and your calculated \(R_A\), state the relationship between the diameter of a wire and its resistance.
(e) Suggest one reason why the switch should be opened (turned off) between taking readings.
(f) The student wants to extend the investigation to see how the resistance of a wire depends on its length. Briefly describe how they would carry out this experiment using a single length of wire. Include: - what they would measure - what they would keep constant - how they would use their measurements to draw a conclusion
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解题
(a) Standard circuit: series circuit containing cell, switch, ammeter, and test wire, with a voltmeter connected in parallel across the test wire. (b) Current \(I_A = 0.35\text{ A}\), potential difference \(V_A = 2.10\text{ V}\). (c) \(R_A = \frac{2.10}{0.35} = 6.0\text{ }\Omega\). (d) Wire A (\(0.25\text{ mm}\)) has \(R = 6.0\text{ }\Omega\). Wire B (\(0.40\text{ mm}\)) has \(R = 4.1\text{ }\Omega\). Wire C (\(0.60\text{ mm}\)) has \(R = 1.8\text{ }\Omega\). As diameter increases, resistance decreases. (e) Current causes heating in the wire, and resistance of metals increases with temperature. Turning it off keeps the temperature constant. (f) Move a crocodile clip to change the length \(L\) of wire in the circuit. Measure \(V\) and \(I\) at each length, calculate \(R\). Keep diameter, material, and temperature constant. Plot a graph of \(R\) against \(L\) to check for direct proportionality.
评分标准
- (a) [2] Diagram: - Correct symbols for cell, switch, ammeter, voltmeter, and resistor/wire [1] - Ammeter in series, voltmeter in parallel across the test wire [1] - (b) [2] Readings: - Current: \(0.35\text{ A}\) (including unit) [1] - Potential difference: \(2.10\text{ V}\) (including unit) [1] - (c) [1] \(R_A = 6.0\text{ }\Omega\) (correct calculation and unit \(\Omega\) or ohm) [1] - (d) [1] Statement: Larger diameter leads to lower resistance / resistance is inversely related to diameter (or cross-sectional area) [1] - (e) [1] Reason: To prevent the wire heating up (which changes its resistance) [1] - (f) [3] Extension: - Method: Use sliding contact/crocodile clip to vary length \(L\), measure \(V\) and \(I\) for each length [1] - Constant: Wire diameter / material / temperature [1] - Conclusion: Calculate \(R\) for each length, plot a graph of \(R\) against \(L\) (expect straight line through origin) [1]
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