An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V3) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
部分 選擇題 (Extended)
Answer all forty questions. For each question there are four possible answers, A, B, C, and D. Choose the one you consider correct.
40 题目 · 40 分
题目 1 · 選擇題
1 分
A cyclist starts from rest and accelerates at a constant rate of \(1.5\text{ m/s}^2\) for \(6.0\text{ s}\). The cyclist then travels at a constant velocity for \(10.0\text{ s}\), before decelerating uniformly to rest in a further \(4.0\text{ s}\).
What is the total distance travelled by the cyclist?
A.108 m
B.126 m
C.135 m
D.153 m
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解题
First, find the maximum velocity reached: \(v = u + at = 0 + (1.5 \times 6.0) = 9.0\text{ m/s}\).
An electric motor is used to lift a load of mass \(4.0\text{ kg}\) vertically through a height of \(15\text{ m}\) in a time of \(6.0\text{ s}\). The electrical power input to the motor is \(140\text{ W}\).
Take the gravitational field strength \(g\) to be \(9.8\text{ N/kg}\).
What is the efficiency of the motor?
A.24%
B.42%
C.70%
D.98%
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解题
The useful work done (gain in gravitational potential energy) is: \(E = mgh = 4.0\text{ kg} \times 9.8\text{ N/kg} \times 15\text{ m} = 588\text{ J}\).
The useful power output of the motor is: \(P_{\text{out}} = \frac{\text{Work done}}{\text{time}} = \frac{588\text{ J}}{6.0\text{ s}} = 98\text{ W}\).
The efficiency of the motor is: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{98\text{ W}}{140\text{ W}} \times 100\% = 70\%\).
评分标准
Award 1 mark for the correct option C.
题目 3 · 選擇題
1 分
A spring has an unstretched length of \(12.0\text{ cm}\). When a load of \(6.0\text{ N}\) is suspended from it, the length of the spring becomes \(15.0\text{ cm}\).
Assuming the limit of proportionality is not exceeded, what load is required to stretch the spring to a total length of \(19.0\text{ cm}\)?
A.8.0 N
B.9.5 N
C.14.0 N
D.38.0 N
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解题
The extension produced by the \(6.0\text{ N}\) load is: \(x_1 = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\).
The spring constant \(k\) is: \(k = \frac{F_1}{x_1} = \frac{6.0\text{ N}}{3.0\text{ cm}} = 2.0\text{ N/cm}\).
To reach a total length of \(19.0\text{ cm}\), the required extension is: \(x_2 = 19.0\text{ cm} - 12.0\text{ cm} = 7.0\text{ cm}\).
The load required is: \(F_2 = k \times x_2 = 2.0\text{ N/cm} \times 7.0\text{ cm} = 14.0\text{ N}\).
评分标准
Award 1 mark for the correct option C.
题目 4 · 選擇題
1 分
A toy railway car of mass \(0.80\text{ kg}\) is moving at a speed of \(3.0\text{ m/s}\) along a straight track. It collides with a stationary toy car of mass \(1.2\text{ kg}\). After the collision, the two cars couple together and move off with a common velocity \(v\).
What is the value of \(v\)?
A.1.2 m/s
B.1.5 m/s
C.2.0 m/s
D.4.5 m/s
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解题
According to the principle of conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
A submerged submarine has a flat hatch of area \(0.60\text{ m}^2\). The maximum force the hatch can withstand from the external water pressure is \(4.41 \times 10^5\text{ N}\).
The density of water is \(1000\text{ kg/m}^3\) and the acceleration of free fall \(g = 9.8\text{ m/s}^2\).
What is the maximum depth to which the submarine can submerge without the hatch failing? (Ignore atmospheric pressure).
A.45 m
B.75 m
C.441 m
D.735 m
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解题
First, find the maximum pressure \(P\) the hatch can withstand: \(P = \frac{F}{A} = \frac{4.41 \times 10^5\text{ N}}{0.60\text{ m}^2} = 7.35 \times 10^5\text{ Pa}\).
Using the liquid pressure formula \(P = \rho g h\): \(7.35 \times 10^5 = 1000 \times 9.8 \times h\)
\(735000 = 9800 h\)
\(h = \frac{735000}{9800} = 75\text{ m}\).
评分标准
Award 1 mark for the correct option B.
题目 6 · 選擇題
1 分
A ray of light travels from glass into air. The refractive index of the glass is \(1.50\). The speed of light in a vacuum is \(3.0 \times 10^8\text{ m/s}\).
What is the critical angle for the glass-to-air boundary and the speed of light in the glass?
2. Speed of light in glass \(v\): \(v = \frac{\text{speed of light in vacuum}}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.50} = 2.0 \times 10^8\text{ m/s}\).
评分标准
Award 1 mark for the correct option A.
题目 7 · 選擇題
1 分
A cell of electromotive force (e.m.f.) \(1.5\text{ V}\) is connected to a resistor. A current of \(0.40\text{ A}\) flows through the resistor for a time of \(5.0\text{ minutes}\).
How much charge passes through the resistor and how much chemical energy is converted in the cell during this time?
A.charge = 2.0 C, energy = 3.0 J
B.charge = 120 C, energy = 3.0 J
C.charge = 120 C, energy = 180 J
D.charge = 2.0 C, energy = 180 J
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解题
First, convert time into seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\).
1. Charge \(Q\): \(Q = I \times t = 0.40\text{ A} \times 300\text{ s} = 120\text{ C}\).
2. Energy \(E\): \(E = V \times Q = 1.5\text{ V} \times 120\text{ C} = 180\text{ J}\).
评分标准
Award 1 mark for the correct option C.
题目 8 · 選擇題
1 分
A sample of a radioactive isotope has an initial count rate of \(800\text{ counts/minute}\). After a time of \(18\text{ hours}\), the count rate has fallen to \(100\text{ counts/minute}\) (after correcting for background radiation).
What is the half-life of the radioactive isotope?
A.2.25 hours
B.4.5 hours
C.6.0 hours
D.9.0 hours
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解题
The fraction of the original activity remaining is: \(\frac{100}{800} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\).
This means that exactly \(3\) half-lives have elapsed in \(18\text{ hours}\).
An object starts from rest and accelerates with a constant acceleration of \(2.5\text{ m/s}^2\) for \(4.0\text{ s}\). It then travels at a constant speed for \(6.0\text{ s}\) before decelerating uniformly to rest in a further \(2.0\text{ s}\).
What is the total distance travelled by the object?
A.\(70\text{ m}\)
B.\(80\text{ m}\)
C.\(90\text{ m}\)
D.\(100\text{ m}\)
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解题
1. Find the maximum speed reached during acceleration: \(v = u + a t = 0 + (2.5\text{ m/s}^2 \times 4.0\text{ s}) = 10.0\text{ m/s}\).
2. Calculate the distance for each phase of motion using the area under the speed-time graph (trapezium): - During acceleration: \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 10.0\text{ m/s} = 20.0\text{ m}\). - During constant speed: \(d_2 = \text{speed} \times \text{time} = 10.0\text{ m/s} \times 6.0\text{ s} = 60.0\text{ m}\). - During deceleration: \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0\text{ s} \times 10.0\text{ m/s} = 10.0\text{ m}\).
3. Calculate the total distance: \(d_{\text{total}} = 20.0\text{ m} + 60.0\text{ m} + 10.0\text{ m} = 90.0\text{ m}\).
评分标准
Award 1 mark for the correct option C.
题目 10 · 選擇題
1 分
A student mixes \(40\text{ cm}^3\) of liquid P (density \(0.80\text{ g/cm}^3\)) with \(60\text{ cm}^3\) of liquid Q (density \(1.2\text{ g/cm}^3\)).
The total volume of the mixture is equal to the sum of the individual volumes of P and Q.
What is the density of the mixture?
A.\(0.96\text{ g/cm}^3\)
B.\(1.00\text{ g/cm}^3\)
C.\(1.04\text{ g/cm}^3\)
D.\(2.00\text{ g/cm}^3\)
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解题
1. Find the mass of each liquid: - Mass of P: \(m_P = \rho_P \times V_P = 0.80\text{ g/cm}^3 \times 40\text{ cm}^3 = 32\text{ g}\). - Mass of Q: \(m_Q = \rho_Q \times V_Q = 1.2\text{ g/cm}^3 \times 60\text{ cm}^3 = 72\text{ g}\).
2. Find the total mass and total volume: - Total mass: \(m_{\text{total}} = 32\text{ g} + 72\text{ g} = 104\text{ g}\). - Total volume: \(V_{\text{total}} = 40\text{ cm}^3 + 60\text{ cm}^3 = 100\text{ cm}^3\).
3. Find the density of the mixture: \(\rho_{\text{mixture}} = \frac{m_{\text{total}}}{V_{\text{total}}} = \frac{104\text{ g}}{100\text{ cm}^3} = 1.04\text{ g/cm}^3\).
评分标准
Award 1 mark for the correct option C.
题目 11 · 選擇題
1 分
A block of mass \(3.0\text{ kg}\) is moving to the right along a frictionless horizontal track at a speed of \(4.0\text{ m/s}\). It collides with a second block of mass \(2.0\text{ kg}\) moving to the left along the same track at a speed of \(3.0\text{ m/s}\).
After the collision, the two blocks stick together.
What is their common speed and direction of travel?
A.\(1.2\text{ m/s}\) to the left
B.\(1.2\text{ m/s}\) to the right
C.\(3.6\text{ m/s}\) to the left
D.\(3.6\text{ m/s}\) to the right
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解题
1. Using the principle of conservation of momentum, taking direction to the right as positive: \(p_{\text{initial}} = (m_1 \times v_1) + (m_2 \times v_2) = (3.0\text{ kg} \times 4.0\text{ m/s}) + (2.0\text{ kg} \times (-3.0\text{ m/s}))\) \(p_{\text{initial}} = 12.0 - 6.0 = +6.0\text{ kg m/s}\) (to the right).
2. Find the total combined mass: \(m_{\text{total}} = 3.0\text{ kg} + 2.0\text{ kg} = 5.0\text{ kg}\).
3. Calculate the common final velocity: \(v_{\text{final}} = \frac{p_{\text{initial}}}{m_{\text{total}}} = \frac{6.0\text{ kg m/s}}{5.0\text{ kg}} = 1.2\text{ m/s}\) (positive, so to the right).
评分标准
Award 1 mark for the correct option B.
题目 12 · 選擇題
1 分
A vertical tube of depth \(1.5\text{ m}\) is filled with a liquid of density \(1200\text{ kg/m}^3\). A piston at the top of the tube exerts an additional pressure of \(5.0\times 10^4\text{ Pa}\) on the liquid surface.
The acceleration of free fall, \(g\), is \(9.8\text{ m/s}^2\).
What is the total pressure at the bottom of the tube?
A.\(1.8\times 10^4\text{ Pa}\)
B.\(5.0\times 10^4\text{ Pa}\)
C.\(6.8\times 10^4\text{ Pa}\)
D.\(1.8\times 10^5\text{ Pa}\)
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解题
1. Calculate the hydrostatic pressure exerted by the liquid column: \(p_{\text{hydrostatic}} = \rho g h = 1200\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 1.5\text{ m} = 17\,640\text{ Pa}\).
2. Find the total pressure at the bottom by adding the surface pressure: \(p_{\text{total}} = p_{\text{surface}} + p_{\text{hydrostatic}} = 5.0\times 10^4\text{ Pa} + 1.764\times 10^4\text{ Pa} = 6.764\times 10^4\text{ Pa}\).
3. Rounding to two significant figures yields \(6.8\times 10^4\text{ Pa}\).
评分标准
Award 1 mark for the correct option C.
题目 13 · 選擇題
1 分
An electric heater of power \(60\text{ W}\) is used to heat a \(0.50\text{ kg}\) block of metal. The heater is switched on for \(5.0\text{ minutes}\). The temperature of the block increases from \(20\text{ }^\circ\text{C}\) to \(65\text{ }^\circ\text{C}\).
Assume there is no thermal energy loss to the surroundings.
What is the specific heat capacity of the metal?
A.\(80\text{ J / (kg }^\circ\text{C)}\)
B.\(800\text{ J / (kg }^\circ\text{C)}\)
C.\(4800\text{ J / (kg }^\circ\text{C)}\)
D.\(48\,000\text{ J / (kg }^\circ\text{C)}\)
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解题
1. Calculate the total energy supplied by the heater: \(E = P \times t = 60\text{ W} \times (5.0 \times 60\text{ s}) = 60 \times 300 = 18\,000\text{ J}\).
2. Find the temperature rise: \(\Delta \theta = 65\text{ }^\circ\text{C} - 20\text{ }^\circ\text{C} = 45\text{ }^\circ\text{C}\).
3. Use the formula \(E = m c \Delta \theta\) to calculatepecific heat capacity \(c\): \(c = \frac{E}{m \Delta \theta} = \frac{18\,000\text{ J}}{0.50\text{ kg} \times 45\text{ }^\circ\text{C}} = \frac{18\,000}{22.5} = 800\text{ J / (kg }^\circ\text{C)}\).
评分标准
Award 1 mark for the correct option B.
题目 14 · 選擇題
1 分
A ray of light in a transparent plastic block is incident on the boundary with air.
The refractive index of the plastic is \(1.45\).
What is the critical angle for the plastic-air boundary?
A.\(34^\circ\)
B.\(44^\circ\)
C.\(46^\circ\)
D.\(90^\circ\)
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解题
1. The formula relating refractive index \(n\) and critical angle \(c\) is: \(\sin(c) = \frac{1}{n}\).
3. Calculate the angle: \(c = \arcsin(0.6897) \approx 43.6^\circ\).
Rounding to the nearest whole degree gives \(44^\circ\).
评分标准
Award 1 mark for the correct option B.
题目 15 · 選擇題
1 分
Two resistors of resistance \(6.0\text{ }\Omega\) and \(12\text{ }\Omega\) are connected in parallel. This combination is connected in series with a third resistor of resistance \(4.0\text{ }\Omega\).
What is the total equivalent resistance of this network?
A.\(4.0\text{ }\Omega\)
B.\(8.0\text{ }\Omega\)
C.\(10\text{ }\Omega\)
D.\(22\text{ }\Omega\)
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解题
1. Find the equivalent resistance \(R_p\) of the two parallel resistors: \(\frac{1}{R_p} = \frac{1}{6.0\text{ }\Omega} + \frac{1}{12\text{ }\Omega} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12}\) \(R_p = \frac{12}{3} = 4.0\text{ }\Omega\).
2. Add the series resistor to find the total resistance: \(R_{\text{total}} = R_p + 4.0\text{ }\Omega = 4.0\text{ }\Omega + 4.0\text{ }\Omega = 8.0\text{ }\Omega\).
评分标准
Award 1 mark for the correct option B.
题目 16 · 選擇題
1 分
A radioactive source has a half-life of \(12\text{ minutes}\). The initial count rate recorded by a detector is \(850\text{ counts/minute}\), which includes a constant background count rate of \(50\text{ counts/minute}\).
What is the count rate recorded by the detector after \(36\text{ minutes}\)?
A.\(106\text{ counts/minute}\)
B.\(150\text{ counts/minute}\)
C.\(250\text{ counts/minute}\)
D.\(263\text{ counts/minute}\)
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解题
1. Calculate the initial corrected count rate due to the source alone: \(\text{Corrected initial count rate} = 850 - 50 = 800\text{ counts/minute}\).
2. Determine the number of half-lives that have elapsed in \(36\text{ minutes}\): \(N = \frac{36\text{ minutes}}{12\text{ minutes}} = 3\text{ half-lives}\).
3. Calculate the corrected count rate after 3 half-lives: \(\text{Final corrected count rate} = \frac{800}{2^3} = \frac{800}{8} = 100\text{ counts/minute}\).
4. Add the background count rate back to find the final reading on the detector: \(\text{Recorded count rate} = 100 + 50 = 150\text{ counts/minute}\).
评分标准
Award 1 mark for the correct option B.
题目 17 · 選擇題
1 分
A train accelerates from rest to a speed of \(24\text{ m/s}\) in \(12\text{ s}\). It then travels at constant speed for \(20\text{ s}\), before decelerating uniformly to rest in a further \(8.0\text{ s}\). What is the average speed of the train for the entire journey?
A trolley of mass \(2.0\text{ kg}\) travelling at \(6.0\text{ m/s}\) collides with a stationary trolley of mass \(4.0\text{ kg}\). After the collision, the two trolleys stick together and move off with a common velocity. What is the loss in total kinetic energy of the trolleys as a result of the collision?
A.\(12\text{ J}\)
B.\(18\text{ J}\)
C.\(24\text{ J}\)
D.\(36\text{ J}\)
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解题
Using the conservation of momentum: \(m_1 u_1 + m_2 v_2 = (m_1 + m_2) v\) \((2.0 \times 6.0) + 0 = (2.0 + 4.0) v \Rightarrow 12 = 6.0 v \Rightarrow v = 2.0\text{ m/s}\). Next, calculate kinetic energy before and after: Initial KE \(= \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2.0 \times 6.0^2 = 36\text{ J}\). Final KE \(= \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 6.0 \times 2.0^2 = 12\text{ J}\). Loss in kinetic energy \(= 36\text{ J} - 12\text{ J} = 24\text{ J}\).
评分标准
Award 1 mark for the correct option C.
题目 19 · 選擇題
1 分
An electric pump lifts \(15\text{ kg}\) of water through a vertical height of \(8.0\text{ m}\) in a time of \(5.0\text{ s}\). The electrical power input to the pump is \(300\text{ W}\). Taking \(g = 9.8\text{ m/s}^2\), what is the efficiency of the pump?
A.\(40\%\)
B.\(50\%\)
C.\(78\%\)
D.\(98\%\)
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解题
First, find the useful power output: \(W_{\text{useful}} = mgh = 15\text{ kg} \times 9.8\text{ m/s}^2 \times 8.0\text{ m} = 1176\text{ J}\). Useful power output \(P_{\text{out}} = \frac{W_{\text{useful}}}{t} = \frac{1176\text{ J}}{5.0\text{ s}} = 235.2\text{ W}\). Efficiency \(= \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{235.2}{300} \times 100\% = 78.4\%\), which is approximately \(78\%\).
评分标准
Award 1 mark for the correct option C.
题目 20 · 選擇題
1 分
A sealed container holds a fixed mass of gas. The volume of the container is decreased to one-third of its original volume, while the temperature of the gas is kept constant. Which statement correctly describes the change in the pressure of the gas and explains it in terms of the behavior of the gas particles?
A.The pressure decreases because the particles collide with the container walls less frequently.
B.The pressure increases because the particles move with a greater average speed.
C.The pressure increases because the particles collide with the container walls more frequently.
D.The pressure remains constant because the temperature is constant.
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解题
According to Boyle's law, at constant temperature, pressure is inversely proportional to volume. Decreasing the volume decreases the space available for gas particles, so the particles collide with the walls of the container more frequently, increasing the pressure.
评分标准
Award 1 mark for the correct option C.
题目 21 · 選擇題
1 分
A ray of monochromatic light travels from air into a transparent plastic block. The angle of incidence is \(45^\circ\) and the angle of refraction is \(28^\circ\). What is the speed of light in the plastic block? (The speed of light in air is \(3.0 \times 10^8\text{ m/s}\).)
A.\(1.5 \times 10^8\text{ m/s}\)
B.\(2.0 \times 10^8\text{ m/s}\)
C.\(2.3 \times 10^8\text{ m/s}\)
D.\(3.0 \times 10^8\text{ m/s}\)
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解题
Calculate the refractive index \(n\) of the plastic using Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(45^\circ)}{\sin(28^\circ)} = \frac{0.7071}{0.4695} \approx 1.506\). Then, calculate the speed of light \(v\) in the plastic: \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.506} \approx 1.99 \times 10^8\text{ m/s}\), which rounds to \(2.0 \times 10^8\text{ m/s}\).
评分标准
Award 1 mark for the correct option B.
题目 22 · 選擇題
1 分
A metal wire of length \(L\) and cross-sectional area \(A\) has a resistance of \(8.0\ \Omega\). What is the resistance of a wire made of the same metal that has a length of \(3L\) and a cross-sectional area of \(2A\)?
A.\(4.0\ \Omega\)
B.\(12\ \Omega\)
C.\(16\ \Omega\)
D.\(24\ \Omega\)
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解题
The resistance of a wire is given by the formula \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity of the metal. For the second wire: \(R_{\text{new}} = \rho \frac{3L}{2A} = 1.5 \left(\rho \frac{L}{A}\right) = 1.5 \times 8.0\ \Omega = 12\ \Omega\).
评分标准
Award 1 mark for the correct option B.
题目 23 · 選擇題
1 分
A radioactive source has an initial measured count rate of \(420\text{ counts/minute}\) in a laboratory where the background count rate is constant at \(20\text{ counts/minute}\). After \(6.0\text{ hours}\), the measured count rate is \(70\text{ counts/minute}\). What is the half-life of the radioactive isotope in the source?
A.\(1.5\text{ hours}\)
B.\(2.0\text{ hours}\)
C.\(3.0\text{ hours}\)
D.\(4.0\text{ hours}\)
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解题
First, subtract background radiation to find the corrected count rates: Corrected initial count rate \(C_0 = 420 - 20 = 400\text{ counts/minute}\). Corrected final count rate \(C_t = 70 - 20 = 50\text{ counts/minute}\). Determine the fraction of remaining active nuclei: \(\frac{C_t}{C_0} = \frac{50}{400} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), \(3\) half-lives have elapsed in \(6.0\text{ hours}\). Half-life \(T_{1/2} = \frac{6.0\text{ hours}}{3} = 2.0\text{ hours}\).
评分标准
Award 1 mark for the correct option B.
题目 24 · 選擇題
1 分
Light from a distant galaxy is observed to be redshifted. The galaxy is moving away from Earth at a speed of \(1.5 \times 10^6\text{ m/s}\). Taking the Hubble constant to be \(2.2 \times 10^{-18}\text{ s}^{-1}\), what is the approximate distance of this galaxy from Earth?
A.\(1.5 \times 10^{12}\text{ m}\)
B.\(3.3 \times 10^{12}\text{ m}\)
C.\(6.8 \times 10^{23}\text{ m}\)
D.\(1.5 \times 10^{24}\text{ m}\)
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解题
According to Hubble's Law, the speed of recession is related to distance by: \(v = H_0 d\). Rearranging to solve for distance: \(d = \frac{v}{H_0}\). Substitute the given values: \(d = \frac{1.5 \times 10^6\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 6.8 \times 10^{23}\text{ m}\).
评分标准
Award 1 mark for the correct option C.
题目 25 · 選擇題
1 分
A car accelerates from rest at a constant rate of \(2.0\text{ m/s}^2\) for \(5.0\text{ s}\). It then travels at a constant speed for \(10\text{ s}\) before decelerating at a constant rate of \(1.0\text{ m/s}^2\) until it comes to a complete stop. What is the total distance traveled by the car?
A.\(125\text{ m}\)
B.\(150\text{ m}\)
C.\(175\text{ m}\)
D.\(200\text{ m}\)
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解题
1. Distance during acceleration: \(s_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 2.0 \times (5.0)^2 = 25\text{ m}\). The final speed reached is \(v = u + at = 0 + 2.0 \times 5.0 = 10\text{ m/s}\). 2. Distance during constant speed: \(s_2 = v \times t = 10 \times 10 = 100\text{ m}\). 3. Distance during deceleration: Time taken to stop is \(t_3 = \frac{v}{a_{dec}} = \frac{10}{1.0} = 10\text{ s}\). The distance is \(s_3 = \frac{1}{2} (v + u) t_3 = \frac{1}{2} (10 + 0) \times 10 = 50\text{ m}\). Total distance = \(25 + 100 + 50 = 175\text{ m}\).
评分标准
C is correct. 1 mark for calculating the correct total distance of 175 m.
题目 26 · 選擇題
1 分
Two trolleys, X (mass \(2.0\text{ kg}\)) and Y (mass \(3.0\text{ kg}\)), travel towards each other on a frictionless horizontal track. Trolley X travels to the right at \(4.0\text{ m/s}\), and trolley Y travels to the left at \(2.0\text{ m/s}\). They collide and stick together. What is their common velocity after the collision?
A.\(0.40\text{ m/s}\) to the left
B.\(0.40\text{ m/s}\) to the right
C.\(2.8\text{ m/s}\) to the left
D.\(2.8\text{ m/s}\) to the right
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解题
Using the conservation of linear momentum: \(m_x u_x + m_y u_y = (m_x + m_y) v\). Taking right as positive: \(2.0 \times 4.0 + 3.0 \times (-2.0) = (2.0 + 3.0) v\) \(8.0 - 6.0 = 5.0 v\) \(2.0 = 5.0 v \Rightarrow v = 0.40\text{ m/s}\). Since the result is positive, the common velocity is to the right.
评分标准
B is correct. 1 mark for calculating the correct common velocity of 0.40 m/s to the right.
题目 27 · 選擇題
1 分
A box of mass \(15\text{ kg}\) slides down a rough slope from a vertical height of \(4.0\text{ m}\). The box reaches the bottom of the slope with a speed of \(6.0\text{ m/s}\). Taking the acceleration of free fall \(g = 9.8\text{ m/s}^2\), how much work is done against friction as the box slides down?
A.\(270\text{ J}\)
B.\(318\text{ J}\)
C.\(588\text{ J}\)
D.\(858\text{ J}\)
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解题
Initial gravitational potential energy: \(E_p = mgh = 15 \times 9.8 \times 4.0 = 588\text{ J}\). Final kinetic energy: \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 15 \times (6.0)^2 = 270\text{ J}\). Work done against friction is the energy lost: \(W = E_p - E_k = 588 - 270 = 318\text{ J}\).
评分标准
B is correct. 1 mark for calculating the correct work done of 318 J against friction.
题目 28 · 選擇題
1 分
A hatch of area \(0.40\text{ m}^2\) on a submarine is submerged in a liquid of density \(1200\text{ kg/m}^3\). The air inside the submarine is at atmospheric pressure. The net inward force on the hatch due to the liquid is \(1.2 \times 10^5\text{ N}\). Taking the acceleration of free fall to be \(9.8\text{ m/s}^2\), what is the depth of the hatch below the surface of the liquid?
A.\(2.1\text{ m}\)
B.\(21\text{ m}\)
C.\(26\text{ m}\)
D.\(31\text{ m}\)
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解题
The net pressure difference causing the net force is \(\Delta p = \frac{F}{A} = \frac{1.2 \times 10^5\text{ N}}{0.40\text{ m}^2} = 3.0 \times 10^5\text{ Pa}\). This net pressure is due entirely to the liquid column: \(\Delta p = \rho g h\). \(3.0 \times 10^5 = 1200 \times 9.8 \times h\) \(h = \frac{3.0 \times 10^5}{11760} \approx 25.5\text{ m}\), which rounds to \(26\text{ m}\) to two significant figures.
评分标准
C is correct. 1 mark for the correct calculation of depth as 26 m.
题目 29 · 選擇題
1 分
A heater of power \(150\text{ W}\) is used to heat a metal block of mass \(0.50\text{ kg}\). The temperature of the block increases from \(20\text{ }^\circ\text{C}\) to \(80\text{ }^\circ\text{C}\) in \(4.0\text{ minutes}\). Assuming no heat is lost to the surroundings, what is the specific heat capacity of the metal?
A.\(300\text{ J/(kg }^\circ\text{C)}
B.\(600\text{ J/(kg }^\circ\text{C)}
C.\(1200\text{ J/(kg }^\circ\text{C)}
D.\(2400\text{ J/(kg }^\circ\text{C)}
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解题
Time in seconds: \(t = 4.0 \times 60 = 240\text{ s}\). Total energy supplied: \(E = P \times t = 150\text{ W} \times 240\text{ s} = 36\,000\text{ J}\). Using \(E = mc\Delta\theta\): \(36\,000 = 0.50 \times c \times (80 - 20)\) \(36\,000 = 30 \times c \Rightarrow c = 1200\text{ J/(kg }^\circ\text{C)}\).
评分标准
C is correct. 1 mark for the correct calculation of specific heat capacity as 1200 J/(kg °C).
题目 30 · 選擇題
1 分
A ray of light in glass of refractive index \(1.50\) is incident on the boundary with water of refractive index \(1.33\). What is the critical angle for total internal reflection at this boundary?
A.\(42^\circ\)
B.\(49^\circ\)
C.\(62^\circ\)
D.\(70^\circ\)
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解题
The critical angle \(c\) is given by: \(\sin(c) = \frac{n_{\text{rare}}}{n_{\text{dense}}} = \frac{1.33}{1.50}\). \(\sin(c) \approx 0.8867 \Rightarrow c = \arcsin(0.8867) \approx 62.5^\circ\), which rounds to \(62^\circ\) to two significant figures.
评分标准
C is correct. 1 mark for the correct calculation of critical angle as 62°.
题目 31 · 選擇題
1 分
A battery of electromotive force (e.m.f.) \(12\text{ V}\) is connected to a resistor of resistance \(8.0\text{ }\Omega\). How much charge flows through the resistor in \(5.0\text{ minutes}\)?
A.\(7.5\text{ C}\)
B.\(60\text{ C}\)
C.\(450\text{ C}\)
D.\(3600\text{ C}\)
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解题
Using Ohm's law: \(I = \frac{V}{R} = \frac{12\text{ V}}{8.0\text{ }\Omega} = 1.5\text{ A}\). Time in seconds: \(t = 5.0 \times 60 = 300\text{ s}\). Total charge: \(Q = I \times t = 1.5\text{ A} \times 300\text{ s} = 450\text{ C}\).
评分标准
C is correct. 1 mark for the correct calculation of charge as 450 C.
题目 32 · 選擇題
1 分
An ideal transformer has a primary coil with \(600\text{ turns}\) and a secondary coil with \(150\text{ turns}\). The primary coil is connected to a \(240\text{ V}\) a.c. supply. A resistor is connected across the secondary coil, producing a current of \(4.0\text{ A}\) in the secondary circuit. What is the current in the primary circuit?
A.\(1.0\text{ A}\)
B.\(2.0\text{ A}\)
C.\(8.0\text{ A}\)
D.\(16\text{ A}\)
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解题
For an ideal transformer, input power equals output power: \(I_p V_p = I_s V_s\). Using \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\), we get: \(I_p = I_s \times \frac{V_s}{V_p} = I_s \times \frac{N_s}{N_p}\). \(I_p = 4.0\text{ A} \times \frac{150}{600} = 4.0 \times 0.25 = 1.0\text{ A}\).
评分标准
A is correct. 1 mark for the correct calculation of primary current as 1.0 A.
题目 33 · 選擇題
1 分
A motorcycle travels along a straight track. The speed-time graph of the motorcycle consists of three phases:
- Uniform acceleration from rest to a speed of \(24\text{ m/s}\) in \(8.0\text{ s}\). - Deceleration at a constant rate of \(1.5\text{ m/s}^2\) for \(4.0\text{ s}\). - Uniform deceleration to rest in another \(6.0\text{ s}\).
What is the total distance travelled by the motorcycle?
A.216 m
B.234 m
C.246 m
D.270 m
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解题
Phase 1: Acceleration from rest \(u = 0\) to \(v = 24\text{ m/s}\) in \(t_1 = 8.0\text{ s}\). Distance \(d_1 = \frac{0 + 24}{2} \times 8.0 = 96\text{ m}\).
Phase 2: Deceleration at \(1.5\text{ m/s}^2\) for \(t_2 = 4.0\text{ s}\). Speed decreases by \(1.5 \times 4.0 = 6.0\text{ m/s}\) from \(24\text{ m/s}\) to \(18\text{ m/s}\). Distance \(d_2 = \frac{24 + 18}{2} \times 4.0 = 84\text{ m}\).
Phase 3: Deceleration from \(18\text{ m/s}\) to rest \(v = 0\) in \(t_3 = 6.0\text{ s}\). Distance \(d_3 = \frac{18 + 0}{2} \times 6.0 = 54\text{ m}\).
Total distance = \(96 + 84 + 54 = 234\text{ m}\).
评分标准
B1: 1 mark for the correct answer of 234 m (Option B).
题目 34 · 選擇題
1 分
An alloy of total mass \(240\text{ g}\) is made by mixing metal X (density \(8.0\text{ g/cm}^3\)) and metal Y (density \(6.0\text{ g/cm}^3\)). The volume of metal X used is twice the volume of metal Y used.
Assuming no change in total volume occurs during mixing, what is the density of the alloy?
A.6.67 g/cm³
B.7.00 g/cm³
C.7.20 g/cm³
D.7.33 g/cm³
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解题
Let the volume of metal Y used be \(V\). The volume of metal X used is \(2V\). Total volume \(V_{\text{total}} = V + 2V = 3V\).
Mass of Y is \(m_Y = 6.0 \times V = 6V\). Mass of X is \(m_X = 8.0 \times 2V = 16V\). Total mass \(m_{\text{total}} = 6V + 16V = 22V\).
The density of the alloy is: \rho = \frac{m_{\text{total}}}{V_{\text{total}}} = \frac{22V}{3V} = \frac{22}{3} \approx 7.33\text{ g/cm}^3\).
评分标准
B1: 1 mark for the correct calculation and option selection (Option D).
题目 35 · 選擇題
1 分
A trolley of mass \(3.0\text{ kg}\) travels to the right at \(4.0\text{ m/s}\). It collides with a second trolley of mass \(2.0\text{ kg}\) travelling to the left at \(1.0\text{ m/s}\). After the collision, the two trolleys stick together and move as a single combined mass.
What is the loss in total kinetic energy of the system due to the collision?
A.5.0 J
B.10.0 J
C.15.0 J
D.25.0 J
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解题
Taking the direction to the right as positive: Initial momentum: \(p_i = (3.0 \times 4.0) + (2.0 \times -1.0) = 12.0 - 2.0 = 10.0\text{ kg m/s}\). Total mass after collision: \(M = 3.0 + 2.0 = 5.0\text{ kg}\). Common velocity after collision: \(v = \frac{10.0}{5.0} = 2.0\text{ m/s}\).
Final kinetic energy: \(E_{kf} = \frac{1}{2}(5.0)(2.0)^2 = 10.0\text{ J}\).
Loss in kinetic energy: \(\Delta E_k = 25.0 - 10.0 = 15.0\text{ J}\).
评分标准
B1: 1 mark for calculating the loss in kinetic energy as 15.0 J (Option C).
题目 36 · 選擇題
1 分
A ray of light in a transparent plastic block is incident on the boundary with air. The refractive index of the plastic is \(1.45\).
The angle of incidence of the ray is increased from \(35^\circ\) to \(50^\circ\). Which statement correctly describes what happens to the ray of light?
A.At 35°, the ray is totally internally reflected. At 50°, it refracts into the air.
B.At 35°, the ray refracts into the air. At 50°, it is totally internally reflected.
C.In both cases, the ray refracts into the air.
D.In both cases, the ray is totally internally reflected.
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解题
First calculate the critical angle \(\theta_c\): \sin \theta_c = \frac{1}{n} = \frac{1}{1.45} \approx 0.6897 \implies \theta_c \approx 43.6^\circ\).
- Since \(35^\circ < 43.6^\circ\), refraction into air occurs at \(35^\circ\). - Since \(50^\circ > 43.6^\circ\), total internal reflection occurs at \(50^\circ\).
Therefore, at \(35^\circ\) the ray refracts into the air, and at \(50^\circ\) it is totally internally reflected.
评分标准
B1: 1 mark for finding the critical angle and correctly determining the behavior at both angles (Option B).
题目 37 · 選擇題
1 分
A battery of electromotive force (e.m.f.) \(V\) drives a current \(I\) through a lamp for a time \(t\). During this time, a charge \(Q\) passes through the lamp and an amount of electrical energy \(E\) is transferred.
Which expression is correct?
A.E = I Q / t
B.V = E / Q
C.I = Q V / t
D.t = E I / V
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解题
By definition, the electromotive force \(V\) is the work done (energy transferred \(E\)) per unit charge \(Q\).
Therefore, \(V = \frac{E}{Q}\). This can also be seen because \(E = V I t = V Q\).
评分标准
B1: 1 mark for identifying the correct relationship between V, E, and Q (Option B).
题目 38 · 選擇題
1 分
Three identical resistors, each of resistance \(R\), are connected together. Which combination of these resistors has the lowest total resistance?
A.three resistors connected in parallel
B.three resistors connected in series
C.two resistors connected in parallel, with the third resistor connected in series with this parallel pair
D.two resistors connected in series, with the third resistor connected in parallel across this series pair
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解题
Let us calculate the total resistance for each combination:
- Three in parallel (Option A): \(R_{\text{total}} = \frac{R}{3} \approx 0.33R\). - Three in series (Option B): \(R_{\text{total}} = 3R\). - Parallel pair in series with third (Option C): \(R_{\text{total}} = \frac{R}{2} + R = 1.5R\). - Series pair in parallel with third (Option D): \(R_{\text{total}} = \frac{2R \times R}{2R + R} = \frac{2}{3}R \approx 0.67R\).
The lowest total resistance is obtained when all three are connected in parallel.
评分标准
B1: 1 mark for analyzing the combinations and choosing the lowest resistance (Option A).
题目 39 · 選擇題
1 分
An ideal transformer has \(1200\text{ turns}\) on its primary coil and \(300\text{ turns}\) on its secondary coil. The primary coil is connected to a \(240\text{ V}\) a.c. mains supply and draws a current of \(0.50\text{ A}\).
What are the output voltage and the output current of the secondary coil?
Answer all questions. Show your working clearly. Give your final answers to 2 or 3 significant figures with units.
11 题目 · 80 分
题目 1 · structured-theory
7 分
A sky-diver of mass 75 kg jumps from an aircraft.
(a) Define acceleration. [2]
(b) Describe and explain, in terms of forces, the motion of the sky-diver from the moment they jump until they reach terminal velocity. [3]
(c) Later, the sky-diver opens a parachute and decelerates at an average rate of \(4.5\text{ m/s}^2\). Taking \(g = 9.8\text{ m/s}^2\), calculate the average upward force exerted by the parachute on the sky-diver during this deceleration. [2]
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解题
(a) Acceleration is defined as the change in velocity per unit time (or the rate of change of velocity).
(b) Initially, the only force acting downwards is weight, so the sky-diver accelerates downwards at \(9.8\text{ m/s}^2\). As speed increases, the upward air resistance increases. This reduces the resultant downward force, meaning the downward acceleration decreases. Eventually, the upward air resistance equals the downward weight, the resultant force becomes zero, and the sky-diver falls at a constant terminal velocity.
(c) First, calculate the downward weight of the sky-diver: \(W = mg = 75 \times 9.8 = 735\text{ N}\)
During deceleration, the net force is directed upwards: \(F_{\text{net}} = F_{\text{up}} - W\) \(F_{\text{net}} = ma = 75 \times 4.5 = 337.5\text{ N}\)
Substitute the values to find the upward force: \(F_{\text{up}} - 735 = 337.5\) \(F_{\text{up}} = 735 + 337.5 = 1072.5\text{ N}\)
Rounding to 2 or 3 significant figures gives \(1100\text{ N}\) (or \(1070\text{ N}\)).
评分标准
(a) - change in velocity [1] - per unit time / time taken (or rate of change of velocity) [1]
(b) - downward force of gravity / weight is initially the only force [1] - air resistance increases with speed, reducing the resultant force and acceleration [1] - at terminal velocity, air resistance equals weight (resultant force is zero) [1]
A student determines the density of an organic solvent.
(a) He measures the mass of an empty beaker as \(45.2\text{ g}\). After pouring \(60\text{ cm}^3\) of the solvent into the beaker, the total mass is \(92.6\text{ g}\).
(i) Calculate the density of the solvent. [3]
(ii) State the name of the apparatus used to measure the volume of the liquid. [1]
(b) Explain why the density of a gas is much less than the density of a liquid, using ideas about particles. [3]
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解题
(a) (i) First, calculate the mass of the solvent: \(m = 92.6\text{ g} - 45.2\text{ g} = 47.4\text{ g}\)
Using the density formula: \(\rho = \frac{m}{V} = \frac{47.4\text{ g}}{60\text{ cm}^3} = 0.79\text{ g/cm}^3\) (or \(790\text{ kg/m}^3\))
(ii) The apparatus used to measure the volume of the liquid is a measuring cylinder.
(b) In a gas, the particles are spaced very far apart with a large amount of empty space between them. In contrast, particles in a liquid are closely packed together and touching. Therefore, for a given volume, there are far fewer particles (and thus significantly less mass) in a gas than in a liquid, resulting in a much lower density.
评分标准
(a) (i) - mass of solvent = \(47.4\text{ g}\) [1] - \(\rho = m / V\) used correctly [1] - final density = \(0.79\text{ g/cm}^3\) (or \(790\text{ kg/m}^3\)) with unit [1]
(a) (ii) - measuring cylinder [1]
(b) - gas particles are far apart / liquid particles are close together [1] - gas has lots of empty space / liquid has very little empty space [1] - fewer particles / less mass in the same volume for gas [1]
题目 3 · structured-theory
8 分
Linear air tracks are used to study collisions in a frictionless environment.
(a) Define impulse. [1]
(b) Glider A of mass \(0.40\text{ kg}\) travels at \(1.5\text{ m/s}\) along a frictionless track. It collides with a stationary glider B of mass \(0.60\text{ kg}\). After the collision, the two gliders stick together and move with a common velocity \(v\).
(i) Calculate the velocity \(v\). [3]
(ii) Calculate the impulse exerted by glider A on glider B during the collision. [2]
(iii) State the direction of this impulse. [1]
(c) Explain whether the collision is elastic or inelastic. [1]
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解题
(a) Impulse is defined as the force acting on an object multiplied by the time for which it acts (or the change in momentum of the object).
(b) (i) Using the principle of conservation of momentum: \(m_A u_A + m_B u_B = (m_A + m_B) v\) \(0.40 \times 1.5 + 0 = (0.40 + 0.60) v\) \(0.60 = 1.00 v \implies v = 0.60\text{ m/s}\)
(ii) Impulse on glider B is equal to its change in momentum: \(\text{Impulse} = m_B (v - u_B) = 0.60 \times (0.60 - 0) = 0.36\text{ N s}\) (or \(\text{kg m/s}\))
(iii) The direction of the impulse on glider B is in the direction of glider A's initial motion (forwards).
(c) The collision is inelastic because the gliders stick together (or kinetic energy is not conserved: initial \(E_k = 0.45\text{ J}\), final \(E_k = 0.18\text{ J}\)).
评分标准
(a) - force \(\times\) time (or change in momentum) [1]
(b) (i) - initial momentum = \(0.60\text{ kg m/s}\) [1] - conservation of momentum equation: \(0.60 = 1.00 \times v\) [1] - \(v = 0.60\text{ m/s}\) with unit [1]
(b) (ii) - \(\text{impulse} = \Delta p\) (or use of B's change in momentum) [1] - \(0.36\text{ N s}\) (or \(\text{kg m/s}\)) [1]
(b) (iii) - in the direction of A's initial motion / forwards / to the right [1]
(c) - inelastic because gliders stick together (or calculation showing kinetic energy is lost) [1]
题目 4 · structured-theory
7 分
A pump is used to lift water from a well to a storage tank on a roof.
(a) The tank is \(15\text{ m}\) above the water level in the well. The pump lifts \(240\text{ kg}\) of water every minute. Take \(g = 9.8\text{ m/s}^2\).
(i) Calculate the useful work done by the pump in lifting the water in one minute. [3]
(ii) Calculate the useful power output of the pump. [2]
(b) The electrical power input to the pump is \(800\text{ W}\). Calculate the efficiency of the pump. [2]
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解题
(a) (i) The work done in lifting the water is equal to the gravitational potential energy gained: \(W = mgh = 240\text{ kg} \times 9.8\text{ m/s}^2 \times 15\text{ m} = 35280\text{ J}\) (or \(3.5 \times 10^4\text{ J}\))
(ii) Power is the rate of doing work: \(P = \frac{W}{t} = \frac{35280\text{ J}}{60\text{ s}} = 588\text{ W}\) (or \(590\text{ W}\))
(b) Efficiency is calculated as: \(\text{Efficiency} = \frac{\text{Useful power output}}{\text{Power input}} \times 100\%\) \(\text{Efficiency} = \frac{588}{800} \times 100\% = 73.5\%\) (which rounds to \(74\%\)).
评分标准
(a) (i) - use of \(W = mgh\) [1] - substitution: \(240 \times 9.8 \times 15\) [1] - \(35280\text{ J}\) (accept \(35000\text{ J}\) or \(3.5 \times 10^4\text{ J}\)) [1]
(a) (ii) - use of \(P = W / t\) with \(t = 60\text{ s}\) [1] - \(588\text{ W}\) (or \(590\text{ W}\)) with unit [1]
(b) - use of \(\text{Efficiency} = (P_{\text{out}} / P_{\text{in}}) \times 100\%\) [1] - \(73.5\%\) or \(74\%\) [1]
题目 5 · structured-theory
7 分
A student carries out an experiment to determine the specific latent heat of fusion of ice.
(a) Define specific latent heat of fusion. [2]
(b) An electric heater of power \(150\text{ W}\) is placed in a funnel containing melting ice at \(0^\circ\text{C}\).
(i) In a time of \(4.0\text{ minutes}\), \(64\text{ g}\) of ice melts. Calculate the specific latent heat of fusion of ice from these results. [3]
(ii) Suggest one reason why the value calculated in (b)(i) might be different from the actual textbook value, and state whether your calculated value is likely to be larger or smaller than the true value. [2]
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解题
(a) Specific latent heat of fusion is the thermal energy required to change unit mass of a solid to liquid with no change in temperature.
(b) (i) Convert time to seconds: \(t = 4.0 \times 60 = 240\text{ s}\)
Calculate the electrical energy supplied by the heater: \(E = P \times t = 150\text{ W} \times 240\text{ s} = 36000\text{ J}\)
(ii) Heat may be absorbed from the surrounding air, which causes additional ice to melt. Because more ice melts than would be caused by the heater alone, the calculated specific latent heat of fusion is smaller than the true value. Alternatively, some thermal energy from the heater may escape to the surrounding air/funnel instead of being transferred to the ice, meaning less ice melts than expected for that energy input, which would make the calculated value larger than the true value.
评分标准
(a) - thermal energy required to change state from solid to liquid [1] - per unit mass at constant temperature [1]
(b) (i) - time conversion to \(240\text{ s}\) and energy calculation: \(E = 150 \times 240 = 36000\text{ J}\) [1] - mass conversion to \(0.064\text{ kg}\) [1] - \(L_f = 5.6 \times 10^5\text{ J/kg}\) (or \(560\text{ J/g}\)) with correct unit [1]
(b) (ii) - identification of heat transfer to/from surroundings [1] - correct deduction of whether calculated value is larger/smaller based on the identified path [1]
题目 6 · structured-theory
8 分
A ray of light is incident on the boundary between glass and air.
(a) (i) State what is meant by critical angle. [2]
(ii) The refractive index of the glass is \(1.52\). Calculate the critical angle \(c\) for this glass-air boundary. [2]
(b) A ray of light enters a right-angled glass prism of refractive index \(1.52\) normally through one of its shorter sides. It meets the hypotenuse of the prism.
(i) Show that the angle of incidence at the hypotenuse is \(45^\circ\). [2]
(ii) State and explain what happens to the ray of light at the hypotenuse. [2]
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解题
(a) (i) The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).
(ii) Using the relation: \(\sin c = \frac{1}{n} = \frac{1}{1.52}\) \(\sin c \approx 0.6579 \implies c \approx 41.1^\circ\) (or \(41^\circ\))
(b) (i) The ray enters normally (at \(90^\circ\) to the surface), so it passes into the prism without refraction. It continues in a straight line until it meets the hypotenuse. Since the prism is a right-angled isosceles triangle, the angle between the hypotenuse and the base is \(45^\circ\). The normal to the hypotenuse makes an angle of \(90^\circ\) with the hypotenuse, so the angle between the ray and the normal is \(90^\circ - 45^\circ = 45^\circ\).
(ii) Since the angle of incidence at the hypotenuse (\(45^\circ\)) is greater than the critical angle (\(41.1^\circ\)), the ray of light undergoes total internal reflection. No light escapes from the hypotenuse; instead, it is reflected back inside the glass prism.
评分标准
(a) (i) - angle of incidence in denser medium [1] - resulting in angle of refraction of \(90^\circ\) in less dense medium [1]
(a) (ii) - formula \(\sin c = 1/n\) seen/used [1] - \(c = 41^\circ\) (or \(41.1^\circ\)) [1]
(b) (i) - ray is undeviated at first surface because it enters normally [1] - geometry showed that the angle with the normal at the hypotenuse is \(45^\circ\) [1]
(b) (ii) - total internal reflection occurs [1] - because angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41^\circ\)) [1]
题目 7 · structured-theory
7 分
A simple circuit contains a cell connected to a copper wire.
(a) Explain what is meant by the electromotive force (e.m.f.) of a cell. [2]
(b) A copper wire has a current of \(0.80\text{ A}\) flowing through it.
(i) Calculate the total charge that passes any point in the wire in \(5.0\text{ minutes}\). [2]
(ii) A cell of e.m.f. \(1.5\text{ V}\) drives this current. Calculate the energy transferred by the cell in these \(5.0\text{ minutes}\). [3]
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解题
(a) The electromotive force (e.m.f.) of a cell is defined as the work done by the source in driving unit charge around a complete circuit (or energy transferred per unit charge).
(b) (i) Convert time to seconds: \(t = 5.0 \times 60 = 300\text{ s}\)
Using the charge formula: \(Q = I \times t = 0.80\text{ A} \times 300\text{ s} = 240\text{ C}\)
(ii) Using the relation for electrical energy: \(E = V \times Q\) (or \(E = V I t\)) \(E = 1.5\text{ V} \times 240\text{ C} = 360\text{ J}\)
评分标准
(a) - work done / energy transferred by the source [1] - per unit charge around a complete circuit [1]
(b) (i) - time conversion to \(300\text{ s}\) [1] - \(Q = 240\text{ C}\) with unit [1]
(b) (ii) - formula \(E = V I t\) or \(E = V Q\) seen/used [1] - substitution: \(1.5 \times 240\) [1] - \(E = 360\text{ J}\) with unit [1]
题目 8 · structured-theory
8 分
Radioactive isotopes are used to study processes in physics and medicine.
(a) A radioactive source contains an isotope of bismuth, \(\text{Bi}-212\). It decays by emitting a beta (\(\beta^-\)) particle to form an isotope of polonium (\(\text{Po}\)).
(i) Complete the decay equation to find \(A\) and \(Z\): $${}_{83}^{212}\text{Bi} \rightarrow {}_{Z}^{A}\text{Po} + {}_{-1}^{0}\beta$$
State the values of \(A\) and \(Z\). [2]
(ii) State the nature of a beta particle. [1]
(b) The initial count rate from a sample of this bismuth isotope is \(800\text{ counts/second}\). After a time of \(3.0\text{ hours}\), the count rate has fallen to \(100\text{ counts/second}\).
(i) Calculate the half-life of the isotope. [3]
(ii) Suggest why a detector placed near the sample still registers a small count rate even after several days when almost all the bismuth has decayed. [2]
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解题
(a) (i) In a beta-minus decay, a neutron decays into a proton and an electron. Thus: - The total nucleon number \(A\) remains unchanged, so \(A = 212\). - The proton number \(Z\) increases by 1, so \(Z = 83 + 1 = 84\).
(ii) A beta particle is a high-speed electron emitted from the nucleus.
(b) (i) Determine the number of half-lives that have passed: \(800 \rightarrow 400\) (1 half-life) \(400 \rightarrow 200\) (2 half-lives) \(200 \rightarrow 100\) (3 half-lives)
So, 3 half-lives have passed in \(3.0\text{ hours}\). \(3 \times T_{1/2} = 3.0\text{ hours} \implies T_{1/2} = 1.0\text{ hour}\) (or \(60\text{ minutes}\)).
(ii) The detector still measures background radiation, which is ionizing radiation always present in the environment from sources like cosmic rays, radioactive rocks (granite), radon gas in the air, or medical/industrial sources.
评分标准
(a) (i) - \(A = 212\) [1] - \(Z = 84\) [1]
(a) (ii) - electron (from nucleus) [1]
(b) (i) - recognizing the count rate halved three times [1] - setting up the equation: \(3 \times T_{1/2} = 3.0\text{ hours}\) [1] - \(T_{1/2} = 1.0\text{ hour}\) (or \(60\text{ minutes}\)) with unit [1]
(b) (ii) - background radiation is present [1] - example source (e.g. cosmic rays, rocks, radon) [1]
题目 9 · structured-theory
7 分
A tennis ball of mass \(0.058\text{ kg}\) travels horizontally at a speed of \(28\text{ m/s}\) towards a tennis racket. The ball is struck by the racket and leaves in the opposite direction at a speed of \(35\text{ m/s}\).
(a) Define impulse in terms of force and time. [1]
(b) Calculate the change in momentum (impulse) of the ball during the collision. [2]
(c) The contact time between the ball and the racket is \(4.5\text{ ms}\). Calculate the average force exerted on the ball by the racket. [2]
(d) Explain, in terms of momentum, why the tennis racket experiences an equal and opposite force to the force on the ball during the impact. [2]
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解题
(a) Impulse is defined as the force multiplied by the time for which it acts, which equals the change in momentum.
(b) Taking the initial direction of the ball's travel as negative and the final direction as positive: Initial momentum, \(p_i = m \times v_i = 0.058 \times (-28) = -1.624\text{ kg m/s}\) Final momentum, \(p_f = m \times v_f = 0.058 \times 35 = 2.030\text{ kg m/s}\) Change in momentum, \(\Delta p = p_f - p_i = 2.030 - (-1.624) = 3.654\text{ kg m/s}\) To two significant figures, \(\Delta p = 3.7\text{ kg m/s}\) (or \(\text{N s}\)).
(c) Average force \(F = \frac{\Delta p}{\Delta t} = \frac{3.654}{4.5 \times 10^{-3}} = 812\text{ N}\) To two significant figures, \(F = 810\text{ N}\).
(d) According to the principle of conservation of momentum, in any closed system, the total momentum remains constant. The momentum lost by the racket must equal the momentum gained by the ball. Therefore, the change in momentum of the racket is equal and opposite to that of the ball. Since force is the rate of change of momentum, the forces are equal and opposite.
评分标准
(a) B1: force \(\times\) time (for which it acts) OR change in momentum
(b) C1: Correct formulation of change in momentum: \(0.058 \times (35 - (-28))\) OR \(0.058 \times 63\) A1: \(3.7\text{ N s}\) (or \(\text{kg m/s}\)), allow \(3.65\text{ N s}\)
(c) C1: \(F = \frac{\text{change in momentum}}{t}\) OR \(\frac{3.654}{0.0045}\) A1: \(810\text{ N}\) (or \(812\text{ N}\)), allow ecf from (b)
(d) B1: Rate of change of momentum of the racket is equal and opposite to that of the ball B1: Newton's third law states action and reaction forces are equal and opposite / rate of change of momentum equals force
题目 10 · structured-theory
7 分
A ray of light in a transparent plastic block travels towards a boundary with air. The refractive index of the plastic is \(1.48\).
(a) State what is meant by the critical angle. [1]
(b) Calculate the critical angle for the boundary between the plastic and the air. [2]
(c) The ray of light meets the boundary at an angle of incidence of \(40^\circ\). State and explain what happens to the ray at this boundary. [2]
(d) Describe how optical fibres are used in medicine to examine internal organs, stating the role of total internal reflection. [2]
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解题
(a) The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).
(b) Using \(\sin(c) = \frac{1}{n}\): \(\sin(c) = \frac{1}{1.48} = 0.6757\) \(c = \sin^{-1}(0.6757) = 42.51^\circ\) To three significant figures, the critical angle is \(42.5^\circ\) (accept \(43^\circ\)).
(c) Since the angle of incidence (\(40^\circ\)) is less than the critical angle (\(42.5^\circ\)), the light ray will be refracted into the air, bending away from the normal. (There will also be a weak, partially reflected ray back inside the plastic).
(d) Flexible bundles of optical fibres (in an endoscope) are inserted into the patient's body. Light travels down one bundle of fibres by total internal reflection to illuminate the internal organ. Reflected light from the organ travels back up a second bundle of fibres, also by total internal reflection, to form an image on a monitor for the doctor to observe.
评分标准
(a) B1: The angle of incidence in the denser medium yielding an angle of refraction of \(90^\circ\)
(c) B1: The light is refracted / exits into the air (bending away from the normal) B1: Because the angle of incidence is less than the critical angle (\(40^\circ < 42.5^\circ\))
(d) B1: Light is guided down flexible fibres to illuminate the organ via total internal reflection B1: Reflected light returns along another bundle of fibres via total internal reflection to a viewer/camera
题目 11 · structured-theory
7 分
Astronomers observe the light from a distant galaxy and compare its spectrum with a reference spectrum from a source on Earth.
(a) Explain what is meant by the redshift of light from a galaxy. [2]
(b) State the formula that relates the recession speed \(v\) of a distant galaxy to its distance \(d\) from Earth. Identify the constant in your formula and state its unit. [2]
(c) A galaxy is detected to be moving away from Earth at a speed of \(4.2 \times 10^6\text{ m/s}\). Using the Hubble constant value of \(2.2 \times 10^{-18}\text{ s}^{-1}\), calculate the distance of this galaxy from Earth. [3]
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解题
(a) Redshift is the increase in the observed wavelength (or decrease in frequency) of electromagnetic radiation emitted by a galaxy, which occurs because the galaxy is moving away from the observer.
(b) The formula is \(v = H_0 \times d\), where \(H_0\) is the Hubble constant. Its SI unit is \(\text{s}^{-1}\).
(c) Rearranging the formula to find the distance \(d\): \(d = \frac{v}{H_0}\) \(d = \frac{4.2 \times 10^6\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 1.909 \times 10^{24}\text{ m}\) To two significant figures, the distance is \(1.9 \times 10^{24}\text{ m}\).
评分标准
(a) B1: Increase in observed wavelength / decrease in frequency of light B1: Caused by the galaxy/source moving away from Earth/observer
(b) B1: \(v = H_0 \times d\) (or equivalent rearrangement) B1: \(H_0\) is the Hubble constant, unit is \(\text{s}^{-1}\) (allow \(\text{km/s per Mpc}\))
Answer all questions. Use your knowledge of practical physics techniques, graphing, and experimental design.
4 题目 · 40 分
题目 1 · practical-structured
10 分
A student is investigating the period of oscillation of a simple pendulum. (a) A digital stopwatch displays the time taken for 20 complete oscillations of a pendulum of length L = 80.0 cm as 00:35.80 (minutes:seconds). State the time t in seconds. (b) The student records the times for other lengths. For L = 40.0 cm, the time for 20 oscillations is t = 25.4 s. Calculate the period T (time for one complete oscillation) and the value of T^2 for L = 40.0 cm and L = 80.0 cm. (c) Explain why timing 20 oscillations is more accurate than timing a single oscillation. (d) Describe, with the aid of a brief explanation, how a student can use a ruler and a set-square to measure the length L of the pendulum bob to the center of the sphere as accurately as possible. (e) State one variable, other than the length of the pendulum, that must be kept constant during this experiment. (f) Suggest why the angle of swing should be kept small.
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解题
Detailed calculations: (b) T = t / 20. For L = 40.0 cm: T = 25.4 / 20 = 1.27 s. T^2 = 1.27^2 = 1.61 s^2. For L = 80.0 cm: T = 35.8 / 20 = 1.79 s. T^2 = 1.79^2 = 3.20 s^2.
评分标准
(a) 1 mark: t = 35.8 s. (b) 2 marks: Correct calculations of T (1 mark) and T^2 (1 mark) to 3 significant figures with units. (c) 1 mark: Mention of reducing the error due to human reaction time. (d) 2 marks: Explaining the use of the set-square to project the center of the bob horizontally to the vertical ruler (1 mark) and ensuring the ruler is vertical (1 mark). (e) 2 marks: Stating two controlled variables (mass of the bob, amplitude/angle of swing). (f) 2 marks: Stating the physical reason for small angles of swing.
题目 2 · practical-structured
10 分
A student investigates the rate of cooling of hot water in two identical beakers under different conditions: Beaker A (uninsulated) and Beaker B (insulated with a layer of cotton wool). (a) A thermometer is used to measure the room temperature before the experiment. The thermometer scale shows a reading between 21 and 22, with the meniscus exactly on the fifth tenth division past 21. State this room temperature \(\theta_R\). (b) Hot water is poured into both beakers. The temperature is recorded every 30 s for 180 s. At t = 0 s, both beakers are at 85.0 \(^\circ\)C. At t = 180 s, the temperature in Beaker A is 61.5 \(^\circ\)C, and the temperature in Beaker B is 72.0 \(^\circ\)C. Calculate the average cooling rate R over the 180 s period for both beakers using the equation R = (\(\theta_{\text{start}}\) - \(\theta_{\text{end}}\)) / t. Include the units. (c) State and explain which beaker had the higher rate of cooling. (d) Describe two precautions that should be taken to ensure that the temperature readings are as accurate as possible. (e) State one variable that must be kept constant to ensure a fair comparison between the two beakers. (f) Explain how the student should position their eye to avoid parallax error when reading the thermometer.
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解题
(a) The thermometer reading is 21 + 0.5 = 21.5 \(^\circ\)C. (b) Beaker A: R_A = (85.0 - 61.5) / 180 = 23.5 / 180 = 0.130 \(^\circ\)C/s. Beaker B: R_B = (85.0 - 72.0) / 180 = 13.0 / 180 = 0.072 \(^\circ\)C/s. (c) Uninsulated beaker A loses thermal energy faster through conduction, convection, and radiation. (d) Precautions include: ensuring the thermometer does not touch the sides or bottom of the beaker; stirring the liquid to ensure uniform temperature. (e) Control volume of water, initial temperature, beaker shape/material. (f) Look horizontally/perpendicularly at the scale.
评分标准
(a) 1 mark: 21.5 \(^\circ\)C. (b) 2 marks: Correct calculation for Beaker A (1 mark) and Beaker B (1 mark) with correct units (\(^\circ\)C/s or \(^\circ\)C/min). (c) 1 mark: Correct identification of Beaker A with explanation. (d) 2 marks: Two valid precautions (e.g. thermometer not touching sides, stirring, reading at eye level). (e) 2 marks: Two controlled variables (initial volume, initial temperature, same beaker type). (f) 2 marks: Clear explanation of eye positioning perpendicular to the meniscus.
题目 3 · practical-structured
10 分
A student determines the refractive index of a rectangular glass block using optical pins. (a) The student traces the outline of the block and draws a normal line. An incident ray is drawn making an angle of incidence i = 45.0\(^\circ\) with the normal. The refracted ray inside the block makes an angle of refraction r = 27.5\(^\circ\) with the normal. Calculate the refractive index n of the glass block using the equation n = sin(i) / sin(r). (b) Explain why the student must place the optical pins as far apart as possible (at least 5.0 cm apart). (c) State one precaution the student must take when placing the pins to ensure accuracy. (d) Describe how the student can locate the emergent ray using the pins. (e) Suggest one advantage of using a thin laser beam instead of optical pins. (f) The student measures the width of the glass block using a micrometer screw gauge instead of a standard ruler. Explain why this choice improves accuracy.
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解题
(a) n = sin(45.0\(^\circ\)) / sin(27.5\(^\circ\)) = 0.7071 / 0.4617 = 1.53. (b) Standard practical precaution: longer distance between pins minimizes alignment errors. (c) Verticality ensures the pin path represents the exact light path at the surface interface. (d) Line-up method of parallax. (e) Laser produces a highly collimated, narrow line of light. (f) Higher precision device decreases experimental uncertainty.
评分标准
(a) 2 marks: Correct substitution (1 mark) and final answer 1.53 to 2 or 3 sig figs (1 mark). (b) 2 marks: Clearly stating that it reduces the error/uncertainty in drawing the ray direction. (c) 1 mark: Valid precaution (e.g. view bases of pins, ensure pins are vertical). (d) 2 marks: Detailed explanation of alignment of images (1 mark) and placement of tracking pins (1 mark). (e) 1 mark: Laser provides a direct, highly visible narrow line. (f) 2 marks: Stating the higher resolution/precision of the micrometer (1 mark) and the reduction of percentage uncertainty (1 mark).
题目 4 · practical-structured
10 分
A student investigates the resistance of a metal wire of uniform cross-sectional area. (a) Draw a circuit diagram showing a cell, a switch, an ammeter, a voltmeter, and a test wire of length L connected so that the potential difference across the wire can be measured. (b) For a wire length of L = 60.0 cm, the ammeter reading is 0.40 A and the voltmeter reading is 2.20 V. Calculate the resistance R of this length of wire. (c) State and explain how the student should use the switch during the experiment to prevent the wire from heating up. (d) Describe the relationship expected between the length L of the wire and its resistance R. (e) State one variable that must be kept constant to ensure a fair test. (f) Explain why sliding a metal contact hard along the resistance wire should be avoided.
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解题
(a) Voltmeter must be in parallel across the test wire, ammeter must be in series in the main loop. (b) R = V / I = 2.20 / 0.40 = 5.50 \(\Omega\). (c) Heating increases the resistance of the metal wire. Open switch when not taking readings to minimize temperature change. (d) R is proportional to L. (e) Constant area/diameter, same material, same room temperature. (f) Physical wear reduces diameter, causing localized high resistance.
评分标准
(a) 2 marks: Correct symbols and ammeter in series (1 mark), voltmeter in parallel with test wire (1 mark). (b) 2 marks: Correct calculation with unit \(\Omega\). (c) 2 marks: Opening switch between readings (1 mark) because heating changes resistance (1 mark). (d) 1 mark: Stating direct proportionality. (e) 2 marks: Two control variables (material, diameter/cross-sectional area, temperature). (f) 1 mark: Stating that sliding damages the wire or changes its cross-sectional area.
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