An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V1) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
卷二 (Extended MCQ)
Forty multiple-choice questions. Four possible answers. Take weight of 1.0 kg to be 9.8 N.
40 题目 · 40 分
题目 1 · 選擇題
1 分
A model rocket is launched vertically upwards. It accelerates uniformly from rest to a speed of \(23.5\text{ m/s}\) in \(3.0\text{ s}\). The fuel is then exhausted, and the rocket decelerates under gravity until it reaches its maximum height.
What is the total time from launch until the rocket reaches its maximum height?
A.3.0 s
B.5.0 s
C.5.4 s
D.7.8 s
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解题
During the first phase of acceleration, the time taken is \(t_1 = 3.0\text{ s}\). During the second phase, the rocket decelerates under gravity with \(g = 9.8\text{ m/s}^2\). The time \(t_2\) taken to reach maximum height (where \(v = 0\text{ m/s}\)) from a speed of \(u = 23.5\text{ m/s}\) is calculated using: \(v = u - gt_2 \Rightarrow 0 = 23.5 - 9.8 \times t_2 \Rightarrow t_2 = \frac{23.5}{9.8} \approx 2.4\text{ s}\).
The total time is \(t_1 + t_2 = 3.0\text{ s} + 2.4\text{ s} = 5.4\text{ s}\).
评分标准
C is the correct answer. 1 mark is awarded for the correct option.
题目 2 · 選擇題
1 分
An electric motor is used to lift a load of mass \(12\text{ kg}\) through a vertical height of \(8.0\text{ m}\) in a time of \(4.0\text{ s}\). The electrical power input to the motor is \(350\text{ W}\).
What is the efficiency of the motor?
A.27%
B.67%
C.84%
D.96%
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解题
First, calculate the weight of the load: \(W = m \times g = 12\text{ kg} \times 9.8\text{ m/s}^2 = 117.6\text{ N}\). Next, calculate the useful work output: \(E_{\text{out}} = W \times h = 117.6\text{ N} \times 8.0\text{ m} = 940.8\text{ J}\). Then, calculate the useful power output: \(P_{\text{out}} = \frac{E_{\text{out}}}{t} = \frac{940.8\text{ J}}{4.0\text{ s}} = 235.2\text{ W}\). Finally, calculate the efficiency: \\text{efficiency} = \\frac{P_{\\text{out}}}{P_{\\text{in}}} \\times 100\\% = \\frac{235.2\\text{ W}}{350\\text{ W}} \\times 100\\% = 67.2\\% \\approx 67\\%.
评分标准
B is the correct answer. 1 mark is awarded for the correct option.
题目 3 · 選擇題
1 分
A U-tube manometer contains water of density \(1000\text{ kg/m}^3\). It is connected to a gas supply. The water level on the side connected to the gas supply is \(15\text{ cm}\) lower than on the side open to the atmosphere.
The atmospheric pressure is \(1.01 \times 10^5\text{ Pa}\).
What is the pressure of the gas supply?
A.1.5 \times 10^3\text{ Pa}
B.9.95 \times 10^4\text{ Pa}
C.1.01 \times 10^5\text{ Pa}
D.1.02 \times 10^5\text{ Pa}
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解题
Since the water level on the side connected to the gas supply is lower, the pressure of the gas supply is greater than atmospheric pressure: \(P_{\text{gas}} = P_{\text{atm}} + \Delta P\) where the pressure difference \(\Delta P\) is given by: \(\Delta P = \rho g h = 1000\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 0.15\text{ m} = 1470\text{ Pa}\).
D is the correct answer. 1 mark is awarded for the correct option.
题目 4 · 選擇題
1 分
A block of copper of mass \(2.0\text{ kg}\) is heated by a \(150\text{ W}\) heater for \(5.0\text{ minutes}\). The temperature of the block increases from \(20\text{ }^\circ\text{C}\) to \(77\text{ }^\circ\text{C}\). Assume no thermal energy is lost to the surroundings.
What is the calculated value for the specific heat capacity of copper from this experiment?
A.120\text{ J/(kg }^\circ\text{C)}
B.390\text{ J/(kg }^\circ\text{C)}
C.790\text{ J/(kg }^\circ\text{C)}
D.2400\text{ J/(kg }^\circ\text{C)}
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解题
First, calculate the thermal energy supplied by the heater: \(E = P \times t = 150\text{ W} \times (5.0 \times 60)\text{ s} = 45000\text{ J}\).
The temperature increase is: \(\Delta \theta = 77\text{ }^\circ\text{C} - 20\text{ }^\circ\text{C} = 57\text{ }^\circ\text{C}\).
Using the formula \(E = m c \Delta \theta\): \(45000 = 2.0\text{ kg} \times c \times 57\text{ }^\circ\text{C}\) \(c = \frac{45000}{114} \approx 394.7\text{ J/(kg }^\circ\text{C)} \approx 390\text{ J/(kg }^\circ\text{C)}\).
评分标准
B is the correct answer. 1 mark is awarded for the correct option.
题目 5 · 選擇題
1 分
A semicircular glass block has a refractive index of \(1.52\). A ray of light is directed towards the centre of the flat face of the block, from inside the glass.
At which angle of incidence at the glass-air boundary will total internal reflection first occur?
A.30^\circ
B.41^\circ
C.49^\circ
D.90^\circ
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解题
Total internal reflection first occurs when the angle of incidence equals the critical angle \(c\). The relationship between critical angle and refractive index \(n\) is: \(\sin(c) = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579\).
B is the correct answer. 1 mark is awarded for the correct option.
题目 6 · 選擇題
1 分
A battery drives a current of \(0.50\text{ A}\) through a lamp for \(3.0\text{ minutes}\). During this time, the battery transfers \(1620\text{ J}\) of electrical energy to the lamp.
What is the electromotive force (e.m.f.) of the battery?
A.1.5 V
B.6.0 V
C.18 V
D.54 V
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解题
First, calculate the total charge \(Q\) that flows through the lamp: \(Q = I \times t = 0.50\text{ A} \times (3.0 \times 60)\text{ s} = 90\text{ C}\).
The e.m.f. \(E_{\text{emf}}\) is the energy transferred per unit charge: \(E_{\text{emf}} = \frac{W}{Q} = \frac{1620\text{ J}}{90\text{ C}} = 18\text{ V}\).
评分标准
C is the correct answer. 1 mark is awarded for the correct option.
题目 7 · 選擇題
1 分
Two resistors, \(R_1 = 4.0\text{ }\Omega\) and \(R_2 = 12.0\text{ }\Omega\), are connected in parallel across a power supply of voltage \(V\). The current in the \(12.0\text{ }\Omega\) resistor is \(1.5\text{ A}\).
What is the total current supplied by the power supply to the circuit?
A.1.5 A
B.2.0 A
C.4.5 A
D.6.0 A
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解题
Since the resistors are in parallel, they have the same potential difference \(V\) across them. The voltage \(V\) is: \(V = I_2 \times R_2 = 1.5\text{ A} \times 12.0\text{ }\Omega = 18.0\text{ V}\).
The current \(I_1\) in the first resistor is: \(I_1 = \frac{V}{R_1} = \frac{18.0\text{ V}}{4.0\text{ }\Omega} = 4.5\text{ A}\).
The total current \(I\) is the sum of the currents in each branch: \(I = I_1 + I_2 = 4.5\text{ A} + 1.5\text{ A} = 6.0\text{ A}\).
评分标准
D is the correct answer. 1 mark is awarded for the correct option.
题目 8 · 選擇題
1 分
Light from a distant galaxy is observed to have its wavelength redshifted. The galaxy is at a distance of \(8.0 \times 10^{24}\text{ m}\) from Earth and is moving away at a speed of \(1.8 \times 10^7\text{ m/s}\).
Using these data, what is the estimated value of the Hubble constant, \(H_0\)?
A.2.3 \times 10^{-18}\text{ s}^{-1}
B.4.4 \times 10^{-17}\text{ s}^{-1}
C.4.4 \times 10^{17}\text{ s}
D.1.4 \times 10^{32}\text{ s}^{-1}
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解题
According to Hubble's Law, the recession speed of a galaxy is proportional to its distance: \(v = H_0 d \Rightarrow H_0 = \frac{v}{d}\).
A is the correct answer. 1 mark is awarded for the correct option.
题目 9 · MCQ
1 分
An electric toy car accelerates from rest to a speed of 3.0 m/s in 4.0 s. It then travels at this constant speed of 3.0 m/s for a further 6.0 s. Finally, it decelerates uniformly to rest in a time of 2.0 s.
What is the average speed of the toy car for the whole journey?
A.1.5 m/s
B.2.3 m/s
C.2.5 m/s
D.3.0 m/s
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解题
The motion consists of three parts: 1. Acceleration from 0 to 3.0 m/s in 4.0 s: $$\text{Distance } d_1 = \frac{1}{2} \times 4.0\text{ s} \times 3.0\text{ m/s} = 6.0\text{ m}$$
2. Constant speed of 3.0 m/s for 6.0 s: $$\text{Distance } d_2 = 3.0\text{ m/s} \times 6.0\text{ s} = 18.0\text{ m}$$
B is correct (1 mark). Award 1 mark for correct calculation of total distance (27 m) and dividing it by total time (12 s) to obtain 2.3 m/s.
题目 10 · MCQ
1 分
A small ball of mass 0.20 kg is attached to a string and whirled in a horizontal circle of radius 0.50 m at a constant speed of 4.0 m/s.
Which statement about the resultant force acting on the ball is correct?
A.The resultant force on the ball is zero because its speed is constant.
B.The resultant force on the ball is 6.4 N directed away from the centre of the circle.
C.The resultant force on the ball is 6.4 N directed towards the centre of the circle.
D.The resultant force on the ball is 1.6 N directed towards the centre of the circle.
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解题
An object moving in a circle at constant speed undergoes centripetal acceleration, which requires a centripetal force directed towards the centre of the circle.
The magnitude of this force is given by: $$F = \frac{m v^2}{r}$$
Substituting the given values: $$F = \frac{0.20\text{ kg} \times (4.0\text{ m/s})^2}{0.50\text{ m}} = \frac{0.20 \times 16.0}{0.50} = 6.4\text{ N}$$
Therefore, the resultant force is 6.4 N directed towards the centre of the circle.
评分标准
C is correct (1 mark). Award 1 mark for correctly calculating the force magnitude (6.4 N) and identifying that its direction is towards the centre.
题目 11 · MCQ
1 分
A trolley of mass 3.0 kg moving at a speed of 4.0 m/s collides with a stationary trolley of mass 5.0 kg. After the collision, the two trolleys stick together and move off with a common velocity v.
Which row correctly identifies the common velocity v and the total kinetic energy after the collision?
A.velocity v = 1.5 m/s, total kinetic energy = 9.0 J
B.velocity v = 1.5 m/s, total kinetic energy = 24 J
C.velocity v = 2.4 m/s, total kinetic energy = 9.0 J
D.velocity v = 2.4 m/s, total kinetic energy = 24 J
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解题
According to the principle of conservation of momentum: $$\text{Total initial momentum} = \text{Total final momentum}$$ $$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$$ $$3.0\text{ kg} \times 4.0\text{ m/s} + 5.0\text{ kg} \times 0 = (3.0\text{ kg} + 5.0\text{ kg}) v$$ $$12.0 = 8.0 v \implies v = 1.5\text{ m/s}$$
The total kinetic energy after the collision is: $$E_k = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 8.0\text{ kg} \times (1.5\text{ m/s})^2 = 4.0 \times 2.25 = 9.0\text{ J}$$
评分标准
A is correct (1 mark). Award 1 mark for the correct combination of common velocity (1.5 m/s) and kinetic energy (9.0 J).
题目 12 · MCQ
1 分
A box of mass 15 kg is pulled up a rough slope of length 12 m to a vertical height of 4.0 m by a constant force of 80 N acting parallel to the slope. (Take the weight of 1.0 kg to be 9.8 N).
What is the efficiency of this process?
A.5.1%
B.54%
C.61%
D.82%
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解题
1. Useful output work (gain in gravitational potential energy): $$E_p = m g h = 15\text{ kg} \times 9.8\text{ m/s}^2 \times 4.0\text{ m} = 588\text{ J}$$
2. Total input work done by the pulling force: $$W_{\text{in}} = F \times d = 80\text{ N} \times 12\text{ m} = 960\text{ J}$$
3. Efficiency: $$\text{Efficiency} = \frac{\text{Useful work output}}{\text{Total work input}} \times 100\% = \frac{588}{960} \times 100\% = 61.25\% \approx 61\%$$
评分标准
C is correct (1 mark). Award 1 mark for calculating both GPE (588 J) and input work (960 J) and determining the efficiency is 61%.
题目 13 · MCQ
1 分
A ray of light travels inside a glass block of refractive index 1.6 towards the boundary with air.
Which row correctly identifies the critical angle c for light in this glass and describes what happens to a ray of light incident at an angle of 40° to the normal inside the glass?
A.critical angle c = 39°; the ray is partially refracted and partially reflected.
B.critical angle c = 39°; the ray is totally internally reflected.
C.critical angle c = 51°; the ray is partially refracted and partially reflected.
D.critical angle c = 51°; the ray is totally internally reflected.
2. Calculate the current I: $$I = \frac{Q}{t} = \frac{6.0\text{ C}}{5.0\text{ s}} = 1.2\text{ A}$$
评分标准
A is correct (1 mark). Award 1 mark for calculating potential difference (7.5 V) and current (1.2 A).
题目 15 · MCQ
1 分
An ideal step-down transformer has a primary coil with 1200 turns and a secondary coil with 300 turns. The primary coil is connected to a 240 V a.c. mains supply, and the power input to the transformer is 48 W.
What is the output voltage and the current in the secondary coil?
A.output voltage = 60 V, secondary current = 0.20 A
B.output voltage = 60 V, secondary current = 0.80 A
C.output voltage = 960 V, secondary current = 0.05 A
D.output voltage = 960 V, secondary current = 0.80 A
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解题
1. Calculate the secondary voltage V_s: $$\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 240\text{ V} \times \frac{300}{1200} = 60\text{ V}$$
2. Since the transformer is ideal, power output = power input = 48 W: $$P_s = V_s \times I_s \implies I_s = \frac{48\text{ W}}{60\text{ V}} = 0.80\text{ A}$$
评分标准
B is correct (1 mark). Award 1 mark for the correct combination of output voltage (60 V) and secondary current (0.80 A).
题目 16 · MCQ
1 分
Light from a distant galaxy is observed to have a longer wavelength than the light emitted by the same elements in a laboratory on Earth.
Which row correctly describes the name of this shift and the direction of the galaxy's motion relative to the Earth?
A.name of shift: redshift; direction of motion: away from Earth
B.name of shift: redshift; direction of motion: towards Earth
C.name of shift: blueshift; direction of motion: away from Earth
D.name of shift: blueshift; direction of motion: towards Earth
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解题
An observed increase in wavelength is called a redshift. This indicates that the light source is moving away from the observer on Earth.
评分标准
A is correct (1 mark). Award 1 mark for identifying both the shift as redshift and the motion as moving away from Earth.
题目 17 · mcq
1 分
A cyclist climbs a steep hill of length 1200 m at a constant speed of 4.0 m/s. The cyclist immediately turns around and descends the same hill at a constant speed of 12.0 m/s.
What is the average speed of the cyclist for the entire double journey?
A.6.0 m/s
B.8.0 m/s
C.9.6 m/s
D.16.0 m/s
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解题
Average speed is defined as total distance divided by total time.
a is correct. Award 1 mark for correct calculation of average speed. b is the simple arithmetic mean (incorrect). c and d are other distractors.
题目 18 · mcq
1 分
A uniform plank of length 3.0 m and weight 80 N is supported by a pivot placed at its centre. A child of weight 300 N sits at a distance of 1.2 m from the pivot.
At what distance from the pivot on the opposite side must a second child of weight 450 N sit to keep the plank horizontal and balanced?
A.0.53 m
B.0.80 m
C.1.2 m
D.1.8 m
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解题
For the plank to be balanced, the sum of clockwise moments must equal the sum of anticlockwise moments about the pivot.
Since the plank is uniform and the pivot is at its centre, the weight of the plank acts through the pivot, producing zero turning moment.
Using the principle of moments: \[\text{Moment of first child} = \text{Moment of second child}\] \[300\text{ N} \times 1.2\text{ m} = 450\text{ N} \times d\] \[360\text{ N m} = 450\text{ N} \times d\] \[d = \frac{360}{450} = 0.80\text{ m}\]
评分标准
b is correct. Award 1 mark for the correct application of the principle of moments leading to 0.80 m.
题目 19 · mcq
1 分
An electric motor with an efficiency of 60% is used to lift a crate of mass 120 kg vertically upwards through a height of 5.0 m. The lift takes 8.0 s.
What is the electrical power input to the motor?
A.441 W
B.735 W
C.1225 W
D.2940 W
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解题
1. Calculate the gravitational potential energy gained by the crate (useful work output): \[\Delta E_{\text{p}} = mgh = 120\text{ kg} \times 9.8\text{ m/s}^2 \times 5.0\text{ m} = 5880\text{ J}\]
2. Calculate the useful power output: \[P_{\text{out}} = \frac{\text{useful work}}{t} = \frac{5880\text{ J}}{8.0\text{ s}} = 735\text{ W}\]
3. Use the efficiency formula to find the electrical power input: \[\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\] \[0.60 = \frac{735\text{ W}}{P_{\text{in}}}\] \[P_{\text{in}} = \frac{735\text{ W}}{0.60} = 1225\text{ W}\]
评分标准
c is correct. Award 1 mark for the correct calculation of input power by dividing output power by efficiency.
题目 20 · mcq
1 分
Two identical copper blocks, one painted dull black and the other polished shiny silver, are heated to a temperature of 90 °C. They are placed in a vacuum in a cold, dark room.
Which block cools down faster, and what is the primary mechanism of thermal energy transfer?
A.The dull black block cools faster because it is a better emitter of infrared radiation.
B.The shiny silver block cools faster because it is a better reflector of infrared radiation.
C.The dull black block cools faster because it is a better conductor of thermal energy.
D.Both blocks cool at the same rate because they are in a vacuum.
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解题
Since the blocks are suspended in a vacuum, no conduction or convection can occur through a surrounding gas. Therefore, the primary mechanism of heat transfer is thermal radiation (infrared radiation).
Dull black surfaces are much better emitters of thermal radiation than shiny silver surfaces. Consequently, the dull black block will radiate heat away much faster and cool down more quickly.
评分标准
a is correct. Award 1 mark for identifying the dull black block as cooling faster because it is a better emitter of infrared radiation.
题目 21 · mcq
1 分
A ray of light in a glass block is incident on the boundary with air. The refractive index of the glass is 1.60.
What is the critical angle for light travelling from the glass into air?
A.24°
B.31°
C.39°
D.51°
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解题
The critical angle \(c\) is related to the refractive index \(n\) by the formula: \[\sin(c) = \frac{1}{n}\]
c is correct. Award 1 mark for correct calculation of the critical angle using the formula \(\sin(c) = 1/n\).
题目 22 · mcq
1 分
A metal wire of length \(L\) and diameter \(d\) has a resistance of \(4.0\ \Omega\). Another wire made of the same metal has a length of \(2L\) and a diameter of \(2d\).
What is the resistance of the second wire?
A.1.0 Α
B.2.0 Α
C.4.0 Α
D.8.0 Α
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解题
The resistance \(R\) of a wire is given by: \[R = \rho \frac{L}{A}\] where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area.
Since \(A = \pi \left(\frac{d}{2}\right)^2 \propto d^2\), we can write the proportionality: \[R \propto \frac{L}{d^2}\]
For the second wire: \[R_2 \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \left(\frac{L}{d^2}\right)\]
Therefore, the resistance of the second wire is half that of the first wire: \[R_2 = \frac{1}{2} \times 4.0\ \Omega = 2.0\ \Omega\]
评分标准
b is correct. Award 1 mark for correct application of proportionality \(R \propto L/d^2\) to find that the resistance is halved to 2.0 \(\Omega\).
题目 23 · mcq
1 分
An ideal transformer has 400 turns on its primary coil and 100 turns on its secondary coil. An alternating voltage of 240 V is applied to the primary coil, and the current in the secondary circuit is 8.0 A.
What is the current in the primary coil?
A.0.50 A
B.2.0 A
C.8.0 A
D.32 A
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解题
For an ideal transformer, the power input to the primary coil is equal to the power output from the secondary coil: \[V_{\text{p}} I_{\text{p}} = V_{\text{s}} I_{\text{s}}\]
This can be rewritten in terms of the turns ratio as: \[\frac{I_{\text{p}}}{I_{\text{s}}} = \frac{N_{\text{s}}}{N_{\text{p}}}\]
Substituting the given values: \[\frac{I_{\text{p}}}{8.0\text{ A}} = \frac{100}{400}\] \[I_{\text{p}} = 8.0\text{ A} \times 0.25 = 2.0\text{ A}\]
评分标准
b is correct. Award 1 mark for correct use of ideal transformer equations to find primary current of 2.0 A.
题目 24 · mcq
1 分
A distant galaxy is observed to have a redshift that corresponds to a recessional speed of \(15\ 000\text{ km/s}\).
Using a Hubble constant value of \(H_0 = 70\text{ km/(s Mpc)}\), estimate the distance of this galaxy from Earth.
A.4.7 Mpc
B.210 Mpc
C.1050 Mpc
D.1 050 000 Mpc
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解题
According to Hubble's Law, the recessional speed \(v\) of a galaxy is proportional to its distance \(d\) from Earth: \[v = H_0 d\]
Rearranging the formula to solve for distance \(d\): \[d = \frac{v}{H_0}\]
Substituting the given values: \[d = \frac{15\ 000\text{ km/s}}{70\text{ km/(s Mpc)}} \approx 214\text{ Mpc}\]
Of the options provided, 210 Mpc is the closest estimate.
评分标准
b is correct. Award 1 mark for correct application of Hubble's Law formula to estimate distance.
题目 25 · MCQ
1 分
A car starts from rest and accelerates at a constant rate of for . It then travels at a constant velocity for , before decelerating uniformly to rest in a further .
What is the total distance travelled by the car?
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解题
The motion can be divided into three parts: 1. Acceleration phase: The final velocity is . The distance travelled is . 2. Constant velocity phase: The distance travelled is . 3. Deceleration phase: The distance travelled is .
Total distance travelled = .
评分标准
C is correct. Award 1 mark for the correct calculation of total distance.
题目 26 · MCQ
1 分
Two identical springs are connected in parallel to support a load of . The extension of the parallel spring system is .
What is the spring constant of one of these springs?
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解题
Since the two identical springs are in parallel, they share the load equally. Each spring supports a force of: .
The extension of each spring is .
Using Hooke's Law, : .
评分标准
B is correct. Award 1 mark for the correct spring constant calculation.
题目 27 · MCQ
1 分
An electric motor has an efficiency of . The motor is used to lift a load vertically upwards through a height of in a time of .
What is the electrical power input to the motor?
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解题
First, calculate the useful work output, which is the gain in gravitational potential energy: .
The useful power output is: .
Using the efficiency formula:
.
评分标准
C is correct. Award 1 mark for the correct calculation of input power.
题目 28 · MCQ
1 分
A metal block of mass is heated by an electric heater rated at . The heater is switched on for , and the temperature of the block rises from to . No thermal energy is lost to the surroundings.
What is the specific heat capacity of the metal?
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解题
First, calculate the total thermal energy supplied by the heater: .
The temperature increase is: .
Using the specific heat capacity formula, : .
评分标准
B is correct. Award 1 mark for the correct specific heat capacity calculation.
题目 29 · MCQ
1 分
A ray of light is travelling inside a transparent plastic block towards the boundary with air. The refractive index of the plastic is .
What is the critical angle for the light in this plastic?
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解题
The critical angle is given by the formula:
where is the refractive index.
.
评分标准
B is correct. Award 1 mark for the correct calculation of the critical angle.
题目 30 · MCQ
1 分
An electric lamp has a constant current of passing through it for a time of .
What is the total charge that passes through the lamp in this time?
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解题
The relationship between charge , current and time is:
First, convert the time from hours to seconds: .
Now, calculate the charge: .
评分标准
C is correct. Award 1 mark for the correct calculation of charge.
题目 31 · MCQ
1 分
A step-up transformer has on its primary coil and on its secondary coil. The primary coil is connected to an alternating current (a.c.) power supply of .
What is the output voltage across the secondary coil?
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解题
The relationship between voltages and number of turns in a transformer is given by:
Rearranging for the secondary voltage : .
评分标准
C is correct. Award 1 mark for the correct transformer voltage calculation.
题目 32 · MCQ
1 分
A distant galaxy is observed to be moving away from Earth at a speed of . The Hubble constant is .
What is the estimated distance from Earth to this galaxy?
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解题
According to Hubble's Law, the recession speed of a galaxy is proportional to its distance:
First, convert the speed from to : .
Now, solve for distance : .
评分标准
C is correct. Award 1 mark for the correct calculation of galaxy distance.
题目 33 · 選擇題
1 分
A vehicle is travelling along a straight road at a velocity of \(24\text{ m/s}\). It then decelerates uniformly at a rate of \(3.0\text{ m/s}^2\) until its velocity decreases to \(12\text{ m/s}\). What distance does the vehicle travel while it is decelerating?
A.\(4.0\text{ m}\)
B.\(36\text{ m}\)
C.\(72\text{ m}\)
D.\(144\text{ m}\)
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解题
We can use the equations of motion for uniform acceleration: \(v^2 = u^2 + 2as\). Here, the initial velocity \(u = 24\text{ m/s}\), the final velocity \(v = 12\text{ m/s}\), and the acceleration (deceleration) \(a = -3.0\text{ m/s}^2\). Substituting these values into the formula gives: \(12^2 = 24^2 + 2(-3.0)s\), which simplifies to \(144 = 576 - 6.0s\). Solving for \(s\): \(6.0s = 576 - 144 = 432 \implies s = 72\text{ m}\).
评分标准
1 mark: Correct calculation of distance to show 72 m.
题目 34 · 選擇題
1 分
A tennis ball of mass \(0.060\text{ kg}\) hits a wall horizontally at \(25\text{ m/s}\) and rebounds in the opposite direction at \(15\text{ m/s}\). The contact time between the ball and the wall is \(0.050\text{ s}\). What is the average force exerted by the wall on the ball?
A.\(12\text{ N}\)
B.\(18\text{ N}\)
C.\(30\text{ N}\)
D.\(48\text{ N}\)
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解题
Using the relationship between impulse and change in momentum: \(F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}\). Taking the initial direction as positive, \(u = 25\text{ m/s}\) and \(v = -15\text{ m/s}\). This gives: \(\Delta p = 0.060 \times (-15 - 25) = -2.4\text{ kg m/s}\). The magnitude of the force is \(F = \frac{2.4\text{ N s}}{0.050\text{ s}} = 48\text{ N}\).
评分标准
1 mark: Correct calculation of average force to show 48 N.
题目 35 · 選擇題
1 分
An electric heater of power \(800\text{ W}\) is used to heat \(2.0\text{ kg}\) of a liquid in a well-insulated container. The temperature of the liquid increases from \(20^\circ\text{C}\) to \(50^\circ\text{C}\) in \(150\text{ s}\). What is the specific heat capacity of the liquid?
A.\(1300\text{ J}/(\text{kg }^\circ\text{C})\)
B.\(2000\text{ J}/(\text{kg }^\circ\text{C})\)
C.\(4000\text{ J}/(\text{kg }^\circ\text{C})\)
D.\(8000\text{ J}/(\text{kg }^\circ\text{C})\)
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解题
First, calculate the thermal energy supplied by the heater: \(E = P \times t = 800\text{ W} \times 150\text{ s} = 120\,000\text{ J}\). The temperature change is \(\Delta \theta = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}\). Using the formula \(E = m c \Delta \theta\), we get \(120\,000\text{ J} = 2.0\text{ kg} \times c \times 30^\circ\text{C}\), which gives \(60 c = 120\,000 \implies c = 2000\text{ J}/(\text{kg }^\circ\text{C})\).
评分标准
1 mark: Correct calculation of specific heat capacity to show 2000 J/(kg °C).
题目 36 · 選擇題
1 分
A ray of light in air is incident on the flat surface of a glass block at an angle of incidence of \(40^\circ\). The refractive index of the glass is \(1.5\). What is the angle of refraction inside the glass block?
A.\(25^\circ\)
B.\(27^\circ\)
C.\(30^\circ\)
D.\(60^\circ\)
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解题
Using Snell's law: \(n = \frac{\sin(i)}{\sin(r)}\), where \(n = 1.5\) and \(i = 40^\circ\). Thus, \(1.5 = \frac{\sin(40^\circ)}{\sin(r)} \implies \sin(r) = \frac{\sin(40^\circ)}{1.5} \approx \frac{0.6428}{1.5} \approx 0.4285\). Taking the inverse sine: \(r = \arcsin(0.4285) \approx 25.4^\circ \approx 25^\circ\).
评分标准
1 mark: Correct calculation of the angle of refraction to show approximately 25°.
题目 37 · 選擇題
1 分
A metal wire of length \(L\) and uniform cross-sectional area \(A\) has a resistance of \(8.0\ \Omega\). A second wire made of the same metal has a length of \(2L\) and a uniform cross-sectional area of \(4A\). What is the resistance of the second wire?
A.\(1.0\ \Omega\)
B.\(4.0\ \Omega\)
C.\(16\ \Omega\)
D.\(64\ \Omega\)
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解题
The resistance of a wire is given by \(R = \rho \frac{L}{A}\). For the second wire: \(R_2 = \rho \frac{2L}{4A} = \frac{1}{2} \left(\rho \frac{L}{A}\right) = \frac{1}{2} R\). Given that \(R = 8.0\ \Omega\), the resistance of the second wire is \(R_2 = 0.5 \times 8.0\ \Omega = 4.0\ \Omega\).
评分标准
1 mark: Correct calculation of resistance to show 4.0 Ω.
题目 38 · 選擇題
1 分
Two resistors, one of \(4.0\ \Omega\) and one of \(6.0\ \Omega\), are connected in parallel. This parallel combination is then connected in series with a \(3.6\ \Omega\) resistor and a \(12\text{ V}\) power supply. What is the current drawn from the power supply?
A.\(0.87\text{ A}\)
B.\(1.2\text{ A}\)
C.\(2.0\text{ A}\)
D.\(3.3\text{ A}\)
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解题
First, calculate the equivalent resistance of the parallel pair: \(R_p = \frac{4.0 \times 6.0}{4.0 + 6.0} = \frac{24}{10} = 2.4\ \Omega\). Next, add the series resistor to find the total resistance of the circuit: \(R_{\text{total}} = R_p + 3.6\ \Omega = 2.4\ \Omega + 3.6\ \Omega = 6.0\ \Omega\). Finally, use Ohm's law to find the total current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).
评分标准
1 mark: Correct calculation of the current drawn from the supply to show 2.0 A.
题目 39 · 選擇題
1 分
An ideal step-down transformer has \(400\) turns on its primary coil and \(100\) turns on its secondary coil. The primary coil is connected to a \(240\text{ V}\) a.c. supply and draws a current of \(0.50\text{ A}\). What are the output voltage and current in the secondary circuit?
A.\(60\text{ V}\) and \(0.125\text{ A}\)
B.\(60\text{ V}\) and \(2.0\text{ A}\)
C.\(960\text{ V}\) and \(0.125\text{ A}\)
D.\(960\text{ V}\) and \(2.0\text{ A}\)
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解题
Using the transformer equation for voltage: \(\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 240\text{ V} \times \frac{100}{400} = 60\text{ V}\). Since it is an ideal transformer, input power equals output power: \(V_p I_p = V_s I_s \implies 240\text{ V} \times 0.50\text{ A} = 60\text{ V} \times I_s \implies I_s = 2.0\text{ A}\).
评分标准
1 mark: Correct determination of both voltage (60 V) and current (2.0 A).
题目 40 · 選擇題
1 分
Light of wavelength \(600\text{ nm}\) emitted by a stationary source is observed to have a wavelength of \(612\text{ nm}\) when detected on Earth from a distant galaxy. The speed of light is \(3.0 \times 10^8\text{ m/s}\). What is the recessional speed of the galaxy?
A.\(2.0 \times 10^6\text{ m/s}\)
B.\(6.0 \times 10^6\text{ m/s}\)
C.\(1.5 \times 10^7\text{ m/s}\)
D.\(3.0 \times 10^7\text{ m/s}\)
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解题
Using the redshift equation: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\), where \(\Delta \lambda = 612\text{ nm} - 600\text{ nm} = 12\text{ nm}\), \(\lambda_0 = 600\text{ nm}\), and \(c = 3.0 \times 10^8\text{ m/s}\). This gives: \(\frac{12}{600} = \frac{v}{3.0 \times 10^8} \implies 0.020 = \frac{v}{3.0 \times 10^8} \implies v = 0.020 \times 3.0 \times 10^8 = 6.0 \times 10^6\text{ m/s}\).
评分标准
1 mark: Correct calculation of the recessional speed to show 6.0 * 10^6 m/s.
Ten structured questions. Show all working and use appropriate units.
10 题目 · 80 分
题目 1 · 結構題
8 分
A skydiver jumps from an airplane.
(a) State the name of the two vertical forces acting on the skydiver as they fall through the air. [2]
(b) At one instant, the skydiver of mass \(75\text{ kg}\) experiences an upward air resistance force of \(220\text{ N}\). Calculate the acceleration of the skydiver at this instant. Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\). [3]
(c) Explain, in terms of forces and acceleration, why the skydiver eventually reaches a constant speed (terminal velocity). [3]
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解题
(a) The two vertical forces are: 1. Weight (or gravitational force acting downwards). 2. Air resistance (or drag force acting upwards).
(b) First, calculate the downward weight \(W\) of the skydiver: \[W = m \times g = 75\text{ kg} \times 9.8\text{ m/s}^2 = 735\text{ N}\]
Next, calculate the resultant force \(F\) acting on the skydiver: \[F = \text{Weight} - \text{Air resistance} = 735\text{ N} - 220\text{ N} = 515\text{ N}\text{ (downwards)}\]
Using Newton's second law, \(F = ma\): \[a = \frac{F}{m} = \frac{515\text{ N}}{75\text{ kg}} \approx 6.87\text{ m/s}^2\] Rounding to two significant figures gives \(6.9\text{ m/s}^2\).
(c) As the skydiver's speed increases, the upward air resistance force also increases. This reduces the resultant downward force acting on the skydiver, which in turn decreases their acceleration. Eventually, the upward air resistance increases to a value equal to the downward weight of the skydiver. At this point, the resultant force becomes zero, the acceleration becomes zero, and the skydiver continues to fall at a constant terminal velocity.
评分标准
(a) - B1: Weight (or gravity / force of gravity) acting downwards. - B1: Air resistance (or drag) acting upwards.
(b) - C1: Calculation of weight: \(75 \times 9.8 = 735\text{ N}\). - C1: Calculation of resultant force: \(735 - 220 = 515\text{ N}\). - A1: Correct calculation of acceleration with units: \(6.9\text{ m/s}^2\) (or \(6.87\text{ m/s}^2\)).
(c) - B1: State that air resistance increases as speed increases. - B1: State that the resultant force (and thus acceleration) decreases. - B1: Explain that terminal velocity is reached when air resistance equals weight, resulting in zero resultant force and zero acceleration.
题目 2 · 結構題
8 分
An electric pump is used to lift water from a well.
(a) The pump lifts \(150\text{ kg}\) of water vertically through a height of \(12\text{ m}\) in a time of \(5.0\text{ s}\). Calculate the gravitational potential energy gained by the water. Take \(g = 9.8\text{ m/s}^2\). [2]
(b) Calculate the useful power output of the pump. [2]
(c) The electrical power input to the pump is \(5.5\text{ kW}\). Calculate the efficiency of the pump. [2]
(d) State what happens to the energy supplied to the pump that is not transferred usefully to the water. [2]
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解题
(a) The gravitational potential energy \(\Delta E_p\) is calculated as: \[\Delta E_p = mgh = 150\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 17640\text{ J}\]
(b) Useful power output \(P\) is: \[P = \frac{\Delta E_p}{t} = \frac{17640\text{ J}}{5.0\text{ s}} = 3528\text{ W}\text{ (or }3.53\text{ kW)}\]
(c) First, express the input power in watts: \(P_{\text{in}} = 5.5\text{ kW} = 5500\text{ W}\). Efficiency is: \[ \text{Efficiency} = \frac{\text{Useful power output}}{\text{Power input}} \times 100\% = \frac{3528}{5500} \times 100\% \approx 64.1\% \] Rounding to two significant figures gives \(64\%\).
(d) The wasted energy is transferred to the surroundings as thermal energy (heat) and sound energy, primarily due to friction in the motor and moving parts of the pump.
评分标准
(a) - C1: Formula \(mgh\) or substitution \(150 \times 9.8 \times 12\). - A1: Correct value with units: \(17640\text{ J}\) (or \(1.8 \times 10^4\text{ J}\)).
(b) - C1: Formula \(P = E/t\) or substitution \(17640 / 5.0\). - A1: Correct power with units: \(3528\text{ W}\) (or \(3500\text{ W}\) or \(3.5\text{ kW}\)).
(c) - C1: Recall of efficiency formula and conversion of \(5.5\text{ kW}\) to \(5500\text{ W}\). - A1: Correct efficiency value: \(64\%\) (or \(64.1\%\) or \(0.64\)).
(d) - B1: Identifies that energy is wasted as thermal energy (or heat). - B1: Explains that this energy is dissipated / transferred to the surroundings (or the pump itself heats up).
题目 3 · 結構題
8 分
A ray of light traveling in air is incident on a flat boundary of a transparent glass block.
(a) State what is meant by the *critical angle* for light in a medium. [2]
(b) The ray of light is incident on the glass block at an angle of incidence of \(42.0^\circ\). The refractive index of the glass is \(1.52\).
(i) Calculate the angle of refraction inside the glass block. [3]
(ii) Calculate the critical angle for the glass-air boundary. [3]
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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)(ii) The formula relating critical angle \(c\) and refractive index \(n\) is: \[\sin c = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579\] \[c = \sin^{-1}(0.6579) \approx 41.1^\circ\]
评分标准
(a) - B1: Angle of incidence in the optically denser medium. - B1: The angle of refraction is \(90^\circ\) (or light travels along the boundary).
(b)(i) - C1: Recall of Snell's law: \(n = \sin i / \sin r\). - C1: Correct substitution: \(\sin r = \sin 42.0^\circ / 1.52\). - A1: Angle of refraction calculated correctly: \(26.1^\circ\) (accept \(26^\circ\)).
(b)(ii) - C1: Recall of formula: \(\sin c = 1/n\). - C1: Correct substitution: \(\sin c = 1 / 1.52\). - A1: Critical angle calculated correctly with units: \(41.1^\circ\) (accept \(41^\circ\)).
题目 4 · 結構題
8 分
A circuit contains a battery and a single resistor.
(a) Define the term *electromotive force (e.m.f.)* of a source. [2]
(b) A battery of e.m.f. \(12.0\text{ V}\) is connected in series with a resistor of resistance \(15.0\ \Omega\).
(i) Calculate the current in the resistor. [2]
(ii) Calculate the total charge that passes through the resistor in \(4.0\text{ minutes}\). [2]
(iii) Calculate the energy transferred by the battery to the charge in this time. [2]
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解题
(a) Electromotive force (e.m.f.) is defined as the electrical work done by a source in moving a unit charge around a complete circuit.
(b)(ii) The time in seconds is \(t = 4.0 \times 60 = 240\text{ s}\). The charge \(Q\) is: \[Q = I \times t = 0.80\text{ A} \times 240\text{ s} = 192\text{ C}\]
(b)(iii) The energy transferred \(E\) is: \[E = V \times Q = 12.0\text{ V} \times 192\text{ C} = 2304\text{ J}\] Rounding to two significant figures gives \(2300\text{ J}\) (or \(2.3\text{ kJ}\)).
评分标准
(a) - B1: Work done (or energy transferred) per unit charge by a source. - B1: In moving charge around a complete circuit.
(b)(i) - C1: Formula \(I = V/R\) or substitution \(12.0 / 15.0\). - A1: Correct current with units: \(0.80\text{ A}\).
(b)(ii) - C1: Convert minutes to seconds: \(4.0\text{ min} = 240\text{ s}\) or formula \(Q = It\). - A1: Correct charge with units: \(192\text{ C}\).
(b)(iii) - C1: Formula \(E = VQ\) or \(E = VIt\). - A1: Correct energy with units: \(2300\text{ J}\) (or \(2304\text{ J}\)).
题目 5 · 結構題
8 分
An electric heater is used to heat a block of copper.
(a) State what is meant by the *specific heat capacity* of a substance. [2]
(b) An electric heater of power \(800\text{ W}\) is used to heat a copper block of mass \(2.5\text{ kg}\). The specific heat capacity of copper is \(385\text{ J/(kg}^\circ\text{C)} . (i) Calculate the thermal energy supplied by the heater in \)3.0\text{ minutes}\). [2]
(ii) Calculate the temperature rise of the copper block, assuming no thermal energy is lost to the surroundings. [2]
(iii) In practice, the actual temperature rise of the block is less than the value calculated in (ii). Explain why. [2]
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解题
(a) Specific heat capacity is the thermal energy required to raise the temperature of \(1\text{ kg}\) of a substance by \(1^\circ\text{C}\).
(b)(i) First, convert time to seconds: \(t = 3.0 \times 60 = 180\text{ s}\). The thermal energy \(E\) supplied is: \[E = P \times t = 800\text{ W} \times 180\text{ s} = 144000\text{ J}\text{ (or }1.44 \times 10^5\text{ J)}\]
(b)(ii) Using \(\Delta E = mc\Delta \theta\): \[144000\text{ J} = 2.5\text{ kg} \times 385\text{ J/(kg}^\circ\text{C)} \times \Delta \theta\] \[ \Delta \theta = \frac{144000}{962.5} \approx 149.6^\circ\text{C} \] Rounding to two significant figures gives \(150^\circ\text{C}\).
(b)(iii) In practice, some thermal energy is lost to the surrounding air and table by conduction, convection, and radiation. Consequently, not all the energy supplied by the heater goes into raising the temperature of the copper block.
评分标准
(a) - B1: Thermal energy / heat required per unit mass (or per \(1\text{ kg}\)). - B1: To raise the temperature by \(1^\circ\text{C}\) (or \(1\text{ K}\)).
(b)(i) - C1: Convert minutes to seconds: \(3.0\text{ min} = 180\text{ s}\) or formula \(E = Pt\). - A1: Correct energy calculation: \(1.44 \times 10^5\text{ J}\) (or \(144000\text{ J}\)).
(b)(ii) - C1: Formula \(\Delta E = mc\Delta \theta\) or rearrangement. - A1: Correct temperature rise with units: \(150^\circ\text{C}\) (or \(149.6^\circ\text{C}\)).
(b)(iii) - B1: States that thermal energy is lost to the surroundings. - B1: Mentions a mechanism of heat transfer (conduction, convection, or radiation) or thermal energy is absorbed by the heater casing.
题目 6 · 結構題
8 分
Transformers are widely used in the distribution of electrical power.
(a) Explain the principle of operation of a basic transformer. [3]
(b) A step-down transformer has a primary coil with \(1200\text{ turns}\) and a secondary coil with \(50\text{ turns}\). The primary voltage is \(240\text{ V}\) a.c.
(i) Calculate the secondary voltage. [2]
(ii) The secondary current is \(4.0\text{ A}\). Assuming the transformer is \(100\%\) efficient, calculate the primary current. [3]
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解题
(a) An alternating current (a.c.) in the primary coil creates a continuously changing magnetic field in the soft iron core. This changing magnetic field is guided by the core through the secondary coil. The changing magnetic field linking with the secondary coil induces an alternating electromotive force (e.m.f.) / voltage across the secondary coil.
(b)(ii) Since the transformer is \(100\%\) efficient, the input power equals the output power: \[P_{\text{in}} = P_{\text{out}} \implies V_p I_p = V_s I_s\] \[240\text{ V} \times I_p = 10\text{ V} \times 4.0\text{ A}\] \[I_p = \frac{40}{240} \approx 0.167\text{ A}\] Rounding to two significant figures gives \(0.17\text{ A}\).
评分标准
(a) - B1: Alternating current in primary coil creates a changing/alternating magnetic field. - B1: Core guides/links the changing magnetic field to the secondary coil. - B1: Changing magnetic field induces an alternating e.m.f. (or voltage) in the secondary coil.
(b)(ii) - C1: Recall of power equation for ideal transformer: \(V_p I_p = V_s I_s\). - C1: Correct substitution: \(240 \times I_p = 10 \times 4.0\). - A1: Correct current calculation with units: \(0.17\text{ A}\) (or \(0.167\text{ A}\)).
题目 7 · 結構題
8 分
Astronomers observe the light from distant galaxies to study the Universe.
(a) State what is meant by *redshift* of light from distant galaxies. [2]
(b) Explain how the observation of redshift provides evidence that the Universe is expanding. [3]
(c) 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. Take 1 year = \(3.16 \times 10^7\text{ s}\). [3]
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解题
(a) Redshift is the observed increase in the wavelength (or decrease in frequency) of light emitted from distant galaxies as they move away from the observer.
(b) When light from distant galaxies is analysed, almost all of it is redshifted, indicating that these galaxies are moving away from us. Furthermore, galaxies that are further away show a greater redshift, meaning they are moving away at a faster speed. This proportional relationship between distance and recession speed implies that the space itself between galaxies is expanding.
(c) The age of the Universe \(t\) is approximately given by the reciprocal of the Hubble constant: \[t \approx \frac{1}{H_0} = \frac{1}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 4.545 \times 10^{17}\text{ s}\]
To convert this time into years: \[t\text{ (years)} = \frac{4.545 \times 10^{17}\text{ s}}{3.16 \times 10^7\text{ s/year}} \approx 1.44 \times 10^{10}\text{ years}\] Rounding to two significant figures gives \(1.4 \times 10^{10}\text{ years}\) (or 14 billion years).
评分标准
(a) - B1: Observed increase in wavelength / shift toward red end of spectrum. - B1: Caused by distant light source moving away from the observer.
(b) - B1: Redshift shows that distant galaxies are moving away (receding) from us. - B1: Further galaxies have a larger redshift, meaning they recede faster. - B1: This indicates that the Universe / space itself is expanding.
(c) - C1: Recall of relation \(t \approx 1/H_0\). - C1: Conversion from seconds to years by dividing by \(3.16 \times 10^7\). - A1: Age calculated correctly with units: \(1.4 \times 10^{10}\text{ years}\) (accept range \(1.4 - 1.5 \times 10^{10}\text{ years}\)).
题目 8 · 結構題
8 分
A student carries out an experiment to measure the speed of sound.
(a) Describe the difference between longitudinal waves and transverse waves. [2]
(b) The student stands at a distance of \(150\text{ m}\) from a tall, vertical wall. The student claps their hands once and hears the echo after \(0.88\text{ s}\).
(i) Calculate the speed of sound in air from these measurements. [3]
(ii) The frequency of the sound wave produced is \(440\text{ Hz}\). Calculate the wavelength of this sound wave. [3]
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解题
(a) In longitudinal waves (such as sound waves), the vibrations of the particles are parallel to the direction of wave travel. In transverse waves (such as light waves), the vibrations are perpendicular to the direction of wave travel.
(b)(i) The sound travels to the wall and back to the student, so the total distance travelled \(d\) is: \[d = 2 \times 150\text{ m} = 300\text{ m}\] The speed \(v\) of sound is: \[v = \frac{d}{t} = \frac{300\text{ m}}{0.88\text{ s}} \approx 340.9\text{ m/s}\] Rounding to three significant figures gives \(341\text{ m/s}\).
(b)(ii) Using the wave equation \(v = f \lambda\): \[340.9\text{ m/s} = 440\text{ Hz} \times \lambda\] \[\lambda = \frac{340.9}{440} \approx 0.775\text{ m}\]
评分标准
(a) - B1: Longitudinal: vibrations are parallel to direction of energy transfer / wave travel. - B1: Transverse: vibrations are perpendicular to direction of energy transfer / wave travel.
(b)(i) - C1: Recognises that sound travels double distance: \(2 \times 150 = 300\text{ m}\). - C1: Use of \(v = d/t\) or substitution \(300 / 0.88\). - A1: Correct speed calculation with units: \(341\text{ m/s}\) (accept range \(340\text{ to }341\text{ m/s}\)).
(b)(ii) - C1: Recall of formula \(v = f\lambda\). - C1: Correct rearrangement or substitution: \(\lambda = 341 / 440\) (allow ecf from (b)(i)). - A1: Correct wavelength with units: \(0.775\text{ m}\) (accept \(0.77\text{ m}\) or \(0.78\text{ m}\)).
题目 9 · 結構題
8 分
A ray of monochromatic light is incident on the flat face of a semi-circular glass block. The angle of incidence in air is \(38^\circ\). The refractive index of the glass is \(1.52\).
(a) Calculate the angle of refraction inside the glass block.
(b) The angle of incidence at the flat face is increased to \(45^\circ\). Explain what happens to the ray of light. Show any necessary calculation to support your answer.
(c) Calculate the speed of light inside the glass block. The speed of light in a vacuum is \(3.0 \times 10^8\text{ m/s}\).
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解题
(a) Using Snell's Law for light entering the glass block: \(n = \frac{\sin(i)}{\sin(r)}\) \(1.52 = \frac{\sin(38^\circ)}{\sin(r)}\) \(\sin(r) = \frac{\sin(38^\circ)}{1.52} = \frac{0.6157}{1.52} \approx 0.4050\) \(r = \arcsin(0.4050) \approx 23.9^\circ\) (or \(24^\circ\)).
(b) First, calculate the critical angle \(c\) for the glass-air boundary: \(\sin(c) = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579\) \(c = \arcsin(0.6579) \approx 41.1^\circ\). Since the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41.1^\circ\)), the light ray undergoes total internal reflection at the flat face instead of refracting out of the block.
(c) Using the definition of refractive index: \(n = \frac{v_{\text{vacuum}}}{v_{\text{glass}}}\) \(1.52 = \frac{3.0 \times 10^8}{v_{\text{glass}}}\) \(v_{\text{glass}} = \frac{3.0 \times 10^8}{1.52} \approx 1.97 \times 10^8\text{ m/s}\) (or \(2.0 \times 10^8\text{ m/s}\)).
评分标准
(a) - C1: State or use \(n = \frac{\sin(i)}{\sin(r)}\) - C1: Correct substitution: \(\sin(r) = \frac{\sin(38^\circ)}{1.52}\) - A1: \(24^\circ\) (accept \(23.9^\circ\))
(b) - C1: State or use \(\sin(c) = \frac{1}{n}\) - A1: Calculate critical angle \(c = 41^\circ\) or \(41.1^\circ\) - B1: Total internal reflection stated and justified because the angle of incidence (\(45^\circ\)) is greater than the critical angle
(c) - C1: State or use \(n = \frac{c}{v}\) or rearranged - A1: \(2.0 \times 10^8\text{ m/s}\) or \(1.97 \times 10^8\text{ m/s}\)
题目 10 · 結構題
8 分
A toy car of mass \(0.80\text{ kg}\) is released from rest at the top of a rough slope. The vertical height of the slope is \(1.5\text{ m}\).
(a) Calculate the gravitational potential energy lost by the car. (Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\))
(b) At the bottom of the slope, the speed of the toy car is \(4.2\text{ m/s}\). (i) Calculate the kinetic energy of the car at the bottom of the slope. (ii) Determine the energy dissipated as thermal energy and sound during the motion down the slope.
(c) The length of the slope is \(2.5\text{ m}\). Calculate the average friction force acting on the car as it travels down the slope.
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解题
(a) Gravitational potential energy lost: \(\Delta E_p = mgh\) \(\Delta E_p = 0.80 \times 9.8 \times 1.5 = 11.76\text{ J}\) (rounds to \(11.8\text{ J}\) or \(12\text{ J}\)).
(ii) Energy dissipated is the difference between potential energy lost and kinetic energy gained: \(E_{\text{dissipated}} = \Delta E_p - E_k\) \(E_{\text{dissipated}} = 11.76 - 7.056 = 4.704\text{ J}\) (rounds to \(4.7\text{ J}\)).
(c) Work done against friction is equal to the dissipated energy: \(W = F \times d\) \(4.704 = F \times 2.5\) \(F = \frac{4.704}{2.5} \approx 1.88\text{ N}\) (rounds to \(1.9\text{ N}\)).
评分标准
(a) - C1: State or use \(\Delta E_p = mgh\) - A1: \(12\text{ J}\) or \(11.8\text{ J}\) (accept \(11.76\text{ J}\))
(b)(i) - C1: State or use \(E_k = \frac{1}{2}mv^2\) - A1: \(7.1\text{ J}\) (accept \(7.06\text{ J}\) or \(7.056\text{ J}\))
(c) - C1: State or use \(W = Fd\) or \(F = \frac{\text{Energy}}{d}\) - A1: \(1.9\text{ N}\) or \(1.88\text{ N}\) (allow ecf from b(ii))
Paper 6 (Alternative to Practical)
Four practical questions covering measurement, circuit configuration, optical ray-traces, and planning.
4 题目 · 40 分
题目 1 · Practical Investigation
10 分
A student investigates the balancing of a metre rule to determine the mass of an unknown object M. Fig. 1.1 shows a metre rule balanced on a pivot at the 50.0 cm mark. An object of unknown mass M is placed at a distance d on one side, and a standard mass m = 100 g is positioned on the other side at distance x to balance the rule. (a) (i) The student records the value of d = 20.0 cm. State the value of d in metres. (ii) The student records the balancing distance x = 24.5 cm. State the value of x in metres. (b) Use the principle of moments to calculate the mass M, using the equation: M = (m * x) / d. Show your working. (c) The student repeats the experiment for several values of d. The values of d and the corresponding balancing distances x are shown in Table 1.1. Table 1.1: [d / cm: 15.0, 20.0, 25.0, 30.0, 35.0] [x / cm: 18.4, 24.5, 30.6, 36.8, 42.9]. Describe the relationship between d and x. Justify your answer with reference to the data. (d) List two precautions that the student should take to obtain accurate results in this experiment. (e) State one reason why it can be difficult to find the exact balance point. (f) The metre rule has a uniform cross-section. State the position of its centre of mass.
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解题
(a) (i) d = 20.0 / 100 = 0.200 m. (ii) x = 24.5 / 100 = 0.245 m. (b) M = (100 * 24.5) / 20.0 = 122.5 g. (c) Calculating the ratio x/d for each pair: 18.4/15.0 = 1.23, 24.5/20.0 = 1.23, 30.6/25.0 = 1.22, 36.8/30.0 = 1.23, 42.9/35.0 = 1.23. Since this ratio is constant, x is directly proportional to d. (d) To improve accuracy, the pivot must be exactly perpendicular to the rule, and the student must look vertically downwards at the scale to avoid parallax error. (e) Friction between the rule and the pivot opposes rotation, creating a range of positions where the rule appears balanced. (f) For a uniform rule, the centre of mass is at its geometric center, which is the 50.0 cm mark.
评分标准
a(i) 0.200 m [1] a(ii) 0.245 m [1] (b) Substitution: (100 * 24.5) / 20.0 [1], Correct calculation: 122.5 g [1] (c) Correct description: x increases as d increases [1], Justification: ratio x/d is constant at 1.23 [1] (d) Any two valid precautions [2] (e) Friction at pivot / width of pivot [1] (f) 50.0 cm mark [1]
题目 2 · Practical Investigation
10 分
A student investigates the resistance of various combinations of resistors. (a) A voltmeter connected across a resistor shows a reading. The voltmeter has a scale from 0 to 5.0 V with 10 major divisions (every 0.5 V) and each major division is divided into 5 small subdivisions. The pointer is on the 4th subdivision after the 2.0 V mark. State the voltmeter reading, V. (b) An ammeter connected in the circuit has a scale from 0 to 1.0 A with 5 major divisions (every 0.2 A) and each major division has 10 small subdivisions. The pointer is on the 8th subdivision after the 0.2 A mark. State the ammeter reading, I. (c) Calculate the resistance R1 of the resistor using the values from (a) and (b) and the equation R1 = V / I. Include the unit. (d) The student connects a second identical resistor in parallel with the first. The new potential difference is Vp = 2.4 V and the new current is Ip = 0.72 A. (i) Calculate the combined resistance Rp of the parallel combination. (ii) State how Rp compares to R1. (e) Describe how you would draw a circuit diagram showing the cell, a switch, two resistors in parallel, an ammeter measuring the total current, and a voltmeter measuring the potential difference across the parallel resistors. (f) The student notices that the resistors become warm during the experiment. State how this might affect the accuracy of the resistance calculation.
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解题
(a) Each subdivision on the voltmeter represents 0.5 V / 5 = 0.1 V. Pointer is on 2.0 V + (4 * 0.1 V) = 2.4 V. (b) Each subdivision on the ammeter represents 0.2 A / 10 = 0.02 A. Pointer is on 0.2 A + (8 * 0.02 A) = 0.36 A. (c) R1 = 2.4 / 0.36 = 6.67 ohms. (d) (i) Rp = 2.4 / 0.72 = 3.33 ohms. (ii) Rp is half of R1. (e) Connect cell, switch, and ammeter in series, then branch the wire into two parallel loops containing the resistors, with a voltmeter wired in parallel across both resistors. (f) Increased temperature increases the resistance of metal wire conductors, leading to higher-than-expected values.
评分标准
(a) 2.4 V [1] (b) 0.36 A [1] (c) Correct calculation: 6.67 (6.7) [1], Unit: ohms [1] (d)(i) Rp = 3.33 ohms [1] (ii) Rp is half of R1 [1] (e) Correct connections described: parallel resistors [1], ammeter in series [1], voltmeter in parallel [1] (f) Temperature rise increases resistance [1]
题目 3 · Practical Investigation
10 分
A student investigates the refraction of light through a transparent rectangular block. (a) The student draws a line to represent the incident ray making an angle of incidence i with the normal. The angle measured using a protractor is 40 degrees. State the angle of incidence, i. (b) The ray enters the block at point P. The student places two pins, P1 and P2, on the incident ray. State the minimum recommended distance between pins P1 and P2, and explain why this distance is necessary. (c) The student views the images of pins P1 and P2 through the opposite side of the block, and places two more pins, P3 and P4, so that they appear to line up with the images. After removing the block, the angle of refraction r inside the block is measured as 25 degrees. (i) State the angle of refraction, r. (ii) Calculate the refractive index n of the block using the formula n = sin(i) / sin(r). (d) Suggest two techniques the student must use when placing the pins to ensure an accurate ray-trace. (e) The student repeats the experiment using a block made of a different material. The angle of refraction is found to be 28 degrees for the same angle of incidence of 40 degrees. State and explain what this indicates about the optical density of the second material compared to the first.
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解题
(a) i = 40 degrees. (b) Pins must be separated by at least 5.0 cm so that minor errors in placing the pins result in negligible angular deviations of the drawn ray. (c) (i) r = 25 degrees. (ii) n = sin(40) / sin(25) = 0.6428 / 0.4226 = 1.52. (d) Ensuring pins are vertical allows more precise alignment. Viewing the bases of the pins avoids errors from bent pins. (e) A larger angle of refraction (28 degrees > 25 degrees) shows that the light bent less upon entering the second medium. Since the bending is less, the second medium is less optically dense.
评分标准
(a) 40 degrees [1] (b) Minimum separation 5.0 cm [1], Explanation: reduces angular error [1] (c)(i) 25 degrees [1] (ii) Correct substitution: sin(40)/sin(25) [1], Correct calculation: 1.52 [1] (d) Any two techniques: view bases of pins / ensure vertical pins / use thin pins [2] (e) Second material has lower optical density [1], Explanation: larger angle of refraction / bends less [1]
题目 4 · Practical Investigation
10 分
Plan an experiment to investigate how the rate of heat loss from a beaker of hot water depends on the surface area of the exposed water. You are provided with: identical beakers, lids with circular holes of different diameters, a supply of hot water, a thermometer, and a stop-watch. In your plan, you should: (a) list any additional apparatus required, (b) explain briefly how you would carry out the investigation, including the measurements you would take, (c) state the key variables to be kept constant, (d) draw a table with column headings to show how you would display your readings (you do not need to enter any data), (e) explain how you would use your results to reach a conclusion.
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解题
(a) A measuring cylinder is required to measure equal volumes of hot water. (b) Fill the beakers with hot water, add lids with different hole sizes. Take the start temperature, start the clock, and record temperature over time. (c) Control variables include the volume of water, initial temperature of the water, the brand/type of beaker used, and surrounding room temperature. (d) Columns in the table should include: Hole Diameter / cm, Time / min, and Temperature / °C. (e) The rate of cooling is calculated as (Initial Temp - Final Temp) / time. A greater rate of cooling for larger hole diameters confirms that higher surface area increases heat loss.
评分标准
(a) Measuring cylinder [1] (b) Pour equal volumes of hot water [1], Record temperature at regular time intervals [1], Repeat for different hole diameters [1] (c) Any two constant variables: initial temperature / room temperature / volume of water / beaker type [2] (d) Columns in table with appropriate units: Diameter / cm, Time / min (or s), Temperature / °C [2] (e) Calculate temperature drop / rate of cooling [1], Compare cooling rates for different surface areas to reach a conclusion [1]
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