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A Level H1 Physics Practice Paper 2
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Questions
TuitionGoWhere Practice Paper - Physics H1 A-Level
TuitionGoWhere Exam Practice (AI)
Subject: Physics H1 (8867) Level: A-Level Paper: Practice Paper 2 (Structured & Free Response) Version: 2 of 5 Duration: 2 hours Total Marks: 80
Name: _________________________ Class: _________________________ Date: _________________________
Instructions to Candidates
- This paper consists of two sections: Section A and Section B.
- Answer all questions in Section A.
- Answer any two questions in Section B.
- Write your answers in the spaces provided.
- Show all working clearly. Marks are awarded for method as well as final answers.
- You may use a scientific calculator.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- Assume g = 9.81 m s⁻² unless otherwise stated.
Section A: Structured Questions (60 marks)
Answer all questions in this section.
Question 1: Kinematics and Projectile Motion
A stone is projected horizontally from the top of a vertical cliff of height 45.0 m above the sea. The initial speed of the stone is 20.0 m s⁻¹. Air resistance may be neglected.
(a) Calculate the time taken for the stone to reach the sea. [2]
(b) Calculate the horizontal distance from the base of the cliff at which the stone enters the sea. [1]
(c) Calculate the magnitude of the velocity of the stone just before it hits the sea. [3]
(d) On the axes below, sketch the variation of the vertical component of the stone's velocity with time from the moment of projection until it reaches the sea. Label any significant values on the axes. [2]
Vertical velocity / m s⁻¹
^
|
|
|
|
|
+---------------------------> Time / s
|
|
|
|
|
Question 2: Forces and Equilibrium
A uniform ladder of length 5.00 m and weight 250 N rests against a smooth vertical wall. The foot of the ladder rests on rough horizontal ground. The ladder makes an angle of 60° with the ground.
(a) Draw a free-body diagram showing all the forces acting on the ladder. Label each force clearly. [3]
(b) By taking moments about the foot of the ladder, calculate the magnitude of the reaction force exerted by the wall on the ladder. [3]
(c) Calculate the magnitude of the frictional force between the ladder and the ground. [2]
(d) State the minimum coefficient of static friction required to prevent the ladder from slipping. [2]
Question 3: Work, Energy, and Power
A car of mass 1200 kg accelerates uniformly from rest to a speed of 25.0 m s⁻¹ in 8.00 s along a straight horizontal road. The total resistive force acting on the car is constant at 600 N.
(a) Calculate the acceleration of the car. [1]
(b) Calculate the distance travelled by the car during the 8.00 s. [2]
(c) Calculate the work done by the engine of the car during the acceleration. [3]
(d) Calculate the average power developed by the engine during the acceleration. [2]
(e) At the instant the car reaches 25.0 m s⁻¹, the engine continues to provide the same driving force. Explain why the car will not continue to accelerate indefinitely. [2]
Question 4: Momentum and Collisions
A trolley A of mass 2.00 kg moves with a velocity of 3.00 m s⁻¹ to the right on a smooth horizontal track. It collides with a stationary trolley B of mass 1.00 kg. After the collision, trolley A moves with a velocity of 1.00 m s⁻¹ to the right.
(a) State the principle of conservation of linear momentum. [2]
(b) Calculate the velocity of trolley B after the collision. [2]
(c) Show that the collision is elastic. [3]
(d) The two trolleys are fitted with magnets so that they repel each other without making physical contact. Explain, with reference to Newton's laws, how momentum is still conserved in this interaction. [3]
Question 5: Current Electricity
A battery of e.m.f. 12.0 V and internal resistance 0.500 Ω is connected to an external resistor of resistance 5.50 Ω.
(a) Calculate the current in the circuit. [2]
(b) Calculate the terminal potential difference across the battery. [1]
(c) Calculate the power dissipated in the external resistor. [2]
(d) Calculate the power wasted in the internal resistance of the battery. [1]
(e) Explain, using the concept of a potential divider, why the terminal potential difference is less than the e.m.f. of the battery. [2]
Question 6: D.C. Circuits
A potential divider circuit consists of two resistors, R₁ = 4.00 kΩ and R₂ = 6.00 kΩ, connected in series across a 10.0 V supply of negligible internal resistance.
(a) Draw the circuit diagram for this potential divider. [2]
(b) Calculate the output voltage across R₂. [2]
(c) A load resistor of resistance 3.00 kΩ is now connected across R₂. Calculate the new output voltage across the load resistor. [4]
(d) Explain why the output voltage changes when the load resistor is connected. [2]
Question 7: Nuclear Physics
The isotope carbon-14 (¹⁴₆C) is radioactive and decays by beta-minus emission.
(a) Write the nuclear equation for the decay of carbon-14. [2]
(b) The half-life of carbon-14 is 5730 years. A sample of ancient wood contains 25% of the carbon-14 activity found in living wood. Calculate the age of the sample. [3]
(c) Explain why carbon-14 dating is not suitable for samples older than about 50,000 years. [2]
Section B: Free Response Questions (20 marks)
Answer any two questions from this section. Each question carries 10 marks.
Question 8: Mechanics – Energy and Momentum
A ball of mass 0.500 kg is dropped from rest from a height of 10.0 m above the ground. Air resistance may be neglected.
(a) Calculate the speed of the ball just before it hits the ground. [2]
(b) The ball rebounds vertically with a speed of 8.00 m s⁻¹. Calculate the impulse exerted on the ball by the ground during the impact. [3]
(c) The impact lasts for 0.0500 s. Calculate the average force exerted on the ball by the ground. [2]
(d) Calculate the maximum height reached by the ball after the rebound. [2]
(e) State and explain one reason why the ball does not reach its original height. [1]
Question 9: Electricity – Circuit Analysis
A student investigates a circuit containing a thermistor and a fixed resistor.
The thermistor has resistance R_T that varies with temperature θ according to the relationship:
R_T = R₀ e^(k/θ)
where R₀ = 2.00 Ω and k = 3000 K. θ is the absolute temperature in kelvin.
The thermistor is connected in series with a fixed resistor of 100 Ω and a 6.00 V battery of negligible internal resistance.
(a) Calculate the resistance of the thermistor at 300 K. [2]
(b) Calculate the current in the circuit at 300 K. [2]
(c) Calculate the potential difference across the thermistor at 300 K. [1]
(d) The temperature of the thermistor increases to 350 K. Without performing a full calculation, explain how the current in the circuit and the potential difference across the thermistor will change. [3]
(e) Suggest one practical application of a thermistor in a potential divider circuit. [2]
Question 10: Waves and Photoelectric Effect
Light of wavelength 450 nm is incident on a clean metal surface. The work function of the metal is 2.00 eV.
(a) Calculate the energy of a photon of this light, giving your answer in joules. [2]
(b) Convert the work function of the metal to joules. [1]
(c) Calculate the maximum kinetic energy of the emitted photoelectrons, in joules. [2]
(d) Calculate the stopping potential required to prevent photoelectrons from reaching the collector. [2]
(e) The intensity of the light is doubled while the wavelength remains the same. State and explain the effect, if any, on: (i) the maximum kinetic energy of the photoelectrons, [1] (ii) the photoelectric current. [2]
END OF PAPER
Answers
TuitionGoWhere Practice Paper - Physics H1 A-Level
Answer Key and Marking Scheme
Paper: Practice Paper 2 (Structured & Free Response) Version: 2 of 5 Total Marks: 80
Section A: Structured Questions (60 marks)
Question 1: Kinematics and Projectile Motion
(a) Time taken for stone to reach the sea [2]
Vertical motion: s = ut + ½at² 45.0 = 0 + ½(9.81)t² [M1] t² = (2 × 45.0) / 9.81 = 9.17 t = 3.03 s [A1]
(b) Horizontal distance [1]
Horizontal distance = horizontal velocity × time = 20.0 × 3.03 = 60.6 m [A1]
(c) Magnitude of velocity just before hitting sea [3]
Vertical component: v_y = u_y + at = 0 + 9.81 × 3.03 = 29.7 m s⁻¹ [M1] Horizontal component: v_x = 20.0 m s⁻¹ (constant) [M1] Resultant speed = √(v_x² + v_y²) = √(20.0² + 29.7²) = √(400 + 882) = √1282 = 35.8 m s⁻¹ [A1]
(d) Graph of vertical velocity vs. time [2]
- Straight line through origin with positive gradient [B1]
- Gradient = 9.81 m s⁻²
- Line extends from t = 0 to t = 3.03 s
- Final vertical velocity labelled as -29.7 m s⁻¹ (or 29.7 m s⁻¹ downward)
- Axes labelled correctly [B1]
Question 2: Forces and Equilibrium
(a) Free-body diagram [3]
Forces to be shown:
- Weight (250 N) acting downward at centre of ladder (2.50 m from either end) [B1]
- Normal reaction from wall (R_W) acting horizontally to the right at top of ladder [B1]
- Normal reaction from ground (R_G) acting vertically upward at foot of ladder
- Frictional force (F) acting horizontally to the left at foot of ladder [B1]
All forces correctly labelled and positioned.
(b) Reaction force from wall [3]
Take moments about foot of ladder: Clockwise moment = anticlockwise moment R_W × (5.00 sin 60°) = 250 × (2.50 cos 60°) [M1] R_W × 4.33 = 250 × 1.25 [M1] R_W = (250 × 1.25) / 4.33 = 72.2 N [A1]
(c) Frictional force [2]
Horizontal equilibrium: F = R_W [M1] F = 72.2 N [A1]
(d) Minimum coefficient of static friction [2]
Vertical equilibrium: R_G = 250 N [M1] μ_min = F / R_G = 72.2 / 250 = 0.289 [A1]
Question 3: Work, Energy, and Power
(a) Acceleration [1]
a = (v - u) / t = (25.0 - 0) / 8.00 = 3.13 m s⁻² [A1]
(b) Distance travelled [2]
s = ut + ½at² = 0 + ½(3.13)(8.00)² [M1] s = 100 m [A1] (Alternative: s = ½(u + v)t = ½(0 + 25.0)(8.00) = 100 m)
(c) Work done by engine [3]
Resultant force = ma = 1200 × 3.13 = 3756 N [M1] Driving force = resultant force + resistive force = 3756 + 600 = 4356 N [M1] Work done = driving force × distance = 4356 × 100 = 4.36 × 10⁵ J [A1]
(d) Average power [2]
Average power = work done / time = 4.36 × 10⁵ / 8.00 [M1] = 5.45 × 10⁴ W [A1] (Alternative: P = Fv_avg = 4356 × 12.5 = 5.45 × 10⁴ W)
(e) Explanation [2]
As speed increases, air resistance (drag force) increases [B1]. The net force (driving force - resistive forces) decreases, so acceleration decreases. Eventually, when driving force equals total resistive forces, net force = 0 and the car reaches terminal velocity / constant maximum speed [B1].
Question 4: Momentum and Collisions
(a) Principle of conservation of linear momentum [2]
The total momentum of a closed/isolated system remains constant [B1] provided no external resultant force acts on the system [B1].
(b) Velocity of trolley B after collision [2]
Total momentum before = total momentum after (2.00 × 3.00) + (1.00 × 0) = (2.00 × 1.00) + (1.00 × v_B) [M1] 6.00 = 2.00 + v_B v_B = 4.00 m s⁻¹ to the right [A1]
(c) Show collision is elastic [3]
Total KE before = ½(2.00)(3.00)² + 0 = 9.00 J [M1] Total KE after = ½(2.00)(1.00)² + ½(1.00)(4.00)² = 1.00 + 8.00 = 9.00 J [M1] Since total KE before = total KE after, the collision is elastic [A1].
(d) Conservation of momentum with magnetic repulsion [3]
Newton's Third Law: The force exerted by A on B is equal and opposite to the force exerted by B on A [B1]. The time of interaction is the same for both trolleys [B1]. Impulse on A = -Impulse on B (since FΔt is equal and opposite). Change in momentum of A = -Change in momentum of B. Therefore, total momentum remains constant [B1].
Question 5: Current Electricity
(a) Current in circuit [2]
Total resistance = R + r = 5.50 + 0.500 = 6.00 Ω [M1] I = EMF / total resistance = 12.0 / 6.00 = 2.00 A [A1]
(b) Terminal potential difference [1]
V = EMF - Ir = 12.0 - (2.00 × 0.500) = 11.0 V [A1] (Alternative: V = IR = 2.00 × 5.50 = 11.0 V)
(c) Power dissipated in external resistor [2]
P = I²R = (2.00)² × 5.50 [M1] = 22.0 W [A1] (Alternative: P = VI = 11.0 × 2.00 = 22.0 W)
(d) Power wasted in internal resistance [1]
P = I²r = (2.00)² × 0.500 = 2.00 W [A1]
(e) Explanation using potential divider [2]
The internal resistance r and the external resistance R form a potential divider in series [B1]. The EMF is divided between r and R in proportion to their resistances. The terminal p.d. (voltage across R) = EMF × [R/(R + r)], which is less than the EMF because some voltage is dropped across the internal resistance [B1].
Question 6: D.C. Circuits
(a) Circuit diagram [2]
- Battery symbol with 10.0 V labelled [B1]
- R₁ (4.00 kΩ) and R₂ (6.00 kΩ) in series
- Output voltage V_out labelled across R₂
- Correct circuit symbols and connections [B1]
(b) Output voltage across R₂ [2]
V_out = V_supply × [R₂/(R₁ + R₂)] [M1] = 10.0 × [6.00/(4.00 + 6.00)] = 10.0 × 0.600 = 6.00 V [A1]
(c) New output voltage with load resistor [4]
R₂ and load (3.00 kΩ) in parallel: 1/R_parallel = 1/6.00 + 1/3.00 = 1/6.00 + 2/6.00 = 3/6.00 [M1] R_parallel = 2.00 kΩ [A1]
New total resistance = R₁ + R_parallel = 4.00 + 2.00 = 6.00 kΩ [M1] New V_out = 10.0 × (2.00/6.00) = 3.33 V [A1]
(d) Explanation of voltage change [2]
The load resistor provides an additional path for current, reducing the effective resistance of the lower arm of the potential divider [B1]. This changes the ratio of resistances in the divider, reducing the fraction of the supply voltage appearing across the output [B1].
Question 7: Nuclear Physics
(a) Nuclear equation for carbon-14 decay [2]
¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄ₑ [B1 for correct products, B1 for balanced atomic and mass numbers]
(b) Age of sample [3]
After n half-lives, fraction remaining = (½)ⁿ 0.25 = (½)ⁿ [M1] n = 2 (since ½² = ¼ = 0.25) [M1] Age = n × t₁/₂ = 2 × 5730 = 11,460 years [A1]
(c) Limitation of carbon-14 dating [2]
After about 50,000 years (approximately 9 half-lives), the activity of carbon-14 becomes too low to measure accurately [B1]. The count rate becomes comparable to background radiation, making reliable dating impossible [B1].
Section B: Free Response Questions (20 marks)
Question 8: Mechanics – Energy and Momentum
(a) Speed just before hitting ground [2]
Using v² = u² + 2as: v² = 0 + 2(9.81)(10.0) = 196.2 [M1] v = 14.0 m s⁻¹ [A1] (Alternative: using energy conservation: mgh = ½mv²)
(b) Impulse on ball [3]
Taking upward as positive: Initial velocity (just before impact) = -14.0 m s⁻¹ Final velocity (just after impact) = +8.00 m s⁻¹ [M1] Change in momentum = m(v - u) = 0.500(8.00 - (-14.0)) = 0.500 × 22.0 [M1] Impulse = 11.0 N s upward [A1]
(c) Average force [2]
F_avg = Impulse / time = 11.0 / 0.0500 [M1] = 220 N upward [A1]
(d) Maximum height after rebound [2]
Using v² = u² + 2as: 0 = (8.00)² + 2(-9.81)h [M1] h = 64.0 / (2 × 9.81) = 3.26 m [A1]
(e) Reason for not reaching original height [1]
Energy is lost during the impact (converted to thermal energy/sound/deformation of the ball and ground) [B1]. The collision is inelastic, so kinetic energy is not conserved.
Question 9: Electricity – Circuit Analysis
(a) Resistance of thermistor at 300 K [2]
R_T = R₀ e^(k/θ) = 2.00 × e^(3000/300) [M1] = 2.00 × e¹⁰ = 2.00 × 22,026 = 4.41 × 10⁴ Ω [A1]
(b) Current at 300 K [2]
Total resistance = R_T + 100 = 44,100 + 100 = 44,200 Ω [M1] I = V / R_total = 6.00 / 44,200 = 1.36 × 10⁻⁴ A [A1]
(c) Potential difference across thermistor at 300 K [1]
V_T = I × R_T = 1.36 × 10⁻⁴ × 44,100 = 6.00 V [A1] (Effectively all the voltage as R_T >> 100 Ω)
(d) Effect of temperature increase to 350 K [3]
As temperature increases, the exponential term e^(k/θ) decreases because k/θ decreases [B1]. Therefore, R_T decreases significantly [B1]. With lower total resistance, the current in the circuit increases. Since R_T decreases relative to the fixed 100 Ω resistor, the potential difference across the thermistor decreases (potential divider effect) [B1].
(e) Practical application [2]
Temperature sensor / electronic thermometer [B1]. The thermistor is placed in one arm of a potential divider. As temperature changes, the resistance changes, altering the output voltage. This voltage can be calibrated to give a temperature reading [B1]. (Other valid applications: fire alarm, overheating protection in circuits, temperature control systems)
Question 10: Waves and Photoelectric Effect
(a) Energy of photon in joules [2]
E = hf = hc/λ [M1] E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (450 × 10⁻⁹) = (1.989 × 10⁻²⁵) / (4.50 × 10⁻⁷) = 4.42 × 10⁻¹⁹ J [A1]
(b) Work function in joules [1]
Φ = 2.00 eV × 1.60 × 10⁻¹⁹ J/eV = 3.20 × 10⁻¹⁹ J [A1]
(c) Maximum kinetic energy [2]
K_max = hf - Φ = 4.42 × 10⁻¹⁹ - 3.20 × 10⁻¹⁹ [M1] = 1.22 × 10⁻¹⁹ J [A1]
(d) Stopping potential [2]
eV_s = K_max [M1] V_s = K_max / e = 1.22 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 0.763 V [A1]
(e) Effect of doubling intensity [3]
(i) Maximum kinetic energy: No change [B1]. The energy of each photon depends only on frequency/wavelength (E = hf), not on intensity. Each photon still transfers the same energy, so K_max remains unchanged.
(ii) Photoelectric current: The current doubles [B1]. Doubling intensity means twice as many photons strike the surface per second. This releases twice as many photoelectrons per second, doubling the current [B1].
END OF ANSWER KEY