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A Level H1 Physics Practice Paper 1

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Questions

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H1
Level: A-Level
Paper: Practice Paper 1 (Version 1 of 5)
Duration: 2 hours
Total Marks: 80
Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

  • Write your name, class, and date in the spaces provided.
  • Answer all questions.
  • The number of marks is given in brackets [ ] at the end of each question or part question.
  • You are advised to spend about 1 hour on Section A and 1 hour on Section B.
  • Use g=9.81 m s−2g = 9.81 \text{ m s}^{-2} unless otherwise stated.

Section A

Answer all questions in this section.

1. A student measures the diameter dd of a steel sphere using a micrometer screw gauge. The reading is 12.45±0.02 mm12.45 \pm 0.02 \text{ mm}. (a) Calculate the percentage uncertainty in the diameter. [1]

(b) The volume VV of the sphere is calculated using V=16πd3V = \frac{1}{6}\pi d^3. Determine the percentage uncertainty in the volume VV. [2]

2. A car travels along a straight horizontal road. The velocity-time graph for the car is shown below.

(Imagine a graph: Velocity starts at 0, increases linearly to 20 m/s in 5s, stays constant at 20 m/s for 10s, then decreases linearly to 0 in 5s.)

(a) Calculate the acceleration of the car during the first 5 seconds. [2]

(b) Determine the total distance travelled by the car during the 20 seconds. [3]

3. State the principle of conservation of linear momentum. [2]

4. Two trolleys, A and B, move on a frictionless horizontal track. Trolley A has mass 2.0 kg2.0 \text{ kg} and moves with velocity 4.0 m s−14.0 \text{ m s}^{-1} to the right. Trolley B has mass 3.0 kg3.0 \text{ kg} and is initially at rest. They collide and stick together. (a) Calculate the common velocity of the trolleys after the collision. [3]

(b) Show that the collision is inelastic. [2]

5. A uniform beam AB of length 4.0 m4.0 \text{ m} and weight 200 N200 \text{ N} is hinged at end A to a vertical wall. The beam is held horizontal by a cable attached to end B and to the wall at a point 3.0 m3.0 \text{ m} vertically above A. (a) Draw a free-body diagram showing all forces acting on the beam. Label the forces clearly. [3]

(Space for diagram)

(b) Calculate the tension in the cable. [4]

6. A box of mass 5.0 kg5.0 \text{ kg} is pulled up a rough inclined plane at a constant speed by a force parallel to the plane. The plane is inclined at 30∘30^\circ to the horizontal. The frictional force acting on the box is 10 N10 \text{ N}. (a) Calculate the component of the weight of the box acting down the slope. [2]

(b) Determine the magnitude of the pulling force. [2]

7. Define the term power. [1]

8. An electric motor lifts a load of mass 50 kg50 \text{ kg} vertically through a height of 12 m12 \text{ m} in 8.0 s8.0 \text{ s}. The motor operates at an efficiency of 60%60\%. (a) Calculate the useful power output of the motor. [3]

(b) Calculate the input power to the motor. [2]

9. A ball is thrown horizontally from the top of a cliff with a speed of 15 m s−115 \text{ m s}^{-1}. It hits the ground 3.0 s3.0 \text{ s} later. Air resistance is negligible. (a) Calculate the height of the cliff. [2]

(b) Calculate the horizontal distance from the base of the cliff to where the ball lands. [1]

10. Explain why the horizontal component of the velocity of the ball in Question 9 remains constant during its flight. [2]

Section B

Answer all questions in this section.

11. A skydiver falls vertically from rest. (a) Describe and explain the variation in the acceleration of the skydiver from the moment he jumps until he reaches terminal velocity. [4]

(b) Sketch a graph of velocity against time for the skydiver. Label the terminal velocity vTv_T. [2]

(Space for graph)

12. A spring obeys Hooke’s Law. When a force of 10 N10 \text{ N} is applied, the extension is 4.0 cm4.0 \text{ cm}. (a) Calculate the spring constant kk. [2]

(b) Calculate the elastic potential energy stored in the spring when the extension is 4.0 cm4.0 \text{ cm}. [2]

13. A car of mass 1200 kg1200 \text{ kg} travels at a constant speed of 25 m s−125 \text{ m s}^{-1} on a level road. The total resistive force acting on the car is 800 N800 \text{ N}. (a) State the driving force produced by the engine. [1]

(b) Calculate the power developed by the engine. [2]

14. Two forces, F1=6.0 NF_1 = 6.0 \text{ N} and F2=8.0 NF_2 = 8.0 \text{ N}, act on a point object. The angle between the two forces is 90∘90^\circ. (a) Calculate the magnitude of the resultant force. [2]

(b) Determine the angle between the resultant force and the 6.0 N6.0 \text{ N} force. [2]

15. A projectile is launched with an initial velocity of 40 m s−140 \text{ m s}^{-1} at an angle of 60∘60^\circ to the horizontal. (a) Calculate the vertical component of the initial velocity. [1]

(b) Calculate the maximum height reached by the projectile. [3]

16. A block of mass 2.0 kg2.0 \text{ kg} slides down a smooth curved track from a height of 5.0 m5.0 \text{ m} and onto a rough horizontal surface. (a) Calculate the speed of the block at the bottom of the curved track. [3]

(b) The block travels 10 m10 \text{ m} on the rough horizontal surface before coming to rest. Calculate the average frictional force acting on the block. [3]

17. Distinguish between scalar and vector quantities, giving one example of each. [3]

18. A uniform ladder of weight WW rests against a smooth vertical wall and on a rough horizontal ground. The ladder makes an angle θ\theta with the ground. (a) Explain why the reaction force from the wall is horizontal. [2]

(b) State the condition for the ladder to be in equilibrium. [2]

19. A student investigates the relationship between the force applied to a spring and its extension. The student obtains the following data:

Force / N0.02.04.06.08.0
Extension / cm0.01.53.04.56.0

(a) Plot a graph of Force (y-axis) against Extension (x-axis). [3]

(Space for graph)

(b) Use the graph to determine the spring constant. [2]

20. A ball of mass 0.5 kg0.5 \text{ kg} is dropped from a height of 2.0 m2.0 \text{ m}. It rebounds to a height of 1.5 m1.5 \text{ m}. (a) Calculate the speed of the ball just before it hits the ground. [2]

(b) Calculate the loss in mechanical energy during the impact. [3]

Answers

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TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key and Marking Scheme (Version 1)

Subject: Physics H1
Paper: Practice Paper 1

Section A

1. (a) Percentage uncertainty = 0.0212.45×100%\frac{0.02}{12.45} \times 100\% [M1]
=0.16%= 0.16\% [A1]

(b) V∝d3V \propto d^3, so % uncertainty in V=3×V = 3 \times (% uncertainty in dd) [M1]
=3×0.16%=0.48%= 3 \times 0.16\% = 0.48\% [A1]

2. (a) Acceleration a=ΔvΔta = \frac{\Delta v}{\Delta t} [M1]
a=20−05=4.0 m s−2a = \frac{20 - 0}{5} = 4.0 \text{ m s}^{-2} [A1]

(b) Distance = Area under graph [M1]
Area = Area of triangle (0-5s) + Area of rectangle (5-15s) + Area of triangle (15-20s)
=(12×5×20)+(10×20)+(12×5×20)= (\frac{1}{2} \times 5 \times 20) + (10 \times 20) + (\frac{1}{2} \times 5 \times 20) [M1]
=50+200+50=300 m= 50 + 200 + 50 = 300 \text{ m} [A1]

3. In a closed system (or isolated system) [B1],
the total linear momentum remains constant (or is conserved) provided no external forces act. [B1]

4. (a) Conservation of momentum: mAuA+mBuB=(mA+mB)vm_A u_A + m_B u_B = (m_A + m_B)v [M1]
(2.0)(4.0)+(3.0)(0)=(2.0+3.0)v(2.0)(4.0) + (3.0)(0) = (2.0 + 3.0)v [M1]
8.0=5.0v⇒v=1.6 m s−18.0 = 5.0v \Rightarrow v = 1.6 \text{ m s}^{-1} [A1]

(b) Initial KE =12mAuA2=12(2.0)(4.0)2=16 J= \frac{1}{2} m_A u_A^2 = \frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J} [M1]
Final KE =12(mA+mB)v2=12(5.0)(1.6)2=6.4 J= \frac{1}{2} (m_A+m_B) v^2 = \frac{1}{2}(5.0)(1.6)^2 = 6.4 \text{ J}
Since Initial KE ≠\neq Final KE (KE is lost), the collision is inelastic. [A1]

5. (a) Diagram should show:

  1. Weight WW acting downwards from the center of the beam. [B1]
  2. Tension TT acting along the cable from B towards the wall. [B1]
  3. Reaction force RR at hinge A (direction can be general, usually up and right). [B1]

(b) Take moments about A. [M1]
Clockwise moment = Anticlockwise moment
Weight acts at 2.0 m2.0 \text{ m} from A.
Perpendicular distance of Tension from A:
Geometry: Triangle with base 4m, height 3m. Hypotenuse =32+42=5 m= \sqrt{3^2+4^2} = 5 \text{ m}.
sin⁡(angle at B)=35=0.6\sin(\text{angle at B}) = \frac{3}{5} = 0.6.
Vertical component of Tension Ty=Tsin⁡θ=0.6TT_y = T \sin \theta = 0.6 T.
Moment of Tension =Ty×4.0=2.4T= T_y \times 4.0 = 2.4 T.
Alternatively, perpendicular distance from A to line of action of T:
d=4sin⁡(angle between beam and cable)d = 4 \sin(\text{angle between beam and cable}). Angle α\alpha where tan⁡α=3/4\tan \alpha = 3/4. sin⁡α=3/5\sin \alpha = 3/5.
d=4×(3/5)=2.4 md = 4 \times (3/5) = 2.4 \text{ m}.
W×2.0=T×2.4W \times 2.0 = T \times 2.4 [M1]
200×2.0=2.4T200 \times 2.0 = 2.4 T
400=2.4T400 = 2.4 T
T=167 NT = 167 \text{ N} (or 166.7 N166.7 \text{ N}) [A1]
(Accept 170 N if 2 s.f. used throughout)

6. (a) Component of weight down slope =mgsin⁡θ= mg \sin \theta [M1]
=5.0×9.81×sin⁡30∘=24.5 N= 5.0 \times 9.81 \times \sin 30^\circ = 24.5 \text{ N} [A1]

(b) Since speed is constant, acceleration is zero, so resultant force is zero.
Pulling Force F=Friction+Weight ComponentF = \text{Friction} + \text{Weight Component} [M1]
F=10+24.5=34.5 NF = 10 + 24.5 = 34.5 \text{ N} [A1]

7. Power is the rate of doing work (or rate of energy transfer). [B1]

8. (a) Useful Work Done =mgh=50×9.81×12=5886 J= mgh = 50 \times 9.81 \times 12 = 5886 \text{ J} [M1]
Useful Power Output Pout=Worktime=58868.0P_{out} = \frac{\text{Work}}{\text{time}} = \frac{5886}{8.0} [M1]
Pout=736 WP_{out} = 736 \text{ W} (or 735.75 W735.75 \text{ W}) [A1]

(b) Efficiency =PoutPin×100%= \frac{P_{out}}{P_{in}} \times 100\% [M1]
0.60=735.75Pin0.60 = \frac{735.75}{P_{in}}
Pin=735.750.60=1226 WP_{in} = \frac{735.75}{0.60} = 1226 \text{ W} (or 1.23 kW1.23 \text{ kW}) [A1]

9. (a) Vertical motion: uy=0u_y = 0, a=g=9.81 m s−2a = g = 9.81 \text{ m s}^{-2}, t=3.0 st = 3.0 \text{ s}.
h=uyt+12gt2h = u_y t + \frac{1}{2}gt^2 [M1]
h=0+12(9.81)(3.0)2=44.1 mh = 0 + \frac{1}{2}(9.81)(3.0)^2 = 44.1 \text{ m} [A1]

(b) Horizontal motion: vx=15 m s−1v_x = 15 \text{ m s}^{-1} (constant).
Distance =vxt=15×3.0=45 m= v_x t = 15 \times 3.0 = 45 \text{ m} [A1]

10. There is no horizontal force acting on the ball (air resistance is negligible). [B1]
According to Newton's First Law, an object continues in its state of uniform motion unless acted upon by a resultant force. Therefore, horizontal velocity is constant. [B1]

Section B

11. (a)

  1. Initially, velocity is zero, so air resistance is zero. Resultant force is weight (mgmg), so acceleration is gg (maximum). [B1]
  2. As velocity increases, air resistance increases. [B1]
  3. Resultant force (mg−air resistancemg - \text{air resistance}) decreases, so acceleration decreases. [B1]
  4. Eventually, air resistance equals weight. Resultant force is zero, acceleration is zero, and velocity becomes constant (terminal velocity). [B1]

(b) Graph:

  • Starts at origin (0,0). [B1]
  • Curve with decreasing gradient, approaching a horizontal asymptote labeled vTv_T. [B1]

12. (a) Hooke's Law: F=kxF = kx [M1]
k=Fx=100.04=250 N m−1k = \frac{F}{x} = \frac{10}{0.04} = 250 \text{ N m}^{-1} [A1]

(b) Elastic Potential Energy E=12kx2E = \frac{1}{2}kx^2 [M1]
E=12(250)(0.04)2=0.20 JE = \frac{1}{2}(250)(0.04)^2 = 0.20 \text{ J} [A1]
(Alternatively E=12Fx=12(10)(0.04)=0.20 JE = \frac{1}{2}Fx = \frac{1}{2}(10)(0.04) = 0.20 \text{ J})

13. (a) Since speed is constant, driving force = resistive force.
Driving Force =800 N= 800 \text{ N} [B1]

(b) Power P=FvP = Fv [M1]
P=800×25=20,000 WP = 800 \times 25 = 20,000 \text{ W} (or 20 kW20 \text{ kW}) [A1]

14. (a) Resultant R=F12+F22R = \sqrt{F_1^2 + F_2^2} [M1]
R=6.02+8.02=36+64=100=10 NR = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N} [A1]

(b) Let α\alpha be the angle with the 6.0 N6.0 \text{ N} force.
tan⁡α=8.06.0\tan \alpha = \frac{8.0}{6.0} [M1]
α=53.1∘\alpha = 53.1^\circ [A1]

15. (a) uy=usin⁡θ=40sin⁡60∘u_y = u \sin \theta = 40 \sin 60^\circ [M1]
uy=34.64 m s−1u_y = 34.64 \text{ m s}^{-1} (or 34.6 m s−134.6 \text{ m s}^{-1}) [A1]

(b) At max height, vy=0v_y = 0.
vy2=uy2+2asv_y^2 = u_y^2 + 2as [M1]
0=(34.64)2+2(−9.81)h0 = (34.64)^2 + 2(-9.81)h
h=120019.62h = \frac{1200}{19.62} [M1]
h=61.2 mh = 61.2 \text{ m} [A1]

16. (a) Conservation of Energy: Loss in GPE = Gain in KE
mgh=12mv2mgh = \frac{1}{2}mv^2 [M1]
v=2gh=2×9.81×5.0v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 5.0} [M1]
v=98.1=9.90 m s−1v = \sqrt{98.1} = 9.90 \text{ m s}^{-1} [A1]

(b) Work Done against friction = Loss in KE
Ff×d=12mv2F_f \times d = \frac{1}{2}mv^2 [M1]
Ff×10=12(2.0)(9.90)2F_f \times 10 = \frac{1}{2}(2.0)(9.90)^2
10Ff=98.110 F_f = 98.1 [M1]
Ff=9.81 NF_f = 9.81 \text{ N} [A1]

17. Scalar quantity has magnitude only. [B1] Example: Mass, Speed, Energy, Distance. [B1]
Vector quantity has magnitude and direction. [B1] Example: Force, Velocity, Displacement, Acceleration. [B1]
(Award max 3 marks. 1 for scalar def, 1 for vector def, 1 for correct examples of both).

18. (a) The wall is smooth, so there is no frictional force parallel to the wall. [B1]
Therefore, the reaction force must be perpendicular (normal) to the wall, which is horizontal. [B1]

(b) The sum of forces acting on the ladder is zero (translational equilibrium). [B1]
AND the sum of moments about any point is zero (rotational equilibrium). [B1]

19. (a) Graph:

  • Axes labeled "Force / N" (y) and "Extension / cm" (x) with units. [B1]
  • Points plotted correctly. [B1]
  • Straight line of best fit passing through the origin. [B1]

(b) Gradient =ΔFΔx= \frac{\Delta F}{\Delta x} [M1]
Using points (0,0)(0,0) and (8.0,6.0)(8.0, 6.0):
Gradient =8.0−06.0−0=1.33 N cm−1= \frac{8.0 - 0}{6.0 - 0} = 1.33 \text{ N cm}^{-1}
Convert to SI: 1.33 N/0.01 m=133 N m−11.33 \text{ N} / 0.01 \text{ m} = 133 \text{ N m}^{-1} [A1]
(Accept 1.33 N/cm1.33 \text{ N/cm} if units stated, but SI preferred).

20. (a) Conservation of Energy (fall): mgh=12mv2mgh = \frac{1}{2}mv^2
v=2gh=2×9.81×2.0v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 2.0} [M1]
v=39.24=6.26 m s−1v = \sqrt{39.24} = 6.26 \text{ m s}^{-1} [A1]

(b) Initial Energy (at drop) =mgh1=0.5×9.81×2.0=9.81 J= mgh_1 = 0.5 \times 9.81 \times 2.0 = 9.81 \text{ J} [M1]
Final Energy (at rebound peak) =mgh2=0.5×9.81×1.5=7.36 J= mgh_2 = 0.5 \times 9.81 \times 1.5 = 7.36 \text{ J} [M1]
Loss in Energy =9.81−7.36=2.45 J= 9.81 - 7.36 = 2.45 \text{ J} [A1]