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A Level H1 Physics Practice Paper 1

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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 $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A

Answer all questions in this section.

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

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

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

Image pending generation for this question.

(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 $2.0 \text{ kg}$ and moves with velocity $4.0 \text{ m s}^{-1}$ to the right. Trolley B has mass $3.0 \text{ kg}$ 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 $4.0 \text{ m}$ and weight $200 \text{ N}$ 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 $3.0 \text{ m}$ vertically above A. (a) Draw a free-body diagram showing all forces acting on the beam. Label the forces clearly. [3]

Drawing space

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

6. A box of mass $5.0 \text{ kg}$ is pulled up a rough inclined plane at a constant speed by a force parallel to the plane. The plane is inclined at $30^\circ$ to the horizontal. The frictional force acting on the box is $10 \text{ N}$ . (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 $50 \text{ kg}$ vertically through a height of $12 \text{ m}$ in $8.0 \text{ s}$ . The motor operates at an efficiency of $60\%$ . (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 $15 \text{ m s}^{-1}$ . It hits the ground $3.0 \text{ s}$ 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 $v_T$ . [2]

Graph space

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

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

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

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

14. Two forces, $F_1 = 6.0 \text{ N}$ and $F_2 = 8.0 \text{ N}$ , act on a point object. The angle between the two forces is $90^\circ$ . (a) Calculate the magnitude of the resultant force. [2]

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

15. A projectile is launched with an initial velocity of $40 \text{ m s}^{-1}$ at an angle of $60^\circ$ 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 $2.0 \text{ kg}$ slides down a smooth curved track from a height of $5.0 \text{ m}$ 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 $10 \text{ m}$ 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 $W$ rests against a smooth vertical wall and on a rough horizontal ground. The ladder makes an angle $\theta$ 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 / N	0.0	2.0	4.0	6.0	8.0
Extension / cm	0.0	1.5	3.0	4.5	6.0

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

Graph space

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

20. A ball of mass $0.5 \text{ kg}$ is dropped from a height of $2.0 \text{ m}$ . It rebounds to a height of $1.5 \text{ m}$ . (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]

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Answers

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 = $\frac{0.02}{12.45} \times 100\%$ [M1]
$= 0.16\%$ [A1]

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

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

(b) Distance = Area under graph [M1]
Area = Area of triangle (0-5s) + Area of rectangle (5-15s) + Area of triangle (15-20s)
$= (\frac{1}{2} \times 5 \times 20) + (10 \times 20) + (\frac{1}{2} \times 5 \times 20)$ [M1]
$= 50 + 200 + 50 = 300 \text{ m}$ [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: $m_A u_A + m_B u_B = (m_A + m_B)v$ [M1]
$(2.0)(4.0) + (3.0)(0) = (2.0 + 3.0)v$ [M1]
$8.0 = 5.0v \Rightarrow v = 1.6 \text{ m s}^{-1}$ [A1]

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

5. (a) Diagram should show:

Weight $W$ acting downwards from the center of the beam. [B1]
Tension $T$ acting along the cable from B towards the wall. [B1]
Reaction force $R$ 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 $2.0 \text{ m}$ from A.
Perpendicular distance of Tension from A:
Geometry: Triangle with base 4m, height 3m. Hypotenuse $= \sqrt{3^2+4^2} = 5 \text{ m}$ .
$\sin(\text{angle at B}) = \frac{3}{5} = 0.6$ .
Vertical component of Tension $T_y = T \sin \theta = 0.6 T$ .
Moment of Tension $= T_y \times 4.0 = 2.4 T$ .
Alternatively, perpendicular distance from A to line of action of T:
$d = 4 \sin(\text{angle between beam and cable})$ . Angle $\alpha$ where $\tan \alpha = 3/4$ . $\sin \alpha = 3/5$ .
$d = 4 \times (3/5) = 2.4 \text{ m}$ .
$W \times 2.0 = T \times 2.4$ [M1]
$200 \times 2.0 = 2.4 T$
$400 = 2.4 T$
$T = 167 \text{ N}$ (or $166.7 \text{ N}$ ) [A1]
(Accept 170 N if 2 s.f. used throughout)

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

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

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

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

(b) Efficiency $= \frac{P_{out}}{P_{in}} \times 100\%$ [M1]
$0.60 = \frac{735.75}{P_{in}}$
$P_{in} = \frac{735.75}{0.60} = 1226 \text{ W}$ (or $1.23 \text{ kW}$ ) [A1]

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

(b) Horizontal motion: $v_x = 15 \text{ m s}^{-1}$ (constant).
Distance $= v_x t = 15 \times 3.0 = 45 \text{ m}$ [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)

Initially, velocity is zero, so air resistance is zero. Resultant force is weight ( $mg$ ), so acceleration is $g$ (maximum). [B1]
As velocity increases, air resistance increases. [B1]
Resultant force ( $mg - \text{air resistance}$ ) decreases, so acceleration decreases. [B1]
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 $v_T$ . [B1]

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

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

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

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

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

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

15. (a) $u_y = u \sin \theta = 40 \sin 60^\circ$ [M1]
$u_y = 34.64 \text{ m s}^{-1}$ (or $34.6 \text{ m s}^{-1}$ ) [A1]

(b) At max height, $v_y = 0$ .
$v_y^2 = u_y^2 + 2as$ [M1]
$0 = (34.64)^2 + 2(-9.81)h$
$h = \frac{1200}{19.62}$ [M1]
$h = 61.2 \text{ m}$ [A1]

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

(b) Work Done against friction = Loss in KE
$F_f \times d = \frac{1}{2}mv^2$ [M1]
$F_f \times 10 = \frac{1}{2}(2.0)(9.90)^2$
$10 F_f = 98.1$ [M1]
$F_f = 9.81 \text{ N}$ [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 $= \frac{\Delta F}{\Delta x}$ [M1]
Using points $(0,0)$ and $(8.0, 6.0)$ :
Gradient $= \frac{8.0 - 0}{6.0 - 0} = 1.33 \text{ N cm}^{-1}$
Convert to SI: $1.33 \text{ N} / 0.01 \text{ m} = 133 \text{ N m}^{-1}$ [A1]
(Accept $1.33 \text{ N/cm}$ if units stated, but SI preferred).

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

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