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A Level H1 Physics Practice Paper 3

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI)
Version: 3 of 5
Subject: Physics H1 (8867)
Level: A-Level
Paper: Structured Questions (Mechanics Focus)
Duration: 1 hour 15 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Answer all questions.
Write your answers in the spaces provided.
You may lose marks if you do not show your working or if you do not use appropriate units.
Take the acceleration of free fall $g = 9.81 \text{ m s}^{-2}$ .

Section A: Kinematics and Dynamics

(Answer all questions in this section.)

1. A car accelerates uniformly from rest along a straight road. It reaches a speed of $24 \text{ m s}^{-1}$ in $8.0 \text{ s}$ . (a) Calculate the acceleration of the car.

Answer: ________________________ $\text{m s}^{-2}$ [2]

(b) Calculate the distance travelled by the car during this $8.0 \text{ s}$ interval.

Answer: ________________________ $\text{m}$ [2]

2. A stone is thrown vertically upwards from the edge of a cliff with an initial velocity of $15 \text{ m s}^{-1}$ . The stone hits the sea $4.0 \text{ s}$ later. Air resistance is negligible. (a) Explain why the acceleration of the stone is constant throughout its motion.

_________________________________________________________________________ [1]

(b) Calculate the height of the cliff above the sea level.

Answer: ________________________ $\text{m}$ [3]

3. Fig. 3.1 shows a velocity-time graph for a toy train moving along a straight track.

(Imagine a graph: Velocity starts at 0, increases linearly to 4 m/s in 2s, stays constant at 4 m/s for 3s, then decreases linearly to 0 in 2s.)

(a) Describe the motion of the train between $t = 5 \text{ s}$ and $t = 7 \text{ s}$ .

_________________________________________________________________________ [1]

(b) Calculate the total distance travelled by the train during the $7 \text{ s}$ .

Answer: ________________________ $\text{m}$ [3]

4. A projectile is fired horizontally from a height of $45 \text{ m}$ with a speed of $20 \text{ m s}^{-1}$ . (a) State the horizontal acceleration of the projectile.

Answer: ________________________ $\text{m s}^{-2}$ [1]

(b) Calculate the time taken for the projectile to hit the ground.

Answer: ________________________ $\text{s}$ [2]

Answer: ________________________ $\text{m}$ [2]

5. A skydiver falls from a stationary helicopter. Fig. 5.1 shows how her vertical velocity $v$ varies with time $t$ .

(Imagine a graph: Curve starts at origin, gradient decreases, becomes horizontal at terminal velocity.)

(a) Explain, in terms of forces, why the gradient of the graph decreases with time.

_________________________________________________________________________ [2]

(b) State and explain the value of the skydiver's acceleration when she reaches terminal velocity.

_________________________________________________________________________ [2]

Section B: Forces, Moments, and Equilibrium

(Answer all questions in this section.)

6. Define the term moment of a force.

_________________________________________________________________________ [2]

7. 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 above $A$ . The cable makes an angle of $30^\circ$ with the beam. (a) Draw a free-body diagram for the beam, showing all forces acting on it. Label the forces clearly.

[2]

(b) By taking moments about point $A$ , calculate the tension $T$ in the cable.

Answer: ________________________ $\text{N}$ [3]

8. A ladder of weight $W$ rests against a smooth vertical wall and on rough horizontal ground. The ladder is in equilibrium. (a) Explain why there must be a frictional force acting on the ladder at the ground.

_________________________________________________________________________ [1]

(b) State two conditions required for the ladder to be in equilibrium.

_____________________________________________________________________ [2]

9. A block of mass $5.0 \text{ kg}$ rests on a rough plane inclined at $30^\circ$ to the horizontal. The block is stationary. (a) Calculate the component of the weight acting down the slope.

Answer: ________________________ $\text{N}$ [2]

(b) Determine the magnitude of the frictional force acting on the block.

Answer: ________________________ $\text{N}$ [1]

Answer: ________________________ $\text{N}$ [2]

10. 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.

Answer: ________________________ $\text{N}$ [2]

(b) Calculate the angle between the resultant force and the $8.0 \text{ N}$ force.

Answer: ________________________ $^\circ$ [2]

Section C: Momentum, Work, Energy, and Power

(Answer all questions in this section.)

11. State the principle of conservation of linear momentum.

_________________________________________________________________________ [2]

12. A trolley $A$ of mass $2.0 \text{ kg}$ moving at $3.0 \text{ m s}^{-1}$ collides with a stationary trolley $B$ of mass $1.0 \text{ kg}$ . After the collision, the two trolleys stick together and move with a common velocity $v$ . (a) Calculate the common velocity $v$ .

Answer: ________________________ $\text{m s}^{-1}$ [3]

(b) Show that the collision is inelastic by comparing the kinetic energy before and after the collision.

[3]

13. A ball of mass $0.20 \text{ kg}$ hits a vertical wall horizontally with a speed of $10 \text{ m s}^{-1}$ and rebounds horizontally with a speed of $8.0 \text{ m s}^{-1}$ . (a) Calculate the change in momentum of the ball.

Answer: ________________________ $\text{N s}$ [3]

(b) If the contact time with the wall is $0.05 \text{ s}$ , calculate the average force exerted by the wall on the ball.

Answer: ________________________ $\text{N}$ [2]

14. Define work done by a force.

_________________________________________________________________________ [2]

15. A crane lifts a load of mass $500 \text{ kg}$ vertically through a height of $20 \text{ m}$ in $10 \text{ s}$ . (a) Calculate the work done against gravity.

Answer: ________________________ $\text{J}$ [2]

(b) Calculate the useful power output of the crane.

Answer: ________________________ $\text{W}$ [2]

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

Answer: ________________________ $\text{N}$ [1]

(b) Calculate the power developed by the engine.

Answer: ________________________ $\text{W}$ [2]

17. A roller-coaster car starts from rest at the top of a hill of height $30 \text{ m}$ . It travels down the track to the bottom. Assume friction and air resistance are negligible. (a) State the principle of conservation of energy as applied to this system.

_________________________________________________________________________ [1]

(b) Calculate the speed of the car at the bottom of the hill.

Answer: ________________________ $\text{m s}^{-1}$ [3]

18. An electric motor lifts a mass of $10 \text{ kg}$ through a height of $2.0 \text{ m}$ . The motor consumes $250 \text{ J}$ of electrical energy. (a) Calculate the useful energy output (gain in gravitational potential energy).

Answer: ________________________ $\text{J}$ [2]

(b) Calculate the efficiency of the motor.

Answer: ________________________ $\%$ [2]

19. A spring obeys Hooke's Law. When a force of $10 \text{ N}$ is applied, the extension is $0.05 \text{ m}$ . (a) Calculate the spring constant $k$ .

Answer: ________________________ $\text{N m}^{-1}$ [2]

(b) Calculate the elastic potential energy stored in the spring at this extension.

Answer: ________________________ $\text{J}$ [2]

20. A student investigates the relationship between force and extension for a rubber band. The loading and unloading curves do not coincide. (a) Name this phenomenon.

Answer: ________________________ [1]

(b) Explain what the area enclosed between the loading and unloading curves represents.

_________________________________________________________________________ [2]

End of Paper

Answers

TuitionGoWhere Practice Paper - Physics H1 A-Level (Answer Key)

Version: 3 of 5
Subject: Physics H1
Topic: Mechanics

Section A: Kinematics and Dynamics

1. (a) $a = \frac{v - u}{t} = \frac{24 - 0}{8.0} = 3.0 \text{ m s}^{-2}$ [2] (b) $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3.0)(8.0)^2 = 96 \text{ m}$ Alternative: $s = \frac{1}{2}(u+v)t = \frac{1}{2}(0+24)(8) = 96 \text{ m}$ [2]

2. (a) The only force acting is gravity (weight), which is constant near the Earth's surface. Therefore, acceleration is constant ( $g$ ). [1] (b) Taking upward as positive: $u = +15 \text{ m s}^{-1}$ , $a = -9.81 \text{ m s}^{-2}$ , $t = 4.0 \text{ s}$ $s = ut + \frac{1}{2}at^2$ $s = (15)(4.0) + \frac{1}{2}(-9.81)(4.0)^2$ $s = 60 - 78.48 = -18.48 \text{ m}$ Displacement is $-18.48 \text{ m}$ (downwards). Height of cliff = $18 \text{ m}$ (to 2 s.f.) [3]

3. (a) The train is decelerating uniformly (constant negative acceleration) until it stops. [1] (b) Distance = Area under graph. Area 1 (triangle): $\frac{1}{2} \times 2 \times 4 = 4 \text{ m}$ Area 2 (rectangle): $3 \times 4 = 12 \text{ m}$ Area 3 (triangle): $\frac{1}{2} \times 2 \times 4 = 4 \text{ m}$ Total distance = $4 + 12 + 4 = 20 \text{ m}$ [3]

4. (a) $0 \text{ m s}^{-2}$ (Horizontal acceleration is zero as air resistance is neglected). [1] (b) Vertical motion: $u_y = 0$ , $s_y = 45 \text{ m}$ , $a_y = 9.81 \text{ m s}^{-2}$ $s = ut + \frac{1}{2}at^2 \Rightarrow 45 = 0 + \frac{1}{2}(9.81)t^2$ $t^2 = \frac{90}{9.81} \approx 9.174$ $t = 3.03 \text{ s}$ (approx $3.0 \text{ s}$ ) [2] (c) Horizontal distance $s_x = u_x t = 20 \times 3.03 = 60.6 \text{ m}$ (approx $61 \text{ m}$ ) [2]

5. (a) As velocity increases, air resistance (drag) increases. The resultant downward force ( $Weight - Drag$ ) decreases. Since $F=ma$ , acceleration decreases. [2] (b) Acceleration is zero. At terminal velocity, air resistance equals weight, so the resultant force is zero. [2]

Section B: Forces, Moments, and Equilibrium

6. The product of the force and the perpendicular distance from the pivot (or axis of rotation) to the line of action of the force. [2]

7. (a) Diagram should show:

Weight ( $200 \text{ N}$ ) acting downwards at the center of the beam ( $2.0 \text{ m}$ from A).
Tension ( $T$ ) acting at B, at $30^\circ$ to the beam (upwards and towards wall).
Reaction force at hinge A (vertical and/or horizontal components). [2] (b) Taking moments about A: Clockwise moment = Anticlockwise moment Weight moment: $200 \text{ N} \times 2.0 \text{ m} = 400 \text{ N m}$ Tension moment: The perpendicular component of Tension is $T \sin(30^\circ)$ . Distance is $4.0 \text{ m}$ . Moment = $(T \sin 30^\circ) \times 4.0$ $400 = T \times 0.5 \times 4.0$ $400 = 2.0 T$ $T = 200 \text{ N}$ [3]

8. (a) The wall is smooth, so it exerts only a normal horizontal force on the ladder. To balance this horizontal force and prevent the ladder from sliding outwards, there must be a horizontal frictional force at the ground acting towards the wall. [1] (b)

The resultant force on the ladder is zero (translational equilibrium).
The resultant moment about any point is zero (rotational equilibrium). [2]

9. (a) Component down slope = $mg \sin \theta = 5.0 \times 9.81 \times \sin(30^\circ) = 24.5 \text{ N}$ [2] (b) Since the block is stationary, forces are balanced. Friction = Component down slope = $24.5 \text{ N}$ (or $25 \text{ N}$ ). [1] (c) Normal contact force = $mg \cos \theta = 5.0 \times 9.81 \times \cos(30^\circ) = 42.5 \text{ N}$ (or $42 \text{ N}$ ). [2]

10. (a) Resultant $R = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N}$ [2] (b) $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{6.0}{8.0} = 0.75$ $\theta = \tan^{-1}(0.75) = 36.9^\circ$ (or $37^\circ$ ) [2]

Section C: Momentum, Work, Energy, and Power

11. In a closed system (no external forces), the total momentum before interaction is equal to the total momentum after interaction. [2]

12. (a) Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ $(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v$ $6.0 = 3.0 v$ $v = 2.0 \text{ m s}^{-1}$ [3] (b) $KE_{before} = \frac{1}{2} m_A u_A^2 = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}$ $KE_{after} = \frac{1}{2} (m_A+m_B) v^2 = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}$ Since $KE_{before} \neq KE_{after}$ ( $9.0 \text{ J} > 6.0 \text{ J}$ ), kinetic energy is not conserved. The collision is inelastic. [3]

13. (a) Taking initial direction as positive: $u = +10 \text{ m s}^{-1}$ , $v = -8.0 \text{ m s}^{-1}$ $\Delta p = m(v - u) = 0.20 (-8.0 - 10) = 0.20 (-18) = -3.6 \text{ N s}$ Magnitude of change in momentum = $3.6 \text{ N s}$ [3] (b) $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{3.6}{0.05} = 72 \text{ N}$ [2]

14. Work done is defined as the product of the force and the displacement moved in the direction of the force. ( $W = Fs \cos \theta$ ) [2]

15. (a) Work done = Gain in GPE = $mgh = 500 \times 9.81 \times 20 = 98,100 \text{ J}$ (or $98 \text{ kJ}$ ) [2] (b) Power = $\frac{\text{Work}}{\text{Time}} = \frac{98,100}{10} = 9,810 \text{ W}$ (or $9.8 \text{ kW}$ ) [2]

16. (a) Since speed is constant, acceleration is zero. Driving force = Resistive force = $800 \text{ N}$ . [1] (b) Power = $Fv = 800 \times 25 = 20,000 \text{ W}$ (or $20 \text{ kW}$ ) [2]

17. (a) The total mechanical energy (sum of kinetic and potential energy) remains constant in the absence of external resistive forces. [1] (b) Loss in GPE = Gain in KE $mgh = \frac{1}{2}mv^2$ $gh = \frac{1}{2}v^2 \Rightarrow v = \sqrt{2gh}$ $v = \sqrt{2 \times 9.81 \times 30} = \sqrt{588.6} = 24.3 \text{ m s}^{-1}$ (or $24 \text{ m s}^{-1}$ ) [3]

18. (a) Useful Energy = $mgh = 10 \times 9.81 \times 2.0 = 196.2 \text{ J}$ (or $196 \text{ J}$ ) [2] (b) Efficiency = $\frac{\text{Useful Output}}{\text{Total Input}} \times 100\%$ Efficiency = $\frac{196.2}{250} \times 100\% = 78.48\%$ (or $78\%$ ) [2]

19. (a) $F = kx \Rightarrow k = \frac{F}{x} = \frac{10}{0.05} = 200 \text{ N m}^{-1}$ [2] (b) $EPE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.05)^2 = 100 \times 0.0025 = 0.25 \text{ J}$ Alternative: $EPE = \frac{1}{2}Fx = \frac{1}{2}(10)(0.05) = 0.25 \text{ J}$ [2]

20. (a) Hysteresis [1] (b) The area represents the energy dissipated (lost as heat/internal energy) during the loading and unloading cycle. [2]

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