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A Level H1 Physics Mechanics Quiz

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Questions

A-Level Physics H1 Quiz - Mechanics

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 45

Duration: 60 minutes
Total Marks: 45
Instructions:

Answer all questions.
Show all working clearly. Marks are awarded for correct method as well as final answers.
Use $g = 9.81 \, \text{m s}^{-2}$ where necessary.
Non-programmable calculators are allowed.

Section A: Kinematics and Dynamics (Questions 1–5)

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.
[2]

(b) Calculate the distance travelled by the car during this time.
[2]

2. A ball is thrown vertically upwards with an initial velocity of $15 \, \text{m s}^{-1}$ . Air resistance is negligible.
(a) Calculate the maximum height reached by the ball.
[2]

(b) Determine the total time taken for the ball to return to its starting position.
[2]

3. The graph below shows the variation of velocity $v$ with time $t$ for a toy car moving in a straight line.

(Imagine a graph: Velocity starts at 0, increases linearly to 10 m/s at t=4s, remains constant at 10 m/s until t=8s, then decreases linearly to 0 at t=12s.)

(a) Describe the motion of the car between $t = 4 \, \text{s}$ and $t = 8 \, \text{s}$ .
[1]

(b) Calculate the total distance travelled by the car during the 12 seconds.
[3]

4. State Newton’s Second Law of Motion in terms of momentum.
[2]

5. A block of mass $5.0 \, \text{kg}$ rests on a rough horizontal surface. A horizontal force of $20 \, \text{N}$ is applied to the block, causing it to accelerate at $2.0 \, \text{m s}^{-2}$ .
Calculate the magnitude of the frictional force acting on the block.
[3]

Section B: Forces, Equilibrium, and Work/Energy (Questions 6–10)

6. A uniform beam AB of length $4.0 \, \text{m}$ and weight $120 \, \text{N}$ is hinged at end A. It is held horizontal by a vertical cable attached at end B.
(a) Draw a free-body diagram for the beam, showing all forces acting on it. Label the forces clearly.
[2]

(b) Calculate the tension in the cable.
[2]

7. Define the term work done by a force.
[2]

8. A crane lifts a load of mass $500 \, \text{kg}$ vertically through a height of $20 \, \text{m}$ at a constant speed. The lift takes $40 \, \text{s}$ .
(a) Calculate the work done by the crane in lifting the load.
[2]

(b) Calculate the useful power output of the crane.
[2]

9. A skier of mass $70 \, \text{kg}$ slides down a slope inclined at $30^\circ$ to the horizontal. The length of the slope is $100 \, \text{m}$ .
(a) Calculate the loss in gravitational potential energy as the skier moves from the top to the bottom of the slope.
[3]

(b) If the skier starts from rest and reaches a speed of $25 \, \text{m s}^{-1}$ at the bottom, calculate the work done against resistive forces (friction and air resistance).
[3]

10. Explain why the efficiency of a machine is always less than 100%.
[2]

Section C: Momentum and Collisions (Questions 11–15)

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 $v$ .
[3]

(b) Determine whether the collision is elastic or inelastic. Show your working.
[3]

13. A ball of mass $0.20 \, \text{kg}$ strikes 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.
[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.
[2]

14. Two ice skaters, initially at rest, push away from each other. Skater X has a mass of $60 \, \text{kg}$ and moves off with a speed of $2.0 \, \text{m s}^{-1}$ . Skater Y has a mass of $80 \, \text{kg}$ .
Calculate the speed of Skater Y.
[3]

15. A rocket accelerates upwards in space by ejecting gas downwards. Explain, using Newton’s laws of motion, how the rocket accelerates.
[3]

Section D: Advanced Mechanics Applications (Questions 16–20)

16. A projectile is fired horizontally from the top of a cliff $45 \, \text{m}$ high with a speed of $20 \, \text{m s}^{-1}$ . Air resistance is negligible.
(a) Calculate the time taken for the projectile to hit the ground.
[2]

(b) Calculate the horizontal distance from the base of the cliff where the projectile lands.
[2]

17. A car of mass $1200 \, \text{kg}$ travels around a circular bend of radius $50 \, \text{m}$ at a constant speed of $15 \, \text{m s}^{-1}$ .
Calculate the centripetal force required to keep the car on the circular path.
[2]

18. The graph shows the force $F$ acting on an object versus its displacement $s$ . The force increases linearly from $0 \, \text{N}$ at $s=0$ to $10 \, \text{N}$ at $s=5 \, \text{m}$ .
Calculate the work done by the force over this displacement.
[2]

19. A student claims that "if an object is moving, there must be a net force acting on it."
Explain why this statement is incorrect, referring to Newton’s First Law.
[2]

20. A box is pushed up a rough inclined plane at a constant speed.
(a) State the direction of the frictional force acting on the box.
[1]

(b) Explain why the work done by the pushing force is greater than the gain in gravitational potential energy of the box.
[2]

[END OF QUIZ]

Answers

A-Level Physics H1 Quiz - Mechanics (Answer Key)

1.
(a) Using $v = u + at$ :
$24 = 0 + a(8.0)$
$a = \frac{24}{8.0} = 3.0 \, \text{m s}^{-2}$
[1 for formula/substitution, 1 for answer]

(b) Using $s = ut + \frac{1}{2}at^2$ :
$s = 0 + \frac{1}{2}(3.0)(8.0)^2$
$s = 0.5 \times 3.0 \times 64 = 96 \, \text{m}$
(Alternatively, $s = \frac{1}{2}(u+v)t = \frac{1}{2}(0+24)(8) = 96 \, \text{m}$ )
[1 for formula/substitution, 1 for answer]

2.
(a) At max height, $v=0$ . Using $v^2 = u^2 + 2as$ :
$0 = (15)^2 + 2(-9.81)s$
$19.62s = 225$
$s = 11.47 \approx 11.5 \, \text{m}$
[1 for formula/substitution, 1 for answer]

(b) Time to reach max height: $v = u + at \Rightarrow 0 = 15 - 9.81t \Rightarrow t = 1.53 \, \text{s}$ .
Total time = $2 \times 1.53 = 3.06 \approx 3.1 \, \text{s}$ .
[1 for time up, 1 for total time]

3.
(a) The car moves with constant velocity (or constant speed in a straight line).
[1]

(b) Distance = Area under graph.
Area 1 (Triangle, 0-4s): $\frac{1}{2} \times 4 \times 10 = 20 \, \text{m}$
Area 2 (Rectangle, 4-8s): $4 \times 10 = 40 \, \text{m}$
Area 3 (Triangle, 8-12s): $\frac{1}{2} \times 4 \times 10 = 20 \, \text{m}$
Total Distance = $20 + 40 + 20 = 80 \, \text{m}$
[1 for each area component or correct method, 1 for final answer]

4.
The rate of change of momentum of a body is directly proportional to the resultant force acting on it [1] and takes place in the direction of the force. [1]
(Or: $F = \frac{dp}{dt}$ )

5.
Resultant Force $F_{net} = ma = 5.0 \times 2.0 = 10 \, \text{N}$ .
$F_{net} = F_{applied} - F_{friction}$
$10 = 20 - F_{friction}$
$F_{friction} = 10 \, \text{N}$
[1 for $F_{net}$ , 1 for equation, 1 for answer]

6.
(a) Diagram should show:

Weight ( $120 \, \text{N}$ ) acting downwards at the center of the beam (2.0 m from A).
Tension ( $T$ ) acting upwards at B (4.0 m from A).
Reaction force at hinge A (vertical component upwards, horizontal component if any, though here only vertical forces exist so reaction is vertical upwards).
[1 for correct positions, 1 for correct labels/directions]

(b) Taking moments about A:
Clockwise Moment = Anticlockwise Moment
$120 \times 2.0 = T \times 4.0$
$240 = 4T$
$T = 60 \, \text{N}$
[1 for moment equation, 1 for answer]

7.
Work done is defined as the product of the force [1] and the displacement moved in the direction of the force. [1]
( $W = Fs \cos \theta$ )

8.
(a) Work Done = Gain in GPE = $mgh$
$W = 500 \times 9.81 \times 20$
$W = 98,100 \, \text{J}$ (or $98.1 \, \text{kJ}$ )
[1 for formula, 1 for answer]

(b) Power = $\frac{\text{Work Done}}{\text{Time}}$
$P = \frac{98,100}{40}$
$P = 2,452.5 \approx 2,450 \, \text{W}$ (or $2.45 \, \text{kW}$ )
[1 for formula, 1 for answer]

9.
(a) Vertical height $h = 100 \sin(30^\circ) = 50 \, \text{m}$ .
Loss in GPE = $mgh = 70 \times 9.81 \times 50$
Loss in GPE = $34,335 \approx 34,300 \, \text{J}$
[1 for height, 1 for formula, 1 for answer]

(b) Gain in KE = $\frac{1}{2}mv^2 = \frac{1}{2}(70)(25)^2 = 21,875 \, \text{J}$ .
Work done against resistance = Loss in GPE - Gain in KE
$W_{res} = 34,335 - 21,875 = 12,460 \approx 12,500 \, \text{J}$
[1 for KE calc, 1 for energy balance concept, 1 for answer]

10.
Some energy is always dissipated/lost as heat or sound due to friction/resistance [1], so useful output energy is always less than total input energy. [1]

11.
For a closed/isolated system [1], the total momentum before collision/interaction is equal to the total momentum after collision/interaction, provided no external forces act. [1]

12.
(a) Conservation of Momentum:
$m_A u_A + m_B u_B = (m_A + m_B)v$
$(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v$
$6.0 = 3.0v$
$v = 2.0 \, \text{m s}^{-1}$
[1 for equation, 1 for substitution, 1 for answer]

(b) Initial KE = $\frac{1}{2}(2.0)(3.0)^2 = 9.0 \, \text{J}$ .
Final KE = $\frac{1}{2}(3.0)(2.0)^2 = 6.0 \, \text{J}$ .
Since KE is not conserved ( $9.0 \neq 6.0$ ), the collision is inelastic.
[1 for initial KE, 1 for final KE, 1 for conclusion]

13.
(a) Let direction towards wall be positive.
$u = +10 \, \text{m s}^{-1}$ , $v = -8.0 \, \text{m s}^{-1}$ .
$\Delta p = m(v - u) = 0.20(-8.0 - 10)$
$\Delta p = 0.20(-18) = -3.6 \, \text{N s}$ (or $\text{kg m s}^{-1}$ ).
Magnitude is $3.6 \, \text{N s}$ .
[1 for correct signs/direction, 1 for formula, 1 for answer]

(b) $F_{avg} = \frac{\Delta p}{\Delta t}$
$F_{avg} = \frac{3.6}{0.05} = 72 \, \text{N}$
[1 for formula, 1 for answer]

14.
Total initial momentum = 0.
$m_X v_X + m_Y v_Y = 0$
$(60)(2.0) + (80)(v_Y) = 0$
$120 + 80v_Y = 0$
$v_Y = -\frac{120}{80} = -1.5 \, \text{m s}^{-1}$ .
Speed = $1.5 \, \text{m s}^{-1}$ .
[1 for conservation principle, 1 for substitution, 1 for answer]

15.
The rocket exerts a force on the gas to eject it downwards (Action) [1]. By Newton’s Third Law, the gas exerts an equal and opposite force on the rocket upwards (Reaction) [1]. This resultant upward force causes the rocket to accelerate upwards (Newton’s Second Law) [1].

16.
(a) Vertical motion: $u_y = 0$ , $s_y = 45 \, \text{m}$ , $a = 9.81 \, \text{m s}^{-2}$ .
$s = ut + \frac{1}{2}at^2 \Rightarrow 45 = 0 + \frac{1}{2}(9.81)t^2$
$t^2 = \frac{90}{9.81} \approx 9.17$
$t = 3.03 \approx 3.0 \, \text{s}$
[1 for formula/sub, 1 for answer]

(b) Horizontal motion: $v_x = 20 \, \text{m s}^{-1}$ (constant).
Distance = $v_x \times t = 20 \times 3.03 = 60.6 \approx 61 \, \text{m}$ .
[1 for formula, 1 for answer]

17.
$F_c = \frac{mv^2}{r}$
$F_c = \frac{1200 \times (15)^2}{50}$
$F_c = \frac{1200 \times 225}{50} = 24 \times 225 = 5,400 \, \text{N}$
[1 for formula, 1 for answer]

18.
Work Done = Area under F-s graph.
Area = Triangle = $\frac{1}{2} \times \text{base} \times \text{height}$
$W = \frac{1}{2} \times 5 \times 10 = 25 \, \text{J}$
[1 for method, 1 for answer]

19.
Newton’s First Law states that an object continues in its state of rest or uniform motion in a straight line unless acted upon by a resultant external force [1]. Therefore, an object moving at constant velocity has zero net force acting on it. [1]

20.
(a) Down the slope (opposite to the direction of motion). [1]

(b) The work done by the pushing force goes into increasing the gravitational potential energy [1] AND overcoming friction/resistive forces (which dissipates energy as heat). [1]