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A Level H1 Physics Mechanics Quiz

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A-Level Physics H1 Quiz - Mechanics

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

Duration: 45 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Numerical answers should be given to an appropriate number of significant figures.
Take the acceleration of free fall $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

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

1. A car accelerates uniformly from rest to a speed of $24 \text{ m s}^{-1}$ in $6.0 \text{ s}$ . It then travels at this constant speed for $10 \text{ s}$ before decelerating uniformly to rest in $4.0 \text{ s}$ .

(a) Calculate the acceleration of the car during the first $6.0 \text{ s}$ .

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

(b) Calculate the total distance travelled by the car.

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

2. A stone is thrown horizontally from the top of a cliff with a speed of $15 \text{ m s}^{-1}$ . The cliff is $45 \text{ m}$ high. Air resistance is negligible.

(a) Calculate the time taken for the stone to reach the ground.

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

(b) Calculate the horizontal distance from the base of the cliff where the stone lands.

Answer: ____________________ $\text{m}$ [1]

3. State Newton’s Second Law of Motion in terms of momentum.

_________________________________________________________________________ [2]

4. 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.

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

5. Two trolleys, A and B, move along a straight frictionless track. Trolley A has mass $2.0 \text{ kg}$ and velocity $3.0 \text{ m s}^{-1}$ to the right. Trolley B has mass $1.0 \text{ kg}$ and velocity $2.0 \text{ m s}^{-1}$ to the left. They collide and stick together.

Calculate the common velocity of the trolleys after the collision. (Take right as positive).

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

Section B: Forces, Equilibrium, and Energy (Questions 6–12)

6. A uniform beam AB of length $4.0 \text{ m}$ and weight $120 \text{ N}$ is hinged at end A to a vertical wall. The beam is held horizontal by a cable attached to end B, which makes an angle of $30^\circ$ with the beam.

(a) Draw a free-body diagram showing all forces acting on the beam. Label the forces clearly.

[2]

(b) Calculate the tension in the cable.

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

7. Define the term work done by a force.

_________________________________________________________________________ [1]

8. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards at a constant speed of $2.0 \text{ m s}^{-1}$ .

Calculate the power developed by the crane motor. (Ignore air resistance).

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

9. A ball of mass $0.20 \text{ kg}$ is dropped from a height of $2.0 \text{ m}$ . It rebounds to a height of $1.5 \text{ m}$ .

(a) Calculate the loss in gravitational potential energy during the fall and rebound process.

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

(b) Suggest what happens to the lost energy.

_________________________________________________________________________ [1]

10. A car of mass $1200 \text{ kg}$ travels up a slope inclined at $5.0^\circ$ to the horizontal at a constant speed of $20 \text{ m s}^{-1}$ . The resistive forces acting on the car total $400 \text{ N}$ .

Calculate the driving force required to maintain this constant speed.

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

11. Explain why the principle of conservation of energy applies to a pendulum swinging in a vacuum, but not to one swinging in air.

_________________________________________________________________________ [2]

12. A spring obeys Hooke’s Law. When a force of $10 \text{ N}$ is applied, the extension is $5.0 \text{ cm}$ .

Calculate the elastic potential energy stored in the spring when the extension is $5.0 \text{ cm}$ .

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

Section C: Momentum, Impulse, and Advanced Applications (Questions 13–20)

13. State the principle of conservation of linear momentum.

_________________________________________________________________________ [2]

14. A golf club strikes a stationary golf ball of mass $0.045 \text{ kg}$ . The club is in contact with the ball for $0.50 \text{ ms}$ . The ball leaves the club with a speed of $50 \text{ m s}^{-1}$ .

Calculate the average force exerted by the club on the ball.

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

15. Distinguish between an elastic collision and an inelastic collision in terms of kinetic energy.

_________________________________________________________________________ [2]

16. A particle of mass $m$ moves in a horizontal circle of radius $r$ with constant speed $v$ .

(a) State the direction of the resultant force acting on the particle.

_________________________________________________________________________ [1]

(b) Derive the expression for the centripetal acceleration $a = \frac{v^2}{r}$ . (You may use vector diagrams or kinematic arguments).

[3]

17. A box of mass $10 \text{ kg}$ is pushed across a horizontal floor by a force of $50 \text{ N}$ acting at an angle of $30^\circ$ below the horizontal. The coefficient of dynamic friction between the box and the floor is $0.20$ .

(a) Calculate the normal reaction force acting on the box.

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

(b) Calculate the acceleration of the box.

Answer: ____________________ $\text{m s}^{-2}$ [3]

18. The graph below shows the variation of velocity $v$ with time $t$ for a falling object subject to air resistance.

(Imagine a graph starting at v=0, curving upwards with decreasing gradient, approaching a horizontal asymptote at $v_T$ )

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

_________________________________________________________________________ [2]

(b) State the condition for the object to reach terminal velocity $v_T$ .

_________________________________________________________________________ [1]

19. A projectile is launched with speed $u$ at an angle $\theta$ to the horizontal. Show that the maximum height $H$ reached is given by: $H = \frac{u^2 \sin^2 \theta}{2g}$

[3]

20. Two spheres, X and Y, undergo a head-on collision on a smooth surface.

Sphere X: mass $2m$ , initial velocity $+u$ .
Sphere Y: mass $m$ , initial velocity $-u$ .

After the collision, sphere X moves with velocity $+0.5u$ .

(a) Calculate the velocity of sphere Y after the collision.

Answer: ____________________ [3]

(b) Determine whether this collision is elastic or inelastic. Show your working.

_________________________________________________________________________ [3]

End of Quiz

Answers

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

1. (a) $a = \frac{v - u}{t} = \frac{24 - 0}{6.0} = 4.0 \text{ m s}^{-2}$ [1] (b) Distance = Area under graph. Area 1 (triangle) = $\frac{1}{2} \times 6.0 \times 24 = 72 \text{ m}$ Area 2 (rectangle) = $10 \times 24 = 240 \text{ m}$ Area 3 (triangle) = $\frac{1}{2} \times 4.0 \times 24 = 48 \text{ m}$ Total Distance = $72 + 240 + 48 = 360 \text{ m}$ [2]

2. (a) Vertical motion: $s = ut + \frac{1}{2}at^2$ . $u_y = 0$ , $a = g = 9.81$ , $s = 45$ . $45 = 0 + \frac{1}{2}(9.81)t^2$ $t^2 = \frac{90}{9.81} = 9.174$ $t = 3.03 \text{ s}$ (approx $3.0 \text{ s}$ ) [2] (b) Horizontal distance $s_x = u_x t = 15 \times 3.03 = 45.45 \text{ m}$ (approx $45 \text{ m}$ ) [1]

3. The resultant force acting on an object is equal to the rate of change of its momentum. [1] Mathematically: $F = \frac{\Delta p}{\Delta t}$ or $F = \frac{d(mv)}{dt}$ . [1]

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

5. Conservation of Momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ $(2.0)(3.0) + (1.0)(-2.0) = (2.0 + 1.0)v$ [1] $6.0 - 2.0 = 3.0v$ $4.0 = 3.0v$ $v = +1.33 \text{ m s}^{-1}$ (to the right) [2]

6. (a) Diagram should show:

Weight $W$ acting downwards from the center of the beam. [1]
Tension $T$ acting from end B at $30^\circ$ to the beam (upwards/left). [1]
Reaction force at hinge A (vertical/horizontal components or resultant). [Accept if implied or not explicitly asked to calculate, but good practice]. Note: Question asks for forces on beam. Weight and Tension are critical. (b) Take moments about hinge A. Clockwise Moment = Anticlockwise Moment $W \times (\text{distance to center}) = T_{\perp} \times (\text{length})$ $120 \times 2.0 = (T \sin 30^\circ) \times 4.0$ [1] $240 = T(0.5)(4.0)$ $240 = 2.0 T$ $T = 120 \text{ N}$ [2]

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

8. Power $P = F v$ . Since speed is constant, Driving Force $F = \text{Weight} = mg$ . [1] $F = 500 \times 9.81 = 4905 \text{ N}$ $P = 4905 \times 2.0 = 9810 \text{ W}$ (or $9.8 \text{ kW}$ ) [1]

9. (a) Initial GPE $= mgh_1 = 0.20 \times 9.81 \times 2.0 = 3.924 \text{ J}$ Final GPE $= mgh_2 = 0.20 \times 9.81 \times 1.5 = 2.943 \text{ J}$ Loss $= 3.924 - 2.943 = 0.981 \text{ J}$ (approx $0.98 \text{ J}$ ) [2] (b) Energy is converted to thermal energy (heat) and sound upon impact with the ground / due to air resistance. [1]

10. Forces acting down the slope: Component of weight ( $mg \sin \theta$ ) + Resistive forces. Driving Force $F_D$ acts up the slope. Since speed is constant, $F_D = mg \sin \theta + F_{resist}$ . [1] $mg \sin 5.0^\circ = 1200 \times 9.81 \times \sin 5.0^\circ = 1200 \times 9.81 \times 0.08716 = 1026 \text{ N}$ [1] $F_D = 1026 + 400 = 1426 \text{ N}$ (approx $1430 \text{ N}$ ) [1]

11. In a vacuum, only gravity (conservative force) does work, so mechanical energy (KE + GPE) is conserved. [1] In air, air resistance (non-conservative force) does negative work, converting mechanical energy into thermal energy/heat, so total mechanical energy decreases. [1]

12. Spring constant $k = \frac{F}{x} = \frac{10}{0.05} = 200 \text{ N m}^{-1}$ . [1] Elastic PE $E = \frac{1}{2} k x^2 = \frac{1}{2} (200) (0.05)^2 = 100 \times 0.0025 = 0.25 \text{ J}$ . (Alternatively $E = \frac{1}{2} F x = \frac{1}{2} (10)(0.05) = 0.25 \text{ J}$ ) [1]

13. In a closed/isolated system (no external forces), [1] the total linear momentum remains constant (or total momentum before collision = total momentum after collision). [1]

14. Impulse $= \Delta p = m(v - u) = 0.045(50 - 0) = 2.25 \text{ N s}$ . [1] Impulse $= F_{avg} \Delta t$ $2.25 = F_{avg} \times (0.50 \times 10^{-3})$ [1] $F_{avg} = \frac{2.25}{0.0005} = 4500 \text{ N}$ [1]

15. In an elastic collision, total kinetic energy is conserved. [1] In an inelastic collision, total kinetic energy is not conserved (some is converted to other forms like heat/sound/deformation). [1]

16. (a) Towards the center of the circle. [1] (b) Consider velocity vectors $\vec{v}_1$ and $\vec{v}_2$ at times $t$ and $t+\Delta t$ . The change in velocity $\Delta \vec{v}$ points towards the center. Using similar triangles for velocity vector triangle and position triangle: $\frac{\Delta v}{v} = \frac{\Delta s}{r}$ $\Delta v = \frac{v \Delta s}{r}$ Divide by $\Delta t$ : $\frac{\Delta v}{\Delta t} = \frac{v}{r} \frac{\Delta s}{\Delta t}$ As $\Delta t \to 0$ , $\frac{\Delta v}{\Delta t} = a$ and $\frac{\Delta s}{\Delta t} = v$ . Therefore, $a = \frac{v^2}{r}$ . [3] (Award marks for logical steps/diagram)

17. (a) Resolve forces vertically. Upward forces = Downward forces. $R = mg + F \sin 30^\circ$ [1] $R = (10 \times 9.81) + (50 \times 0.5) = 98.1 + 25 = 123.1 \text{ N}$ (approx $123 \text{ N}$ ) [1] (b) Frictional force $F_f = \mu R = 0.20 \times 123.1 = 24.62 \text{ N}$ . Horizontal component of applied force $F_x = F \cos 30^\circ = 50 \cos 30^\circ = 43.30 \text{ N}$ . Resultant horizontal force $F_{net} = F_x - F_f = 43.30 - 24.62 = 18.68 \text{ N}$ . [1] $a = \frac{F_{net}}{m} = \frac{18.68}{10} = 1.87 \text{ m s}^{-2}$ (approx $1.9 \text{ m s}^{-2}$ ) [2]

18. (a) As speed increases, air resistance (drag) increases. [1] The resultant downward force ( $Weight - Drag$ ) decreases. Since $F=ma$ , acceleration (gradient) decreases. [1] (b) Air resistance equals weight (Resultant force is zero). [1]

19. At maximum height, vertical velocity $v_y = 0$ . Initial vertical velocity $u_y = u \sin \theta$ . Using $v^2 = u^2 + 2as$ with $a = -g$ and $s = H$ : $0 = (u \sin \theta)^2 - 2gH$ [1] $2gH = u^2 \sin^2 \theta$ [1] $H = \frac{u^2 \sin^2 \theta}{2g}$ [1]

20. (a) Conservation of Momentum: $P_{initial} = (2m)(u) + (m)(-u) = 2mu - mu = mu$ . $P_{final} = (2m)(0.5u) + (m)(v_Y) = mu + m v_Y$ . $mu = mu + m v_Y \Rightarrow m v_Y = 0 \Rightarrow v_Y = 0$ . [3] (b) Check Kinetic Energy. $KE_{initial} = \frac{1}{2}(2m)u^2 + \frac{1}{2}(m)(-u)^2 = mu^2 + 0.5mu^2 = 1.5mu^2$ . $KE_{final} = \frac{1}{2}(2m)(0.5u)^2 + \frac{1}{2}(m)(0)^2 = m(0.25u^2) = 0.25mu^2$ . $KE_{initial} \neq KE_{final}$ ( $1.5mu^2 > 0.25mu^2$ ). Therefore, the collision is inelastic. [3]

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