A Level H1 Physics Mechanics Quiz

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]