A Level H1 Physics Practice Paper 2

TuitionGoWhere Practice Paper - Physics H1 A-Level (Answers)

Version 2 of 5

Section A

1. (a) Mean reading = $(12.42 + 12.44 + 12.41 + 12.43) / 4 = 12.425 \text{ mm}$.
Corrected mean = Mean - Zero Error = $12.425 - 0.02 = 12.405 \text{ mm}$.
Rounding to 2 decimal places (precision of instrument): 12.41 mm (or 12.40 mm depending on rounding convention, usually keep extra digit for intermediate). Let's use 12.41 mm.
[1 for mean, 1 for correction]

(b) Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\frac{12.41 \times 10^{-3}}{2})^3 = 9.99 \times 10^{-7} \text{ m}^3$.
Density $\rho = \frac{m}{V} = \frac{7.80 \times 10^{-3}}{9.99 \times 10^{-7}} = 7808 \text{ kg m}^{-3}$.
Uncertainty:
% unc in $m = \frac{0.05}{7.80} \times 100 = 0.64\%$.
% unc in $d = \frac{0.01}{12.41} \times 100 \approx 0.08\%$ (using range/2 or least count). Let's assume least count 0.01mm.
% unc in $V = 3 \times \% \text{ unc in } d = 3 \times 0.08 = 0.24\%$.
Total % unc in $\rho = 0.64 + 0.24 = 0.88\%$.
Absolute unc = $0.0088 \times 7808 \approx 69 \text{ kg m}^{-3}$.
Answer: $7810 \pm 70 \text{ kg m}^{-3}$.
[1 for density, 1 for % unc method, 1 for final answer]

2. (a) Acceleration $a = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{5} = \mathbf{4.0 \text{ m s}^{-2}}$.
[1]

(b) Distance = Area under graph.
Area = Area of triangle (0-5s) + Area of rectangle (5-15s) + Area of triangle (15-20s).
$= (0.5 \times 5 \times 20) + (10 \times 20) + (0.5 \times 5 \times 20)$
$= 50 + 200 + 50 = \mathbf{300 \text{ m}}$.
[1 for method, 1 for answer]

(c) At constant velocity, acceleration is zero. According to Newton's First Law, if acceleration is zero, the resultant force is zero. Therefore, the driving force equals the resistive forces (friction/air resistance).
[1 for zero resultant force, 1 for explanation]

3. (a) $v_x = 30 \cos 40^\circ = \mathbf{23.0 \text{ m s}^{-1}}$.
[1]

(b) Vertical component $v_y = 30 \sin 40^\circ = 19.28 \text{ m s}^{-1}$.
At max height, $v_y = 0$.
$v^2 = u^2 + 2as \Rightarrow 0 = (19.28)^2 + 2(-9.81)h$.
$h = \frac{371.7}{19.62} = \mathbf{18.9 \text{ m}}$.
[1 for vertical component, 1 for substitution, 1 for answer]

(c) Air resistance opposes motion. It reduces the horizontal velocity over time (horizontal deceleration). Thus, the horizontal range is reduced.
[1 for reduced, 1 for explanation]

4. (a) From graph, max static friction is the peak value before it drops. 25 N.
[1]

(b) Dynamic friction $f_d = 20 \text{ N}$. Normal reaction $R = mg = 5.0 \times 9.81 = 49.05 \text{ N}$.
$\mu_d = \frac{f_d}{R} = \frac{20}{49.05} = \mathbf{0.41}$.
[1 for formula, 1 for answer]

(c) Net force $F_{net} = F_{applied} - f_d = 30 - 20 = 10 \text{ N}$.
$a = \frac{F_{net}}{m} = \frac{10}{5.0} = \mathbf{2.0 \text{ m s}^{-2}}$.
[1 for net force, 1 for answer]

5. (a) In a closed/isolated system (no external forces), the total momentum before collision equals the total momentum after collision.
[1 for "closed system/no external forces", 1 for "momentum constant"]

(b) $m_A u_A + m_B u_B = (m_A + m_B)v$.
$(2.0)(4.0) + (3.0)(0) = (2.0 + 3.0)v$.
$8.0 = 5.0v \Rightarrow v = \mathbf{1.6 \text{ m s}^{-1}}$.
[1 for equation, 1 for substitution, 1 for answer]

(c) $KE_{initial} = \frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J}$.
$KE_{final} = \frac{1}{2}(5.0)(1.6)^2 = 6.4 \text{ J}$.
Since $KE_{initial} \neq KE_{final}$, the collision is inelastic.
[1 for KE init, 1 for KE final, 1 for conclusion]

Section B

6. (a) Forces:

1. Weight ($200 \text{ N}$) acting downwards at the center of the beam (2.0 m from A).
2. Tension ($T$) acting along the cable from B towards the wall.
3. Reaction force at hinge A ($R$), with horizontal and vertical components.
   [1 for each correct force vector]

(b) Angle of cable with horizontal: $\tan \theta = \frac{3.0}{4.0} \Rightarrow \theta = 36.9^\circ$.
Take moments about A:
Clockwise moment = Anticlockwise moment.
Weight moment: $200 \times 2.0 = 400 \text{ Nm}$.
Tension moment: Vertical component of $T$ is $T \sin \theta$. Perpendicular distance is 4.0 m.
$T \sin(36.9^\circ) \times 4.0 = 400$.
$T \times 0.6 \times 4.0 = 400 \Rightarrow 2.4 T = 400$.
$T = \mathbf{167 \text{ N}}$ (or 166.7 N).
[1 for moment equation, 1 for angle/component, 1 for calculation, 1 for answer]

(c) Horizontal equilibrium: $R_x = T \cos \theta = 166.7 \cos(36.9^\circ) = 133.3 \text{ N}$ (to the left).
Vertical equilibrium: $R_y + T \sin \theta = 200$.
$R_y + 100 = 200 \Rightarrow R_y = 100 \text{ N}$ (upwards).
Magnitude $R = \sqrt{133.3^2 + 100^2} = \mathbf{167 \text{ N}}$.
Direction: $\tan \alpha = \frac{100}{133.3} \Rightarrow \alpha = 36.9^\circ$ above horizontal.
[1 for Rx, 1 for Ry, 1 for magnitude, 1 for direction]

7. (a) Gravitational force $F_g = \frac{GMm}{r^2}$. Centripetal force $F_c = \frac{mv^2}{r}$. For circular orbit, $F_g$ provides $F_c$.
[1 for stating forces, 1 for equality]

(b) $\frac{GMm}{r^2} = \frac{mv^2}{r}$.
Cancel $m$ and one $r$: $\frac{GM}{r} = v^2$.
$v = \mathbf{\sqrt{\frac{GM}{r}}}$.
[1 for algebraic steps, 1 for final expression]

(c) $v \propto \frac{1}{\sqrt{r}}$. If $r$ doubles, $v$ becomes $\frac{1}{\sqrt{2}}$ times the original speed.
Speed decreases by factor of $\sqrt{2}$.
[1 for relationship, 1 for comparison]

8. (a) Work done against gravity $W = mgh = 500 \times 9.81 \times 20 = \mathbf{98100 \text{ J}}$ (or 98.1 kJ).
[1 for formula, 1 for answer]

(b) Input Energy = Power $\times$ Time = $8000 \times 15 = 120000 \text{ J}$.
Efficiency = $\frac{\text{Useful Output}}{\text{Total Input}} \times 100\% = \frac{98100}{120000} \times 100\% = \mathbf{81.8\%}$.
[1 for input energy, 1 for ratio, 1 for answer]

(c) Energy is lost as heat due to friction in the motor parts and air resistance, and sound.
[1 for heat/friction, 1 for sound/other]

9. (a) $v^2 = u^2 + 2gh$. $u=0$. $v = \sqrt{2 \times 9.81 \times 2.0} = \sqrt{39.24} = \mathbf{6.26 \text{ m s}^{-1}}$.
[1 for formula, 1 for answer]

(b) Rebound height $h' = 1.5 \text{ m}$. At max height $v=0$.
$0 = v_{rebound}^2 - 2gh' \Rightarrow v_{rebound} = \sqrt{2 \times 9.81 \times 1.5} = \sqrt{29.43} = \mathbf{5.42 \text{ m s}^{-1}}$.
[1 for formula, 1 for answer]

(c) Impulse $I = \Delta p = m(v_{final} - v_{initial})$.
Take upward as positive. $v_{final} = +5.42$, $v_{initial} = -6.26$.
$I = 0.20 (5.42 - (-6.26)) = 0.20 (11.68) = \mathbf{2.34 \text{ N s}}$.
[1 for delta p concept, 1 for sign convention, 1 for answer]

10. (a) Component of weight down slope $= mg \sin \theta = 1200 \times 9.81 \times \sin 5^\circ = 1200 \times 9.81 \times 0.0872 = \mathbf{1026 \text{ N}}$.
[1 for formula, 1 for answer]

(b) Constant speed means equilibrium. Driving Force $D = \text{Resistive Force} + \text{Weight Component}$.
$D = 400 + 1026 = \mathbf{1426 \text{ N}}$.
[1 for equilibrium concept, 1 for answer]

(c) Power $P = Fv = 1426 \times 20 = 28520 \text{ W} = \mathbf{28.5 \text{ kW}}$.
[1 for formula, 1 for answer]

Section C

11. (a) Hooke's Law: The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.
[1]

(b) Gradient = Spring Constant $k$. $k = 50 \text{ N m}^{-1}$.
[1]

(c) $E = \frac{1}{2}kx^2 = 0.5 \times 50 \times (0.10)^2 = 25 \times 0.01 = \mathbf{0.25 \text{ J}}$.
[1 for formula, 1 for answer]

(d) Spring constant is inversely proportional to length ($k \propto 1/L$). If length is halved, the spring constant doubles.
[1 for doubles, 1 for explanation]

12. (a) Vector diagram: Vertical vector (4.0) and Horizontal vector (3.0) added head-to-tail. Resultant is hypotenuse.
[1 for correct vectors, 1 for resultant]

(b) Magnitude $v = \sqrt{3.0^2 + 4.0^2} = \sqrt{9+16} = \sqrt{25} = \mathbf{5.0 \text{ m s}^{-1}}$.
[1 for Pythagoras, 1 for answer]

(c) Direction $\theta$ from North: $\tan \theta = \frac{3.0}{4.0} = 0.75$.
$\theta = \mathbf{36.9^\circ}$ East of North.
[1 for tan ratio, 1 for answer]

13. (a) Resultant $R = \sqrt{10^2 + 10^2} = \sqrt{200} = \mathbf{14.1 \text{ N}}$.
[1 for formula, 1 for answer]

(b) For equilibrium, $F_3$ must be equal and opposite to the resultant of $F_1$ and $F_2$.
Magnitude = 14.1 N.
Direction: $45^\circ$ below the horizontal (or $225^\circ$ from positive x-axis).
[1 for magnitude, 1 for direction logic, 1 for specific direction]

14. (a) Velocity is a vector (speed + direction). Since the direction is constantly changing, the velocity is changing. Acceleration is the rate of change of velocity.
[1 for vector nature, 1 for changing direction]

(b) $a = \frac{v^2}{r} = \frac{10^2}{50} = \frac{100}{50} = \mathbf{2.0 \text{ m s}^{-2}}$.
[1 for formula, 1 for answer]

(c) Friction between tires and road.
[1]

15. (a) Weight $W = mg = 1000 \times 9.81 = \mathbf{9810 \text{ N}}$.
[1]

(b) Net Force $F_{net} = \text{Thrust} - \text{Weight} = 15000 - 9810 = 5190 \text{ N}$.
$a = \frac{F_{net}}{m} = \frac{5190}{1000} = \mathbf{5.19 \text{ m s}^{-2}}$.
[1 for net force, 1 for formula, 1 for answer]

(c) Mass decreases. Since $F_{net}$ is constant (Thrust constant, Weight decreases slightly but Thrust >> Weight, actually Net Force increases as Weight drops), but primarily $a = F/m$. As $m$ decreases, $a$ increases.
[1 for increases, 1 for explanation]

16. (a) $GPE = mgh = 0.5 \times 9.81 \times 0.2 = \mathbf{0.981 \text{ J}}$.
[1 for formula, 1 for answer]

(b) Conservation of energy: $GPE_{lost} = KE_{gained}$.
$0.981 = \frac{1}{2}mv^2 = 0.5 \times 0.5 \times v^2 = 0.25 v^2$.
$v^2 = \frac{0.981}{0.25} = 3.924$.
$v = \sqrt{3.924} = \mathbf{1.98 \text{ m s}^{-1}}$.
[1 for conservation principle, 1 for substitution, 1 for answer]

(c) Energy is dissipated as heat and sound due to air resistance and friction at the pivot.
[1 for heat/air resistance, 1 for sound/pivot friction]

17. (a) Diagram: Weight (down), Normal Reaction (perpendicular to slope). No friction (smooth).
[1 for Weight, 1 for Normal]

(b) Component of weight down slope $= mg \sin 30^\circ$.
$ma = mg \sin 30^\circ \Rightarrow a = g \sin 30^\circ = 9.81 \times 0.5 = \mathbf{4.91 \text{ m s}^{-2}}$.
[1 for component, 1 for Newton's 2nd law, 1 for answer]

(c) $F_{net} = ma = 2.0 \times 2.0 = 4.0 \text{ N}$.
Driving force (weight component) $= mg \sin 30^\circ = 2.0 \times 9.81 \times 0.5 = 9.81 \text{ N}$.
$F_{net} = \text{Weight Component} - \text{Friction}$.
$4.0 = 9.81 - f \Rightarrow f = 9.81 - 4.0 = \mathbf{5.81 \text{ N}}$.
[1 for net force, 1 for weight component, 1 for friction]

18. (a) Constant speed means equilibrium. Tension $T = \text{Weight} = mg = 2000 \times 9.81 = \mathbf{19620 \text{ N}}$.
[1 for equilibrium, 1 for answer]

(b) Power $P = Fv = 19620 \times 0.5 = \mathbf{9810 \text{ W}}$ (or 9.81 kW).
[1 for formula, 1 for answer]

(c) Tension is 19620 N (same). Since speed is constant, acceleration is zero, so resultant force is zero. $T = mg$ still holds.
[1 for same value, 1 for explanation]

19. (a) Total Momentum $P = m_X u_X + m_Y u_Y = m(2v) + 2m(-v) = 2mv - 2mv = \mathbf{0}$.
[1 for substitution, 1 for answer]

(b) Conservation of momentum: $P_{initial} = P_{final}$.
$0 = m_X v_X' + m_Y v_Y'$.
$0 = m(-v) + 2m(v_Y')$.
$mv = 2m v_Y' \Rightarrow v_Y' = \mathbf{0.5v}$ (in the original direction of X).
[1 for equation, 1 for substitution, 1 for answer]

(c) $KE_{initial} = \frac{1}{2}m(2v)^2 + \frac{1}{2}(2m)(-v)^2 = 2mv^2 + mv^2 = 3mv^2$.
$KE_{final} = \frac{1}{2}m(-v)^2 + \frac{1}{2}(2m)(0.5v)^2 = 0.5mv^2 + 0.25mv^2 = 0.75mv^2$?
Wait. Let's recheck.
$KE_{final} = \frac{1}{2}m v^2 + \frac{1}{2}(2m)(\frac{v}{2})^2 = \frac{1}{2}mv^2 + m(\frac{v^2}{4}) = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
$KE_{initial} = 3mv^2$.
They are NOT equal. The question asks to "Show that kinetic energy is conserved". My calculation shows it is NOT.
Let's re-read the prompt. "Sphere X moves with velocity -v".
If $v_X' = -v$, then $v_Y' = 0.5v$.
$KE_i = 3mv^2$. $KE_f = 0.75mv^2$.
Kinetic energy is NOT conserved. The collision is inelastic.
Correction for Answer Key: The question likely implies an elastic collision scenario or I must state it is NOT conserved. Given the phrasing "Show that...", usually implies it IS. Did I make a mistake?
Momentum: $2mv - 2mv = 0$. Final: $-mv + 2m(v_Y') = 0 \rightarrow v_Y' = v/2$.
KE Initial: $0.5(m)(4v^2) + 0.5(2m)(v^2) = 2mv^2 + mv^2 = 3mv^2$.
KE Final: $0.5(m)(v^2) + 0.5(2m)(v^2/4) = 0.5mv^2 + 0.25mv^2 = 0.75mv^2$.
Energy is lost.
Answer: Calculate both. State they are unequal. Conclude KE is not conserved. (If the question strictly says "Show it is conserved", the question premise might be flawed for these specific numbers, but in an exam, you show the working and state the fact).
[1 for KE init, 1 for KE final, 1 for conclusion "Not Conserved"]

20. (a) $a = \frac{30 - 0}{10} = \mathbf{3.0 \text{ m s}^{-2}}$.
[1]

(b) $F = ma = 1000 \times 3.0 = \mathbf{3000 \text{ N}}$.
[1 for formula, 1 for answer]

(c) $s = ut + \frac{1}{2}at^2 = 0 + 0.5 \times 3.0 \times 100 = \mathbf{150 \text{ m}}$.
[1 for formula, 1 for answer]

(d) Work Done $W = Fs = 3000 \times 150 = \mathbf{450,000 \text{ J}}$ (or 450 kJ).
Alternatively, $W = \Delta KE = \frac{1}{2}(1000)(30^2) = 500 \times 900 = 450,000 \text{ J}$.
[1 for formula, 1 for answer]