A Level H2 Physics Practice Paper 1

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key and Marking Scheme Version 1

Total Marks: 60

Section A: Structured Questions

1. Car Motion with Resistive Force

(a) At terminal speed, acceleration is zero, so driving force equals resistive force.
$F_{drive} = F_{resist} = kv$
$2400 = k(40)$
$k = \frac{2400}{40} = 60 \text{ N s m}^{-1}$ (or $\text{kg s}^{-1}$)
[2 marks]: 1 for correct equation, 1 for correct answer with units.

(b) At start ($v=0$), resistive force $F_{resist} = k(0) = 0$.
Resultant force $F_{net} = F_{drive} = 2400 \text{ N}$.
$F_{net} = ma \Rightarrow 2400 = 1200 a$
$a = 2.0 \text{ m s}^{-2}$
[2 marks]: 1 for identifying net force, 1 for correct answer.

(c) As speed $v$ increases, the resistive force $F_{resist} = kv$ increases.
The driving force is constant.
Therefore, the resultant force ($F_{drive} - F_{resist}$) decreases.
Since $a = F_{net}/m$, the acceleration decreases.
[2 marks]: 1 for linking speed to increased resistive force, 1 for linking reduced net force to reduced acceleration.

2. Vertical Projectile Motion

(a) Using conservation of energy or kinematics:
$v^2 = u^2 + 2as$
At max height, $v=0$, $a = -9.81 \text{ m s}^{-2}$, $u = 12 \text{ m s}^{-1}$.
$0 = 12^2 + 2(-9.81)h$
$19.62 h = 144$
$h = 7.34 \text{ m}$
[2 marks]: 1 for correct substitution, 1 for correct answer (7.34 or 7.35 m).

(b) Time to reach max height: $v = u + at \Rightarrow 0 = 12 - 9.81 t_{up} \Rightarrow t_{up} = 1.223 \text{ s}$.
Total time $T = 2 \times t_{up} = 2.45 \text{ s}$.
(Alternatively using $s = ut + \frac{1}{2}at^2$ with $s=0$).
[2 marks]: 1 for method, 1 for correct answer (2.45 s).

(c) Graph shape: Parabola opening upwards (since $E_k \propto v^2$ and $v$ varies linearly with $t$, $E_k$ is quadratic).

• Starts at max value at $t=0$.
• Zero at $t = 1.22 \text{ s}$ (halfway).
• Returns to max value at $t = 2.45 \text{ s}$.
• Symmetric about the time axis midpoint.
  Max $E_k = \frac{1}{2}mv^2 = 0.5(0.15)(12^2) = 10.8 \text{ J}$.
  [3 marks]: 1 for correct shape (U-shaped/parabolic), 1 for touching zero axis at midpoint, 1 for labeling max energy approx 10.8 J.

3. Inelastic Collision

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

(b) Conservation of momentum:
$m_A u_A + m_B u_B = (m_A + m_B) v$
$(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0) v$
$6.0 = 3.0 v$
$v = 2.0 \text{ m s}^{-1}$
[2 marks]: 1 for correct equation, 1 for correct answer.

(c) Initial KE: $KE_i = \frac{1}{2} m_A u_A^2 = 0.5(2.0)(3.0^2) = 9.0 \text{ J}$.
Final KE: $KE_f = \frac{1}{2} (m_A+m_B) v^2 = 0.5(3.0)(2.0^2) = 6.0 \text{ J}$.
Loss in KE $= 9.0 - 6.0 = 3.0 \text{ J}$.
[3 marks]: 1 for initial KE, 1 for final KE, 1 for correct difference.

4. Satellite Orbit

(a) Gravitational force provides centripetal force:
$\frac{GMm}{R^2} = \frac{mv^2}{R}$
Cancel $m$ and one $R$:
$\frac{GM}{R} = v^2$
$v = \sqrt{\frac{GM}{R}}$
[3 marks]: 1 for equating forces, 1 for algebraic steps, 1 for final result.

(b) (i) Orbital speed decreases. Since $v \propto \frac{1}{\sqrt{R}}$, as $R$ increases, $v$ decreases.
[2 marks]: 1 for "decreases", 1 for explanation using formula.

(ii) Total mechanical energy increases (becomes less negative).
Work must be done against gravity to move to a higher orbit, increasing potential energy more than kinetic energy decreases.
$E_{total} = -\frac{GMm}{2R}$. As $R$ increases, $E_{total}$ becomes closer to 0 (increases).
[2 marks]: 1 for "increases", 1 for explanation.

5. Simple Pendulum

(a) Vertical height change $h = L - L \cos \theta = L(1 - \cos \theta)$.
$h = 1.2 (1 - \cos 30^\circ) = 1.2 (1 - 0.866) = 1.2(0.134) = 0.1608 \text{ m}$.
$\Delta PE = mgh = 0.50 \times 9.81 \times 0.1608 = 0.789 \text{ J}$.
[3 marks]: 1 for height calculation, 1 for formula, 1 for answer (0.79 J).

(b) Conservation of energy: $\Delta PE = \Delta KE$.
$0.789 = \frac{1}{2} m v^2$
$v^2 = \frac{2 \times 0.789}{0.50} = 3.156$
$v = 1.78 \text{ m s}^{-1}$
[2 marks]: 1 for substitution, 1 for answer.

(c) At lowest point, forces are Tension ($T$) up and Weight ($mg$) down. Resultant force is centripetal.
$T - mg = \frac{mv^2}{L}$
$T = mg + \frac{mv^2}{L} = (0.50)(9.81) + \frac{0.50(1.78^2)}{1.2}$
$T = 4.905 + \frac{1.584}{1.2} = 4.905 + 1.32 = 6.225 \text{ N}$.
[3 marks]: 1 for correct force equation, 1 for substitution, 1 for answer (6.23 N).

6. Circular Motion (Cyclist)

(a) Frictional force between tires and road.
[1 mark]

(b) Max friction provides centripetal force:
$F_{fric} = \mu R = \mu mg$
$\frac{mv^2}{r} = \mu mg$
$v^2 = \mu g r$
$v = \sqrt{0.80 \times 9.81 \times 25} = \sqrt{196.2} = 14.0 \text{ m s}^{-1}$.
[3 marks]: 1 for equation setup, 1 for substitution, 1 for answer.

(c) Leaning creates a horizontal component of the normal reaction force (or torque balance) that provides the necessary centripetal force/moment to prevent toppling outwards. It aligns the resultant force of gravity and normal reaction through the center of mass.
[2 marks]: 1 for mentioning torque/moment or component of force, 1 for stability/preventing toppling.

Section B: Data-Based and Contextual Questions

7. Spring Experiment

(a) Graph:

• Points plotted correctly.
• Straight line drawn through origin up to (8.0, 4.8).
• Curve drawn for points beyond 8.0 N.
  [4 marks]: 1 for axes labels/units, 1 for all points correct, 1 for linear fit, 1 for curve at end.

(b) Gradient of linear section:
$k = \frac{\Delta F}{\Delta x} = \frac{8.0 - 0}{0.048 - 0} = 166.67 \text{ N m}^{-1}$.
$k \approx 167 \text{ N m}^{-1}$.
[2 marks]: 1 for gradient method, 1 for answer (167 N/m). Note: x must be in meters.

(c) Energy = Area under graph up to 8.0 N.
$E = \frac{1}{2} F x = 0.5 \times 8.0 \times 0.048 = 0.192 \text{ J}$.
[2 marks]: 1 for formula/area concept, 1 for answer.

(d) The limit of proportionality has been exceeded. The spring undergoes plastic deformation or the material structure is changing, so Hooke's Law no longer applies.
[2 marks]: 1 for "limit of proportionality exceeded", 1 for elaboration.

8. Projectile from Cliff

(a) Vertical motion:
$s_y = -45 \text{ m}$ (down is negative), $u_y = 20 \sin 30^\circ = 10 \text{ m s}^{-1}$, $a_y = -9.81 \text{ m s}^{-2}$.
$s = ut + \frac{1}{2}at^2$
$-45 = 10t - 4.905t^2$
$4.905t^2 - 10t - 45 = 0$
Using quadratic formula: $t = \frac{10 \pm \sqrt{100 - 4(4.905)(-45)}}{2(4.905)}$
$t = \frac{10 \pm \sqrt{100 + 882.9}}{9.81} = \frac{10 \pm 31.35}{9.81}$
Positive root: $t = \frac{41.35}{9.81} = 4.215 \text{ s}$.
[4 marks]: 1 for resolving $u_y$, 1 for correct equation, 1 for quadratic handling, 1 for answer (4.22 s).

(b) Horizontal motion:
$u_x = 20 \cos 30^\circ = 17.32 \text{ m s}^{-1}$.
Distance $= u_x \times t = 17.32 \times 4.215 = 73.0 \text{ m}$.
[2 marks]: 1 for $u_x$, 1 for distance calculation.

(c) Vertical velocity at impact:
$v_y = u_y + at = 10 + (-9.81)(4.215) = 10 - 41.35 = -31.35 \text{ m s}^{-1}$.
Horizontal velocity $v_x = 17.32 \text{ m s}^{-1}$ (constant).
Magnitude $v = \sqrt{v_x^2 + v_y^2} = \sqrt{17.32^2 + (-31.35)^2}$
$v = \sqrt{299.98 + 982.82} = \sqrt{1282.8} = 35.8 \text{ m s}^{-1}$.
[4 marks]: 1 for $v_y$ calc, 1 for stating $v_x$, 1 for Pythagorean combination, 1 for answer.

9. Elastic Collision (Billiards)

(a) 1. Kinetic Energy. 2. Linear Momentum.
[2 marks]: 1 for each.

(b) For equal masses in elastic collision, the angle between final velocity vectors is $90^\circ$.
Alternatively, use conservation equations:
x-mom: $mu = mv_1 \cos 30 + mv_2 \cos \theta$
y-mom: $0 = mv_1 \sin 30 - mv_2 \sin \theta$
KE: $u^2 = v_1^2 + v_2^2$
From vector triangle, since $u^2 = v_1^2 + v_2^2$, the triangle is right-angled.
Thus $30^\circ + \theta = 90^\circ \Rightarrow \theta = 60^\circ$.
[3 marks]: 1 for citing equal mass/elastic property or equations, 1 for logical deduction of 90 deg separation, 1 for final angle.

(c) From the right-angled velocity vector triangle:
$v_{red} = u \sin 30^\circ$ (component opposite to white ball's angle? No, geometry: $v_{white}$ is adjacent to 30, $v_{red}$ is opposite? Let's check).
Vector sum $\vec{u} = \vec{v}_1 + \vec{v}_2$.
Angle of $\vec{v}_1$ is 30 to $\vec{u}$. Angle of $\vec{v}_2$ is 60 to $\vec{u}$.
$v_{red} = u \cos 60^\circ$ or $u \sin 30^\circ$.
$v_{red} = 2.0 \times 0.5 = 1.0 \text{ m s}^{-1}$.
[3 marks]: 1 for correct trigonometric relation, 1 for substitution, 1 for answer.

10. Inclined Plane

(a) Component down slope $= mg \sin \theta$.
$W_{parallel} = 5.0 \times 9.81 \times \sin 20^\circ = 49.05 \times 0.342 = 16.78 \text{ N}$.
[2 marks]: 1 for formula, 1 for answer (16.8 N).

(b) Normal reaction $R = mg \cos \theta$.
$R = 5.0 \times 9.81 \times \cos 20^\circ = 49.05 \times 0.940 = 46.1 \text{ N}$.
[2 marks]: 1 for formula, 1 for answer.

(c) Constant speed means equilibrium. Forces up slope = Forces down slope.
$P = W_{parallel} + F_{friction}$
$F_{friction} = \mu R = 0.30 \times 46.1 = 13.83 \text{ N}$.
$P = 16.78 + 13.83 = 30.61 \text{ N}$.
[3 marks]: 1 for friction calc, 1 for equilibrium equation, 1 for answer (30.6 N).

(d) Newton's 2nd Law: $F_{net} = ma$.
$P' - W_{parallel} - F_{friction} = ma$
$P' - 16.78 - 13.83 = 5.0 \times 1.5$
$P' - 30.61 = 7.5$
$P' = 38.11 \text{ N}$.
[3 marks]: 1 for net force equation, 1 for substitution, 1 for answer (38.1 N).