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A Level H2 Physics Practice Paper 1

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H2
Level: A-Level
Paper: Practice Paper 1 (Version 1 of 5)
Topic Focus: Mechanics (Newtonian Mechanics, Energy, Momentum, Circular Motion)
Duration: 1 hour 30 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Answer all questions.
Write your answers in the spaces provided.
All working must be clearly shown. Numerical answers should be given to 3 significant figures unless otherwise stated.
Take the acceleration of free fall $g = 9.81 \text{ m s}^{-2}$ .

Section A: Structured Questions (40 Marks)

1. A car of mass $1200 \text{ kg}$ travels along a straight horizontal road. The engine provides a constant driving force of $2400 \text{ N}$ . The total resistive force acting on the car is proportional to its speed $v$ , given by $F_{resist} = kv$ , where $k$ is a constant.

(a) The car reaches a terminal speed of $40 \text{ m s}^{-1}$ . Calculate the value of the constant $k$ .
[2]

(b) Calculate the initial acceleration of the car when it starts from rest.
[2]

2. A ball of mass $0.15 \text{ kg}$ is thrown vertically upwards with an initial speed of $12 \text{ m s}^{-1}$ from ground level. Air resistance is negligible.

(a) Calculate the maximum height reached by the ball.
[2]

(b) Determine the time taken for the ball to return to the ground.
[2]

(c) On the axes below, sketch the variation with time $t$ of the kinetic energy $E_k$ of the ball from the moment it is thrown until it returns to the ground. Label the maximum value on the energy axis.
[3]

      E_k / J
        |
        |
        |
        |
        |
        |
        |________________________ t / s
        0

3. Two trolleys, A and B, move along a frictionless horizontal track. Trolley A has a mass of $2.0 \text{ kg}$ and moves with a velocity of $3.0 \text{ m s}^{-1}$ to the right. Trolley B has a mass of $1.0 \text{ kg}$ and is initially at rest. They collide and stick together.

(a) State the principle of conservation of linear momentum.
[2]

(b) Calculate the common velocity of the trolleys after the collision.
[2]

4. A satellite orbits the Earth in a circular orbit of radius $R$ . The mass of the Earth is $M$ and the mass of the satellite is $m$ .

(a) Show that the orbital speed $v$ of the satellite is given by $v = \sqrt{\frac{GM}{R}}$ , where $G$ is the gravitational constant.
[3]

(b) The satellite moves to a higher circular orbit. State and explain the effect on: (i) the orbital speed of the satellite,
[2]

(ii) the total mechanical energy of the satellite.  
[2]

5. A simple pendulum consists of a bob of mass $0.50 \text{ kg}$ attached to a light inextensible string of length $1.2 \text{ m}$ . The bob is pulled aside until the string makes an angle of $30^\circ$ with the vertical and is released from rest.

(a) Calculate the change in gravitational potential energy of the bob as it moves from the release point to the lowest point of its swing.
[3]

(b) Calculate the speed of the bob at the lowest point of its swing.
[2]

6. A cyclist travels around a horizontal circular bend of radius $25 \text{ m}$ . The coefficient of static friction between the tires and the road is $0.80$ .

(a) Identify the force that provides the centripetal acceleration.
[1]

(b) Calculate the maximum speed at which the cyclist can travel around the bend without skidding.
[3]

Section B: Data-Based and Contextual Questions (20 Marks)

7. A student investigates the relationship between the extension $x$ of a spring and the load $F$ applied to it. The spring obeys Hooke's Law up to its limit of proportionality. The student obtains the following data:

Load $F$ / N	Extension $x$ / cm
0.0	0.0
2.0	1.2
4.0	2.4
6.0	3.6
8.0	4.8
10.0	6.5
12.0	9.0

(a) Plot a graph of $F$ (y-axis) against $x$ (x-axis) on the grid below. Draw the line of best fit for the linear region.
[4]

      F / N
   12 |
      |
   10 |
      |
    8 |
      |
    6 |
      |
    4 |
      |
    2 |
      |
    0 |________________________________________ x / cm
      0    2    4    6    8    10   12

(b) Determine the spring constant $k$ of the spring.
[2]

(d) Explain why the graph deviates from a straight line for loads greater than $8.0 \text{ N}$ .
[2]

8. A projectile is launched from the top of a cliff $45 \text{ m}$ high with an initial velocity of $20 \text{ m s}^{-1}$ at an angle of $30^\circ$ above the horizontal. Air resistance is negligible.

(a) Calculate the time taken for the projectile to hit the ground.
[4]

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

9. In a game of billiards, a white ball of mass $m$ moving with speed $u$ collides elastically with a stationary red ball of equal mass $m$ . After the collision, the white ball moves off at an angle of $30^\circ$ to its original direction.

(a) State two quantities that are conserved in this elastic collision.
[2]

(b) Show that the red ball moves off at an angle of $60^\circ$ to the original direction of the white ball.
[3]

10. A block of mass $5.0 \text{ kg}$ rests on a rough inclined plane that makes an angle of $20^\circ$ with the horizontal. A force $P$ acts parallel to the plane and up the slope. The coefficient of dynamic friction between the block and the plane is $0.30$ .

(a) Calculate the component of the weight of the block acting down the slope.
[2]

(b) Calculate the normal reaction force acting on the block.
[2]

(d) The force $P$ is increased so that the block accelerates up the slope at $1.5 \text{ m s}^{-2}$ . Calculate the new value of $P$ .
[3]

Answers

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).