A Level H2 Physics Mechanics Quiz

A-Level Physics H2 Quiz - Mechanics: Answer Key

Section A: Short-Answer and Conceptual Questions

1. [2 marks]
The principle of conservation of linear momentum states that the total momentum of a closed system remains constant (is conserved) provided that no net external force acts on the system. Equivalently: in an isolated system, the total momentum before an interaction equals the total momentum after the interaction.

• 1 mark for stating that total momentum remains constant / is conserved.
• 1 mark for specifying the condition: no external force / closed (isolated) system.
  Common mistake: Stating only "momentum is conserved" without mentioning the condition of no external force. This is incomplete and would score only 1 mark.

2. [2 marks]
Using $a = \frac{v - u}{t}$:
$a = \frac{28 - 0}{7.0} = 4.0 \text{ m s}^{-2}$

• 1 mark for correct formula or method.
• 1 mark for correct answer with unit.
  Teaching note: Uniform acceleration means the velocity increases by the same amount each second. The equation $a = \frac{\Delta v}{\Delta t}$ is one of the key kinematic equations for constant acceleration.

3. [2 marks]
Impulse is defined as the product of the force acting on an object and the time for which it acts: $F \Delta t$. Equivalently, impulse equals the change in momentum of the object: $\Delta p = mv - mu$.
The SI unit of impulse is newton second (N s), which is equivalent to kg m s⁻¹.

• 1 mark for correct definition.
• 1 mark for correct SI unit.
  Teaching note: Impulse and change in momentum are equivalent quantities. This is why the impulse-momentum theorem $F\Delta t = \Delta p$ is so useful — it connects force and motion.

4. [2 marks]
Newton's second law states that the net force acting on an object is equal to the rate of change of its momentum:
$F_{net} = \frac{\Delta p}{\Delta t} = \frac{mv - mu}{\Delta t}$
For constant mass, this reduces to $F_{net} = ma$.

• 1 mark for stating force equals rate of change of momentum.
• 1 mark for correct mathematical expression.
  Common mistake: Writing only $F = ma$ without the momentum form. While $F = ma$ is correct for constant mass, the more general statement in terms of momentum is what the question asks for.

5. [2 marks]
At maximum height, the final velocity $v = 0$. Using $v^2 = u^2 - 2gh$:
$0 = (15)^2 - 2(9.81)h$
$h = \frac{225}{2 \times 9.81} = \frac{225}{19.62} = 11.5 \text{ m}$

• 1 mark for correct substitution.
• 1 mark for correct answer (accept 11.4–11.5 m depending on $g$ value used).
  Teaching note: At the highest point, the ball momentarily stops before falling back down. The deceleration is $g$ throughout the upward motion.

6. [2 marks]

• Elastic collision: Both total kinetic energy and total momentum of the system are conserved. The objects separate after collision.
• Inelastic collision: Total momentum is conserved, but total kinetic energy is not conserved (some is converted to other forms such as heat, sound, or deformation energy). In a perfectly inelastic collision, the objects stick together and move with a common velocity after impact.
• 1 mark for each correct distinction.
  Common mistake: Thinking that momentum is not conserved in inelastic collisions. Momentum is always conserved in collisions (provided no external forces act); it is kinetic energy that is not conserved in inelastic collisions.

7. [3 marks]
The tension in the string provides the centripetal force for circular motion:
$T = \frac{mv^2}{r} = \frac{0.50 \times (4.0)^2}{1.2} = \frac{0.50 \times 16}{1.2} = \frac{8.0}{1.2} = 6.7 \text{ N}$

• 1 mark for correct formula $T = \frac{mv^2}{r}$.
• 1 mark for correct substitution.
• 1 mark for correct answer with unit.
  Teaching note: In horizontal circular motion, the tension is the only horizontal force and it provides the centripetal force directed towards the centre of the circle. The weight is balanced by other forces (or is perpendicular to the plane of motion).

8. [3 marks]
A body is in equilibrium when it is at rest or moving with constant velocity (i.e., no acceleration).
The two conditions for equilibrium are:

1. The net (resultant) force acting on the body is zero (translational equilibrium): $\sum F = 0$.
2. The net (resultant) moment about any point is zero (rotational equilibrium): $\sum \tau = 0$.

• 1 mark for definition of equilibrium.
• 1 mark for condition 1 (net force = 0).
• 1 mark for condition 2 (net moment = 0).
  Teaching note: Equilibrium does not mean no forces act on the body — it means all forces balance out. A book resting on a table is in equilibrium because its weight is exactly balanced by the normal reaction force.

9. [2 marks]
The gravitational force (weight) on an object is related to the gravitational field strength by:
$W = mg$
where $W$ is the weight (gravitational force), $m$ is the mass of the object, and $g$ is the gravitational field strength at that point. The gravitational field strength $g$ is defined as the gravitational force per unit mass: $g = \frac{W}{m}$.

• 1 mark for stating $W = mg$ or equivalent.
• 1 mark for explaining that $g$ is force per unit mass.
  Teaching note: Gravitational field strength $g$ has units of N kg⁻¹, which is equivalent to m s⁻². Near Earth's surface, $g \approx 9.81 \text{ N kg}^{-1}$.

Section B: Structured and Calculation Questions

10. [7 marks]

(a) [2 marks] Using conservation of energy (the track is smooth, so no energy is lost to friction):
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 1.8} = \sqrt{35.316} = 5.94 \text{ m s}^{-1}$

• 1 mark for using conservation of energy.
• 1 mark for correct answer.

(b) [2 marks] Using conservation of linear momentum (perfectly inelastic collision):
$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$
$0.20 \times 5.94 + 0.80 \times 0 = (0.20 + 0.80)v_f$
$v_f = \frac{1.188}{1.0} = 1.19 \text{ m s}^{-1}$

• 1 mark for correct momentum conservation equation.
• 1 mark for correct answer.

(c) [2 marks]
Kinetic energy before collision:
$KE_{before} = \frac{1}{2}(0.20)(5.94)^2 = \frac{1}{2}(0.20)(35.3) = 3.53 \text{ J}$
Kinetic energy after collision:
$KE_{after} = \frac{1}{2}(1.0)(1.19)^2 = \frac{1}{2}(1.0)(1.416) = 0.708 \text{ J}$
Kinetic energy lost:
$\Delta KE = 3.53 - 0.708 = 2.82 \text{ J}$

• 1 mark for calculating both KE values.
• 1 mark for correct energy lost.

(d) [1 mark] The collision is not elastic because kinetic energy is not conserved — a significant amount of kinetic energy ($2.82 \text{ J}$) is lost during the collision, converted into other forms of energy such as heat, sound, and deformation energy. Additionally, the two objects stick together after impact, which is characteristic of a perfectly inelastic collision.

11. [6 marks]

(a) [2 marks] At the top of the circle, the minimum speed occurs when the tension in the string is zero (only weight provides the centripetal force):
$mg = \frac{mv_{min}^2}{r}$
$v_{min} = \sqrt{gr} = \sqrt{9.81 \times 0.90} = \sqrt{8.829} = 2.97 \text{ m s}^{-1}$

• 1 mark for setting $mg = \frac{mv^2}{r}$ (tension = 0 at minimum speed).
• 1 mark for correct answer.

(b) [3 marks] At the bottom of the circle, both tension and weight act. The net centripetal force (towards the centre, which is upward) is:
$T - mg = \frac{mv^2}{r}$
$T = mg + \frac{mv^2}{r} = 0.30 \times 9.81 + \frac{0.30 \times (6.0)^2}{0.90}$
$T = 2.943 + \frac{0.30 \times 36}{0.90} = 2.943 + 12.0 = 14.9 \text{ N}$

• 1 mark for correct equation $T - mg = \frac{mv^2}{r}$.
• 1 mark for correct substitution.
• 1 mark for correct answer (accept 14.9 N or 15 N).

(c) [1 mark] The tension at the top is zero. At the minimum speed, the weight of the object alone provides the necessary centripetal force, so the string is just barely taut with no tension. If the speed were any less, the string would go slack and the object would not complete the circular path.

12. [9 marks]

(a) [2 marks] Using Newton's second law:
$F_{net} = F_{driving} - F_{resistive} = ma$
$3600 - 1200 = 1200a$
$a = \frac{2400}{1200} = 2.0 \text{ m s}^{-2}$

• 1 mark for correct net force.
• 1 mark for correct answer with unit.

(b) [2 marks] Using $s = ut + \frac{1}{2}at^2$ with $u = 0$:
$s = 0 + \frac{1}{2}(2.0)(10.0)^2 = \frac{1}{2}(2.0)(100) = 100 \text{ m}$

• 1 mark for correct equation.
• 1 mark for correct answer.

(c) [2 marks] Speed after 10.0 s: $v = u + at = 0 + 2.0 \times 10.0 = 20.0 \text{ m s}^{-1}$
$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(20.0)^2 = 600 \times 400 = 2.4 \times 10^5 \text{ J}$

• 1 mark for finding final speed.
• 1 mark for correct KE value.

(d) [3 marks] On the slope, the component of weight acting down the slope is $mg\sin\theta$:
$F_{net} = F_{driving} - F_{resistive} - mg\sin\theta = ma'$
$3600 - 1200 - 1200 \times 9.81 \times \sin 5.0^{\circ} = 1200a'$
$2400 - 1200 \times 9.81 \times 0.0872 = 1200a'$
$2400 - 1024.6 = 1200a'$
$a' = \frac{1375.4}{1200} = 1.15 \text{ m s}^{-2}$

• 1 mark for including the $mg\sin\theta$ component.
• 1 mark for correct substitution.
• 1 mark for correct answer (accept 1.1–1.2 m s⁻²).

Section C: Data Interpretation and Multi-Concept Questions

13. [7 marks]

(a) [1 mark]
$W = mg = 0.10 \times 9.81 = 0.98 \text{ N}$

• 1 mark for correct answer with unit.

(b) [3 marks] The tension in the string accelerates both the trolley and the hanging mass. For the system:
The net force accelerating the system is the weight of the hanging mass $mg$. The total mass being accelerated is $M + m$.
$a = \frac{mg}{M + m} = \frac{0.10 \times 9.81}{0.50 + 0.10} = \frac{0.981}{0.60} = 1.64 \text{ m s}^{-2}$

• 1 mark for identifying net force = weight of hanging mass.
• 1 mark for total mass = $M + m$.
• 1 mark for correct answer.

(c) [2 marks] Using $v^2 = u^2 + 2as$ with $u = 0$:
$a = \frac{v^2}{2s} = \frac{(0.88)^2}{2 \times 0.40} = \frac{0.7744}{0.80} = 0.968 \text{ m s}^{-2}$

• 1 mark for correct equation.
• 1 mark for correct answer.

(d) [1 mark] The experimental acceleration is lower than the theoretical value because friction acts on the trolley (between the trolley and the track), which opposes the motion and reduces the net accelerating force. Other acceptable answers: air resistance, the pulley is not perfectly smooth (friction at the pulley), the string has mass.

14. [5 marks]

(a) [1 mark]
$F = mg_{field} = 2.0 \times 10^4 \times 8.2 = 1.64 \times 10^5 \text{ N}$

• 1 mark for correct answer with unit.

(b) [2 marks] The spacecraft is accelerating because its direction of motion is continuously changing. Even though the speed (magnitude of velocity) is constant, velocity is a vector quantity. A change in direction constitutes a change in velocity, which means there is acceleration. This acceleration is directed towards the centre of the circular orbit (centripetal acceleration).

• 1 mark for identifying that direction changes.
• 1 mark for stating that velocity is a vector / acceleration is towards the centre.

(c) [2 marks] The work done by the gravitational force during one complete orbit is zero. The gravitational force is always directed towards the centre of the orbit (radially inward), while the displacement of the spacecraft at any instant is tangential to the orbit (perpendicular to the radial direction). Since the force is always perpendicular to the displacement, $W = Fd\cos 90° = 0$ at every point along the orbit. Over a complete orbit, the total work done is therefore zero.

• 1 mark for stating work done = 0.
• 1 mark for correct explanation (force perpendicular to displacement / tangential motion).

15. [7 marks]

(a) [2 marks]
Horizontal component: $v_x = 25\cos 35° = 25 \times 0.819 = 20.5 \text{ m s}^{-1}$
Vertical component: $v_y = 25\sin 35° = 25 \times 0.574 = 14.3 \text{ m s}^{-1}$

• 1 mark for each correct component.

(b) [2 marks] At maximum height, vertical velocity = 0:
$v_y = u_y - gt$
$0 = 14.3 - 9.81t$
$t = \frac{14.3}{9.81} = 1.46 \text{ s}$

• 1 mark for correct equation.
• 1 mark for correct answer.

(c) [3 marks] Total time of flight = $2t = 2 \times 1.46 = 2.92 \text{ s}$
Horizontal range:
$R = v_x \times T = 20.5 \times 2.92 = 59.9 \text{ m}$

• 1 mark for total time of flight.
• 1 mark for using range = horizontal velocity × time of flight.
• 1 mark for correct answer (accept 59–60 m).

16. [5 marks]

(a) [1 mark]
$W = mg = 12 \times 9.81 = 117.7 \text{ N} \approx 118 \text{ N}$

• 1 mark for correct answer.

(b) [3 marks] Taking moments about A (clockwise = anticlockwise):
The weight acts at the centre, $1.50 \text{ m}$ from A. Support B is $2.40 \text{ m}$ from A.
$R_B \times 2.40 = W \times 1.50$
$R_B = \frac{117.7 \times 1.50}{2.40} = \frac{176.6}{2.40} = 73.6 \text{ N}$

• 1 mark for correct moment equation.
• 1 mark for correct distances.
• 1 mark for correct answer.

(c) [1 mark] Using vertical equilibrium:
$R_A + R_B = W$
$R_A = 117.7 - 73.6 = 44.1 \text{ N}$

• 1 mark for correct answer.

17. [4 marks]

(a) [1 mark] Free-body diagram should show:

• Weight ($W = mg$) acting vertically downward from the centre of mass of the person.
• Normal reaction force ($R$) acting vertically upward from the floor on the person's feet.
• The reaction force arrow should be longer than the weight arrow (since the lift is accelerating upward, $R > mg$).
• 1 mark for correct diagram with both forces labelled.

(b) [2 marks] Using Newton's second law (upward positive):
$R - mg = ma$
$R = m(g + a) = 70(9.81 + 1.5) = 70 \times 11.31 = 791.7 \text{ N} \approx 792 \text{ N}$

• 1 mark for correct equation.
• 1 mark for correct answer.

(c) [1 mark] When the lift moves at constant velocity, acceleration = 0, so:
$R = mg = 70 \times 9.81 = 686.7 \text{ N} \approx 687 \text{ N}$
The reaction force equals the person's weight because there is no net force (Newton's first law — constant velocity means zero net force).

• 1 mark for correct value and explanation.

18. [6 marks]

(a) [1 mark]
$p_{total} = m_{car} \times v_{car} + m_{van} \times v_{van} = 1000 \times 20 + 3000 \times 0 = 2.0 \times 10^4 \text{ kg m s}^{-1}$

• 1 mark for correct answer with unit.

(b) [2 marks] Conservation of momentum:
$m_{car}v_{car} = (m_{car} + m_{van})v_f$
$2.0 \times 10^4 = (1000 + 3000)v_f$
$v_f = \frac{2.0 \times 10^4}{4000} = 5.0 \text{ m s}^{-1}$

• 1 mark for correct equation.
• 1 mark for correct answer.

(c) [2 marks]
$KE_{before} = \frac{1}{2}(1000)(20)^2 = 2.0 \times 10^5 \text{ J}$
$KE_{after} = \frac{1}{2}(4000)(5.0)^2 = 5.0 \times 10^4 \text{ J}$

• 1 mark for each correct KE value.

(d) [1 mark] The collision is inelastic because kinetic energy is not conserved ($KE_{after} < KE_{before}$). Kinetic energy has decreased from $2.0 \times 10^5 \text{ J}$ to $5.0 \times 10^4 \text{ J}$, with the difference converted to other forms of energy (deformation, heat, sound).

19. [4 marks]

(a) [2 marks]
$\vec{p}_{total} = m_1\vec{v}_1 + m_2\vec{v}_2$
$= 0.050(3\mathbf{i} + 4\mathbf{j}) + 0.030(-2\mathbf{i} + 1\mathbf{j})$
$= (0.15\mathbf{i} + 0.20\mathbf{j}) + (-0.06\mathbf{i} + 0.03\mathbf{j})$
$= 0.09\mathbf{i} + 0.23\mathbf{j} \text{ kg m s}^{-1}$

• 1 mark for correct calculation of each sphere's momentum.
• 1 mark for correct total.

(b) [2 marks] By conservation of momentum, total momentum after = total momentum before:
$0.050 \times \vec{0} + 0.030 \times \vec{v}_2' = 0.09\mathbf{i} + 0.23\mathbf{j}$
$\vec{v}_2' = \frac{0.09\mathbf{i} + 0.23\mathbf{j}}{0.030} = 3.0\mathbf{i} + 7.67\mathbf{j} \text{ m s}^{-1}$

• 1 mark for applying conservation of momentum.
• 1 mark for correct answer.

20. [5 marks]

Let $t$ be the time when the two balls meet. Let the meeting height be $h$ above the ground.

For the dropped ball (Ball 1, from height 45 m):
$h_1 = 45 - \frac{1}{2}gt^2$

For the projected ball (Ball 2, from ground):
$h_2 = 30t - \frac{1}{2}gt^2$

At the meeting point, $h_1 = h_2$:
$45 - \frac{1}{2}gt^2 = 30t - \frac{1}{2}gt^2$
$45 = 30t$
$t = 1.5 \text{ s}$

• 2 marks for setting up displacement equations for both balls.
• 1 mark for solving to get $t = 1.5$ s.

(b) [2 marks] Substituting $t = 1.5$ s into the equation for Ball 2:
$h = 30(1.5) - \frac{1}{2}(9.81)(1.5)^2$
$h = 45 - 11.04 = 33.96 \text{ m} \approx 34.0 \text{ m}$

• 1 mark for correct substitution.
• 1 mark for correct answer (accept 33.9–34.0 m).

Teaching note: The $\frac{1}{2}gt^2$ terms cancel because both balls experience the same gravitational acceleration. This means the distance fallen by Ball 1 plus the distance risen by Ball 2 always equals 45 m, making the problem simpler than it first appears.

End of Answer Key