## PHY 001: MECHANICS AND PROPERTIES OF MATTER
**1. (a)**
(i) Distinguish between static and dynamic friction. **[2 marks]**
(ii) Give one example each of a contact force and a force field. **[1 mark]**
**(b)** Sketch a graph showing how the kinetic energy and potential energy of a mass-spring system varies with displacement. Label your graph to show the amplitude, region of minimum and maximum kinetic and potential energies with any two relevant assumptions stated. **[4 marks]**
**(c)** A spinning wheel initially has an angular velocity of 60 rad/s east; 20 s later its angular velocity is 80 rad/s west. If the angular acceleration is constant; calculate:
(i) the magnitude and direction of the angular acceleration.
(ii) the angular displacement over 20 s. **[3 marks]**
---
**2. (a)** Define the following terms:
(i) Hooke's law
(ii) Elastic limit **[2 marks]**
**(b)** The position of a particle moving along an x-axis is given by x = 2 + 4t + 12t² − 2t³, where x is in metres and t is in seconds. Determine:
(i) the position, velocity and acceleration of the particles at t = 3.0 s.
(ii) determine the average velocity of the particle between t = 2 s and t = 5 s **[5 marks]**
**(c)**
(i) State the principle of conservation of angular momentum. **[1 mark]**
(ii) List 2 factors affecting pressure in fluid. **[2 marks]**
---
## PHY 002: HEAT, WAVES AND OPTICS
**3. (a)**
(i) Define the 'coefficient of performance' of a refrigerator. **[1 mark]**
(ii) On a hot day, an oil truck was loaded with 33000 L of diesel fuel. A cold weather was encountered where the diesel was to be delivered which lowered the temperature by 28.3 K. Determine the amount of diesel delivered. (β_diesel = 9.50 × 10⁻⁴°C⁻¹, α_oil truck = 11 × 10⁻⁶°C⁻¹) **[3 marks]**
**(b)**
(i) Define 'Doppler effect' as applicable to sound waves. **[2 marks]**
(ii) What distinguishes a sound wave from a light wave? **[1 mark]**
**(c)**
(i) State Huygen's principle. **[2 marks]**
(ii) What is a diffraction grating? **[1 mark]**
---
**4. (a)** Define the following terms:
(i) Lateral inversion
(ii) Critical angle
(iii) Total internal reflection **[3 marks]**
**(b)**
(i) Differentiate between Myopia and Hyperopia. **[2 marks]**
(ii) A film of oil of refractive index 1.25 lies on liquid of refractive index 1.33. If the ray of light is incident at an angle of 60° in the oil on the liquid boundary. Calculate the angle of refraction in the liquid. **[2 marks]**
**(c)** A copper calorimeter of mass 0.20 kg contains 0.40 kg of water. The calorimeter and water were initially at 25°C. A 3.0 kg block of metal was heated to 95°C and placed in the water in the calorimeter. If the final steady temperature of the mixture is 60°C. Calculate the specific heat capacity of the metal. (specific heat capacity of water, c_w = 4200 J/kgK, specific heat capacity of copper, c_cu = 400 J/kgK). **[3 marks]**
---
## PHY 003: ELECTRICITY AND MAGNETISM
**5. (a)**
(i) Three point charges Q₁ = 3.0μC, Q₂ = −3.0μC and Q₃ = 2.0μC are located at −0.3 i, 0.3 i and 0.4 j, respectively. Find the resultant force on charge Q₃. **[5 marks]**
(ii) Write down the mathematical expression of Biot-Savart law. **[1 mark]**
**(b)**
(i) Calculate the induced e.m.f of a coil of 15 loops at the time t = 5 s if the magnetic flux through each loop is (1.8t − 2.21t³) × 10⁻² T.m². **[2 marks]**
(ii) Describe how to charge a neutral rod positively by induction. **[2 marks]**
---
**6.** *(Capacitor circuit diagram: two 2μf capacitors in series in the top branch, one 4μf capacitor in the bottom branch, both branches connected in parallel across a 120 V supply)*
**(a)** Using the diagram above, calculate:
(i) The amount of charge stored by the 2μf capacitors.
(ii) The potential difference across each of the 2μf capacitor.
(iii) The effective energy stored by the system. **[3½ marks]**
**(b)**
(i) Why are electric dielectric materials needed in the fabrication of capacitors?
(ii) A copper wire has a diameter of 0.2 mm. if current of 6 μA flows through it at the rate of 1.56 × 10⁸ m/s, how many electrons are involved in the flow and the electron density.
(iii) State Kirchhoff's junction and loop rules. **[4½ marks]**
**(c)** An electric motor is connected to 5 A, 110 V source. Determine the power input and the energy consumed by the motor in 1 hr. **[2 marks]**
---
## PHY 004: MODERN PHYSICS
**7. (a)** List at least two uses of the following:
(i) Ultrasound
(ii) Nuclear Magnetic Resonance
(iii) X – rays **[3 marks]**
**(b)** Calculate the wavelength of light emitted when a hydrogen atom makes a transition from n = 6 to the n = 2 energy level according to the Bohr model. **[5 marks]**
**(c)** State Moseley's law as applicable to X-rays. **[2 marks]**
---
**8. (a)** X-rays of wavelength 3.0 × 10⁻¹⁰ m are reflected off a crystal. When the angle between the x-ray beam and face of the crystal is increased from 0°, a strong reflected beam is detected when the angle becomes 30°. Calculate the spacing of the crystal planes responsible for this reflection. **[3 marks]**
# SECTION B: COMPLETE SOLUTIONS
---
## PHY 001: MECHANICS AND PROPERTIES OF MATTER
### Question 1
---
**1(a)(i) Static vs Dynamic Friction**
| | Static Friction | Dynamic Friction |
|---|---|---|
| Definition | Friction acting on a body **at rest**, preventing motion | Friction acting on a body **in motion** |
| Magnitude | Greater (higher coefficient) | Lesser (lower coefficient) |
| Nature | Self-adjusting up to maximum | Approximately constant |
---
**1(a)(ii) Examples**
- **Contact force:** Normal reaction / Tension / Friction
- **Force field:** Gravitational force / Magnetic force / Electric force
---
**1(b) Kinetic & Potential Energy Graph (Mass-Spring System)**
```
Energy
| KE PE
| ___/ \___ ___/ \___
| / \ / \
|/ \/ \
+-----|-------|-------|-------→ Displacement
-A 0 +A
```
**Assumptions:**
1. No energy is lost to friction (system is conservative)
2. Simple harmonic motion is assumed
**Labels:**
- At x = 0 (equilibrium): **KE is maximum, PE is minimum**
- At x = ±A (amplitude): **KE is minimum (zero), PE is maximum**
- Total Mechanical Energy = KE + PE = constant
---
**1(c) Angular Motion Problem**
Given:
- ω₁ = +60 rad/s (east = positive)
- ω₂ = −80 rad/s (west = negative)
- t = 20 s
**(i) Angular Acceleration:**
$$\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{-80 - 60}{20} = \frac{-140}{20}$$
$$\boxed{\alpha = -7 \ \text{rad/s}^2 \ \text{(westward direction)}}$$
**(ii) Angular Displacement:**
$$\theta = \omega_1 t + \frac{1}{2}\alpha t^2$$
$$\theta = (60)(20) + \frac{1}{2}(-7)(20)^2$$
$$\theta = 1200 - \frac{1}{2}(7)(400)$$
$$\theta = 1200 - 1400$$
$$\boxed{\theta = -200 \ \text{rad (i.e., 200 rad westward)}}$$
---
### Question 2
**2(a) Definitions**
**(i) Hooke's Law:**
The extension (or compression) of a spring is **directly proportional** to the applied force, provided the elastic limit is not exceeded.
$$F = ke$$
**(ii) Elastic Limit:**
The maximum stress/force beyond which a material **does not return** to its original shape or size after the force is removed. Permanent deformation occurs beyond this point.
---
**2(b) Position: x = 2 + 4t + 12t² − 2t³**
Finding velocity and acceleration by differentiation:
$$v = \frac{dx}{dt} = 4 + 24t - 6t^2$$
$$a = \frac{dv}{dt} = 24 - 12t$$
**(i) At t = 3.0 s:**
$$x = 2 + 4(3) + 12(3)^2 - 2(3)^3$$
$$x = 2 + 12 + 108 - 54 = \boxed{68 \ \text{m}}$$
$$v = 4 + 24(3) - 6(3)^2 = 4 + 72 - 54 = \boxed{22 \ \text{m/s}}$$
$$a = 24 - 12(3) = 24 - 36 = \boxed{-12 \ \text{m/s}^2}$$
**(ii) Average velocity between t = 2 s and t = 5 s:**
At t = 2:
$$x_2 = 2 + 4(2) + 12(4) - 2(8) = 2 + 8 + 48 - 16 = 42 \ \text{m}$$
At t = 5:
$$x_5 = 2 + 4(5) + 12(25) - 2(125) = 2 + 20 + 300 - 250 = 72 \ \text{m}$$
$$\bar{v} = \frac{x_5 - x_2}{t_5 - t_2} = \frac{72 - 42}{5 - 2} = \frac{30}{3}$$
$$\boxed{\bar{v} = 10 \ \text{m/s}}$$
---
**2(c)(i) Principle of Conservation of Angular Momentum:**
The total angular momentum of a system remains **constant** if no external torque acts on it.
$$L = I\omega = \text{constant}$$
**2(c)(ii) Two factors affecting pressure in a fluid:**
1. **Depth** of the fluid (P = ρgh)
2. **Density** of the fluid
---
## PHY 002: HEAT, WAVES AND OPTICS
### Question 3
**3(a)(i) Coefficient of Performance (COP) of a Refrigerator:**
$$COP = \frac{Q_L}{W} = \frac{\text{Heat removed from cold reservoir}}{\text{Work input}}$$
It measures how efficiently a refrigerator removes heat per unit of work done.
---
**3(a)(ii) Volume of Diesel Delivered**
Given:
- V₀ = 33,000 L
- ΔT = −28.3 K (temperature drop)
- β_diesel = 9.50 × 10⁻⁴ °C⁻¹
- α_oil truck = 11 × 10⁻⁶ °C⁻¹ → γ_truck = 3α = 33 × 10⁻⁶ °C⁻¹
Change in volume of diesel:
$$\Delta V_{diesel} = V_0 \beta_{diesel} \Delta T = 33000 \times 9.50 \times 10^{-4} \times (-28.3)$$
$$\Delta V_{diesel} = -887.7 \ \text{L}$$
Change in volume of tank:
$$\Delta V_{tank} = V_0 \gamma_{truck} \Delta T = 33000 \times 33 \times 10^{-6} \times (-28.3)$$
$$\Delta V_{tank} = -30.81 \ \text{L}$$
Net volume delivered = V₀ + ΔV_diesel − ΔV_tank (diesel contracts more):
$$V_{delivered} = 33000 + (-887.7) - (-30.81)$$
$$\boxed{V_{delivered} \approx 32,143 \ \text{L}}$$
---
**3(b)(i) Doppler Effect:**
The **apparent change in frequency** of a sound wave observed when there is **relative motion** between the source of sound and the observer.
**3(b)(ii) Sound wave vs Light wave:**
| Sound Wave | Light Wave |
|---|---|
| Mechanical (needs medium) | Electromagnetic (no medium needed) |
| Longitudinal | Transverse |
| Slower (~343 m/s in air) | Faster (3×10⁸ m/s in vacuum) |
---
**3(c)(i) Huygen's Principle:**
Every point on a wavefront acts as a **secondary source** of spherical wavelets. The new wavefront at a later time is the **tangential surface (envelope)** to all these secondary wavelets.
**3(c)(ii) Diffraction Grating:**
An optical device consisting of a large number of **equally spaced parallel slits** that diffracts light into several beams, used to measure wavelengths of light.
---
### Question 4
**4(a) Definitions**
**(i) Lateral Inversion:**
The phenomenon where the **left and right sides** of an image are interchanged relative to the object (as seen in a plane mirror).
**(ii) Critical Angle:**
The **angle of incidence** in the denser medium for which the angle of refraction in the less dense medium is **exactly 90°**.
**(iii) Total Internal Reflection:**
When light travels from a denser to a less dense medium and the angle of incidence **exceeds the critical angle**, all light is **completely reflected** back into the denser medium — none is refracted.
---
**4(b)(i) Myopia vs Hyperopia:**
| Myopia (Short-sightedness) | Hyperopia (Long-sightedness) |
|---|---|
| Can see near objects clearly | Can see distant objects clearly |
| Image forms in front of retina | Image forms behind retina |
| Corrected with concave lens | Corrected with convex lens |
---
**4(b)(ii) Angle of Refraction (Snell's Law)**
Given:
- n₁ = 1.25 (oil), n₂ = 1.33 (liquid)
- θ₁ = 60°
$$n_1 \sin\theta_1 = n_2 \sin\theta_2$$
$$1.25 \times \sin 60° = 1.33 \times \sin\theta_2$$
$$1.25 \times 0.8660 = 1.33 \sin\theta_2$$
$$\sin\theta_2 = \frac{1.0825}{1.33} = 0.8139$$
$$\boxed{\theta_2 = \sin^{-1}(0.8139) \approx 54.5°}$$
---
**4(c) Specific Heat Capacity of Metal**
Given:
- m_cu = 0.20 kg, c_cu = 400 J/kgK
- m_w = 0.40 kg, c_w = 4200 J/kgK
- m_metal = 3.0 kg
- T_initial (water + calorimeter) = 25°C
- T_metal = 95°C
- T_final = 60°C
Heat gained by water + calorimeter = Heat lost by metal
$$[m_w c_w + m_{cu} c_{cu}](T_f - T_i) = m_{metal} \cdot c_{metal}(T_{metal} - T_f)$$
$$[(0.40 \times 4200) + (0.20 \times 400)](60 - 25) = 3.0 \times c_{metal}(95 - 60)$$
$$[1680 + 80](35) = 3.0 \times c_{metal} \times 35$$
$$1760 \times 35 = 105 \cdot c_{metal}$$
$$61600 = 105 \cdot c_{metal}$$
$$\boxed{c_{metal} = \frac{61600}{105} \approx 586.7 \ \text{J/kgK}}$$
---
## PHY 003: ELECTRICITY AND MAGNETISM
### Question 5
**5(a)(i) Resultant Force on Q₃**
Given:
- Q₁ = 3.0 μC at position **r₁** = −0.3**i** m
- Q₂ = −3.0 μC at position **r₂** = +0.3**i** m
- Q₃ = 2.0 μC at position **r₃** = 0.4**j** m
- k = 9 × 10⁹ N·m²/C²
**Force F₁₃ (Q₁ on Q₃):**
$$\vec{r}_{13} = \vec{r}_3 - \vec{r}_1 = (0)\hat{i} + (0.4)\hat{j} - (-0.3\hat{i}) = 0.3\hat{i} + 0.4\hat{j}$$
$$|\vec{r}_{13}| = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \ \text{m}$$
$$F_{13} = \frac{kQ_1Q_3}{r_{13}^2} = \frac{9\times10^9 \times 3\times10^{-6} \times 2\times10^{-6}}{(0.5)^2} = \frac{0.054}{0.25} = 0.216 \ \text{N}$$
Unit vector: $\hat{r}_{13} = \frac{0.3\hat{i} + 0.4\hat{j}}{0.5} = 0.6\hat{i} + 0.8\hat{j}$
$$\vec{F}_{13} = 0.216(0.6\hat{i} + 0.8\hat{j}) = 0.1296\hat{i} + 0.1728\hat{j} \ \text{N}$$
**Force F₂₃ (Q₂ on Q₃):**
$$\vec{r}_{23} = \vec{r}_3 - \vec{r}_2 = -0.3\hat{i} + 0.4\hat{j}$$
$$|\vec{r}_{23}| = \sqrt{0.09 + 0.16} = 0.5 \ \text{m}$$
$$F_{23} = \frac{9\times10^9 \times 3\times10^{-6} \times 2\times10^{-6}}{0.25} = 0.216 \ \text{N}$$
Q₂ is negative → force is **attractive** (toward Q₂):
Unit vector toward Q₂: $\frac{-\vec{r}_{23}}{|\vec{r}_{23}|} = \frac{0.3\hat{i} - 0.4\hat{j}}{0.5} = 0.6\hat{i} - 0.8\hat{j}$
$$\vec{F}_{23} = 0.216(0.6\hat{i} - 0.8\hat{j}) = 0.1296\hat{i} - 0.1728\hat{j} \ \text{N}$$
**Total Force on Q₃:**
$$\vec{F}_{total} = \vec{F}_{13} + \vec{F}_{23}$$
$$= (0.1296 + 0.1296)\hat{i} + (0.1728 - 0.1728)\hat{j}$$
$$\boxed{\vec{F}_{total} = 0.2592\hat{i} \ \text{N} \approx 0.26\hat{i} \ \text{N (in the +x direction)}}$$
---
**5(a)(ii) Biot-Savart Law:**
$$d\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I(d\vec{l} \times \hat{r})}{r^2}$$
Where I = current, dl = length element, r = distance, μ₀ = 4π × 10⁻⁷ T·m/A
---
**5(b)(i) Induced EMF of Coil**
Given:
- N = 15 loops
- Φ = (1.8t − 2.21t³) × 10⁻² T·m²
- t = 5 s
$$\frac{d\Phi}{dt} = (1.8 - 6.63t^2) \times 10^{-2}$$
At t = 5 s:
$$\frac{d\Phi}{dt} = (1.8 - 6.63 \times 25) \times 10^{-2} = (1.8 - 165.75) \times 10^{-2} = -163.95 \times 10^{-2}$$
$$\mathcal{E} = -N\frac{d\Phi}{dt} = -15 \times (-163.95 \times 10^{-2})$$
$$\boxed{\mathcal{E} = 24.59 \ \text{V}}$$
---
**5(b)(ii) Charging a Rod Positively by Induction:**
1. Bring a **negatively charged** object near (not touching) the neutral rod
2. Electrons in the rod are **repelled** to the far end
3. **Ground** the far end — electrons flow to earth
4. **Remove the ground** connection
5. **Remove** the charged object
6. The rod is left with a **net positive charge** (deficit of electrons)
---
### Question 6
**Circuit Analysis:**
The two 2μF capacitors are in **series** (top branch), and the 4μF is in **bottom branch**; both branches are in **parallel** across 120 V.
**Series combination of the two 2μF:**
$$\frac{1}{C_{series}} = \frac{1}{2} + \frac{1}{2} = 1 \ \Rightarrow \ C_{series} = 1 \ \mu F$$
**Total effective capacitance:**
$$C_{total} = C_{series} + C_{4\mu F} = 1 + 4 = 5 \ \mu F$$
---
**6(a)(i) Charge stored by 2μF capacitors:**
The series branch (1 μF equivalent) is across 120 V:
$$Q_{2\mu F} = C_{series} \times V = 1 \times 10^{-6} \times 120$$
$$\boxed{Q = 1.2 \times 10^{-4} \ \text{C} = 120 \ \mu\text{C}}$$
*(Both 2μF capacitors store the same charge = 120 μC)*
---
**6(a)(ii) Potential difference across each 2μF capacitor:**
$$V_{each} = \frac{Q}{C} = \frac{120 \times 10^{-6}}{2 \times 10^{-6}} = \boxed{60 \ \text{V each}}$$
*(Total = 60 + 60 = 120 V ✓)*
---
**6(a)(iii) Effective energy stored:**
$$U = \frac{1}{2}C_{total}V^2 = \frac{1}{2} \times 5 \times 10^{-6} \times (120)^2$$
$$U = \frac{1}{2} \times 5 \times 10^{-6} \times 14400$$
$$\boxed{U = 0.036 \ \text{J} = 36 \ \text{mJ}}$$
---
**6(b)(i) Why dielectric materials are needed in capacitors:**
- Dielectrics **increase capacitance** (C = εA/d) by reducing the electric field
- They act as **insulators**, preventing charge from flowing between plates
- They **increase the breakdown voltage**, allowing more charge to be stored safely
---
**6(b)(ii) Electrons in copper wire:**
Given:
- Diameter = 0.2 mm = 2 × 10⁻⁴ m → radius = 1 × 10⁻⁴ m
- I = 6 μA = 6 × 10⁻⁶ A
- drift velocity v_d = 1.56 × 10⁸ m/s
Cross-sectional area:
$$A = \pi r^2 = \pi (1 \times 10^{-4})^2 = 3.14 \times 10^{-8} \ \text{m}^2$$
Using I = nev_dA:
$$n = \frac{I}{ev_d A} = \frac{6 \times 10^{-6}}{1.6 \times 10^{-19} \times 1.56 \times 10^8 \times 3.14 \times 10^{-8}}$$
$$n = \frac{6 \times 10^{-6}}{7.83 \times 10^{-19}}$$
$$\boxed{n \approx 7.66 \times 10^{12} \ \text{electrons/m}^3}$$
Number of electrons flowing (per second):
$$N = \frac{I}{e} = \frac{6 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 3.75 \times 10^{13} \ \text{electrons/s}$$
---
**6(b)(iii) Kirchhoff's Rules:**
- **Junction Rule (KCL):** The sum of currents entering a junction equals the sum of currents leaving it. (Conservation of charge)
$$\sum I_{in} = \sum I_{out}$$
- **Loop Rule (KVL):** The algebraic sum of all potential differences around any closed loop is zero. (Conservation of energy)
$$\sum V = 0$$
---
**6(c) Electric Motor Power and Energy:**
Given: I = 5 A, V = 110 V, t = 1 hr = 3600 s
$$P = IV = 5 \times 110 = \boxed{550 \ \text{W}}$$
$$E = Pt = 550 \times 3600 = \boxed{1,980,000 \ \text{J} = 1.98 \ \text{MJ}}$$
---
## PHY 004: MODERN PHYSICS
### Question 7
**7(a) Uses:**
**(i) Ultrasound:**
1. Medical imaging (pregnancy scans, organ imaging)
2. Treatment of kidney stones (lithotripsy)
3. Non-destructive testing of materials
**(ii) Nuclear Magnetic Resonance (NMR/MRI):**
1. Medical imaging of soft tissues (MRI scans)
2. Chemical structure determination in chemistry/spectroscopy
**(iii) X-rays:**
1. Medical diagnosis (bone fractures, chest X-rays)
2. Security scanning at airports
3. Crystallography (crystal structure determination)
---
**7(b) Wavelength of Hydrogen Transition (n=6 to n=2)**
Using Rydberg formula:
$$\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
R_H = 1.097 × 10⁷ m⁻¹, n_f = 2, n_i = 6
$$\frac{1}{\lambda} = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{36}\right)$$
$$= 1.097 \times 10^7\left(\frac{9 - 1}{36}\right) = 1.097 \times 10^7 \times \frac{8}{36}$$
$$= 1.097 \times 10^7 \times 0.2222 = 2.438 \times 10^6 \ \text{m}^{-1}$$
$$\lambda = \frac{1}{2.438 \times 10^6}$$
$$\boxed{\lambda = 4.10 \times 10^{-7} \ \text{m} = 410 \ \text{nm (violet light, Balmer series)}}$$
---
**7(c) Moseley's Law:**
The square root of the frequency of characteristic X-ray lines emitted by an element is **directly proportional** to the atomic number (Z) of the element:
$$\sqrt{f} = a(Z - b)$$
Where a and b are constants. This law confirmed the concept of atomic number.
---
### Question 8
**8(a) Crystal Plane Spacing (Bragg's Law)**
Given:
- λ = 3.0 × 10⁻¹⁰ m
- Glancing angle = 30° (angle between beam and crystal face)
- First strong reflection → n = 1
Using **Bragg's Law:**
$$2d\sin\theta = n\lambda$$
$$d = \frac{n\lambda}{2\sin\theta} = \frac{1 \times 3.0 \times 10^{-10}}{2 \times \sin 30°}$$
$$d = \frac{3.0 \times 10^{-10}}{2 \times 0.5} = \frac{3.0 \times 10^{-10}}{1}$$
$$\boxed{d = 3.0 \times 10^{-10} \ \text{m} = 0.3 \ \text{nm}}$$
---