Immediate Summary: Physical Sciences Exam Structure
In the NSC examinations, Physics (Paper 1 · 3 Hours · 150 Marks) covers Momentum & Impulse (15±3 marks), Vertical Projectile Motion (15±3 marks), Work, Energy & Power (15±3 marks), Doppler Effect (15±3 marks), Electrostatics & Electric Circuits (35±3 marks), and Electrodynamics & Photoelectric Effect (25±3 marks). Chemistry (Paper 2 · 3 Hours · 150 Marks) covers Organic Molecules (50±5 marks), Rates of Reaction (25±3 marks), Chemical Equilibrium (25±3 marks), Acids & Bases (20±3 marks), and Electrochemical Cells (30±3 marks).
Examiner Formula & Unit Penalties (DBE Guidelines)
Before attempting calculations, note the strict DBE marking rules that cost students tens of marks every year:
- Formula Mark: You must write the exact formula as listed on the official NSC data sheet (e.g., \( F_{net} = ma \), or \( W_{nc} = \Delta E_k + \Delta E_p \)). If you write a rearranged version or substitute immediately without writing the formula, you forfeit the formula mark.
- Unit Penalties: An answer without a unit or with an incorrect unit (e.g., writing \( m/s \) instead of \( m \cdot s^{-1} \)) incurs a 1-mark penalty. Direction is compulsory for vector quantities (Velocity, Acceleration, Force, Momentum).
Worked Example 1: Newton's Second Law with Connected Bodies (Paper 1 · 6 Marks)
Physics: Two Connected Blocks on a Rough Incline
6 Marks (Paper 1)
Exam Scenario: Block A (mass 4 kg) is on a rough surface inclined at 30° to the horizontal. Block A is connected to Block B (mass 6 kg) by a light inextensible string passing over a frictionless pulley. Block B hangs vertically. The coefficient of kinetic friction between Block A and the incline is 0.25. Calculate the acceleration of the system when released.
Step 1: Calculate frictional force on Block A
Normal Force N = m_A × g × cos(30°)
N = 4 × 9.8 × cos(30°) = 33.948 N
f_k = μ_k × N = 0.25 × 33.948 = 8.487 N (acting down the slope)
N = 4 × 9.8 × cos(30°) = 33.948 N
f_k = μ_k × N = 0.25 × 33.948 = 8.487 N (acting down the slope)
Step 2: Calculate parallel component of gravity on Block A
F_{g//} = m_A × g × sin(30°) = 4 × 9.8 × 0.5 = 19.6 N (acting down slope)
Step 3: Write Newton's Second Law equation for Block A (moving up slope)
F_{net} = m_A × a
T - f_k - F_{g//} = 4a
T - 8.487 - 19.6 = 4a
T - 28.087 = 4a ---- [Equation 1]
T - f_k - F_{g//} = 4a
T - 8.487 - 19.6 = 4a
T - 28.087 = 4a ---- [Equation 1]
Step 4: Write Newton's Second Law equation for Block B (moving downward)
F_{net} = m_B × a
w_B - T = 6a
(6 × 9.8) - T = 6a
58.8 - T = 6a ---- [Equation 2]
w_B - T = 6a
(6 × 9.8) - T = 6a
58.8 - T = 6a ---- [Equation 2]
Step 5: Add Equation 1 and Equation 2 (Tension cancels)
(T - 28.087) + (58.8 - T) = 4a + 6a
30.713 = 10a
a = 3.07 m · s&supmin;²
30.713 = 10a
a = 3.07 m · s&supmin;²
Mark Breakdown:
• Normal force & kinetic friction calculation (1 Mark)
• \(F_{g//}\) calculation (1 Mark)
• Correct equation for Block A with \(F_{net} = ma\) (1 Mark)
• Correct equation for Block B (1 Mark)
• Simultaneous algebraic solution (1 Mark)
• Final value with unit \(3.07 \text{ m} \cdot \text{s}^{-2}\) (1 Mark)
• Normal force & kinetic friction calculation (1 Mark)
• \(F_{g//}\) calculation (1 Mark)
• Correct equation for Block A with \(F_{net} = ma\) (1 Mark)
• Correct equation for Block B (1 Mark)
• Simultaneous algebraic solution (1 Mark)
• Final value with unit \(3.07 \text{ m} \cdot \text{s}^{-2}\) (1 Mark)
Worked Example 2: Doppler Effect with Approaching Source (Paper 1 · 5 Marks)
Physics: Sound Waves & Doppler Frequency Shifts
5 Marks (Paper 1)
Question: An ambulance siren emits sound waves of frequency 650 Hz as it travels towards a stationary pedestrian at a constant speed of 25 m · s-1. The speed of sound in air is 340 m · s-1. Calculate the frequency of sound detected by the pedestrian.
Step 1: Write official formula from Data Sheet
f_L = [ (v ± v_L) / (v ± v_s) ] × f_s
Step 2: Identify variables and correct signs
1. Listener is stationary: \( v_L = 0 \text{ m} \cdot \text{s}^{-1} \).
2. Source approaches listener: Distance decreases, frequency increases → Choose minus sign in denominator: \( v - v_s \).
f_L = [ v / (v - v_s) ] × f_s
f_L = [ 340 / (340 - 25) ] × 650
f_L = [ 340 / 315 ] × 650
f_L = 1.079365 × 650 = 701.59 Hz
f_L = [ 340 / (340 - 25) ] × 650
f_L = [ 340 / 315 ] × 650
f_L = 1.079365 × 650 = 701.59 Hz
Examiner Rule: Never omit the formula mark. Writing down the general formula from the data sheet secures the first mark before substitution.
Worked Example 3: Chemical Equilibrium Table & Kc (Paper 2 · 7 Marks)
Chemistry: RICE Table & Equilibrium Constant Kc
7 Marks (Paper 2)
Question: In a closed 2 dm³ container at 450°C, 4.0 mol of N&sub2; and 10.0 mol of H&sub2; are sealed and allowed to reach equilibrium according to the equation:
\( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \).
At equilibrium, it is found that 4.0 mol of NH&sub3; are present. Calculate the equilibrium constant \( K_c \) at this temperature.
\( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \).
At equilibrium, it is found that 4.0 mol of NH&sub3; are present. Calculate the equilibrium constant \( K_c \) at this temperature.
Step 1: Set up RICE (Ratio, Initial, Change, Equilibrium) Table
| Mole Calculations | N&sub2; | 3H&sub2; | 2NH&sub3; |
|---|---|---|---|
| Molar Ratio | 1 | 3 | 2 |
| Initial Moles (n) | 4.0 mol | 10.0 mol | 0 mol |
| Change in Moles (Δn) | -2.0 mol | -6.0 mol | +4.0 mol |
| Equilibrium Moles (n_eq) | 2.0 mol | 4.0 mol | 4.0 mol |
| Equilibrium Conc. [c = n/V] (V = 2 dm³) | 2.0 / 2 = 1.0 mol·dm&supmin;³ | 4.0 / 2 = 2.0 mol·dm&supmin;³ | 4.0 / 2 = 2.0 mol·dm&supmin;³ |
Step 2: Write Kc Expression & Substitute Equilibrium Concentrations
K_c = [NH&sub3;]² / ( [N&sub2;] × [H&sub2;]³ )
K_c = (2.0)² / [ (1.0) × (2.0)³ ]
K_c = 4.0 / [ 1.0 × 8.0 ] = 4.0 / 8.0 = 0.50
K_c = (2.0)² / [ (1.0) × (2.0)³ ]
K_c = 4.0 / [ 1.0 × 8.0 ] = 4.0 / 8.0 = 0.50
Marking Rubric:
• Ratio deduction: Using ratio to find \( \Delta n(N_2) = -2.0 \) and \( \Delta n(H_2) = -6.0 \) (2 Marks)
• Equilibrium moles calculation for all three species (1 Mark)
• Dividing by container volume (2 dm³) to get concentrations (1 Mark)
• Correct \(K_c\) expression formula (1 Mark)
• Correct substitution into \(K_c\) (1 Mark)
• Final answer \(0.5\) with no units (Kc has no units) (1 Mark)
• Ratio deduction: Using ratio to find \( \Delta n(N_2) = -2.0 \) and \( \Delta n(H_2) = -6.0 \) (2 Marks)
• Equilibrium moles calculation for all three species (1 Mark)
• Dividing by container volume (2 dm³) to get concentrations (1 Mark)
• Correct \(K_c\) expression formula (1 Mark)
• Correct substitution into \(K_c\) (1 Mark)
• Final answer \(0.5\) with no units (Kc has no units) (1 Mark)