Grade 13 — Term 3
14 study notes · 45 practice questions · Aligned with NIE syllabus
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This term in the Idasara Academy app has 14 lessons (NIE Units 3, 6, 7, 8). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
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Lessons in This Term
Velocity and acceleration in circular motion
What you must know
- Centripetal Force requirement: By Newton's second law, circular motion requires an inward net resultant force F_c = m * a_r = m*v^2 / r = m*r*ω^2 provided by tension, friction, gravity, or normal reaction.
- Period and Frequency: Period T = 2π / ω = 2πr / v; Frequency f = 1 / T = ω / (2π).
- Velocity vector is ALWAYS tangential: At every instant, the linear velocity vector v is directed along the tangent to the circular path; perpendicular to the position vector r (v · r = 0).
Try one question
A particle moves along a circular path of radius r = 2.0 m with a constant speed v = 6.0 m s⁻¹. What are the magnitudes of its tangential acceleration a_t and centripetal (radial) acceleration a_r?
- A. a_t = 18 m s⁻², a_r = 0
- B. a_t = 0, a_r = 36 m s⁻²
- C. a_t = 0, a_r = 18 m s⁻²
- D. a_t = 3 m s⁻², a_r = 18 m s⁻²
Show the answer
Answer: C. a_t = 0, a_r = 18 m s⁻². Since the speed is constant (uniform circular motion), the rate of change of speed is zero: a_t = dv/dt = 0. The centripetal acceleration directed toward the center is a_r = v² / r = (6.0 m s⁻¹)² / (2.0 m) = 36 / 2 = 18 m s⁻².
A model question written by Idasara on the exam pattern. It is not from a past paper.
Motion in a horizontal circle
What you must know
- Friction on flat unbanked curved road: Maximum safe turning speed without skidding on flat curve of radius r: friction force f ≤ μ N = μ mg => m*v^2/r ≤ μ mg => v_max = √(μ g r).
- Banked track with friction: Maximum speed before slipping up: tan(θ + λ) = v_max^2 / (rg) where λ is angle of friction (tan λ = μ); minimum speed before slipping down: tan(θ - λ) = v_min^2 / (rg).
- Rotor / Wall of Death ride: Rider of mass m pressed against vertical cylinder wall of radius r spinning at ω: normal reaction N = m*r*ω^2; friction balances weight f = mg; no slipping if f ≤ μ N => mg ≤ μ(m*r*ω^2) => ω_min = √(g / (μ r)).
Try one question
A conical pendulum consists of a small bob of mass m attached to a light inextensible string of length L rotating in a horizontal circle such that the string makes a constant angle θ with the downward vertical. What is the tension T in the string?
- A. T = mg / cos θ
- B. T = mg sin θ
- C. T = mg / sin θ
- D. T = mg cos θ
Show the answer
Answer: A. T = mg / cos θ. Resolving forces on the bob vertically: since there is no vertical acceleration, the vertical component of string tension balances gravity: T cos θ = mg => T = mg / cos θ. (The horizontal component T sin θ provides the centripetal force m v² / r).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Motion in a vertical circle
What you must know
- Motion inside a smooth vertical hollow sphere: Normal reaction R replaces tension T; particle leaves the surface where R = 0 (cos θ = -(u^2 - 2ga)/(3ga)); if u ≤ √(2ga), particle oscillates in lower hemisphere; if u ≥ √(5ga), completes full loops; if √(2ga) < u < √(5ga), leaves surface in upper hemisphere and moves as a free parabola under gravity.
- Particle moving on a light rigid rod: A rigid rod can support thrust (compression) as well as tension; particle completes loop provided speed at top v_top > 0, requiring speed at lowest point u > √(4ga) = 2√(ga).
- Tension difference between bottom and top: In any complete vertical circular revolution: T_bottom - T_top = 6mg, completely INDEPENDENT of initial speed u.
Try one question
A particle of mass m attached to the end of a light inextensible string of length r is whirled in a vertical circle. What is the minimum speed v_top the particle must have at the highest point of the circle so that the string does not become slack (tension T ≥ 0)?
- A. √(5 g r)
- B. √(g r)
- C. √(2 g r)
- D. 2 √(g r)
Show the answer
Answer: B. √(g r). At the topmost point, both tension T and weight mg act vertically downwards toward the center: T + mg = m v_top² / r => T = m (v_top² / r - g). For the string not to slacken, T ≥ 0 => v_top² / r ≥ g => v_top ≥ √(g r). Hence, the minimum speed is √(g r).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Simple harmonic motion
What you must know
- Auxiliary Circle representation: SHM along diameter can be modeled as projection of uniform circular motion of radius a with angular speed ω on the auxiliary circle; displacement x = a cos θ, phase angle θ = ωt.
- Energy in SHM: Kinetic energy E_k = (1/2)m v^2 = (1/2)m ω^2 (a^2 - x^2); Potential energy E_p = (1/2)m ω^2 x^2; Total mechanical energy E = E_k + E_p = (1/2)m ω^2 a^2 (constant).
- Time to travel between positions: Time from center (x = 0) to x = x1 is t = (1/ω) arcsin(x1/a); Time from extremity (x = a) to x = x1 is t = (1/ω) arccos(x1/a).
Try one question
A particle moves in simple harmonic motion (SHM) defined by ẍ = -16x. If the amplitude of oscillation is A = 0.50 m, what is the maximum speed v_max of the particle?
- A. 1.0 m s⁻¹
- B. 8.0 m s⁻¹
- C. 2.0 m s⁻¹
- D. 4.0 m s⁻¹
Show the answer
Answer: C. 2.0 m s⁻¹. Comparing ẍ = -16x with the standard SHM differential equation ẍ = -ω² x gives angular frequency ω = √16 = 4 rad s⁻¹. The maximum speed occurs at the equilibrium position (x = 0): v_max = ω A = (4 rad s⁻¹) × 0.50 m = 2.0 m s⁻¹.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Simple harmonic motion on a horizontal line
What you must know
- Time calculation for composite string-slack motion: Total periodic time = (Time in SHM where string is taut) + (Time in constant speed where string is slack): T_total = 2 * (2 * t_SHM) + 2 * (distance / v_max).
- Elastic Potential Energy: Work done in stretching an elastic string from natural length to extension x is E_e = (1/2) * (λ / l) * x^2 = (1/2) * T * x.
- Energy Conservation in horizontal SHM: (1/2)m v^2 + (1/2)(λ/l)x^2 = (1/2)(λ/l)a^2 = (1/2)m v_max^2.
Try one question
A particle of mass m = 0.20 kg attached to an ideal horizontal spring of stiffness k = 20 N m⁻¹ oscillates without friction on a smooth horizontal table. What is the period T of oscillation?
- A. 0.20π s (approx 0.628 s)
- B. 1.0 s
- C. 0.40π s
- D. 0.10π s
Show the answer
Answer: A. 0.20π s (approx 0.628 s). For a mass-spring system in horizontal SHM, the restoring force is F = -kx = m ẍ => ẍ = -(k/m) x. Angular frequency is ω = √(k/m) = √(20 / 0.20) = √100 = 10 rad s⁻¹. The period of oscillation is T = 2π / ω = 2π / 10 = 0.20π s ≈ 0.628 s.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Simple harmonic motion on a vertical line
What you must know
- Complete vertical oscillation without slackness: The particle executes pure SHM for its entire cycle if and only if amplitude a ≤ e = mgl / λ.
- Maximum height reached when string goes slack (a > e): When particle rises above slack point (x = -e), it has upward speed v_slack = ω√(a^2 - e^2); additional height climbed as free projectile under gravity is h_add = v_slack^2 / (2g) = (a^2 - e^2)/(2e).
- Total Mechanical Energy in vertical SHM: Sum of kinetic energy, gravitational potential energy, and elastic potential energy is conserved throughout the entire motion.
Try one question
A light elastic spring hanging vertically is extended by an equilibrium length e = 0.098 m when a mass m is gently attached to it. The mass is then pulled down slightly and released. Taking g = 9.8 m s⁻², what is the period of the resulting vertical oscillations?
- A. 0.50 s
- B. 1.0 s
- C. 0.10π s
- D. 0.20π s (approx 0.628 s)
Show the answer
Answer: D. 0.20π s (approx 0.628 s). At vertical equilibrium, string tension equals weight: k e = m g => k / m = g / e. The angular frequency of small vertical oscillations about this equilibrium point is ω = √(k/m) = √(g/e) = √(9.8 / 0.098) = √100 = 10 rad s⁻¹. Period T = 2π / ω = 2π / 10 = 0.20π s ≈ 0.628 s.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Events of a random experiment
What you must know
- De Morgan's Laws for Events: (A ∪ B)' = A' ∩ B' (neither A nor B occurs); (A ∩ B)' = A' ∪ B' (at least one does not occur).
- Relative complement (Difference of events): A - B = A ∩ B' (event A occurs but B does not); P(A ∩ B') = P(A) - P(A ∩ B).
- Venn diagram visualization: Always sketch overlapping Venn circles for events A and B to partition sample space into four disjoint subsets: A ∩ B', A ∩ B, A' ∩ B, and A' ∩ B'.
Try one question
Two events A and B in a sample space S are mutually exclusive (disjoint). If P(A) = 0.35 and P(B) = 0.45, what is the probability P(A ∪ B)?
- A. 0.1575
- B. 0.6425
- C. 0.10
- D. 0.80
Show the answer
Answer: D. 0.80. For mutually exclusive events, they cannot occur simultaneously: A ∩ B = ∅ => P(A ∩ B) = 0. By the addition rule of probability: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.35 + 0.45 - 0 = 0.80.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Probability models for random events
What you must know
- Mutually Exclusive vs Independent distinction: Mutually exclusive means A and B CANNOT happen together (A ∩ B = ∅ => P(A ∩ B) = 0); Independent means P(A ∩ B) = P(A)*P(B) > 0 for non-zero events; two events with non-zero probability CANNOT be both mutually exclusive and independent!
- Probability of at least one occurrence: For independent events A1, A2, ..., An: P(at least one occurs) = 1 - P(none occurs) = 1 - [P(A1') * P(A2') * ... * P(An')].
- Exactly one of A or B occurs: P(exactly one) = P(A ∩ B') + P(A' ∩ B) = P(A ∪ B) - P(A ∩ B) = P(A) + P(B) - 2P(A ∩ B).
Try one question
Two events A and B are independent with P(A) = 0.40 and P(B) = 0.50. What is the probability that at least one of the two events occurs, P(A ∪ B)?
- A. 0.60
- B. 0.20
- C. 0.70
- D. 0.90
Show the answer
Answer: C. 0.70. For independent events, P(A ∩ B) = P(A) · P(B) = 0.40 × 0.50 = 0.20. By the general addition rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.40 + 0.50 - 0.20 = 0.70.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Conditional probability
What you must know
- P(A | B) ≠ P(B | A) asymmetry: Conditional probability is strictly asymmetric; e.g., P(cloudy | raining) = 1, but P(raining | cloudy) is far less than 1.
- With vs Without Replacement in urn models: Drawing balls without replacement makes successive draws DEPENDENT (probabilities change); drawing with replacement preserves INDEPENDENCE (probabilities remain constant).
- Law of Total Probability: If B1, B2, ..., Bn form a partition of S (mutually exclusive and exhaustive): P(A) = ∑_{i=1}^n P(Bi) * P(A | Bi).
Try one question
For two events A and B, P(B) = 0.50 and P(A ∩ B) = 0.15. What is the conditional probability of A given that B has occurred, P(A | B)?
- A. 0.35
- B. 0.075
- C. 0.30
- D. 0.65
Show the answer
Answer: C. 0.30. By the definition of conditional probability: P(A | B) = P(A ∩ B) / P(B). Substituting values: P(A | B) = 0.15 / 0.50 = 0.30.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Bayes' theorem
What you must know
- Medical screening diagnostic paradox: False positive rates can cause posterior probability P(Disease | Test Positive) to be counterintuitively low when disease prevalence (prior P(Disease)) is extremely rare in the general population.
- Denominator is the Law of Total Probability: The denominator in Bayes' formula is precisely P(A) evaluated over all possible mutually exclusive alternative paths.
- Two-hypothesis binary case: P(B | A) = [P(B) P(A | B)] / [P(B) P(A | B) + P(B') P(A | B')].
Try one question
Two factories X and Y produce identical light bulbs. Factory X produces 60% of the total output with a 2% defect rate, while Factory Y produces 40% with a 5% defect rate. If a randomly inspected bulb is found to be defective, what is the probability that it was manufactured by Factory X?
- A. 3/8 (0.375)
- B. 1/2 (0.500)
- C. 3/5 (0.600)
- D. 5/8 (0.625)
Show the answer
Answer: A. 3/8 (0.375). Let D be defective. Prior probabilities: P(X) = 0.60, P(Y) = 0.40. Conditional probabilities: P(D|X) = 0.02, P(D|Y) = 0.05. Total probability of defect: P(D) = P(X)P(D|X) + P(Y)P(D|Y) = (0.60 × 0.02) + (0.40 × 0.05) = 0.012 + 0.020 = 0.032. By Bayes' Theorem: P(X|D) = P(X)P(D|X) / P(D) = 0.012 / 0.032 = 12 / 32 = 3/8 = 0.375.
A model question written by Idasara on the exam pattern. It is not from a past paper.
The nature of statistics
What you must know
- Class boundaries vs Class limits: For grouped data with gaps (e.g., 10-19, 20-29), True Class Boundaries are 9.5 - 19.5, 19.5 - 29.5 (class width = upper boundary - lower boundary = 10).
- Class Midpoint (Mark) x: x = (Lower Limit + Upper Limit) / 2 = (Lower Boundary + Upper Boundary) / 2.
- Histogram frequency density: When class widths are unequal, bar height must represent Frequency Density = Frequency / Class Width, so that area remains proportional to frequency.
Try one question
Which of the following variables in statistical data collection is correctly classified as a discrete quantitative variable?
- A. The height of a competitor measured in meters
- B. The eye color of an applicant
- C. The number of students present in a classroom
- D. The body temperature of a patient measured in °C
Show the answer
Answer: C. The number of students present in a classroom. A discrete quantitative variable takes distinct, separated count values (non-negative integers, e.g., number of students). Temperature and height are continuous variables (can assume any value in a continuum), while eye color is a qualitative (categorical) variable.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Measures of central tendency
What you must know
- Sensitivity to Outliers: Mean is heavily distorted by extreme values / outliers; Median is ROBUST against extreme outliers and preferred for skewed income/economic data.
- Empirical relationship in skewed distributions: Mean - Mode ≈ 3 * (Mean - Median).
- Algebraic property of Mean: The algebraic sum of deviations of all observations from their arithmetic mean is identically ZERO: ∑ (x_i - x̄) = 0.
Try one question
In a moderately skewed unimodal frequency distribution, the mean is 24 and the median is 22. Using the empirical Pearson relationship between central tendency measures, what is the estimated mode?
- A. 20
- B. 16
- C. 26
- D. 18
Show the answer
Answer: D. 18. Pearson's empirical formula relating the three measures of central tendency is: Mode ≈ 3 × Median - 2 × Mean. Substituting the given values: Mode ≈ 3(22) - 2(24) = 66 - 48 = 18.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Measures of relative position
What you must know
- Outlier detection using 1.5 * IQR rule: An observation x is considered an Outlier if x < Q1 - 1.5*IQR or x > Q3 + 1.5*IQR.
- Estimating quantiles from Ogive: Plot cumulative frequency polygon (Ogive); read x-value corresponding to N/4 on y-axis for Q1, N/2 for Median, and 3N/4 for Q3.
- Skewness from Quartiles (Bowley's coefficient of skewness): Sk = [(Q3 - Q2) - (Q2 - Q1)] / (Q3 - Q1) = (Q3 + Q1 - 2*Q2) / (Q3 - Q1); Sk = 0 for symmetric, Sk > 0 for right-skewed, Sk < 0 for left-skewed.
Try one question
In a dataset of examination marks, the lower quartile is Q₁ = 45 and the upper quartile is Q₃ = 75. What are the Interquartile Range (IQR) and the Semi-interquartile Range (Quartile Deviation)?
- A. IQR = 15, Quartile Deviation = 30
- B. IQR = 30, Quartile Deviation = 15
- C. IQR = 120, Quartile Deviation = 60
- D. IQR = 30, Quartile Deviation = 30
Show the answer
Answer: B. IQR = 30, Quartile Deviation = 15. Interquartile Range is defined as IQR = Q₃ - Q₁ = 75 - 45 = 30. The Semi-interquartile Range (Quartile Deviation) is half of IQR: QD = (Q₃ - Q₁) / 2 = 30 / 2 = 15.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Measures of dispersion
What you must know
- Effect of Linear Transformations: If y_i = a * x_i + b: New mean ȳ = a * x̄ + b; New variance σ_y^2 = a^2 * σ_x^2 (constant b has zero effect on spread); New standard deviation σ_y = |a| * σ_x.
- Coefficient of Variation (CV): Relative dispersion metric independent of measurement units: CV = (σ / x̄) * 100%; lower CV indicates greater consistency / reliability.
- Combined Variance of two samples: For sample 1 (size n1, mean x̄1, variance s1^2) and sample 2 (size n2, mean x̄2, variance s2^2): Combined variance s^2 = [n1(s1^2 + d1^2) + n2(s2^2 + d2^2)] / (n1 + n2) where d1 = x̄1 - x̄_c, d2 = x̄2 - x̄_c.
Try one question
For a sample of n = 5 observations {2, 4, 6, 8, 10}, what is the population variance (σ²)?
- A. 10.0
- B. 6.0
- C. 2.83
- D. 8.0
Show the answer
Answer: D. 8.0. Mean μ = (2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6. Deviations from mean: (2-6)=-4, (4-6)=-2, (6-6)=0, (8-6)=2, (10-6)=4. Squared deviations: 16, 4, 0, 4, 16. Sum of squares = 40. Variance σ² = ∑(x - μ)² / n = 40 / 5 = 8.0.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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