The problems are on the uncertainty and vectors worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Homogeneity means every term of an equation has the same base units, which lets you test formulae. These worked answers correct the 20 problems of the Physics 11-12 worksheet on uncertainty and vectors, from base units to crossing a river: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result in bold. Intermediate values are rounded only at the end, so a small difference in the last digit is acceptable. Compare your method as well as your numbers.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 1, IB A
Problem 1 Base units
1. F = ma, so N = kg × m s⁻² = kg m s⁻².2. J = N m = kg m s⁻² × m = kg m² s⁻².
3. W = J s⁻¹ = kg m² s⁻² × s⁻¹ = kg m² s⁻³.
Problem 2 Prefix conversions
1. 4.5 mm = 4.5 × 10⁻³ m = 0.0045 m.2. 72 ÷ 3.6 = 20 m/s.
3. 0.25 g/cm³ = 0.25 × 1000 kg/m³ = 250 kg/m³.
4. 1.2 GW = 1.2 × 10⁹ W.
Problem 3 Scalar or vector
1. Scalar. 2. Vector. 3. Scalar. 4. Vector. 5. Scalar. Speed, kinetic energy and time are scalars; displacement and force are vectors.Problem 4 Percentage of one reading
1. 0.1 ÷ 24.3 × 100 = 0.41 %.2. 0.2 ÷ 1.6 × 100 = 12.5 %, so about 13 %.
Problem 5 Components of a pull
1. Fx = 120 cos 40° = 120 × 0.766 = 91.9 N.2. Fy = 120 sin 40° = 120 × 0.643 = 77.1 N.
Problem 6 Spot the wrong equation
1. v and u are in m s⁻¹. a t² is (m s⁻²)(s²) = m.2. A term in m cannot be added to terms in m s⁻¹, so the equation is not homogeneous. It is dimensionally wrong (the correct term is at).
Problem 7 Check v² = u² + 2as
1. v² and u²: (m s⁻¹)² = m² s⁻². 2as: (m s⁻²)(m) = m² s⁻².2. All three terms match, so the equation is homogeneous.
Problem 8 Unit of a drag constant
1. k = F ÷ v².2. Units: kg m s⁻² ÷ m² s⁻² = kg m⁻¹.
Problem 9 Sum and difference
1. 15.2 + 8.7 = 23.9 cm; uncertainty 0.1 + 0.1 = 0.2 cm: 23.9 ± 0.2 cm.2. 15.2 − 8.7 = 6.5 cm, uncertainty again 0.2 cm: 6.5 ± 0.2 cm.
3. Sum: 0.2 ÷ 23.9 = 0.84 %. Difference: 0.2 ÷ 6.5 = 3.1 %. The difference has the larger percentage uncertainty.
Problem 10 Area of a garden
1. 0.5 ÷ 45.0 = 1.11 % and 0.5 ÷ 30.0 = 1.67 %.2. Area = 45.0 × 30.0 = 1350 m². Percentage uncertainty = 1.11 + 1.67 = 2.78 %, so 0.0278 × 1350 = 37.5 m². Area = 1350 ± 40 m².
Problem 11 Speed with uncertainty
1. v = 100.0 ÷ 12.5 = 8.00 m/s.2. 0.5 ÷ 100.0 = 0.50 % and 0.1 ÷ 12.5 = 0.80 %, total 1.3 %.
3. 0.013 × 8.00 = 0.104 m/s, so v = 8.0 ± 0.1 m/s.
Problem 12 Volume of a sphere
1. V = (4/3) × π × 2.50³ = 4.189 × 15.625 = 65.4 cm³.2. 0.05 ÷ 2.50 = 2.0 %.
3. Power of 3: 3 × 2.0 = 6.0 %. 0.060 × 65.4 = 3.9 cm³. V = 65 ± 4 cm³.
Problem 13 Perpendicular forces
1. R = √(8.0² + 6.0²) = √100 = 10.0 N.2. tan θ = 6.0 ÷ 8.0 = 0.750, so θ = 36.9° from the 8.0 N force.
Problem 14 Suitcase handle
1. 55 cos 28° = 55 × 0.883 = 48.6 N.2. 55 sin 28° = 55 × 0.469 = 25.8 N.
3. The upward component of 25.8 N lifts the suitcase slightly and reduces the push on the floor.
Problem 15 Repeated timings
1. Sum = 11.60 s, mean = 11.60 ÷ 5 = 2.32 s.2. Range = 2.35 − 2.29 = 0.06 s, so uncertainty = ±0.03 s.
3. 0.03 ÷ 2.32 = 1.3 %.
Problem 16 Finding g with a pendulum
1. g = 39.48 × 0.840 ÷ 1.84² = 33.16 ÷ 3.386 = 9.79 m/s².2. L: 0.002 ÷ 0.840 = 0.24 %. T: 0.02 ÷ 1.84 = 1.09 %.
3. T is squared, so 2 × 1.09 = 2.17 %. Total = 0.24 + 2.17 = 2.41 %, and 0.0241 × 9.79 = 0.24 m/s².
4. g = 9.8 ± 0.2 m/s².
Problem 17 Gradient uncertainty
1. Best: 42 ÷ 10.0 = 4.2 m/s. Steepest: (47 − 0.5) ÷ 10.0 = 4.65 m/s. Shallowest: (38 + 0.5) ÷ 10.0 = 3.85 m/s.2. Uncertainty = (4.65 − 3.85) ÷ 2 = 0.40 m/s.
3. speed = 4.2 ± 0.4 m/s.
Problem 18 Two tugs on a barge
1. 500 + 350 cos 60° = 500 + 175 = 675 N.2. 350 sin 60° = 303 N.
3. R = √(675² + 303²) = √(455 625 + 91 875) = √547 500 = 740 N.
4. tan θ = 303 ÷ 675 = 0.449, so θ = 24.2° north of east.
Problem 19 Dimensional analysis
1. F: kg m s⁻². m: kg. v: m s⁻¹. r: m.2. kg: 1 = a. s: −2 = −b. m: 1 = b + c.
3. a = 1, b = 2, c = 1 − 2 = −1, so F = k m v² ÷ r. Dimensions cannot give k itself.
Problem 20 Crossing a river
1. t = 90 ÷ 3.0 = 30 s.2. Drift = 1.2 × 30 = 36 m.
3. v = √(3.0² + 1.2²) = √10.44 = 3.2 m/s.
4. tan θ = 1.2 ÷ 3.0 = 0.40, so θ = 21.8° downstream.
