Scalars and Vectors: Practice Questions (Physics C: Mechanics 1.1)

Answer: (C). Position, displacement and acceleration all need a direction to describe them fully, so all three are vectors.

• (A) includes distance, a path length with no direction. It is a scalar.
• (B) includes speed, the size of a velocity without its direction. It is a scalar.
• (D) includes time. Asking “which way is 4 s?” has no answer, so time is a scalar.

Answer: (B). The angle is measured from south (−y). The component next to the angle is along y: v_y = −8.0 cos 40° = −6.1 m/s (south, so negative). The component opposite the angle is along x: v_x = −8.0 sin 40° = −5.1 m/s (west, so negative). Check: √(5.14² + 6.13²) = 8.0 m/s.

• (A) uses cosine for x and sine for y, as if the angle were measured from the x-axis. The magnitude is still 8.0 m/s, so a magnitude check alone does not catch this; a sketch does.
• (C) gives the x-component the wrong sign. West is −x.
• (D) gives the y-component the wrong sign. South is −y.

Answer: (D). |d| = √(36 + 4.0 + 9.0) = √49 = 7.0 m. Divide each component by 7.0 m: 6.0/7.0 = 0.86, 2.0/7.0 = 0.29, −3.0/7.0 = −0.43. Check: 0.857² + 0.286² + 0.429² = 1.00, and the metres cancel.

• (A) divides by 11, the sum of the sizes of the components. Its magnitude is 0.64, not 1.
• (B) keeps only the signs. Its magnitude is √3 ≈ 1.7, and it points in a different direction from d.
• (C) divides by |d|² = 49. Its magnitude is 0.14, not 1.

1. Components of d₂. The angle is measured from west (−x). x-component −300 cos 60.0° = −150 m; y-component +300 sin 60.0° = +259.8 m.
2. Components of d₃. d₃ = (0 î − 150 ĵ) m.
3. (a) R_x = 250 − 150 + 0 = 100 m; R_y = 100 + 259.8 − 150 = 209.8 m. R = (100 î + 210 ĵ) m.
4. (b) |R| = √(100² + 209.8²) = 232 m. Both components are positive, so θ = tan⁻¹(209.8 / 100) = 64.5° north of east.
5. (c) |d₁| = √(250² + 100²) = 269.3 m. Distance = 269.3 + 300 + 150 = 719 m.

Suggested mark points (4): 1 for both components of d₂ with correct signs; 1 for R in unit vector notation; 1 for the magnitude and a direction with a reference line (for example “north of east”); 1 for the distance as the sum of the three magnitudes.

Common error: adding the three magnitudes and calling 719 m the displacement. That is the distance; the walks point different ways.

(a) |A| = √(p² + (−2p)² + (2p)²) = √(9p²) = 3p (p is positive, so no modulus sign is needed).

(b)  = A / |A| = (p î − 2p ĵ + 2p k̂) / 3p = ⅓ î − ⅔ ĵ + ⅔ k̂. Magnitude: √(1/9 + 4/9 + 4/9) = √1 = 1. The metres in p cancel with the metres in 3p, so  has no unit.

(c) B = −L Â = −(L/3) î + (2L/3) ĵ − (2L/3) k̂. Check: |B| = L × |Â| = L.

(d) The z-component of A is |A| cos θ, where θ is the angle from +z. So cos θ = 2p / 3p = ⅔ and θ = 48.2°.

Alternative method. For (d), the dot product with k̂ gives the same result, but it belongs to Topic 3.2 and is not needed here.

(a) Largest 11 m, when A and B point the same way (θ = 0°). Smallest 3.0 m, when they point opposite ways (θ = 180°). Along one axis this is +4.0 m + 7.0 m = +11 m, or +4.0 m + (−7.0 m) = −3.0 m.

(b) B = 7.0 cos θ î + 7.0 sin θ ĵ. Then R = (4.0 + 7.0 cos θ) î + 7.0 sin θ ĵ, so |R|² = (4.0 + 7.0 cos θ)² + (7.0 sin θ)² = 16 + 56 cos θ + 49(cos² θ + sin² θ) = 65 + 56 cos θ (in m²).

(c) 64 = 65 + 56 cos θ, so cos θ = −1/56 = −0.0179 and θ = 91.0°.

(d) The claim is false, though close. For perpendicular vectors, cos θ = 0 and |R| = √65 = 8.06 m, not 8.0 m. A resultant of exactly 8.0 m needs θ = 91.0°, slightly more than a right angle. The right-triangle rule is a special case of the result in (b).

Alternative method. For (b), the law of cosines in the form |R|² = |A|² + |B|² + 2|A||B| cos θ earns both (b) points if θ is clearly the angle between the vectors’ directions. Using the angle inside the triangle needs a minus sign, |R|² = |A|² + |B|² − 2|A||B| cos(180° − θ), which gives the same result.

(a) Choose a scale such as 1 cm : 1 m. Draw the four arrows tip to tail in order, each with its length to scale and pointing in its compass direction. Draw the resultant from the tail of the first arrow to the tip of the last, and label every arrow with its magnitude.

(b) Add components: x: 6.0 − 3.0 = 3.0 m; y: 8.0 − 4.0 = 4.0 m. D = (3.0 î + 4.0 ĵ) m, magnitude √(9.0 + 16) = 5.0 m, direction tan⁻¹(4.0 / 3.0) = 53.1° north of east.

(c) Distance = 6.0 + 8.0 + 3.0 + 4.0 = 21 m.

(d) Average speed = 21 m ÷ 30.0 s = 0.70 m/s. Average velocity = D ÷ 30.0 s = (0.10 î + 0.133 ĵ) m/s, magnitude 5.0 m ÷ 30.0 s = 0.167 m/s. Speed uses distance, a scalar that counts every metre of the path. Velocity uses displacement, a vector, so the parts of the path that point opposite ways cancel. Here the robot doubled back west and south, so the average speed is about 4.2 times the size of the average velocity.