Scalars and vectors

1. (a)(i) Scalars: e.g. mass, time; vectors: e.g. force, acceleration. B1 + B1
2. (a)(ii) magnitude B1 and unit B1
3. (b)(i) North: 13 cos 35° = 11 m s⁻¹ A1; east: 13 sin 35° = 7.5 m s⁻¹ A1 The June 2021 report: some candidates swapped them.
4. (b)(ii) The current is to the west, so subtract from east: 7.5 − 2.7 = 4.8 m s⁻¹. A1
5. (b)(iii) The current has no north component, so north is still 11. \(v = \sqrt{11^2 + 4.8^2}\) C1 = 12 m s⁻¹ A1
6. (b)(iv) Angle from north: \(\tan^{-1}(4.8/11)\) C1 = 24° A1 Opposite (east) over adjacent (north). Only a minority chose the right trig function (examiner report).

The current pushes west, so the ship's heading swings from 35° towards north, and 24° is less than 35°. The speed drops a little, 13 → 12 m s⁻¹.

1. (c)(i) 7.5 cos 40° = 5.7 N A1 (ii) 7.5 sin 40° = 4.8 N A1 The scheme accepts 5.75 or 5.7, but not 5.8: it was rounded wrongly.
2. (d) Resolve horizontally: \(T_2\cos 40^\circ = T_1\cos 50^\circ\). B1
3. Resolve vertically: \(T_1\sin 50^\circ + T_2\sin 40^\circ = 7.5\). B1
4. Make T₂ the subject of the horizontal equation: \(T_2 = T_1 \times \dfrac{\cos 50^\circ}{\cos 40^\circ} = T_1 \times \dfrac{0.643}{0.766} = 0.839\,T_1\).
5. Substitute it into the vertical equation: \(T_1 \sin 50^\circ + (0.839\,T_1)\sin 40^\circ = 7.5\), i.e. \(0.766\,T_1 + (0.839 \times 0.643)\,T_1 = 7.5\).
6. Collect the T₁ terms: \(0.766\,T_1 + 0.539\,T_1 = 1.305\,T_1 = 7.5\), so \(T_1 = \dfrac{7.5}{1.305} = 5.75\) → T₁ = 5.7 N A1
7. Put T₁ back in: \(T_2 = 0.839 \times 5.75 = 4.82\) → T₂ = 4.8 N A1
8. Check in the horizontal equation: \(4.82 \cos 40^\circ = 3.69\) and \(5.75 \cos 50^\circ = 3.69\) ✓

Vector triangle: the strings are at 50° and 40°, so they are perpendicular (50° + 40° = 90°). The closed triangle of forces is right-angled with the 7.5 N weight as the hypotenuse: T₁ = 7.5 cos 40° = 5.7 N and T₂ = 7.5 sin 40° = 4.8 N. That's the same as part (c), and it explains why. (Allow ± 0.2 N for a scale drawing.)

1. East components cancel: \(800\sin\theta = 200\) → \(\sin\theta = 0.25\) → θ = 14°. sin, not tan: 800 is the hypotenuse. The March 2019 report on a similar question: candidates used tan (option B there) instead of sin.
2. Resultant (north) = \(800\cos 14.5^\circ = 775 \approx 770\ \text{km h}^{-1}\). Or \(\sqrt{800^2 - 200^2} = 775\). → A 1

Some of the aircraft's speed is "spent" cancelling the wind, so the speed over the ground must be less than 800. That rules out B and D straight away.

Step 1 · The obvious way: horizontal and vertical

1. The 5 N force is horizontal (to the left). The angle between the forces is 150°, so the 10 N force is 180° − 150° = 30° above the horizontal.
2. Horizontal: \(10\cos 30^\circ - 5 = 3.66\) N. Vertical: \(10\sin 30^\circ = 5.00\) N.
3. Look for \((10\cos 30^\circ - 5,\ 10\sin 30^\circ)\) in the options. It isn't there. A has the right horizontal part but 10 cos 30° for the vertical. So the question used different axes.

Step 2 · The clever way: axes along and across the 10 N force

1. Axis 1 along the 10 N force, axis 2 perpendicular to it.
2. The 10 N force lies on axis 1: it gives 10 along axis 1 and nothing along axis 2.
3. The 5 N force is 150° from axis 1, which is 30° from the backwards direction. It gives \(5\cos 30^\circ\) backwards along axis 1 and \(5\sin 30^\circ\) along axis 2.
4. Add along each axis: axis 1: \(10 - 5\cos 30^\circ\); axis 2: \(5\sin 30^\circ\). → C 1

Step 3 · Check every option against the resultant

The resultant is the same whatever axes you use. Cosine rule: \(R^2 = 10^2 + 5^2 + 2(10)(5)\cos 150^\circ = 38.4\), so \(R = 6.20\) N. A correct pair must give \(\sqrt{a^2 + b^2} = 6.20\) N: