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9702 · 1.4

Scalars and vectors — practice questions

Practice and worked examples for 9702 Scalars and vectors. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A boat is travelling at 3.0 m/s due East, and simultaneously, a current pushes it 4.0 m/s due North. Calculate the magnitude and direction of the boat's resultant velocity.

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  1. Identify components: We have a horizontal velocity component Vx=3.0 m/sV_x = 3.0 \text{ m/s} (East) and a vertical velocity component Vy=4.0 m/sV_y = 4.0 \text{ m/s} (North). These are perpendicular.
  2. Calculate resultant magnitude: Use Pythagoras' theorem. R=(3.0)2+(4.0)2=9.0+16.0=25.0=5.0 m/sR = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0 \text{ m/s}
  3. Calculate resultant direction: Use trigonometry. Let θ\theta be the angle North of East. tan⁡θ=VyVx=4.03.0=1.333\tan \theta = \frac{V_y}{V_x} = \frac{4.0}{3.0} = 1.333 θ=tan⁡−1(1.333)=53.1∘\theta = \tan^{-1}(1.333) = 53.1\circ (to 1 d.p.)
  4. State the resultant velocity: The boat's resultant velocity is 5.0 m/s5.0 \text{ m/s} at 53.1∘53.1\circ North of East.

Worked example 2

Two forces, F1=60F_1 = 60 N and F2=40F_2 = 40 N, act on a small object P. F1F_1 acts horizontally to the right. F2F_2 acts at an angle of 60∘60^\circ above the horizontal. Calculate the magnitude and direction of the resultant force acting on P.

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  1. Strategy: Resolve each force into horizontal (x) and vertical (y) components. Sum the components, then recombine them to find the resultant. Let's define 'right' and 'up' as positive directions.
  2. Resolve F1: F1x=60 NF_{1x} = 60 \text{ N} (since it's purely horizontal) F1y=0 NF_{1y} = 0 \text{ N}
  3. Resolve F2: F2x=F2cos⁡(60∘)=40×0.5=20 NF_{2x} = F_2 \cos(60^\circ) = 40 \times 0.5 = 20 \text{ N} F2y=F2sin⁡(60∘)=40×32≈34.64 NF_{2y} = F_2 \sin(60^\circ) = 40 \times \frac{\sqrt{3}}{2} \approx 34.64 \text{ N}
  4. Sum components: Resultant horizontal component Rx=F1x+F2x=60+20=80 NR_x = F_{1x} + F_{2x} = 60 + 20 = 80 \text{ N} Resultant vertical component Ry=F1y+F2y=0+34.64=34.64 NR_y = F_{1y} + F_{2y} = 0 + 34.64 = 34.64 \text{ N}
  5. Calculate resultant magnitude (R): Use Pythagoras' theorem. R=Rx2+Ry2=(80)2+(34.64)2R = \sqrt{R_x^2 + R_y^2} = \sqrt{(80)^2 + (34.64)^2} R=6400+1200=7600≈87.2 NR = \sqrt{6400 + 1200} = \sqrt{7600} \approx 87.2 \text{ N} (to 3 s.f.)
  6. Calculate resultant direction (θ\theta): Use trigonometry. Let θ\theta be the angle the resultant makes with the horizontal. tan⁡θ=RyRx=34.6480=0.433\tan \theta = \frac{R_y}{R_x} = \frac{34.64}{80} = 0.433 θ=tan⁡−1(0.433)≈23.4∘\theta = \tan^{-1}(0.433) \approx 23.4^\circ (to 1 d.p.)
  7. Final Answer: The resultant force is 87.287.2 N at an angle of 23.4∘23.4^\circ above the horizontal.
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