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WAEC 2018 · Paper 2 · Q14

  1. (a)

    Three forces 6 N6\text{ N}, 4.5 N4.5\text{ N} and 8 N8\text{ N} act on a body AA of mass 1.2 kg1.2\text{ kg} as shown in the diagram: the 6 N6\text{ N} force acts due north, the 4.5 N4.5\text{ N} force at 120∘120^\circ clockwise from it, and the 8 N8\text{ N} force at 150∘150^\circ anticlockwise from it. Calculate the magnitude of the: (i) resultant force; (ii) acceleration of the body AA.

    6 N4.5 N8 N120°150°A

    Separate values with commas, e.g. 3, −2

  2. (b)

    A stone is dropped from the top of a building 80 m80\text{ m} high. Find, in m s−1\text{m s}^{-1}, the velocity with which it hits the ground. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

Worked solution (try it first)

(a)(i)

  1. The forces are on bearings 000∘000^\circ (6 N), 120∘120^\circ (4.5 N) and 210∘210^\circ (8 N).
  2. East: 0+4.5sin⁡120∘+8sin⁡210∘=3.897−40 + 4.5\sin120^\circ + 8\sin210^\circ = 3.897 - 4
    =−0.103= -0.103.
  3. North: 6+4.5cos⁡120∘+8cos⁡210∘=6−2.25−6.9286 + 4.5\cos120^\circ + 8\cos210^\circ = 6 - 2.25 - 6.928
    =−3.178= -3.178.
  4. ∣R∣=0.1032+3.1782|\mathbf R| = \sqrt{0.103^2 + 3.178^2}
    ≈3.18 N\approx 3.18\text{ N}.

(ii)

  1. a=Fma = \dfrac{F}{m}
    =3.181.2= \dfrac{3.18}{1.2}
    ≈2.65 m s−2\approx 2.65\text{ m s}^{-2}.

(b)

  1. v2=u2+2gh=0+2(10)(80)=1600v^2 = u^2 + 2gh = 0 + 2(10)(80) = 1600, so v=40 m s−1v = 40\text{ m s}^{-1}.

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