WAEC 2019 · Paper 2 · Q15

  1. (a)

    A body of mass 4 kg4\text{ kg} placed at the top of a smooth plane inclined at an angle of 35∘35^\circ to the horizontal slides from rest down the plane. If the plane is 300 m300\text{ m} long, calculate, correct to two significant figures, the speed of the body when it has travelled half the length of the plane. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

  2. (b)

    An object falling from a tree passes through points PP and QQ with speeds 5.5 m s−15.5\text{ m s}^{-1} and 30.5 m s−130.5\text{ m s}^{-1} respectively. Find the: (i) distance PQPQ; (ii) average speed of the object while falling from PP to QQ. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

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

Worked solution (try it first)

(a)

  1. The plane is smooth: a=gsin⁡35∘a = g\sin35^\circ
    =10×0.5736= 10 \times 0.5736
    =5.736 m s−2= 5.736\text{ m s}^{-2} down the plane.
  2. Half the plane is 150 m150\text{ m}: v2=0+2(5.736)(150)≈1720.7v^2 = 0 + 2(5.736)(150) \approx 1720.7.
  3. v≈41 m s−1v \approx 41\text{ m s}^{-1} (2 s.f.).

(b)(i)

  1. v2=u2+2gsv^2 = u^2 + 2gs: 30.52=5.52+20s30.5^2 = 5.5^2 + 20s, so s=90020=45 ms = \dfrac{900}{20} = 45\text{ m}.

(ii)

  1. 30.5=5.5+10t30.5 = 5.5 + 10t, so t=2.5 st = 2.5\text{ s}.
  2. Average speed =452.5=18 m s−1= \dfrac{45}{2.5} = 18\text{ m s}^{-1}.

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