WAEC 2017 · Paper 2 · Q14

  1. (a)

    A body PP of mass q kgq\text{ kg} is suspended by two light inextensible strings ABAB and DBDB attached to a horizontal ceiling. The strings are inclined at 30∘30^\circ and 60∘60^\circ respectively to the horizontal and the tension in ABAB is 48 N48\text{ N}. If the system is in equilibrium: (i) sketch a diagram to represent the information; (ii) calculate the tension in DBDB; (iii) find the value of qq. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

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

  2. (b)

    A ball is thrown vertically upwards with a velocity of 20 m s−120\text{ m s}^{-1}. Find, correct to two decimal places, the: (i) maximum height reached by the ball; (ii) time taken to reach the maximum height. [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)(i)

  1. Draw the body hanging from BB, with ABAB at 30∘30^\circ and DBDB at 60∘60^\circ to the ceiling, and its weight 10q10q downwards.

(ii)

  1. Across: 48cos⁡30∘=Tcos⁡60∘48\cos30^\circ = T\cos60^\circ.
  2. T=48cos⁡30∘cos⁡60∘T = \dfrac{48\cos30^\circ}{\cos60^\circ}
    =483= 48\sqrt3
    ≈83.14 N\approx 83.14\text{ N}.

(iii)

  1. Up: 48sin⁡30∘+Tsin⁡60∘=10q48\sin30^\circ + T\sin60^\circ = 10q, so 24+72=10q24 + 72 = 10q and q=9.6q = 9.6.

(b)(i)

  1. At the top v=0v = 0: 0=202−2(10)h0 = 20^2 - 2(10)h, so h=20.00 mh = 20.00\text{ m}.

(ii)

  1. 0=20−10t0 = 20 - 10t, so t=2.00 st = 2.00\text{ s}.

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