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WAEC 2009 · Paper 2 · Q18✱

  1. (a)

    A body of mass 8 kg is placed on a smooth plane inclined at an angle of 40∘40^\circ to the horizontal. Find the magnitude of the force: (i) acting perpendicular to the plane; (ii) acting along the plane, required to keep the body in equilibrium. [Take g=10 m s−2g = 10\text{ m s}^{-2}]

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

  2. (b)(i)

    A particle of mass 800 g is moving in a straight line with a velocity of (5i+3j)(5\mathbf i + 3\mathbf j) m s−1^{-1}. It is acted upon by a force that changes the velocity to (5i+12j)(5\mathbf i + 12\mathbf j) m s−1^{-1}. Find the impulse.

  3. (b)(ii)

    If the force acted for 2 s, find the acceleration of the particle.

Worked solution (try it first)

(a)

  1. The weight is mg=8×10=80mg = 8 \times 10 = 80 N, acting vertically down.
  2. Resolve it along and perpendicular to the plane.

(i)

  1. Perpendicular to the plane: 80cos⁡40∘=61.2880\cos 40^\circ = 61.28 N.
  2. This is balanced by the normal reaction.

(ii)

  1. Down the plane: 80sin⁡40∘=51.4280\sin 40^\circ = 51.42 N.
  2. The plane is smooth, so a force of 51.4251.42 N up the plane is needed for equilibrium.

(b)(i)

  1. Impulse = change in momentum =m(v−u)= m(\mathbf v - \mathbf u), with m=0.8m = 0.8 kg.
  2. v−u=(5i+12j)−(5i+3j)\mathbf v - \mathbf u = (5\mathbf i + 12\mathbf j) - (5\mathbf i + 3\mathbf j)
    =9j= 9\mathbf j, so the impulse is 0.8×9j=7.2j0.8 \times 9\mathbf j = 7.2\mathbf j N s.

(ii)

  1. Impulse =Ft= Ft, so F=7.22=3.6F = \dfrac{7.2}{2} = 3.6 N (in the j\mathbf j direction).
  2. a=Fm=3.60.8=4.5a = \dfrac{F}{m} = \dfrac{3.6}{0.8} = 4.5 m s−2^{-2}, in the direction of j\mathbf j.

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