Question

6(a). The force of attraction, F, between two bodies varies directly as the product of their masses, m₁ and m₂, and inversely as the square of the distance, d, between them. Given that F = 20N, when m₁ = 25kg, m₂ = 10kg and d = 5m, find:
(i) an expression for F in terms of m₁, m₂ and d;
(ii) the distance, d, when F = 30N, m₁ = 7.5kg and m₂ = 4kg.
(b) Find the value of x in the diagram (angles labelled x°, (x+20)°, (x+60)°, (x+40)°, (x+80)° in a pentagon).

Accepted Answer

(a)(i) F ∝ (m₁×m₂) and F ∝ 1/d², so combining: F ∝ (m₁×m₂)/d², i.e. F = k(m₁×m₂)/d², where k is the constant of proportionality.
Substituting F=20N, m₁=25kg, m₂=10kg, d=5m: k = Fd²/(m₁m₂) = (20×5²)/(25×10) = 500/250 = 2. So F = 2m₁m₂/d².
(ii) F×d² = 2m₁m₂ ⇒ d² = 2m₁m₂/F. With F=30N, m₁=7.5kg, m₂=4kg: d² = (2×7.5×4)/30 = 60/30 = 2 ⇒ d = √2 ≈ 1.414m.

(b) Sum of interior angles of a pentagon (n=5) = (n−2)×180° = 3×180° = 540°. So x+(x+20)+(x+60)+(x+40)+(x+80) = 540 ⇒ 5x+200 = 540 ⇒ 5x = 340 ⇒ x = 68°.

WAEC Mathematics 2019 Theory — Question 8

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