The force of attraction F, between two bodies,... | WAEC SSCE - General Mathematics - 2019

Question 1 Report

The force of attraction F, between two bodies, varies directly as the product of their masses, \(m_1\) and m\(_2\) and inversely as the square of the distance, d, between them. Given that F = 20N, when m\(_1\) = 25kg, m\(_2\) = 10kg and d = 5m, find:

(I) an expression for F in terms of m\(_1\), m\(_2\) and d;

(ii) the distance, d for F = 30N, m\(_1\) = 7.5kg and m\(_2\) = 4kg

(b) Find the value of x in the diagram

Answer

(a) Setting up the joint variation

F varies directly as \(m_1 m_2\) and inversely as \(d^2\):

\[F = \frac{k\, m_1 m_2}{d^2}\]

(i) Finding k and the expression

Substitute \(F = 20,\ m_1 = 25,\ m_2 = 10,\ d = 5\):

\[20 = \frac{k(25)(10)}{5^2} = \frac{250k}{25} = 10k\]

\[k = 2\]

Therefore the required expression is

\[F = \frac{2\,m_1 m_2}{d^2}\]

(ii) Finding d

Substitute \(F = 30,\ m_1 = 7.5,\ m_2 = 4\):

\[30 = \frac{2(7.5)(4)}{d^2} = \frac{60}{d^2}\]

\[d^2 = \frac{60}{30} = 2\]

\[d = \sqrt{2} \approx 1.41\ \text{m}\]

(b) Finding x in the diagram

The figure is a five-sided polygon (pentagon) whose interior angles, read from the diagram, are:

\[x,\quad (x+20^\circ),\quad (x+40^\circ),\quad (x+80^\circ),\quad (x+60^\circ)\]

The sum of the interior angles of a pentagon is

\[(5-2)\times 180^\circ = 540^\circ\]

So:

\[x + (x+20^\circ) + (x+40^\circ) + (x+80^\circ) + (x+60^\circ) = 540^\circ\]

\[5x + 200^\circ = 540^\circ\]

\[5x = 340^\circ\]

\[x = 68^\circ\]

Read lesson note on Variation (JAMB)

Read lesson note on Simple Operations On Algebraic Expressions (WAEC)

Read lesson note on Quadratic Equations (WAEC)

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