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Q.

A man running has half the KE of that of a boy of half his mass. The man speeds up by $1 {ms}^{- 1}$ and then has the KE as that of the boy. What were the original speeds of the man and the boy?

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a

$(\sqrt{2} - 1); 2 (\sqrt{2} - 1)$

b

$\sqrt{2}; 2 \sqrt{2}$

c

$(\sqrt{2} + 1); 2 (\sqrt{2} + 1)$

d

$(\sqrt{2} + 1); 2 (\sqrt{2} - 1)$

answer is C.

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Detailed Solution

${KF}_{m} = \frac{{KF}_{b}}{2}$
${KF}_{b} = \frac{1}{2} \frac{m}{2} U_{2}^{2}$
${KF}_{m} = \frac{1}{2} {mv}_{1}^{2}$
If v₁ is inversed to v₁ + 1
$\begin{array}{l} \frac{1}{2} m {(v_{1} + 1)}^{2} = \frac{1}{2} \frac{m}{2} v_{2}^{2} \\ v_{1} + 1 = \frac{v_{2}}{\sqrt{2}} \end{array}$
But $v_{2} = 2 v_{1}$
$\begin{array}{l} ∴ v_{1} + 1 = \sqrt{2} v_{1} \\ v_{1} = (\sqrt{2} + 1) \\ v_{2} = 2 (\sqrt{2} + 1) \end{array}$

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