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The kinetic energy of a man is half that of a boy whose mass is half that of the man. When the man speeds up by $5 {ms}^{- 1}$ , his kinetic energy is $100 %$ more than that of the boy. The initial velocity of the man is,

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5 m/s

(\sqrt{2} + 1) m/s

2 m/s

2 (\sqrt{2} - 1) m/s

answer is B.

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Detailed Solution

{KE}_{M} = \frac{1}{2} {KE}_{B}

$\frac{1}{2} M_{m} V_{m}^{2} = \frac{1}{2} (\frac{1}{2} \frac{Mm}{2} V_{B}^{2})$

$V_{B}^{2} = 4 V_{M}^{2}$

$V_{B} = 2 V_{m}$

$V_{1}^{m} = V_{m} + 5$

${KE}_{M}^{1} = 2 {KE}_{B}$

$\frac{1}{2} M_{m} {(V_{m} + 5)}^{2} = 2 \frac{1}{2} \frac{M_{m}}{2} V_{B}^{2}$

${(V_{m} + 5)}^{2} = 4 V_{m}^{2}$

$V_{m} + 5 = 2 V_{m}$

$V_{m} = 5 m/s$

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