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Question

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

Moderate

A

$2 (\sqrt{2} - 1) m s^{- 1}$
B

$2 (\sqrt{2} + 1) m s^{- 1}$
C

${2ms}^{-1}$
D

${5ms}^{-1}$

Solution

$\begin{array}{l} {KE}_{man} = \frac{1}{2} {KE}_{boy} \\ \frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} m_{2} v_{2}^{2} \\ m_{1} v_{1}^{2} = \frac{1}{2} \times \frac{m_{1}}{2} \times v_{2}^{2} \\ v_{2} = 2 v_{1} \to 1 \\ {KE}_{man} = 2 {KE}_{b} \\ \frac{1}{2} m_{1} {(v_{1} + 5)}^{2} = 2 \times \frac{1}{2} \frac{m_{1}}{2} v_{2}^{2} \\ v_{1} + 5 = 2 v_{1} \\ v_{1} = 5 {ms}^{- 1} \end{array}$

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