A proton (mass $1 u$ ) collides elastically head on with a stationary deuteron (mass $2 u$ ). What fraction of its kinetic energy does it transfer to the deuteron ?

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$\frac{3}{4}$

$\frac{8}{9}$

$\frac{2}{3}$

$\frac{6}{7}$

answer is B.

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Detailed Solution

$K_{i} = \frac{1}{2} m {v_{0}}^{2} v_{d} = \frac{(1 + 1) m v_{0}}{3 m} = \frac{2 v_{0}}{3} K_{i} = \frac{1}{2} \times 2 m \times {(\frac{2 v_{0}}{3})}^{2} = \frac{8}{9} \times K_{i}$

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