A particle of mass m has half the kinetic energy of another particle of mass $m / 2$ . If the speed of the heavier particle is increased by $2 m s^{- 1}$ , its new kinetic energy equals the original kinetic energy of the lighter particle. The ratio of the original speeds of the lighter and heavier particles is $1 : 2$ . What is the original speed of the heavier particle?

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

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

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

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

answer is A.

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Detailed Solution

When the heavier particle is speeded up by $2.0 m s^{- 1}$ , its kinetic energy becomes $\frac{1}{2} m {(v + 2)}^{2} .$ Since this equals the original kinetic energy of the lighter particle, we have

$\frac{1}{2} m {(v + 2)}^{2} = \frac{1}{2} (m / 2) (4 v^{2}) v^{2} + 4 + 4 v = 2 v^{2} o r v^{2} - 4 v - 4 = 0 v = \frac{4 \pm \sqrt{16 + 16}}{2} = \frac{4 \pm 2 \sqrt{8}}{2} = 2 \pm 2 \sqrt{2}$