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Q.

The kinetic energy of a moving body of linear momentum P is doubled. The final momentum of the body will be

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a

$2 P$

b

$\frac{P}{2}$

c

$\sqrt{2} P$

d

$\frac{P}{\sqrt{2}}$

answer is C.

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Detailed Solution

$K_{1} = K_{1}, K_{2} = 2 K_{1}, p_{1} = p p_{2} = ? K_{1} = \frac{{p_{1}}^{2}}{2 m} \Rightarrow {p_{1}}^{2} = 2 m K_{1} ∴ {p_{1}}^{2} = \sqrt{2 m k_{1}} p_{2} = \sqrt{2 m K_{2}} = \sqrt{2 m 2 K_{1}} = \sqrt{4 m K_{1}} = \sqrt{2 p_{1}} ∴ p_{2} = \sqrt{2} p$

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