Kinetic energy of a particle is increased by $4$ times. What will be the relation between initial and final momentum.

P_{2} = 2 P_{1}

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P_{2} = \frac{P_{1}}{2}

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P_{2} = P_{1}

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P_{2} = 4 P_{1}

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Solution

The correct option is D $P_{2} = 2 P_{1}$
$P = \sqrt{2 m K}$
So $\frac{P_{1}}{P_{2}} = \sqrt{\frac{K_{1}}{K_{2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
So $P_{2} = 2 P_{1}$
Momentum will be doubled.

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Kinetic energy of a particle is increased by 4 times. What will be the relation between initial and final momentum.

The correct option is D P2=2P1P=√2mKSo P1P2=√K1K2=√14=12So P2=2P1Momentum will be doubled.

Kinetic energy of a particle is increased by $4$ times. What will be the relation between initial and final momentum.

The correct option is D $P_{2} = 2 P_{1}$
$P = \sqrt{2 m K}$
So $\frac{P_{1}}{P_{2}} = \sqrt{\frac{K_{1}}{K_{2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
So $P_{2} = 2 P_{1}$
Momentum will be doubled.