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Instantaneous Force and Impulse

An impulse ...

Question

An impulse $\to 1$ changes the velocity of a particle from ${\to v}_{1}$ to ${\to v}_{2}$ . Kinetic energy gained by the particle is

A

\frac{1}{2} \to I ({\to v}_{1} + {\to v}_{2})

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B

\frac{1}{2} \to I ({\to v}_{1} - {\to v}_{2})

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C

\to 1 \cdot ({\to v}_{1} - {\to v}_{2})

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D

\to I ({\to v}_{1} + {\to v}_{2})

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Solution

The correct option is A $\frac{1}{2} \to I ({\to v}_{1} + {\to v}_{2})$
Kinetic energy of any moving particle is $\frac{1}{2} m v^{2}$
Impulse of any particle is change in momentum.
Kinetic energy gained by particle is,
$\frac{1}{2} m {v_{f}}^{2} - \frac{1}{2} m {v_{i}}^{2} = \frac{1}{2} m ({v_{f}}^{2} - {v_{i}}^{2}) = \frac{1}{2} m (v_{f} - v_{i}) (v_{f} + v_{i}) = \frac{1}{2} \to I (v_{2} + v_{1})$
Hence, correct option is $(A)$

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