Correct Answer: Option B

Explanation

Given \(K_{1}\) = \(\frac{mv^{2}}{2}\).

When v is halved, we have

\(K_{2}\) = \(\frac{m\times (\frac{v}{2})^{2}}{2}\)

= \(\frac{K_{1}}{4}\)

Hence, the initial kinetic energy is quartered.

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