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Question
What happens to the kinetic energy when mass of body is doubled, but velocity is reduced to half?
Derivation
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Solution
The kinetic energy is halved (becomes \(\frac{1}{2}\) times).
Mathematical Derivation:
Original Kinetic Energy:
\[ K = \frac{1}{2}mv^{2} \]
Substituting new mass \(m' = 2m\) and new velocity \[ v' = \frac{v}{2} \]:
\[ K' = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^{2} \]
\[ = m\left(\frac{v^{2}}{4}\right) \]
\[ = \frac{1}{2} \times \left(\frac{1}{2}mv^{2}\right) \]
\[ = {\frac{1}{2}\,K} \]
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