When the speed of a car is halved, its kinetic energy is
doubled
quartered
halved
quadrupled
Correct answer is B
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.