What happens to the kinetic energy when
$(i)$ the mass of the body is doubled at constant velocity?
$(ii)$ the velocity of the body is doubled at constant mass?
$(iii)$ the mass of the body is doubled but the velocity is reduced to half?

(N/A) The kinetic energy $(K)$ of a body of mass $m$ moving with velocity $v$ is given by the formula: $K = \frac{1}{2}mv^2$.
$(i)$ If the mass is doubled $(m' = 2m)$ and velocity remains constant $(v' = v)$:
$K' = \frac{1}{2}(2m)v^2 = 2 \times (\frac{1}{2}mv^2) = 2K$.
The kinetic energy becomes double the original value.
$(ii)$ If the velocity is doubled $(v' = 2v)$ and mass remains constant $(m' = m)$:
$K' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times (\frac{1}{2}mv^2) = 4K$.
The kinetic energy becomes four times the original value.
$(iii)$ If the mass is doubled $(m' = 2m)$ and velocity is reduced to half $(v' = v/2)$:
$K' = \frac{1}{2}(2m)(v/2)^2 = \frac{1}{2}(2m)(v^2/4) = \frac{1}{2} \times (\frac{1}{2}mv^2) = K/2$.
The kinetic energy becomes half of the original value.