If the momentum of an object becomes double without any change in its mass, then what will be the change in its kinetic energy?

Final momentum, p' = 2p
Final Kinetic energy, K.E' = ?

We know that,
KE = $\frac{1}{2}{\mathrm{mv}}^2$           .... (1)
 $\mathrm{p}= \mathrm{mv}$                  .... (2)

Where, m is the mass body and v is the velocity of the body.

Solving equations 1 and 2, we get,

$\text{∴ KE = }\frac{{\mathrm{p}}^2}{2\mathrm{m}}$
$\text{∴ KE' = }\frac{\mathrm{p}{\text{'}}^2}{2\mathrm{m}}$
Since there is no change in the mass, m will remain the same.
$\text{KE'= }\frac{{\text{(2P)}}^2}{2\mathrm{m}}\text{= 4}\frac{{\mathrm{p}}^2}{2\mathrm{m}}\text{= 4 KE}$

Therefore, kinetic energy will become 4 times.