A light and a heavy body have equal K.E. and  Which one  has greater momentum

The K.E. is the energy possessed by a body due to its state of motion. The expression for K.E. is given as

$K=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$

Here m represents the mass of the body while v represents the speed with which the body is moving and p is used to signify the momentum of the body. The K.E. can be expressed in terms of momentum by using the following expression where velocity can be represented in terms of momentum and mass

$p=mv$

$v=\frac{p}{m}$

Now we've a light body and a heavy body which have the same kinetic energy. If m1 is that the mass of the lighter body and m2 is the mass of the heavier body then we have

$m_1<m_2$

If K1 is that the kinetic energy of the lighter body and K2 is the kinetic energy of the heavier body then we have

$K_1=K_2$

Now we will write the kinetic energies as follows using the expression for kinetic energy

$\frac{p_1^2}{2m_1}=\frac{p_2^2}{2m_2}$

$\frac{p_1}{p_2}=\frac{\sqrt{m_1}}{\sqrt{m_2}}$

$m_1<m_2$

$p_1<p_2$

This expression means the lighter body has momentum less than the momentum of the heavier body.

Hence, option A is correct.