Relate kinetic energy and momentum.

The formula for kinetic energy ($KE$) is $KE = \frac{1}{2}mv^2$, where $m$ is mass and $v$ is velocity.
The formula for momentum ($p$) is $p = mv$.

From the momentum formula, we can express velocity as $v = \frac{p}{m}$.
Substitute this expression for $v$ into the kinetic energy formula:

$$KE = \frac{1}{2}m \left(\frac{p}{m}\right)^2$$ $$KE = \frac{1}{2}m \frac{p^2}{m^2}$$ $$KE = \frac{p^2}{2m}$$

Let the two bodies be body 1 and body 2.
Let their momenta be $p_1$ and $p_2$, and their kinetic energies be $KE_1$ and $KE_2$.
Assuming the masses of the two bodies are the same ($m_1 = m_2 = m$), we can write:

$$KE_1 = \frac{p_1^2}{2m}$$ $$KE_2 = \frac{p_2^2}{2m}$$

Now, form the ratio of their kinetic energies:

$$\frac{KE_1}{KE_2} = \frac{\frac{p_1^2}{2m}}{\frac{p_2^2}{2m}}$$

The $2m$ terms cancel out:

$$\frac{KE_1}{KE_2} = \frac{p_1^2}{p_2^2}$$ $$\frac{KE_1}{KE_2} = \left(\frac{p_1}{p_2}\right)^2$$

The problem states that the momentum of the two bodies are in the ratio of $4:1$.
So, $\frac{p_1}{p_2} = \frac{4}{1}$.

Substitute this ratio into the equation for the ratio of kinetic energies:

$$\frac{KE_1}{KE_2} = \left(\frac{4}{1}\right)^2$$ $$\frac{KE_1}{KE_2} = \frac{16}{1}$$

The ratio of the kinetic energies of these bodies is $\text{16:1}$.