First this may help - Toppling of a cylinder on a block.

Let us consider a different object. Take equal two masses connected together with a long rigid rod of length $l$.

Case 1 - Apply a force $F$ to the center of the rod so that both masses are accelerated equally. After you push for a short distance $d$, you have given the system energy $E = Fd$. You can figure out the velocity from $E = 1/2(2m)v_1^2$

Case 2 - Do it again, but apply the force to one of the masses for a distance $d$. This time that mass accelerates. The other mass is a long way off sideways. The rod pulls a little bit, but that mass accelerates very little. We will approximate its speed as $0$ just to make calculation easier. It will be a little off, but not much. I just want to illustrate the point.

The system is rotating because one mass is moving faster and getting ahead of the other. Again, you have given the system energy $e = Fd$.

One way to understand the energy in this case is to add up the kinetic energies of each mass.

$$E = 1/2mv_2^2 + 1/2m0^2$$

You can calculate that $v_2 = \sqrt 2 v_1$.

Another way to calculate the energy is

$$E = E_{translation} + E_{rotation}$$

where

$$E_{translation} = 1/2 (2m) v_{COM}^2$$

and

$$E_{rotation} = (1/2) I \omega^2$$

$v_{COM}^2$ is the velocity of the center of the rod. You can verify that

$$v_{COM}^2 = 1/2 v_2$$

You can verify that

$$I = 2 m (l/2)^2$$

and

$$\omega = \frac{v_2/2}{l/2}$$

If you add it all up,

$$E_{translation} + E_{rotation} = 1/4 m v_2^2 \space + \space 1/4 m v_2^2$$

So both ways come out the same.

Also the total energy is the same for case 1 and case 2, even though it is distributed differently among the masses.

You should think about how long it took. In case 1, you pushed $2m$ a distance $d$. In case 2, you pushed $m$ a distance $d$. What does that tell you about momentum?

First this may help - Toppling of a cylinder on a block.

Let us consider a different object. Take equal two masses connected together with a long rigid rod of length $l$.

Case 1 - Apply a force $F$ to the center of the rod so that both masses are accelerated equally. After you push for a short distance $d$, you have given the system energy $E = Fd$. You can figure out the velocity from $E = 1/2(2m)v_1^2$

Case 2 - Do it again, but apply the force to one of the masses for a distance $d$. This time that mass accelerates. The other mass is a long way off sideways. The rod pulls a little bit, but that mass accelerates very little. We will approximate its speed as $0$ just to make calculation easier. It will be a little off, but not much. I just want to illustrate the point.

The system is rotating because one mass is moving faster and getting ahead of the other. Again, you have given the system energy $E = Fd$.

One way to understand the energy in this case is to add up the kinetic energies of each mass.

$$E = 1/2mv_2^2 + 1/2m0^2$$

You can calculate that $v_2 = \sqrt 2 v_1$.

Another way to calculate the energy is

$$E = E_{translation} + E_{rotation}$$

where

$$E_{translation} = 1/2 (2m) v_{COM}^2$$

and

$$E_{rotation} = (1/2) I \omega^2$$

$v_{COM}$ is the velocity of the center of the rod. You can verify that

$$v_{COM} = 1/2 v_2$$

You can verify that

$$I = 2 m (l/2)^2$$

and

$$\omega = \frac{v_2/2}{l/2}$$

If you add it all up,

$$E_{translation} + E_{rotation} = 1/4 m v_2^2 \space + \space 1/4 m v_2^2$$

So both ways come out the same.

Also the total energy is the same for case 1 and case 2, even though it is distributed differently among the masses.

You should think about how long it took. In case 1, you pushed $2m$ a distance $d$. In case 2, you pushed $m$ a distance $d$. What does that tell you about momentum?