So let's do your experiment 2 correctly. One thing to notice is that as soon as object 2 starts rotating away from the vertical is that the forces will no longer be purely horizontal. Both objects 1 and 2 will end up with non-zero vertical components in their velocity vectors. So right off the bat experiment 2 is going to differ from experiment 1. However, this effect will be very small if the extended lengths of the springs are much smaller than the length of object 2.
I'm going to generalize experiment 2 so that it covers experiment 1 as a special case. Rather than applying the force from the end of object 2, let's put the spring on object 2 at some distance
r from object 2's center of mass. Experiment 1 results when
r=0 while experiment 2 results when
r=L/2. I'm also going to make the masses of objects 1 and 2,
m[sub]1[/sub] and
m[sub]2[/sub], not necessarily be the same. Your experiments 1 and 2 arise as special cases where
m[sub]1[/sub]=
m[sub]2[/sub].
One way to look at the result of this generalized experiment is from the perspective of the conservation laws. In the case where the spring lengths are very small compared to
l[sub]2[/sub], the length of the second rod, we have object 1 moving with a velocity
v[sub]1[/sub] in the -x direction, object 2 moving with a velocity
v[sub]2[/sub], and object 2 rotating about the -z axis, with x,y, and z forming a right-handed system. (Aside: you have a left-handed coordinate system, Alex. The z-axis should be coming out of the page.)
For convenience, I am going to use a coordinate system in which the origin is at the initial position of object 2's center of mass. The initial position of object 1's center of mass in this system is $$-x_0\hat x + r\hat y$$, where
x[sub]0[/sub] is the sum of the lengths of the compressed springs and
r is the distance between the center of mass of object 2 and the location of the spring on object 2. Ignoring the small $$\hat y$$ velocity components (which we can make very small by making the springs very small), the final configuration will have
- Object 1 moving at a velocity of $$-v_1\hat x$$ along the line $$y=r$$,
- Object 2 moving at a velocity of $$v_2\hat x$$ along the x-axis, and
- Object 2 rotating with an angular velocity $$-\omega\hat z$$.
Note that I have defined the scalars $$v_1$$ and $$\omega$$ so that they will have positive values.
Conservation of linear and angular momentum apply assuming no other forces act on the objects. Assuming that all of the potential energy in the compressed springs went into kinetic energy. So what do the conservation laws say about this final state?
Conservation of linear momentum. Since both objects were initially at rest, the initial linear momentum of the system was zero. Conservation of linear momentum says
$$m_1\vec v_1 + m_2 \vec v_2 = 0$$ or
$$m_1v_1 = m_2 v_2$$
Conservation of angular momentum. Since both objects were initially at rest, the initial angular momentum of the system was zero. Conservation of angular momentum says the total angular momentum of the system will also be zero. In the final state, object 1 has angular momentum about the origin due to its translational motion along a line that does not intersect the origin, $$\vec L_1=r\hat y \times (-m_1 v_1 \hat x) = rm_1v_1\hat z$$. Object 2 is moving along the line
y=0, so it has no angular momentum due to translational motion, but it does have angular momentum due to rotation about its center of mass: $$\vec L_2 = \mathbf{I}_2 \vec{\omega}_2 = -\frac{m_2{l_2}^2}{12}\omega \hat z$$. Applying conservation of angular momentum yields
$$rm_1v_1 = \frac{m_2{l_2}^2}{12}\omega$$
Combining the results of the conservation of momentum laws lets us eliminate all but one of $$v_1$$, $$v_2$$, and $$\omega$$. For example, in terms of $$v_2$$,
$$v_1 = \frac{m_2}{m_1} v_2$$
$$\omega = \frac{12\,r}{{l_2}^2}v_2$$
Conservation of energy. The initial energy of the system is in the form of potential energy in the compressed springs. The final energy is in the form of kinetic energy. Applying conservation of energy yields
$$\frac 1 2\left(m_1 {v_1}^2 + m_2 {v_2}^2 + I \omega^2\right) =
\frac 1 2 \left(1+\left(\frac {m_2}{m_1}\right)^2 + 12\left(\frac {r}{l_2}\right)^2\right)m_2v_2^2 = E_{\text{springs}}$$
In experiment 1,
m[sub]1[/sub]=
m[sub]2[/sub]=
m and
r=0. Thus in this case the two objects will have equal but opposite velocity vectors with magnitude given by
$$v_1 = v_2 = \sqrt{\frac{E_{\text{springs}}}{2m}}$$
In experiment 2,
m[sub]1[/sub]=
m[sub]2[/sub]=
m and
r=
l[sub]2[/sub]/2. Thus in this case the two objects will have equal but opposite velocity vectors with magnitude given by
$$v_1 = v_2 = \sqrt{\frac{E_{\text{springs}}}{5m}}$$
Note that in this case, most (3/5) of the energy is in the form of object 2's rotation.$$$$