First note that a rigid body is an idealization. Some of the questions you ask here therefore actually implicitly also ask "but if it wasn't an ideal rigid body...", well, but often we just assume it is. A rigid body is defined to be one whose every pair of locations $\vec{r}_1, \vec{r}_2$ satisfy $|\vec{r}_1-\vec{r}_2|=\text{constant}$.
This idealized definition is a useful one because it allows us to build simple models that make accurate predictions in many practical situations. With that in mind, let's review your questions.
1️⃣ Despite applying the force only to $B_1$, the entire block $B$ (both $B_1$ and $B_2$) moves together with acceleration $a$.
- I assume internal forces act between particles of the block to maintain rigidity and ensure $B_2$ follows $B_1$ with the same acceleration.
- Is this assumption correct, or are there additional factors ensuring this rigid motion?
When you apply an external force $\vec{F}_{\text{net}}$, with the point of application being anywhere on the rigid body, two things happen:
- It acts as though it was applied directly to the center of mass, by assumption of the body being a rigid body. So $\vec{F}_{\text{net}} = m\vec{a}_C$ where $\vec{a}_C$ denotes the linear acceleration of the body's center of mass.
- A torque is generated, with magnitude and direction depending on the moment of inertia of the body and the distance of the point of application from the CoM.
For a more detailed treatment, see for example this answer and links therein. Or for that matter, any decent classical mechanics textbook.
2️⃣ Assuming internal forces are responsible for rigidity, how do we know they do not alter the net acceleration?
- The standard explanation is that internal forces sum to zero by Newton’s Third Law.
The internal forces can't contribute to the net acceleration induced by the external force, because of the conservation of momentum. In the absence of external forces, we have in inertial frames $d\vec{p}/dt = 0$, so the momentum $\vec{p}$ is conserved. This can also be seen as a consequence of Newton's first law. What happens inside the system during that time, may be all kinds of complicated interactions, but indeed due to Newton's third law we have $\dot{\vec{p}} = \sum \vec{F}_{\text{internal}} = \vec{0}$. The internal forces must cancel in action-reaction pairs and that guarantees the conservation of total momentum.
Now, when you apply the external force, then $d\vec{p}/dt = \vec{F}_{\text{net}}$ where $\vec{F}_{\text{net}}$ is the net external force you've applied to the rigid body, so no "extra" force can be applied to the system due to any kind of internal "response", or otherwise momentum conservation would be violated.
- However, if internal forces act via force fields rather than direct contact, then:
- A particle $P_i$ exerts a force on $P_j$ via a field.
- But $P_j$ does not “know” where $P_i$ is—how does it return an equal and opposite force?
- if $P_j$ created a field to extert force on $P_i$ instead . Could this cause a delay in force transmission, leading to internal energy loss and a net acceleration different from $a$ (i.e., $a \pm \delta$)?
There's no need to get down to the level of fields, because again, the whole point of introducing rigid bodies is so we don't have to think of such stuff. But even if the body isn't rigid, and your system is some wobbly thing connected by springs, with some parts interacting by electric and magnetic fields and whatnot, momentum is still conserved. Recall that what you choose to define as your system is completely arbitrary, so this will hold true for any arbitrary volume you pick as long as no external forces are applied to anything within it.
There are cases where momentum may appear to be not conserved, but that's only if you forget to account for the fact that some of the momentum can be stored in fields.
Since force transmission is not always instantaneous, some energy might get converted into other forms (heat, vibrations, etc.), potentially leading to a slight change in net acceleration.
Don't confuse energy and momentum! For example, in an inelastic collision, energy is lost to heat/environment, but momentum is conserved! Consider two identical bodies approaching each other and "sticking" together upon impact: $KE_i = \frac{1}{2}mv^2+\frac{1}{2}mv^2$ goes to $KE_f = 0$ but $\vec{p}_i = m\vec{v}-m\vec{v} = \vec{0} = \vec{p}_f$. You can generalize this for bodies of different mass and velocity, in which case only some of the KE will be "lost to heat", but momentum will still be perfectly conserved. The fact that momentum is conserved as a vector quantity plays a crucial role in this.
Final Question:
How do we know for certain that these effects do not cause any deviation from the expected acceleration $a$ that is if these effects take place at all? Is there an experimental or theoretical way to confirm this?
As the preceding remarks should have clarified, the theoretical basis is essentially the conservation of linear momentum in inertial frames. Experimentally this law was never observed to be violated.
