Acceleration of two bodies in contact with each other Let us place two blocks having mass $m_1$ and $m_2$ in contact with each other on a frictionless surface. Let us assume that one of the blocks is pushed by a force $F$ which in turn pushes the other block in the same direction in which it is being accelerated.
In such a scenario why do we always assume that both the blocks will move with same acceleration which is $a=F/(m1+m2)$? Can't the acceleration of second body be larger than the first body momentarily such that it loses contact with the first body for a while? If not then how to prove it?
 A: 
Can't the acceleration of second body be larger than the first body
momentarily such that it loses contact with the first body for a
while?

Not the acceleration, but the velocity. For example, if the initial push is not gentle, but sudden. The vibration can lead to a momentarily loss of contact. But just after losing contact, the second body by inertia keeps the same velocity. And as the first body is accelerating, the contact is made again, with another kick. The process can be repeated for a little while, until permanent contact is reached.
We can use the equivalence principle of relativity to see the ground as an accelerated body upwards with "g" acceleration. A ball kicking on the ground is similar to the second body kicking on the first one. Air resistance and damping effects gradually take energy of the ball, until it is at rest.
A: Yes, it can happen but the block on the left will catch up very soon, so for any practical purposes you can assume both move together, even if there is a small periodic motion between them.
To see this, imagine that the contact force between the two blocks is like a spring (but only when they push into each other, because the block on the left can never attract the other block). We have for the two blocks:
$F+kd=m_1a_1$
$-kd=m_2a_2$
where $d=x_2-x_1-l$ is the compression from the equilibrium position $l$.
From the equations we see that $a_1$ decreases linearly with $|d|$ (d is negative when the spring is compressed) from the value $F/m_1$ , and $a_2$ grows linearly with $|d|$. At $|d|=m_2F/(m_1+m_2)$ the two accelerations are equal, and $v_1>v_2$, so the spring keeps compressing for a while, until both speeds are equal and $a_2>a_1$. After this $m_2$ moves away from $m_1$. At some point after the separation  the contact force will disappears, so $m_1$ will quickly catch up. And so for eternity. This oscillation should be very small, I imagine invisible for any practical purposes.
A: Okay let's say you're pushing a ball, that ball itself can be composed of numerous more balls, but you only take the net force an the entire system i.e the ball as a whole.
Similarly, you can assume the blocks in contact to be a singular block whose net force is $F$ and acceleration is $F/(m_1+m_2)$
