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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?

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    $\begingroup$ As soon as it looses the contact the net force on the block two become zero,Thus there is no force to accelerate it in horizontal direction . Is this the kind of proof you are looking for ? $\endgroup$
    – Bhavay
    Commented Nov 8, 2020 at 0:27
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    $\begingroup$ Related physics.stackexchange.com/q/452276/267970 $\endgroup$
    – Prateek Mourya
    Commented Nov 8, 2020 at 2:28

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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.

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  • $\begingroup$ But isn't the cause of greater initial velocity of 2nd body is the greater acceleration experienced by 2nd body as compared to 1st body? And yes I do understand that the 2nd body will keep losing contact and hence will keep the same velocity after losing contact while the first body will keep accelerating but can you explain how did you arrive at the conclusion that permanent contact will be reached? Couldn't this cycle of kicking and accelerating keep on going? $\endgroup$
    – user14598090
    Commented Nov 8, 2020 at 19:49
  • $\begingroup$ Suppose the first body is fixed. But it is hit in the other end. When the elastic wave comes to the boundary with the second body, that can move away. An elastic wave is a periodic accelerated movement, and it is the cause of the displacement of the second body. But it is different from the "bulk" acceleration of the first body. $\endgroup$
    – Claudio Saspinski
    Commented Nov 8, 2020 at 20:28
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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.

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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 $F/(m_1+m_2)$

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  • $\begingroup$ Okay I understand your point but my doubt is that in the ball example you gave, all the particles in the ball are connected to each because they are a part of the same solid whereas two blocks in contact with each other do not have to be connected to each other always. So my question is can those 2 blocks have different acceleration momentarily? $\endgroup$
    – user14598090
    Commented Nov 8, 2020 at 14:45
  • $\begingroup$ Yes good question , should have also added that in my answer .When one block is on top of another it may not necessarily have the same acceleration ,In that case yes ,you have to treat them as different bodies . $\endgroup$
    – Glowingbluejuicebox
    Commented Nov 8, 2020 at 15:14

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