Suppose that there is a wooden block on a smooth horizontal plane. It is hit by some force and it starts moving along a straight line with a constant velocity. Now, cut the block into two slices horizontally and place them one on another. Assume that the surfaces that touch each other are very smooth, so there is no friction. Then again hit it with the same force. Let's imagine that the force hit the bottom part. So that part only starts moving. As there is no friction the upper part stands still and when the it lose the contact with the bottom slice it will fall down. We can think that the centre of the mass(of two blocks) moves with the same speed which it had, before cut into two slices. But after the upper part fell down the COM also descends. Thus why the path of the COM at the second situation differs from the first situation though we have provided same conditions?
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$\begingroup$ If the impulse imparted in both conditions are same, then speed of COM will be same in both the cases. By cutting we have removed the internal friction of the block... initially the internal(molecular) force kept the block together, later that force wasn't present. $\endgroup$– gobbledy-gookCommented Jul 2, 2021 at 14:57
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Thus why the path of the COM at the second situation differs from the first situation though we have provided same conditions?
In short, because there was no net external force acting in either the horizontal or vertical direction on the single block, thus per Newton's first law the COM does not accelerate.
In the second example, when the top block leaves the bottom block there is now a net external force acting downward on the top block (its weight). So while the COM of the bottom block does not accelerate, the COM of the top block accelerates downward because it is subjected to a net external force, per Newton's second law. Consequently the COM of the combination accelerates downward.
Hope this helps.
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$\begingroup$ According to my knowledge weight is always an external force for any object. In the first situation the weight acts downward on the block and also the same net weight acts on the two blocks seperatedly downwards in the second situation as we did not removed any part of the object. So it seems that the net force acting at the second situation is similar to that acting at the first situation. That is why this question occured. I am still confused. $\endgroup$– ACBCommented Jul 2, 2021 at 14:23
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$\begingroup$ But there is no NET external force in the case of single block because of the reaction force of the horizontal surface. When the top block leaves the bottom block there’s no upward reaction on the top block and therefore there is a net horizontal force downward on the top block. What is it you don’t understand about that? $\endgroup$– Bob DCommented Jul 2, 2021 at 14:59
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$\begingroup$ Thank you for your explanation. It solved my problem. $\endgroup$– ACBCommented Jul 2, 2021 at 15:42
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