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Let there be a block of mass $M$ sitting on top of a table. there is a hole in the table right under the center of mass of the block.

A bullet of mass $m$ is shot through this hole and it hits the block and passes through it, losing some velocity.

The impulse imparted in the block due to the bullet passing through it is $I = Fdt = \Delta P = m\Delta v = M\Delta V$. Now there is some velocity $\Delta V$ imparted into the Block due to this impulse.

Now I wanna find the minimum mass of the block such that it doesn't leave contact with the surface.

I have 2 thoughts of how the situation could theoretically proceed :

  1. velocity is imparted into the block regardless due to conservation of momentum and the block will leave the table for any mass of the block, albeit it might be for a negligible duration as the mass gets higher.

  2. something like an impulse imparted by the gravitation force $I_g = Mgdt$. Now we can equate the impulses to get $Mgdt = Fdt \implies Mg = F$. And from here we can find the critical value of $M$.

Are any of these thoughts right or is there a different ways to go about this problem?

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    $\begingroup$ If the force of the bullet on the mass is smaller than the weight of the mass, it never moves. What's so strange about that? $\endgroup$ Commented Jan 19 at 19:28
  • $\begingroup$ @FlatterMann, So my second thought is right? $\endgroup$ Commented Jan 19 at 19:33
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    $\begingroup$ @gandalf61 If we shoot the bullet at a wall of infinite mass, does it move? The problem here is that the table is implicitly assumed as rigid. In reality it's not. So the body may not lift up, but there would be a slight deformation of the table. No paradox, just an insufficient model for the table system. $\endgroup$ Commented Jan 19 at 19:52
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    $\begingroup$ There is no reason to think that the block must lift, even by a small amount. Much of the energy of the bullet will be lost in the deformation of the bullet as it hits the block. More in the sound made when hitting the block, and any other shock wave created by the hit. And the block itself will have an elasticity related to the speed of sound within the block. $\endgroup$ Commented Jan 19 at 19:56
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    $\begingroup$ Momentum is conserved in an isolated system, where no forces are external. Since gravity is one of the forces, the Earth must be included in the system for it to be isolated. The bullet loses momentum, The Earth/table/block gains it. Your second thought is right. $\endgroup$
    – mmesser314
    Commented Jan 19 at 20:29

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Unfortunately, impulse is not the correct way to approach this problem. You need to work in forces and momentum. If you try to think in just impulses, you will find that it is impossible to have a block that doesn't move. No matter how big you make the block, it will have a positive upward velocity after the interaction. This means it will leave the table. Or at least it looks like it does.

What's missing from the equation is a time period over which the impulse is being delivered. To consider that time period meaningfully, we should be talking in terms of continuous momentum transfers, not impulses. Your bullet is applying a force over a period of time, while the bullet deforms. As long as that force is lower than the force of gravity, the block will not move upward.

While you can technically use the concepts of impulse in any situation, we usually use them in scenarios where the momentum transfer is fast enough that we don't really care how the transfer happens. We just care that it does happen. In your case, you really do care about how the transfer happens, so the impulse equations simply don't give you the whole story.

So what does the impulse story tell you? It tells you that if you transfer momentum fast enough (over a short enough period of time), you can generate arbitrarily high forces and overcome stabilizing forces. When designing devices, we often have to subject them to "shock and vibe," which is testing where we intentionally expose it to movement to see if anything gives. A shock, in particular, is very high frequency. It can occur over microseconds. As such, a shock can often take a joint which is stable, using static friction, and temporarily make it unstable, with kinematic friction (because that small delta-V can result in very high forces that start the parts moving with respect to one another). This can cause a part to fail if one was dependent on the high coefficient of friction that comes from static friction.

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  • $\begingroup$ well then. how exactly do we find the minimum mass of the block such that it doesn't leave contact with the table. You have said that "While you can technically use the concepts of impulse in any situation, we usually use them in scenarios where the momentum transfer is fast enough that we don't really care how the transfer happens.", So if I don't care about the analysis of the situation and just care about the final answer, is the calculation I have provided in the 2nd point of my question correct? $\endgroup$ Commented Jan 20 at 12:24
  • $\begingroup$ Approach number 2 works. It's just confusing because it went through the concept of impulse when it did not have to. If you did it using forces, you would reach mg=F instantly, without having to go through the earlier steps. And this points to why you care about how the transfer happens. Instead of a bullet hitting a block, think about a car crash. Modern cars are designed for safety. They have crush zones which try to spread the impulse over the most time possible, but eventually you run out of crush zone, and you face the reality of an engine block that's about to smack into you. $\endgroup$
    – Cort Ammon
    Commented Jan 21 at 4:09
  • $\begingroup$ In such a case, much of the impulse is transmitted by low forces as the crumple zone does its job, spreading things out over time. However, once the engine block strikes you, its a very high force. In such an example, you can see it isn't the impulse that matters. Most of the impulse can be delivered at low force. It's the high maximum force that really drives the behaviors you're looking for. $\endgroup$
    – Cort Ammon
    Commented Jan 21 at 4:10
  • $\begingroup$ Thanks for the explanation. It really helped $\endgroup$ Commented Jan 21 at 7:38

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