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Bullet Block Experiment


CramBoom

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In this veritasium's newest video, you can see that he shot 2 blocks with a bullet. One in the center, other on the side. Therefore, one block was spinning when it went up, whilst the other wasn't. But, both blocks clearly reach the same height. Also, they had the same amount of energy applied to them (one gun shot), but one appeared to have more energy, since it reached the same height but was spinning. Can anyone explain this?

Edited by CramBoom
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1. How much energy does it require to spin the block?

2. How much kinetic energy does it have?

3. What is the margin of error in "measuring" (crudely estimating) the height each block reaches?

3a. How much variation is there in speed, mass, etc of each bullet?

3b. How much variation is there in the mass and size of the blocks?

 

I can't be bothered to estimate 1 and 2 (we would need more information, anyway) but I am fairly confident that the proportion of energy used to spin the block is sufficiently small that it is dwarfed by the errors in 3.

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Answer me this: is kinetic energy conserved in a (totally) inelastic collision? What quantity is conserved?

I am not sure, sorry. All I know is the information from this video, and the previous one. Here is the previous one:

 

 

I believe he said the bullet lodges itself inside the block with both shots.

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I haven't watched the videos, but you would also need to account for manufacturing disparities in the bullets. Even bullets of the same caliber can have varying weights and load amounts, which will result in varying amount of kinetic energy on impact.

 

You would actually need to weigh each of the bullets (sans powder and cartridge) and get an accurate measure of the speed of the projectile as it impacted the block in order to accurately estimate the amount of kinetic energy imparted to each block.

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No. The linear and rotational kinetic energy comparisons are a sleight of hand. KE is not conserved in an inelastic collision, and in a totally inelastic collision, you have the maximum loss of KE. IOW, the bullet+block system has very little of the KE that the bullet had prior to the collision. No energy has to be "manufactured" to rotate the nlock. All you had to do was lose a little less to heating and sound and deforming the block.

 

What is conserved is momentum. And because KE = p2/2m, they must also have the same linear kinetic energy. The amount of rotational energy will depend on the distance from the center; L = r X p = Iw, and similarly, the energy is L2/2I.

 

Edit to add: via twitter, it was pointed out that the bullet/nail probably penetrated less in the rotational case. It remains to be seen if they measure this in the followup video.

Edited by swansont
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I haven't watched the videos, (...)

You should.

 

(...)

 

In this veritasium's newest video, you can see that he shot 2 blocks with a bullet. One in the center, other on the side. Therefore, one block was spinning when it went up, whilst the other wasn't. But, both blocks clearly reach the same height. Also, they had the same amount of energy applied to them (one gun shot), but one appeared to have more energy, since it reached the same height but was spinning. Can anyone explain this?

I cannot really explain this but it means that a spinning block falling over your head has the same KE than a non-spinning block. From what I know from Laws of Motion, that makes sense.

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I haven't watched the videos...

 

You should.

My point was that, in the context of the original post, the idea that they had the same initial KE is dependent on a lot more than them both having been shot once.

 

Although as swansont points out, the initial KE is irrelevant anyway, since it's not conserved in this type of collision. What matters is the initial momentum (though I think my point that "one gunshot" being an invalid measure of the force applied to the block is still relevant).

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My point was that, in the context of the original post, the idea that they had the same initial KE is dependent on a lot more than them both having been shot once.

 

Although as swansont points out, the initial KE is irrelevant anyway, since it's not conserved in this type of collision. What matters is the initial momentum (though I think my point that "one gunshot" being an invalid measure of the force applied to the block is still relevant).

 

The results are consistent with the impacts being essentially identical. In a collision like this, with the kind of mass ratios you would have, around 99% of the KE is lost — it's given by m/(m+M) so if the nail is 1% of the mass of the block, then the final system has about 1% of the KE. If the block rotates, all that means is that a slightly smaller amount of the KE is lost.

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No. The linear and rotational kinetic energy comparisons are a sleight of hand. KE is not conserved in an inelastic collision, and in a totally inelastic collision, you have the maximum loss of KE. IOW, the bullet+block system has very little of the KE that the bullet had prior to the collision. No energy has to be "manufactured" to rotate the nlock. All you had to do was lose a little less to heating and sound and deforming the block.

 

What is conserved is momentum. And because KE = p2/2m, they must also have the same linear kinetic energy. The amount of rotational energy will depend on the distance from the center; L = r X p = Iw, and similarly, the energy is L2/2I.

 

Edit to add: via twitter, it was pointed out that the bullet/nail probably penetrated less in the rotational case. It remains to be seen if they measure this in the followup video.

 

I don't have any degrees in science, I'm barely in highschool, so bear with me if I didn't understand you correctly.

 

Basically, the momentum from the impact is conserved, so no matter what, the same amount of upward force is applied to the block, no matter how far from the center of the block the bullet hits. This is because you lose less energy because there is less heat, due to penetrating the block less when impacted on the side (assumed because the block is easier to move when it can rotate about an axis, so it turns before more heat can be produced by friction). The second equation explains the amount of rotational energy (angular momentum).

 

Did I get that right, or am I compeletly off? Thanks for your help.

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I don't have any degrees in science, I'm barely in highschool, so bear with me if I didn't understand you correctly.

 

Basically, the momentum from the impact is conserved, so no matter what, the same amount of upward force is applied to the block, no matter how far from the center of the block the bullet hits. This is because you lose less energy because there is less heat, due to penetrating the block less when impacted on the side (assumed because the block is easier to move when it can rotate about an axis, so it turns before more heat can be produced by friction). The second equation explains the amount of rotational energy (angular momentum).

 

Did I get that right, or am I compeletly off? Thanks for your help.

 

That's pretty much it. Most of the energy (around 99% in a case like this) of the bullet goes into spreading/compacting the wood, which also heats it up, and there's also sound.

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Alright, I want to make a video response, and I'll be sure to credit you guys. But is this a simple yet effective explanation (so most people understand)?:

 

So, basically, both blocks went the same height because more energy was lost in the first impact. No matter how where you hit the block, there will be an equal amount of upward force. When the bullet hit in the middle, most of the kinetic energy is lost due to sound, heat and friction. This is because the center of mass was where the bullet was shot, so the block had to stay straight when the bullet made contact. This meant the bullet could lodge itself deep inside the block without the block moving out of its way. This compression and splitting of the block used a lot of the bullet’s kinetic energy. But, when the bullet hit the side of the block, the block rotated slightly, since the center of mass was not where the bullet was shot. This meant the bullet could only lodge itself so deep before the block rotated too much for the bullet. Therefore the kinetic energy that was previously used to lodge itself deep inside the block was now used to rotate the block. This allows both blocks to reach the same height, with the same amount of energy.

Edited by CramBoom
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It's not how I would (or did) explain it. You imply that there's a big difference in the penetration, and that's not true, and also make it sound like the rotation requires a large amount of energy. The problem I see in this kind of explanation is how easy it is to misapply it to a different problem. Sticking with conservation of linear and angular momentum is an approach that is less likely to cause other problems.

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It's not how I would (or did) explain it. You imply that there's a big difference in the penetration, and that's not true, and also make it sound like the rotation requires a large amount of energy. The problem I see in this kind of explanation is how easy it is to misapply it to a different problem. Sticking with conservation of linear and angular momentum is an approach that is less likely to cause other problems.

That's true, but I want to try to approach this problem without using equations. What if I added this sentence to the end of the explanation?

 

You have to remember that the difference in penetration is quite tiny, and that the amount of energy required to rotate the block of that size is also very small, which is why a small factors can affect the outcome so much.

Edited by CramBoom
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