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This question is about this video on YouTube, in which a bullet is fired vertically into the center of a wooden block from below, sending the block up into the air. Next, a bullet is fired vertically but off-center into a similar block from below, again causing the block to rise into the air, but simultaneously to rotate. The video asks for a prediction as to which block will rise higher.

Please post your guess/prediction/solution along with answer or in comments.

My first guess was that air friction decreases when object is rotating, but now I think that's not the reason.

Someone please shed some light on this.

EDIT

Is it possible that the first bullet went deeper inside the first block and thus the first block had lesser kinetic energy than expected?

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  • $\begingroup$ Wow, Derek Muller must be a magician! I thought the rotating one would go to a lesser height $\endgroup$ – udiboy1209 Aug 20 '13 at 17:48
  • $\begingroup$ I am sorry for not being active on my own question but I seriously have a little (if any) idea about what you are talking about. Its been really long since I studied physics. :P but I just love your enthusiasm. All of you are awesome. I really admire physics and Physicists $\endgroup$ – Sourabh Aug 28 '13 at 16:55
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The system needs to conserve momentum. In both cases, the momentum is whatever m*v is for the bullet. Since it's the same in both cases, the bullet and block have the same vertical velocity.

Mechanical energy is not conserved. The reason the block hit on the side has more kinetic energy is that the bullet converted less of its kinetic energy into heat upon impact. We would find that this bullet penetrated less far into the block.

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  • $\begingroup$ I don't understand why linear momentum is conserved in second case. I mean, some of the energy is gone into rotating and system and this less energy for translation. I agree the being inelastic collision, energy is not conserved but what about what about that missing 0.5*I*w^2 energy that didn't go into transnational kinetic energy? $\endgroup$ – Sourabh Aug 21 '13 at 12:55
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    $\begingroup$ Momentum is conserved because that's completely general in mechanics. The rest of your comment does not make sense to me. $\endgroup$ – Mark Eichenlaub Aug 21 '13 at 13:20
  • $\begingroup$ So if the block was pinned in it's center where would the bullet momentum go? Linear momentum would not be conserved even if the block was spinning as a result. So what makes this case different? I think Isaac's 2nd law applies, and not a blanket statement about linear momentum. $\endgroup$ – ja72 Aug 21 '13 at 14:25
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    $\begingroup$ If the block were pinned, the pin would exert a force on the block, so momentum wouldn't be conserved. Obviously, momentum isn't conserved when there's an outside force on the system. Momentum would go into the pin and whatever the pin is attached to. $\endgroup$ – Mark Eichenlaub Aug 21 '13 at 14:42
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    $\begingroup$ @ja72 You're completely off base here. If you accept Newton's Second Law for point particles, then it follows immediately that if the net external force on a system of particles is zero in a particular direction, then the momentum of the system in that direction is conserved. There is really nothing else that you need mathematically here. $\endgroup$ – joshphysics Aug 22 '13 at 22:08
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Here is the mathematical treatment of this experiment. A block of mass $m_{block}$ is at rest when struck by a bullet of mass $m_{bullet}$ traveling with $v_{bullet}$ at a distance $x$ from the center of the block.

If the impact is perfectly plastic then the linear velocity of the block at the impact point should equal the velocity of the bullet after the impact. If the momentum transferred from the bullet to the block is $J$ then the change in speed for the bullet during the impact is

$$ \Delta v_{bullet} = - \frac{J}{m_{bullet}} $$

The same momentum $J$ affects the block by imparting linear and angular velocity at the center of gravity of

$$ \begin{aligned} \Delta v_{block} & = \frac{J}{m_{block}} \\ \Delta \omega_{block} & = \frac{x\, J}{I_{block}} \end{aligned} $$

where $I_{block}$ is the mass moment of inertia of the block.

For a purely plastic contact setting the final relative speed equal to zero means

$$ \begin{aligned} \Delta v_{block} + x \, \Delta \omega_{block} & = v_{bullet} + \Delta v_{bullet} \\ \frac{J}{m_{block}} + \frac{x^2\, J}{I_{block}} & = v_{bullet} - \frac{J}{m_{bullet}} \end{aligned} $$

which is solved for: $$ J = \frac{v_{bullet}}{ \frac{1}{m_{bullet}} + \frac{1}{m_{block}} + \frac{x^2}{I_{block}} } $$

Now for the trick. The combined center of gravity is at $x_{cg} = x \frac{ m_{bullet}} { m_{block} +m_{bullet} } $ so the take off speed of the new cg is

$$ \begin{aligned} v_{cg} & = \Delta v_{block} + x_{cg} \Delta \omega_{block} \\ & = \frac{J}{m_{block}} + x \left( \frac{ m_{bullet}} { m_{block} +m_{bullet} } \right) \frac{x\, J}{I_{block}} \\ v_{cg} & =\left( \frac{ m_{bullet}} { m_{block} +m_{bullet} } \right) v_{bullet} \end{aligned} $$

The above expression means that the take of speed (and hence the height) of the cg of the combined block and bullet does not depend on the impact location $x$.

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Maybe I think I got it.

You cannot apply mechanical energy conservation to an inelastic collision, which is happening when the bullet is getting stuck into the block.

So in the first case the bullet is losing more mechanical energy by being fired into the center of the block than the second case, so when they reach the same height the block in the second case has more mechanical energy(because it must have lost lesser energy).

The fact that they will go to the same height is fairly simple to prove, by momentum conservation which can be applied for any kind of collision, elastic or inelastic.

The bullet has the same momentum initially in both cases, and the mass of the final system(Block + bullet) is same in both the cases so we get the same initial linear velocity for both cases, which implies they will both go to the same height.

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  • $\begingroup$ But how are you sure that bullet went deeper in first case? And case II also has some angular momentum. Can you prove this by conservation of angular momentum? $\endgroup$ – Sourabh Aug 20 '13 at 18:44
  • $\begingroup$ You will only get the angular velocity by conserving angular momentum. And thats exactly what I'm saying, the bullet must have gone deeper in the first case than second leading to more energy loss in the first. Note that this does not mean it will apply lesser impulse in the second case. The equation of momentum is same in both cases. $\endgroup$ – udiboy1209 Aug 20 '13 at 18:47
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I was thinking the same thing you were: spinning reduced the air resistance. Since the spinning trick helps bullets travel faster with barrels that have riffling.

A lot of people are considering that it is probably that the block hit dead center because by hitting the block dead-center the collision was supposedly more inelastic.

Although I am going to be honest here and claim that there argument makes sense, I have a suggestion for testing both explanations:

1) Conduct a similar experiment but instead of using a wooden block use a material that will have a more elastic collision. If the results are slightly different it probably was due to heat loss. If not air-resistance was the cause.

2) Conduct a similar experiment with a wooden block but in a vacuum, [I don't know where one can find a vacuum as large other than space, but no matter]. If they travel the same distance then it was due to energy loss in heat, otherwise air-resistance.

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  • $\begingroup$ I don't think we can perform your two experiments because I really don't have a gun :P. And Dr. Derek will tell the solution to this on Tuesday. I will post the link to that video under the question. $\endgroup$ – Sourabh Aug 21 '13 at 12:44
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The answer is: less

Some amount of the momentum (energy) transferred results in spin of the block, so less momentum (energy) is left for the rise. WRT thw idea that off-center hit will transfer more/less momentum: nonsense! The time needed to stop the bullet in the block is much shorter than any movement of the block. That is why that arrangement is called "ballistic". The original version is a wooden Block (or a chest filled with cotton or wet clay) hinged as a pendulum bob. This is the traditional equipment used to measure momentum of rifle bullets depicted in every basic mechanics textbook.

http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html

https://www.google.de/search?q=ballistic+pendulum&tbm=isch&tbo=u&source=univ&sa=X&ei=CxAeUtirOoeZhQfn94DADw&sqi=2&ved=0CDQQsAQ&biw=1338&bih=911

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  • $\begingroup$ I'm interested to hear how you think linear momentum is converted into angular momentum... $\endgroup$ – Michael Brown Aug 28 '13 at 14:59
  • $\begingroup$ @Michael Take a "beam" (eg a ruler) and rest it at its ends on two chairs maybe. Then hit it from below with a thin stick, this will represent the bullet. First in the middle, then close to one end. Nothing superseeds experiment. $\endgroup$ – Georg Aug 28 '13 at 15:05
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    $\begingroup$ This is nonsense. Linear momentum cannot be transferred into angular momentum about the center of mass. Linear momentum is exactly equal to mass times velocity of center of mass. $\endgroup$ – Mark Eichenlaub Aug 28 '13 at 20:19
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    $\begingroup$ The links you posted do not support your answer. You are simply unambiguously wrong. The linear momentum is exactly the same in both cases, not less in one of them. This is not an arguable point - it is completely clear to anyone with the most basic knowledge of introductory physics. $\endgroup$ – Mark Eichenlaub Aug 28 '13 at 20:23
  • $\begingroup$ @Georg Please please try to understand that angular momentum does not mean "rotation about the centre." Compute the angular momentum in both cases before and after. Linear and angular momentum are seperately conserved in this problem. You never convert one into the other. They are seperate conservations laws. They even have different dimensions! $\endgroup$ – Michael Brown Aug 29 '13 at 4:00

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