Veritasium - Firing bullet in block - along center and away from center 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?
 A: 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 $x$.
A: 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.
A: 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.
A: 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
A: 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.
