I've recently seen a bullet (.22, vertically oriented, block rises against gravity) and block experiment on YouTube. There are two parts the first being an on CG impact and the second being an off CG impact. The conclusion after some fiddling with the video runs which were not conclusive in MHO, is that though the off-CG strike rotates the block, the linear momentum imparted to all blocks no matter where impacted, are the same...the block always rises the same amount. I'm still wondering about this and wish to ask about it. Is this correct?
I think there are too many uncontrolled parameters in the backyard experiment including different wood densities in different blocks. I'm assuming the ammunition is consistent in energy imparting a known mass with the same velocity at the muzzle and impacting the block.
I suggest an alternative experiment, but haven't set it up and run it: a very long block is struck by a dart, the dart launched by a consistent system such as a spring, the dart able to strike without damaging the target. First it strikes the target cross wise on it's CG, second, it strikes the block near the end of it most assuredly imparting considerable spin. Is the linear momentum of the block/dart combo always the same? Reason (whatever that is) suggests to me that, no, one would fly forward fast and the second would spin and not fly forward as quickly.
With the bullet and block, a huge amount of kinetic energy (bullet) is lost in the interaction (friction? Breaking wood fibers? distorting the lead bullet? all of these?)...40 grain bullet, 1255 f/s velocity...1# block, that block winding up with something like 1/70th of the kinetic energy of the bullet... so perhaps some of it could be put into spinning the block while the linear momentum transfer remains the same. Since total energy will be conserved, I'd expect a sophisticated experiment has been done that measures these kinds of energy exchanges and shows them adding up to the same initially and finally in the closed system.
A corollary: The bullet hits a 1# block and the block moves off at a velocity, p = mv. Then the block is replaced with a cardboard box filled with sawdust total weight is the same as the block. This time the bullet is slowly captured and does damage to untold fibers along the way dissipating it's energy as it does so. Does the box move off at the same velocity? I think it must. p = mv no matter what happens to the bullet.
In real life (the collapse of the WTC1 on 9/11/01, for example) a small block of (12) floors collapses onto a large block of floors (98) and they crush each other, The falling block pushes down on the remainder of the building and the remainder of the building pushes back slowing the falling block (Newton's third law.) And the crushing (steel being distorted) absorbing energy. There is a p = mv aspect to this. And also mass is lost, tossed aside so it can't participate in subsequent driving of the event downward. The event should lose huge energy even with the first impact, the drop of the block the 12 or 24 feet to the awaiting top floor of the lower section. Conservation of linear momentum says that the event couldn't have continued downward. An egg dropped on a stack of eggs can't crush all the eggs in the stack. In fact, it hardly does much to the first egg it hits.