# Bullet and block w/ (way) off-CG impact [duplicate]

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.

• there's an interesting question here, but you have obfuscated it with too much extra information. Try and condense your question down to a single paragraph. – cms Jan 13 '18 at 20:51
• Consider that the conclusion reached in the YouTube video is likely referring to the gross, approximate result, vs the details down to the 5th decimal place. – Hot Licks Jan 13 '18 at 23:09
• Conservation of momentum does not apply to objects that are connected to the ground (like buildings) because the earth can absorb or emit momentum as needed without affecting its motion (noticeably). – John Alexiou Dec 16 '19 at 13:10

One's intuition might suggest that when an impulse imparts rotational motion to a rigid object, giving it angular momentum, that "uses up" some of the impulse, leaving less to change the linear momentum of the object. This isn't so. It's one of those cases where common sense leads you astray.

There are some factors in a backyard experiment that might lead you to conclude otherwise. The block must be supported somehow, and in the initial motion after impact, this support mechanism may impart some force to the block, the force depending on the details of the motion. The bullet also presumably drags some of the wood fibers along with it as it moves through the block, and the bullet itself is denser than the block, so there may be some uncertainty as to the precise location of the center of mass of the bullet+block system after impact.

To get around these complications, imagine the experiment is carried out in space, where there is no need for any support system for the block, and where you can let the bullet+block drift along for as long as you like after impact to determine their exact speed. If you do this, sitting in the rest frame of the block and watching how fast it moves after being hit with the bullet, you will find that it moves exactly the same speed whether the bullet hit in line with the center of mass or near one edge. The only difference is, in one case the block is spinning.

You can imagine viewing the experiment in the center of mass frame: the block drifts in from the left, with momentum $M$, and the bullet slams into it from the right, also with momentum $M$. The bullet stops the block in its tracks regardless of the point of impact as long as the bullet gets lodged in the block. If the bullet hits in line with the center of mass, the block is motionless. If it hits off center, the block is stationary in space, but it rotates.

Here is an analysis of the situation that came up in the comments--a long narrow block is partly hanging off a table, and is shot with an upward travelling bullet, which lodges in the very end of the block. In order not to have to deal with the constraint that the far end of the block can't drop below the level of the table after impact (in that case you'd have the end of the block sliding across the table for some period), I will assume the block is only supported at its ends. Immediately after it begins to move, it raises off the support where the bullet hits it, and slips off the one at the other end (though that support does prevent any initial downward motion).

$m$ : Mass of the bullet

$M$ : Mass of the block

$l$ : Length of the block

$v$ : Speed of bullet

$\omega$ : Angular velocity of the block

Angular momentum about the far end of the block is the same before and after the impact, and the moment of inertia of a stick about one end is $Ml^2/3$, so: $$mvl=\frac{M\omega l^2}{3}+m\omega l^2.$$ Solving for the angular velocity of the block, $$\omega=\frac{3mv}{L(M+3m)}.$$ After impact, the initial upward velocity of the center of mass of the block is $\frac{\omega l}{2}$, and that of the bullet is $\omega l$. Multiplying these by their respective masses and adding them together gives the total upward momentum of the block+bullet immediately after impact: $$\frac{(M+2m)v}{1+\frac{M}{3m}}.$$ This is to be compared with the initial upward momentum when the block is shot at its center of mass, which is just $mv$. Depending on the ratio of masses of the block and bullet, the increase in momentum due to the table varies: for a very light block, the initial momentum is twice as large than without the table, and for an very heavy block, it's three times as large. The height to which the block will rise goes as the square root of the momentum, so the height reached increases by a factor of $\sqrt 2$ for a very light block, and by $\sqrt 3$ for a very heavy one.

• I have come to the same conclusion. Thank you for thinking about this and taking the time to write it down. Related question: What would happen if the block is long, set overhanging a table, hit at the overhanging end by the bullet? A similar set up is to have the long block balanced a hair above the table so that any rotation is nearly immediately countered by the end over the table hitting the table as it rotates. The table hits it back stopping the spin. Does that just cancel the spin and do nothing to the linear momentum? I think that's what must happen. – Paul K Jan 23 '18 at 14:52
• When the block hits the table, the table hits the block: it imparts upward momentum. The bullet applies an upward force during its collision, and then the table applies an additional upward force during its collision, and you get more overall initial upward momentum of the block+bullet. – Ben51 Jan 23 '18 at 15:04
• Wow! Can we do the math on that? What you have just written is actually what I initially thought would happen but it seems to violate the conservation of momentum law. Perhaps some of that otherwise lost kinetic energy--and there's a lot of it--captured and added to the momentum imparted to the block by the bullet? These things, already non-intuitive, can get complicated. – Paul K Jan 24 '18 at 15:29
• We can certainly do the math. Need some assumptions to start with, such as: long narrow block, bullet hits block right at one end, etc. Energy and momentum are not really the same currency--you can't trade one for the other. But the final energy will be a bit greater in the table case; still much less than the initial energy. – Ben51 Jan 24 '18 at 15:37
• Given that energy can be neither created nor destroyed, I think the initial energy and the final energy must be the same, just in different forms, i.e. kinetic energy vs heat. The most basic bullet and block situation, with the bullet ( a 4o grain .22 at 1255 ft/sec) having far less mass than the block (1.0 pounds...sorry about the non SI units,) and hitting it on the CG, winds up with something like 97% of the initial kinetic energy being turned into heat in the block. Perhaps the bullet doesn't penetrate as far into the long 1.0 lb block when it hits it (way) off center. – Paul K Jan 25 '18 at 19:41

1. Linear momentum is the product of mass and the velocity of the center of mass. $$\boldsymbol{p } = m \boldsymbol{v}_{\rm cm}$$ Any rotation about the center of mass is irrelevant here as far as momentum goes.
2. Force is the rate of change of momentum. Newton's 2nd law clearly connects the net force applied on a body with the motion of the center of mass only. $$\boldsymbol{F} = \frac{\rm d}{{\rm d}t} \boldsymbol{p} = m \boldsymbol{a}_{\rm cm}$$ nothing here about where the force is applied.
3. An off center impulse $$\boldsymbol{J}$$ will affect the velocity of the center of mass regardless of where the impulse is applied $$\Delta \boldsymbol{v}_{\rm cm} = \frac{1}{m} \boldsymbol{J}$$