What is your prediction of the results of this experiment (without googling)

http://youtu.be/vWVZ6APXM4w

Basically.

You fire a bullet vertically up into a stationary block of wood, right in the middle. It will accelerate up.

You then alter the rifle so that instead of firing directly vertically up, it fires off to the corner of the block. Will the block

A) go higher than the original?
B) same height ?
C) lower height?

Why?

Higher? Some sort of lever thingy?

Assuming that the bullet becomes embedded in the wood in both cases, and that we are talking about the increase in height of the centre of mass of the wood block, I would expect the second case to rise less because:

1) The vertical component of the bullets velocity will be less.
3) If the wood rotates, the bullet will rise higher, and hence the centre of mass of the wood will rise less, for equal increase in height of the combined mass.

Well, I’m bad at physics…

OK, now looked at parts 1 and 2 of the video, which are QI.

I may make further comment at a later date.

Is it a later date yet?

My point 1 was not valid because the bullet was fired vertically in both cases.

I should also point out that my point 3 was in fact point 2, and there was no point 3.

I would guess lower.

But the video shows a sample size of one, and has no clearly define scale behind to show if there is a small difference. Slightly
different charge size and variations in block density need to be accounted for.

Probably best not to give it away if you go see the part 2.

Part 2 not out yet

Dropbear said:

Probably best not to give it away if you go see the part 2.

Part 2 not out yet

?

The answer is out there.

The Rev Dodgson said:

Dropbear said:

Probably best not to give it away if you go see the part 2.

Part 2 not out yet

?

The answer is out there.

Sorry part 3 not out there yet

Dropbear said:

Sorry part 3 not out there yet

There’s a lot of 2/3 confusion in this thread.

My reasoning is that the vertical part of the acceleration vector would be reduced when the gun is fired at an angle – basic trig really..

The vertical component at 90 degrees would be 0.

The Rev Dodgson said:

Dropbear said:

Sorry part 3 not out there yet

There’s a lot of 2/3 confusion in this thread.

The other third seems OK but.

>>My reasoning is that the vertical part of the acceleration vector would be reduced when the gun is fired at an angle –

The gun is not fired at an angle, the block is moved off centre so the bullet hits it asymmetrically.

Dropbear said:

My reasoning is that the vertical part of the acceleration vector would be reduced when the gun is fired at an angle – basic trig really..

The vertical component at 90 degrees would be 0.

That’s what I said!

But the gun is fired vertical in both cases; it is just moved sideways (or the target is).

Feck. Good point…

Should pay more attention.

Dropbear said:

Feck. Good point…

Should pay more attention.

At least to your own threads :P

like the challenge btw but will have to catch up after coming in late

I’ve heard the explanation here but I don’t think I’ll remember it before it is explained again

aah now you have a problem I don’t know enough about to solve stuck in my head :(

Riff-in-Thyme said:

aah now you have a problem I don’t know enough about to solve stuck in my head :(

Muhaahaha :)

Dropbear said:

What is your prediction of the results of this experiment (without googling)

http://youtu.be/vWVZ6APXM4w

Basically.

You fire a bullet vertically up into a stationary block of wood, right in the middle. It will accelerate up.

You then alter the rifle so that instead of firing directly vertically up, it fires off to the corner of the block. Will the block

A) go higher than the original?
B) same height ?
C) lower height?

Why?

C, lower.
And the further from the centre of the block it gets the lower the block will go.

I went with not as high, it’ll be interesting to see the answer why it did what it did.

I wouldn’t have guessed this guy grew up in Traralgon. http://www.youtube.com/watch?v=myh94hpFmJY

Spiny Norman said:

Dropbear said:

What is your prediction of the results of this experiment (without googling)

http://youtu.be/vWVZ6APXM4w

Basically.

You fire a bullet vertically up into a stationary block of wood, right in the middle. It will accelerate up.

You then alter the rifle so that instead of firing directly vertically up, it fires off to the corner of the block. Will the block

A) go higher than the original?
B) same height ?
C) lower height?

Why?

C, lower.
And the further from the centre of the block it gets the lower the block will go.

Unfortunately I stated the problem incorrectly.

The block is moved off centre, the rifle remains pointing upwards.
so the impact point of the bullet is on the edge instead of middle

Soso said:

I went with not as high, it’ll be interesting to see the answer why it did what it did.

Seems pretty easy to me – The further from the CoG the bullet hits, the less effect it has on moving that CoG. If the block was long enough, even at the same mass, it would not move very far at all.

Remember with a collision between two objects the combined CoG of the two objects moves in a steady path.

By the way, subscribe to the youtoob channel, it’s awesome

Dropbear said:

Spiny Norman said:

Dropbear said:

What is your prediction of the results of this experiment (without googling)

http://youtu.be/vWVZ6APXM4w

Basically.

You fire a bullet vertically up into a stationary block of wood, right in the middle. It will accelerate up.

You then alter the rifle so that instead of firing directly vertically up, it fires off to the corner of the block. Will the block

A) go higher than the original?
B) same height ?
C) lower height?

Why?

C, lower.
And the further from the centre of the block it gets the lower the block will go.

Unfortunately I stated the problem incorrectly.

The block is moved off centre, the rifle remains pointing upwards.
so the impact point of the bullet is on the edge instead of middle

That’s how I understood it.

OK, just looked at the thread.

My questions is, define higher?

Is it the centre of the mass that needs to be higher, or is a point of the wood can be higher?

Spiny Norman said:

The further from the CoG the bullet hits, the less effect it has on moving that CoG. If the block was long enough, even at the same mass, it would not move very far at all.

Remember with a collision between two objects the combined CoG of the two objects moves in a steady path.

Did you see the outcome?

And aren’t those statements contradictory? The CoG would be moving upwards (before impact) at the same rate no matter how off centre the bullet was.

sibeen said:

OK, just looked at the thread.

My questions is, define higher?

Is it the centre of the mass that needs to be higher, or is a point of the wood can be higher?

COM

Dropbear said:

What is your prediction of the results of this experiment (without googling)

http://youtu.be/vWVZ6APXM4w

Basically.

You fire a bullet vertically up into a stationary block of wood, right in the middle. It will accelerate up.

You then alter the rifle so that instead of firing directly vertically up, it fires off to the corner of the block. Will the block

A) go higher than the original?
B) same height ?
C) lower height?

Why?

B. Conservation of linear momentum.

The second block will spin, which requires energy, so the bullet won’t penetrate it as deeply as the bullet in the first block.

I started the experiment but couldn’t complete it cos the police arrived.

PM 2Ring said:

Dropbear said:

What is your prediction of the results of this experiment (without googling)

http://youtu.be/vWVZ6APXM4w

Basically.

You fire a bullet vertically up into a stationary block of wood, right in the middle. It will accelerate up.

You then alter the rifle so that instead of firing directly vertically up, it fires off to the corner of the block. Will the block

A) go higher than the original?
B) same height ?
C) lower height?

Why?

B. Conservation of linear momentum.

Isn’t some energy converted to angular momentum?

A. Higher.

The block will act as a lever at the opposite end…

Dropbear said:

PM 2Ring said:

B. Conservation of linear momentum.

Isn’t some energy converted to angular momentum?

Sort of. Some kinetic energy goes into lifting the block, some into making it spin. But conservation of energy and conservation of momentum are separate (though related) things.

I’ve had a rethink.

The off-centre case will go higher.

In the case with the central bullet the wood will lift off the support almost instantaneously, but with the offset bullet one end of the wood will remain in contact with the support for a short time, and for this period the wood will have an additional upward force, so the deceleration of the combined centre of mass will be less.

The Rev Dodgson said:

I’ve had a rethink.

The off-centre case will go higher.

In the case with the central bullet the wood will lift off the support almost instantaneously, but with the offset bullet one end of the wood will remain in contact with the support for a short time, and for this period the wood will have an additional upward force, so the deceleration of the combined centre of mass will be less.

wouldn’t the twisting of the block relative to the bullets trajectory counteract that?

Dropbear said:

What is your prediction of the results of this experiment (without googling)

http://youtu.be/vWVZ6APXM4w

Basically.

You fire a bullet vertically up into a stationary block of wood, right in the middle. It will accelerate up.

You then alter the rifle so that instead of firing directly vertically up, it fires off to the corner of the block. Will the block

A) go higher than the original?
B) same height ?
C) lower height?

Why?

lower

i’m thinking that in the first case the bullet and the block travel upwards together, all of the energy is used to push it upwards

if the bullet strikes the block off centre then i’d say the block has two motions, spinning and travelling upwards, in this case because the block has been struck off centre. the two motions have split the original energy of the bullet.

Riff-in-Thyme said:

The Rev Dodgson said:

I’ve had a rethink.

The off-centre case will go higher.

In the case with the central bullet the wood will lift off the support almost instantaneously, but with the offset bullet one end of the wood will remain in contact with the support for a short time, and for this period the wood will have an additional upward force, so the deceleration of the combined centre of mass will be less.

wouldn’t the twisting of the block relative to the bullets trajectory counteract that?

No, that’s what causes it.

The Rev Dodgson said:

Riff-in-Thyme said:

The Rev Dodgson said:

I’ve had a rethink.

The off-centre case will go higher.

In the case with the central bullet the wood will lift off the support almost instantaneously, but with the offset bullet one end of the wood will remain in contact with the support for a short time, and for this period the wood will have an additional upward force, so the deceleration of the combined centre of mass will be less.

wouldn’t the twisting of the block relative to the bullets trajectory counteract that?

No, that’s what causes it.

I mean, once the block starts to spin it is acting as a lever against the bullet and losing the energy that would make it go higher

wookiemeister said:

lower

i’m thinking that in the first case the bullet and the block travel upwards together, all of the energy is used to push it upwards

if the bullet strikes the block off centre then i’d say the block has two motions, spinning and travelling upwards, in this case because the block has been struck off centre. the two motions have split the original energy of the bullet.

A reasonablre hypothesis.

Unfortunately, it doesn’t take into account linear momentum, which is always conserved. Energy can be converted into different forms, but momentum cannot, although it can be shared among multiple entities.

Riff-in-Thyme said:

I mean, once the block starts to spin it is acting as a lever against the bullet and losing the energy that would make it go higher

Forget about energy. Nearly all the energy of the bullet is lost in deforming the wood, and it would be very difficult to work out exactly where it all goes in the two cases.

Fortunately we don’t need to, because we know that momentum is conserved, and that before the bullet hits, and after the wood has completely lifted off the support the external forces on the wood+bullet are almost identical (there is a small difference in air resistance, but we don’t know which way that will go, so we’ll ignore it).

So the only difference is for the time when the wood with the off-centre bullet still has some support, and the wood with central bullet has completely lifted off. The support provides an upward force, so for this very short period the wood with off-centre bullet will have slightly less loss of momentum.

ahh so this bullet is actually not pointing directly in each instance

if the bullet is being fired at an angle then the bullet has two directions of force as it acts on the block

wookiemeister said:

ahh so this bullet is actually not pointing directly in each instance

if the bullet is being fired at an angle then the bullet has two directions of force as it acts on the block

It was my mistake. The rifle is firing directly up in both instances

I’ve just watched the you tube of the experiment

I call shenanigans

the rotational energy wasn’t that much as it turns out so i’m not sure that you could really see that much difference

you don’t know if both bullets had EXACTLY the same energy when they left the barrel

they might have just fudged by showing footage that confirms the same height

Thanks for posting the answer.. Everyone else managed not to..

seconds thoughts

the wooden block and bullet in 1 are hotter

the wooden block and bullet in 2 are cooler

however in block 2 the heat not present in block 2 is translated into angular motion

hence they both rise to the same height

Lower than the original, from basic energy considerations.

wookiemeister said:

seconds thoughts

the wooden block and bullet in 1 are hotter

the wooden block and bullet in 2 are cooler

however in block 2 the heat not present in block 2 is translated into angular motion

hence they both rise to the same height

this bloke seems to agree with me

http://www.youtube.com/watch?v=qaIqanZ00uo&feature=youtu.be

PM 2Ring said:

wookiemeister said:

lower

i’m thinking that in the first case the bullet and the block travel upwards together, all of the energy is used to push it upwards

if the bullet strikes the block off centre then i’d say the block has two motions, spinning and travelling upwards, in this case because the block has been struck off centre. the two motions have split the original energy of the bullet.

A reasonablre hypothesis.

Unfortunately, it doesn’t take into account linear momentum, which is always conserved. Energy can be converted into different forms, but momentum cannot, although it can be shared among multiple entities.

Are you sure PM? If you were to repeat the experiment with a 50m long block of carbon fibre/aerogel (so it is about the same weight), you would expect the centre shot to make the same lift, but would an end shot still give the same lift???

Stealth said:

PM 2Ring said:

wookiemeister said:

lower

i’m thinking that in the first case the bullet and the block travel upwards together, all of the energy is used to push it upwards

if the bullet strikes the block off centre then i’d say the block has two motions, spinning and travelling upwards, in this case because the block has been struck off centre. the two motions have split the original energy of the bullet.

A reasonablre hypothesis.

Unfortunately, it doesn’t take into account linear momentum, which is always conserved. Energy can be converted into different forms, but momentum cannot, although it can be shared among multiple entities.

Are you sure PM? If you were to repeat the experiment with a 50m long block of carbon fibre/aerogel (so it is about the same weight), you would expect the centre shot to make the same lift, but would an end shot still give the same lift???

I think so, but I’m prepared to be corrected. :)

I expect that it’d work the same with big aerogel blocks if we did it in a vacuum. I’m not so sure what’d happen if we did it in air.

Stealth said:

Are you sure PM? If you were to repeat the experiment with a 50m long block of carbon fibre/aerogel (so it is about the same weight), you would expect the centre shot to make the same lift, but would an end shot still give the same lift???

In the case of a 50 m long block the effect of the additional upward thrust on the off centre bullet would become significant (if this 50 m long block of aerogel was rigid), so the centre of mass of the off centre block would rise significantly higher.

For the small blocks used the effect would be very small.

Unfortunately, it doesn’t take into account linear momentum, which is always conserved.

—-

It can be transferred to the air.

dv said:

Unfortunately, it doesn’t take into account linear momentum, which is always conserved.

—-

It can be transferred to the air.

Since we don’t know what the air resistance effects are we are assuming they are the same for both.

Where “we” = “I”.

Since we don’t know what the air resistance effects are we are assuming they are the same for both.
-
I’m not. I am expecting air resistance to be proportional to the square of speed, and I am expecting in the off-centre case the mean square speed of surfaces on this block to be higher.

PM 2Ring said:

I expect that it’d work the same with big aerogel blocks if we did it in a vacuum. I’m not so sure what’d happen if we did it in air.

Or maybe not, since the combined CoM of the block + bullet hit near the edge would be shifted a fair bit. But I assuming that the bullet mass is quite a bit smaller than the mass of the block, I don’t think it’d matter too much.

dv said:

Unfortunately, it doesn’t take into account linear momentum, which is always conserved.

—-

It can be transferred to the air.

Good point.

I figure both blocks push roughly the same amount of air up, but while one edge of the spinning block gives some additional upwards momentum to the air the opposite edge gives an equal amount of momentum to the air its pushing down, and so the additional upwards momentum is cancelled out by the that downwards momentum.

I think its got to do more with the heat missing in the spinning block being translated into the spinning motion

I bet if you measured the heat of the spinning block it would be cooler than the non spinning block

dv said:

Since we don’t know what the air resistance effects are we are assuming they are the same for both.
-
I’m not. I am expecting air resistance to be proportional to the square of speed, and I am expecting in the off-centre case the mean square speed of surfaces on this block to be higher.

But they will be greater in both directions

Why is everyone ignoring the greater support contact time for the rotating block?

I’m assuming that air resistance for both blocks (in the original version using wooden blocks) is fairly small and roughly equal, because the blocks are compact, much denser than air, and the air speed is low.

I bet that if you measured the bullet depth between the blocks the spinning block would have a shallower bullet on the spinner

The Rev Dodgson said:

Why is everyone ignoring the greater support contact time for the rotating block?

I’m not ignoring it. I’m just neglecting it as insignificant. :)

PM 2Ring said:

The Rev Dodgson said:

Why is everyone ignoring the greater support contact time for the rotating block?

I’m not ignoring it. I’m just neglecting it as insignificant. :)

It wouldn’t be in a 50 m long block.

wookiemeister said:

I bet that if you measured the bullet depth between the blocks the spinning block would have a shallower bullet on the spinner

would also probably turn towards the outside of the block before stopping

The Rev Dodgson said:

PM 2Ring said:

The Rev Dodgson said:

Why is everyone ignoring the greater support contact time for the rotating block?

I’m not ignoring it. I’m just neglecting it as insignificant. :)

It wouldn’t be in a 50 m long block.

Fair enough.

But they will be greater in both directions

—

Yes, but this all just means there are greater losses

PM 2Ring said:

The Rev Dodgson said:

PM 2Ring said:

I’m not ignoring it. I’m just neglecting it as insignificant. :)

It wouldn’t be in a 50 m long block.

Fair enough.

50m by 50m? (wasn’t it a square block they used?)

I might be getting tired but Rev’s explanation seems like another way of illustrating PM’s, in that both would result in less penetration to the outside of the block

dv said:

But they will be greater in both directions

—

Yes, but this all just means there are greater losses

Not for momentum.

If the upward momentum increase of the air on one side is balanced by the downward momentum increase on the other side, the path of the block will not be affected.

I think that’s a dodgy way of looking at it. The left hand side will have on average lower speed in the second case than in the first, but the root mean speed will be higher.

c’mon guys. it can’t be that hard. maybe if you used those squiggly things you might do better cos words sure aren’t working.

I dunno, fluid dynamics is hard…

Obviously the solution is to do the experiment in a vacuum.

The Rev Dodgson said:

Obviously the solution is to do the experiment in a vacuum.

With point sized, frictionless cows…

Dropbear said:

What is your prediction of the results of this experiment (without googling)

A) go higher than the original?
B) same height ?
C) lower height?

Why?

I understand that you adjusted your post so that both bullets are fired vertically but one aimed and impacting on the centre and the other on the side?

If you are, say, measuring height by the first part of the block that breaks a beam I reckon the bullet that impacts on a corner, it is not trying to lift an entire block, just rotate a corner which I think would take less energy.

If you are, say, measuring height by the first part of the block that breaks a beam I reckon the bullet that impacts on a corner, it is not trying to lift an entire block, just rotate a corner which I think would take less energy.
—————————
The Fosbury Flop means that a high jumper can keep their CoM below the height of the bar while actually clearing the bar, but I don’t think that that is relevant here.

dv said:

I dunno, fluid dynamics is hard…

well I smiled

Isn’t this a conservation of angular momentum question? The angular momentum present at point of contact in a) is 0 and in b) it is 45degrees? Probably not correctly worded, but…….

By an energy balance, the height it will rise is given in 1/2 mV^2 = (m+M)gh +Ef +Er, where Ef is the frictional loss and Er is the rotational energy. m and M are the two masses.

Ef should remain constant in each case.
Er will be zero in the first case and non-zero in the second case.

h cannot increase, or the energy balance will be violated.
In the second case, h must decrease, even if by an almost infinitely small amount, as there will be some rotational energy imparted.

My 2 c worth.

morrie said:

By an energy balance, the height it will rise is given in 1/2 mV^2 = (m+M)gh +Ef +Er, where Ef is the frictional loss and Er is the rotational energy. m and M are the two masses.

Ef should remain constant in each case.
Er will be zero in the first case and non-zero in the second case.

h cannot increase, or the energy balance will be violated.
In the second case, h must decrease, even if by an almost infinitely small amount, as there will be some rotational energy imparted.

My 2 c worth.

The problem is that not all the kinetic energy of the bullet is converted into potential energy of bullet+wood, in fact nearly all of it is converted into heat, rather than potential energy, and working out the split between residual (rotational) kinetic energy, potential energy, and thermal energy, by just looking at conservation of energy, is next to impossible.

But momentum also needs to be conserved, and working out how that works is much simpler, because (in the vacuum assumed for the experiment :)) after the block has fully lifted off the only external force is gravity, which is the same in both cases.

I have in Inventor a dynamic simulation package that will probably solve this

Here’s an example
http://www.youtube.com/watch?v=jeyqQdbR9EE

However I’ve never used it in anger, I’ll have a crack at it on the weekend if I get time.
Inventor has some cool stuff that most people never use.

Peak Warming Man said:

I have in Inventor a dynamic simulation package that will probably solve this

Here’s an example
http://www.youtube.com/watch?v=jeyqQdbR9EE

However I’ve never used it in anger, I’ll have a crack at it on the weekend if I get time.
Inventor has some cool stuff that most people never use.

How to make a simple problem difficult (and probably end up with the wrong answer :))

Peak Warming Man said:

I have in Inventor a dynamic simulation package that will probably solve this

Here’s an example
http://www.youtube.com/watch?v=jeyqQdbR9EE

However I’ve never used it in anger, I’ll have a crack at it on the weekend if I get time.
Inventor has some cool stuff that most people never use.

What does Inventor cost?

The Rev Dodgson said:

Peak Warming Man said:

I have in Inventor a dynamic simulation package that will probably solve this

Here’s an example
http://www.youtube.com/watch?v=jeyqQdbR9EE

However I’ve never used it in anger, I’ll have a crack at it on the weekend if I get time.
Inventor has some cool stuff that most people never use.

What does Inventor cost?

About 8k for Inventor Professional.

You must be rich as a weather girl to afford that

Peak Warming Man said:

About 8k for Inventor Professional.

I might stick with Strand7 for now.

The Rev Dodgson said:

Peak Warming Man said:

I have in Inventor a dynamic simulation package that will probably solve this

Here’s an example
http://www.youtube.com/watch?v=jeyqQdbR9EE

However I’ve never used it in anger, I’ll have a crack at it on the weekend if I get time.
Inventor has some cool stuff that most people never use.

How to make a simple problem difficult (and probably end up with the wrong answer :))

Like asking a engineer to sign of on it????

Stealth said:

The Rev Dodgson said:

Peak Warming Man said:

I have in Inventor a dynamic simulation package that will probably solve this

Here’s an example
http://www.youtube.com/watch?v=jeyqQdbR9EE

However I’ve never used it in anger, I’ll have a crack at it on the weekend if I get time.
Inventor has some cool stuff that most people never use.

How to make a simple problem difficult (and probably end up with the wrong answer :))

Like asking a engineer to sign of on it????

wooooooooooooooooooooooooooooooooooooooooooooooooooooooooh….

stands back

:)

I would have liked the experiement to have been performed a few times, but oh well…

you’re not going to get perfectly accurate burn rates etc in ammunition anyway, and i believe any difference in height will probably get lost in ‘noise’

Dropbear said:

Stealth said:

The Rev Dodgson said:

How to make a simple problem difficult (and probably end up with the wrong answer :))

Like asking a engineer to sign of on it????

wooooooooooooooooooooooooooooooooooooooooooooooooooooooooh….

stands back

:)

I would have liked the experiement to have been performed a few times, but oh well…

you’re not going to get perfectly accurate burn rates etc in ammunition anyway, and i believe any difference in height will probably get lost in ‘noise’

Engineers don’t always get it right of course, but we do try to simplify things as far as possible (but no further).

I thought they did do the experiment a few times, or did they just play it back several times?