## Eight problems for intuition in basic classical mechanics

A forum for good logic/math puzzles.

Moderators: jestingrabbit, Moderators General, Prelates

Blatm
Posts: 638
Joined: Mon Jun 04, 2007 1:43 am UTC

### Re: Eight problems for intuition in basic classical mechanic

9. (from Mythbusters) Are two identical cars hitting each other head on at 50 mph equivalent to one car hitting a wall at 100 mph?

10. (Mythbusters inspired) Imagine a rigid beam on a pivot (see saw), with a light mass on one end, and nothing on the other. A large mass is dropped from a height hM onto the other end of the see saw. Give the height the small mass reaches, hm, as a function of hM.

11. What would happen if you put a portal (à la Aperture Science) on the moon, and another on Earth?

phlip
Restorer of Worlds
Posts: 7573
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Eight problems for intuition in basic classical mechanic

9.
Spoiler:
When this episode came on I immediately thought it was identical, but when they said that people said it wasn't, I realised why. I really I wish they'd done a fourth variant on the show: a 100mph car hitting a stationary car, rather than a wall. Because that should be equivalent to the two 50mph cars.

But a 100mph car hitting a stationary car is different to a 100mph car hitting a stationary wall - the former target will collapse a bunch, taking its share of the impact, while the latter will stand resolute, and reflect all the impact back at the moving car.

10.
Spoiler:
Hmm, not sure... first thought is that the lever will equalise speeds... the falling heavy mass won't be impeded much, and the light mass will be accelerated up to match its speed... once their speeds match, the lever is so much dead weight. So the light mass will be launched up at about the same velocity the heavy mass lands, and will thus reach about the same height, modulo air resistance and whatnot. And meanwhile, the heavy mass will hit the ground with very little difference in speed to if the lever hadn't been there - it still has most of its KE.

The fact that you say to find hm in terms of h[/sub]M[/sub], and don't mention their masses, suggests that this is probably the answer you're after.

11.
Spoiler:
They'd close instantly, 'cause you can't have portals open in motion.

But if that wasn't the case, I imagine all the air would get sucked through first... once it got to the moon it would mostly dissipate into space, since the moon isn't big enough to hold it. Once the pressure equalised (at near-0) it would just be any other portal.

In the game, gravitational fields don't go through portals (down is always down, and is discontinuous through the portal) but if that wasn't the case in this presumed real-life version of a portal, I have no idea what the effects would be. Alternatively, if it's not some spacetime-warping deal, there'd possibly be some kind if lightspeed delay between the two portals, in which case it could be dangerous to walk through it... who knows what kind of problems would happen if there was a 3 second lag time between two neighbouring neurons that happen to be on either side of the portal...

But ultimately, the portal rampant speculation thread is that way.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

juststrange
Posts: 296
Joined: Wed Jul 23, 2008 3:57 pm UTC

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:
juststrange wrote:Stuff

It seems like you're making up your mind before doing any physics, and then trying to use physics to justify your (often incorrect) guess (as are a number of other posters in this thread). All of phlip's answers are correct and well justified.

No. First, philp and I had most of the same answers. Two, I staunchly disagree with his take on problem #1. Cycloids do not apply here, atleast in my understanding of the problem.

Spoiler:
Problem #1: Total energy of the system X , Summation of kinetic (KE) and potential (PE). Potential energy (mgh) is controlled by the blocks altitude, kinetic by its velocity (0.5mv^2). If you don't add any energy to the system, then the velocity the block is moving at a given elevation is always the same. If A+B = C and D+B = C, then A==D.
Last edited by juststrange on Tue Aug 17, 2010 11:29 am UTC, edited 1 time in total.

jaap
Posts: 2094
Joined: Fri Jul 06, 2007 7:06 am UTC
Contact:

### Re: Eight problems for intuition in basic classical mechanic

phlip wrote:10.
Spoiler:
Hmm, not sure... first thought is that the lever will equalise speeds... the falling heavy mass won't be impeded much, and the light mass will be accelerated up to match its speed... once their speeds match, the lever is so much dead weight. So the light mass will be launched up at about the same velocity the heavy mass lands, and will thus reach about the same height, modulo air resistance and whatnot. And meanwhile, the heavy mass will hit the ground with very little difference in speed to if the lever hadn't been there - it still has most of its KE.

The fact that you say to find hm in terms of h[/sub]M[/sub], and don't mention their masses, suggests that this is probably the answer you're after.

I agree, if the light mass is negligable compared to the heavy mass, and the see-saw is perfectly rigid and also has negligable mass.
Spoiler:
If the light mass m is not negligable compared to M, and the see-saw is rigid and massless, then it is just like a perfectly inelastic collision. The speed of the large mass (and of the small one) once the see-saw moves will be M/(M+m) times the speed that the large mass approaches the see-saw with. The ratio of the heights will then be the square root of that.

In practice however the see-saw will not be rigid, and the collision will not be elastic, so the small mass will I think tend to reach higher.

(edit: removed "root")
Last edited by jaap on Wed Aug 18, 2010 2:39 pm UTC, edited 1 time in total.

Blatm
Posts: 638
Joined: Mon Jun 04, 2007 1:43 am UTC

### Re: Eight problems for intuition in basic classical mechanic

phlip wrote:10.
Spoiler:
Hmm, not sure... first thought is that the lever will equalise speeds... the falling heavy mass won't be impeded much, and the light mass will be accelerated up to match its speed... once their speeds match, the lever is so much dead weight. So the light mass will be launched up at about the same velocity the heavy mass lands, and will thus reach about the same height, modulo air resistance and whatnot. And meanwhile, the heavy mass will hit the ground with very little difference in speed to if the lever hadn't been there - it still has most of its KE.

The fact that you say to find hm in terms of h[/sub]M[/sub], and don't mention their masses, suggests that this is probably the answer you're after.

Spoiler:
I agree with this as well, but this YouTube video seems to show otherwise. I'd chalk it up to the fact that the piece of wood isn't rigid, but I think it's surprising that it has such a dramatic effect.

12. A ping pong ball hits a stationary bowling ball. After the collision, which of the two has more momentum? Which has more kinetic energy?

phlip
Restorer of Worlds
Posts: 7573
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:
Spoiler:
I agree with this as well, but this YouTube video seems to show otherwise. I'd chalk it up to the fact that the piece of wood isn't rigid, but I think it's surprising that it has such a dramatic effect.

Spoiler:
Yeah, if the wood is springy at all (and, as you can see in the high-speed shot, it very much is) then I'd think more energy would be transferred. 'Cause it won't stop once the speed of the two is equalised - when that happens, it just means there's no more force bending the wood, so it'll spring back, and add a bit more kick to the object being launched.

At least, I think that's how it'd happen.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Dopefish
Posts: 855
Joined: Sun Sep 20, 2009 5:46 am UTC
Location: The Well of Wishes

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:12. A ping pong ball hits a stationary bowling ball. After the collision, which of the two has more momentum? Which has more kinetic energy?

Spoiler:
The bowling ball would have more momentum.

Conservation of momentum requires there to be some net momentum pointed in the direction of the bowling ball, initially from the ping pong ball. The ping pong ball is almost certainly going to bounce back in the other direction, so the bowling ball would have to offset that momentum as well as have whatever the inital momentum of the ping pong ball was.

As for which has more kinetic energy...intuition says the ping pong ball, based on v^2 reliance. Calculations say... the ratio of the bowling balls final KE over the ping pongs KE is less than 1 (getting the velocities from rearranging the momentum stuff), so yeah, ping pong ball has more KE.

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:10. (Mythbusters inspired) Imagine a rigid beam on a pivot (see saw), with a light mass on one end, and nothing on the other. A large mass is dropped from a height hM onto the other end of the see saw. Give the height the small mass reaches, hm, as a function of hM.

What is the length of the beam?
Where is the pivot located along the beam?
What height is the pivot off the ground?
What is the mass of the beam?
Can we neglect air resistance?
Can we assume constant gravity at all altitudes?
wee free kings

Blatm
Posts: 638
Joined: Mon Jun 04, 2007 1:43 am UTC

### Re: Eight problems for intuition in basic classical mechanic

Qaanol wrote:What is the length of the beam?

l
Where is the pivot located along the beam?

x0
What height is the pivot off the ground?

y0
What is the mass of the beam?

mbeam
Can we neglect air resistance?

If you like.
Can we assume constant gravity at all altitudes?

Probably.

TheSkyMovesSideways
Posts: 589
Joined: Thu Oct 02, 2008 8:36 am UTC
Location: Melbourne, Australia
Contact:

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:9. (from Mythbusters) Are two identical cars hitting each other head on at 50 mph equivalent to one car hitting a wall at 100 mph?

10. (Mythbusters inspired) Imagine a rigid beam on a pivot (see saw), with a light mass on one end, and nothing on the other. A large mass is dropped from a height hM onto the other end of the see saw. Give the height the small mass reaches, hm, as a function of hM.

11. What would happen if you put a portal (à la Aperture Science) on the moon, and another on Earth?

9.
Spoiler:
No, two people would die instead of one. If your answer is different to this, I can guess what you do for a living.

10.
Spoiler:
Their speeds will be equal just before the dropped mass hits the ground. If the small mass is negligibly small, it won't slow the larger mass down at all, and so it will reach the height that the larger was dropped from. Otherwise, conservation of momentum tells us that M|vM1| = M|vM2| - m|vM2|
|vm2| = |vM2| = M|vM1|/(M-m)
hm = (M/(M-m))2hM since kinetic energy is proportional to v2 which gravitational potential energy is proportional to h.

No, wait, that's not right, as conservation of momentum doesn't apply here, since forces are applied from the ground. If the masses were equal, then their total momentum just before M hits the ground must be zero, as they're equal masses travelling at the same speed in opposite directions, so the momentum of M must have gone into the ground. If they were travelling in the same direction (ignoring the fact that gravity doesn't work that way), it'd be hm = (M/(M+m))2hM, which intuitively feels like the right answer to me, but I can't be stuffed doing any more maths at this time of night.

11.
Spoiler:
You violate the first law of thermodynamics?
I had all kinds of plans in case of a zombie attack.
I just figured I'd be on the other side.
~ASW

brandon309
Posts: 10
Joined: Wed Aug 18, 2010 4:35 am UTC

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:9. (from Mythbusters) Are two identical cars hitting each other head on at 50 mph equivalent to one car hitting a wall at 100 mph?

Spoiler:
I think this depends on perspective. To the cars, no the impact is not equivalent at all (two cars hitting head on at 50 mph is like one hitting a wall at 50 mph).

But I think if your perspective is outside of the vehicle - let's say you are sandwiched between the two cars - that the crashes are equivalent. Is this correct?

juststrange wrote:No. First, philp and I had most of the same answers. Two, I staunchly disagree with his take on problem #1. Cycloids do not apply here, atleast in my understanding of the problem.

[spoiler]Problem #1: Total energy of the system X , Summation of kinetic (KE) and potential (PE). Potential energy (mgh) is controlled by the blocks altitude, kinetic by its velocity (0.5mv^2). If you don't add any energy to the system, then the velocity the block is moving at a given elevation is always the same. If A+B = C and D+B = C, then A==D.
[/spoiler]

Spoiler:
You are correct that the total energy of the system does not change, which is why the two balls will arrive at the same speed. Same final height = same final kinetic energy.

However, a ball that dips downward and comes back up will indeed arrive first. Despite the longer path, the ball has a higher average speed which lets it get ahead. There are some youtube videos demonstrating this exact problem (unfortunately I can't post links at the moment).

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:10. (Mythbusters inspired) Imagine a rigid beam on a pivot (see saw), with a light mass on one end, and nothing on the other. A large mass is dropped from a height hM onto the other end of the see saw. Give the height the small mass reaches, hm, as a function of hM.

Spoiler:
If I’m thinking this through correctly, both angular momentum and kinetic energy should be conserved by the collision. Angular momentum is conserved because the only external force is by the ground, which produces no torque around the pivot. Kinetic energy should be conserved because the ground does no work. Linear momentum, however, is not conserved.

Supposing the beam to be massless and initially horizontal at height 0, with distance d from the pivot to the mass m and distance D from the pivot to where the mass M strikes, then the small mass should reach height
$h_m = \left(\frac{2dM^2}{DM^2+dm^2}\right)^2h_M$
This can also be written in terms of the ratios [imath]a = \frac{m}{M}[/imath] and [imath]b = \frac{d}{D}[/imath] as
$h_m = \left(\frac{2b}{1+ab^2}\right)^2h_M$
wee free kings

dedalus
Posts: 1169
Joined: Fri Apr 24, 2009 12:16 pm UTC
Location: Dark Side of the Moon.

### Re: Eight problems for intuition in basic classical mechanic

juststrange wrote:
Blatm wrote:
juststrange wrote:Stuff

It seems like you're making up your mind before doing any physics, and then trying to use physics to justify your (often incorrect) guess (as are a number of other posters in this thread). All of phlip's answers are correct and well justified.

No. First, philp and I had most of the same answers. Two, I staunchly disagree with his take on problem #1. Cycloids do not apply here, atleast in my understanding of the problem.

Spoiler:
Problem #1: Total energy of the system X , Summation of kinetic (KE) and potential (PE). Potential energy (mgh) is controlled by the blocks altitude, kinetic by its velocity (0.5mv^2). If you don't add any energy to the system, then the velocity the block is moving at a given elevation is always the same. If A+B = C and D+B = C, then A==D.

Spoiler:
Consider the case where initial velocity=0. Obviously the block will never get to the end on a uniform flat track. However, if we have a track that starts sloped, the block will have some velocity and will eventually reach the end. The case doesn't change much when you give the block a starting velocity.

If you need more convincing, there's a fair few youtube videos that have been linked in the thread demonstrating this. But yeah, phlip is quite right. Also, you might want to check the working from other people for q.5 - an answer has been found that's a bit more qualitative then 'fast'.

Oh, and for people thinking about the Dyson sphere (q.7) - the easiest analogy here is to electromagnetism - like gravity it's an inverse square law so the maths is the same. Gauss's law gives that the integral of the dot product over any closed surface of the electric field and the surface normal is equal to the charge contained by that surface divided by the permeability of free space. If we assume a uniform spherical surface of charge, and take a spherical surface of smaller radius, then we can immediately say the integral is equal to zero, which means by symmetry arguments the electric field is equal to zero. Obviously, if you can't make those symmetry arguments, then you can't claim this, so it's not true that any system is unperturbed by an external electric charge (or else there'd be none whatsoever). But for a spherical shell, then it works out quite well.
doogly wrote:Oh yea, obviously they wouldn't know Griffiths from Sakurai if I were throwing them at them.

Charlie!
Posts: 2035
Joined: Sat Jan 12, 2008 8:20 pm UTC

### Re: Eight problems for intuition in basic classical mechanic

3 is slightly trickier than it appears:
Spoiler:
If our block's velocity is v and the bullet velocity is V, then when v = V/2 the rubber bullets and the magic stopping foam bullets are equivalent. After that the foam bullets actually do better. Indeed, due to fact that the rubber bullets end up with a sequence of ever-decreasing pushes while the magic stopping bullets deliver the same amount of push each time, and are therefore capable of reaching the top speed and even surpassing it if conditions are right, I would say that the magic stopping bullets are the "best."
Some people tell me I laugh too much. To them I say, "ha ha ha!"

phlip
Restorer of Worlds
Posts: 7573
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Eight problems for intuition in basic classical mechanic

For 3, I (and some others) have been reading "stop" as relative to the target, not relative to some stationary observer. So if the foam bullets hit a moving target, they only slow down to the same speed as the target, and then fall.

Yes, if they somehow, by magic, slow down to stationary relative to a reference observer, then it'll be a lot more complicated. Also, I think the collisions will be super-elastic in some cases.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Blatm
Posts: 638
Joined: Mon Jun 04, 2007 1:43 am UTC

### Re: Eight problems for intuition in basic classical mechanic

13. Two equally long rows of dominoes are set up, one with the dominoes placed close together, and the other with the dominoes far apart. Which row topples fastest?

14. On a flat train car (i.e. no sides or top) is a speaker emitting a sound at one end and an observer at the other. How does the pitch the observer perceives change with the speed of the cart?

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:13. Two equally long rows of dominoes are set up, one with the dominoes placed close together, and the other with the dominoes far apart. Which row topples fastest?

Spoiler:
If the dominoes are extremely close together, say less than half the thickness of a domino apart, then the center of mass of each one will rise and not fall before striking the next one. Thus the net energy transferred will diminish until it ceases to be enough to topple the next domino. Thus in the limit of very long rows, extremely close tiles will fail.

On the other hand if the dominoes are extremely far apart, say less than half the thick of a domino shy of being a full domino height apart, then the impact of one striking the next will be a glancing blow that does not tend to topple the next one. So this too fails.

My intuition says there will be an optimal spacing as a function of domino height and thickness, which should be feasible to calculate if we neglect air resistance. But it is not so simple as “closer is always better/worse”. This is a decidedly nonlinear problem. It also matters whether we assume the base of the leading edge of each domino is fixed down like a hinge, or if the dominoes are allowed to “bounce up”.

Blatm wrote:14. On a flat train car (i.e. no sides or top) is a speaker emitting a sound at one end and an observer at the other. How does the pitch the observer perceives change with the speed of the cart?

Spoiler:
The pitch does not change. However, if the speaker is at the back of the car and the train reaches the speed of sound in air, then the person at the front will hear nothing from the speaker.
wee free kings

TheNorm05
Posts: 73
Joined: Wed Aug 04, 2010 11:36 pm UTC
Location: Dallas, TX

### Re: Eight problems for intuition in basic classical mechanic

TheNorm05 wrote:
Spoiler:
Since the foam bullets stop immediately on impact, we can safely assume that ALL of the energy is transferred into the block.

Spoiler:
This assumption is not valid. We have no reason to believe that no heat or sound is produced or that the foam bullet does not squash itself flat. The correct reasoning does not require any assumptions not listed in the OP.

on 3.
Spoiler:
We weren't instructed to include it, so it didn't make sense to assume this, or my assertion for that matter. If you ignore the energy "losses" to heat, sound, and friction(which is mentioned in the problem which led me to believe we were to leave out related losses), then it's really just a conservation of energy problem. In each case you can look at is as a simple concept of how much energy does the block alone contain?, and how much do the projectiles contain? Energy is scalar so while the conservation of momentum still generally applies, it's easier to see why I ranked the projectiles in the order I did. I suppose as far as the rubber bullets are concerned it would make some difference as to the ratio of energy they contained after the collision.

The idea is that the first hit will generally provide the largest marginal increase to the block's velocity, and since the foam bullet retains the least energy it would follow that the most energy is transferred from the foam bullet. The rubber bullet confuses the hell out of me simply because we don't have any idea how fast it bounces back. If it retained it's speed, it would resemble a tennis ball hitting a brick wall.

phlip
Restorer of Worlds
Posts: 7573
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Eight problems for intuition in basic classical mechanic

TheNorm05 wrote:on 3.
Spoiler:
We weren't instructed to include it, so it didn't make sense to assume this, or my assertion for that matter. If you ignore the energy "losses" to heat, sound, and friction(which is mentioned in the problem which led me to believe we were to leave out related losses), then it's really just a conservation of energy problem. In each case you can look at is as a simple concept of how much energy does the block alone contain?, and how much do the projectiles contain? Energy is scalar so while the conservation of momentum still generally applies, it's easier to see why I ranked the projectiles in the order I did. I suppose as far as the rubber bullets are concerned it would make some difference as to the ratio of energy they contained after the collision.

The idea is that the first hit will generally provide the largest marginal increase to the block's velocity, and since the foam bullet retains the least energy it would follow that the most energy is transferred from the foam bullet. The rubber bullet confuses the hell out of me simply because we don't have any idea how fast it bounces back. If it retained it's speed, it would resemble a tennis ball hitting a brick wall.

Spoiler:
But if you assume conservation of kinetic energy, then you have perfectly elastic collision. If you have perfectly elastic collision, then you have the rubber bullet. Because in the elastic case, both energy and momentum are preserved... and from the initial velocities and the masses, in 1 dimension, that's enough to figure out the final velocities. There isn't room for a perfectly-elastic collision where the bullet comes to a stop and falls to the ground, and a perfectly-elastic collision where the bullet bounces off and goes back the way it came... there's only one solution. Either one situation is elastic, or the other is (or neither is). And for the one that isn't elastic, you can't assume kinetic energy is conserved, because it isn't.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

redrogue
Posts: 116
Joined: Tue Dec 15, 2009 9:17 pm UTC

### Re: Eight problems for intuition in basic classical mechanic

Blatm wrote:13. Two equally long rows of dominoes are set up, one with the dominoes placed close together, and the other with the dominoes far apart. Which row topples fastest?

Spoiler:
The line of closely-spaced dominoes will topple faster.

The "wave" imparted down the line travels faster through dominoes than it does through air. The further they are apart, the more air is between them. If the line of dominoes is spaced with no gaps, all of the dominoes will topple instantly. As the gaps increase, so will the time it takes for the last one to fall.

Blatm wrote:14. On a flat train car (i.e. no sides or top) is a speaker emitting a sound at one end and an observer at the other. How does the pitch the observer perceives change with the speed of the cart?

Spoiler:
There will be changes in pitch the cart accelerates, but for experiments performed at constant speeds, the pitch would be the same as if the cart were stationary. The Doppler effect imparted into the sound waves by the speaker will be canceled out by the speed of the observer's ears.