## Eight problems for intuition in basic classical mechanics

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aleph_one
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### Eight problems for intuition in basic classical mechanics

1. Is there a frictionless track on which a block sent sliding at some initial horizontal velocity on the track will reach a point that is at the starting elevation faster than if the track were level?

2. A block, a ball, and a hoop of the same mass are released down a slope. The block slides down frictionlessly, whereas the ball and hoop roll without slipping. In what order do they reach the bottom of the slope? How do their speeds on reaching the bottom of the slope compare?

3. A steady stream of bullets is shot at a target resting on a frictionless surface, accelarating it. The bullets are either foam bullets that stop and fall after hitting the target, lead bullets that embed into the target, or rubber bullets that bounce off the target in the opposite direction. In each case, the same number of bullets are fired, and the bullets the same mass and are fired at the same rate and velocity. How do the final speeds of the target compare in each case?

4. A pulley consists of a inelastic rope with two unequal weights attached to its ends draped over a frictionless wheel of negligible mass. As the heavier weight falls and lighter weight rises, the pulley is weighted by a spring scale attached to the center of the pulley's wheel. How does the scale's reading compare to the total weight of the pulley's components?

5. A very massive block sliding at velocity v collides elastically straight-on with a light block that is at rest. With approximately what velocity does the light block travel afterwards?

6. A boat holding a rock floats in a swimming pool. The rock is removed from the boat and sinks to the bottom of the pool. How does the new water level compare to the original water level?

7. You travel down a straight tunnel to the center of a planet that has uniform density. How does the gravity you experience change throughout your trip?

8. A planet orbits a star in a circular orbit. The star collapses to one tenth of its former radius, but keeps the same mass. What happens to the planet's orbit? What if the star were instantaneously replaced by an object ten times as dense?

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

My mechanics is not excellent, but I think I know these.

6.
Spoiler:
It goes down. Floaty rock displaces a volume of water with weight equal to the weight of the rock. Sinky rock displaces a volume of water equal to the volume of the rock. Since the unaided rock sinks, it must have a greater density than water, so the former displaced volume is greater.

7.
Spoiler:
The gravitational force you experience goes as the cube of your distance from the centre. This is because the part of the planet that is "above" you (that is, at a greater distance from the centre) contributes zero resultant gravitational force, so you experience the same gravity as you would if you were standing on the surface of a smaller planet, with the same centre but radius such that you'd be on the surface. Thus the mass is as the cube of your distance, which is divided by the square of the distance, so it's linear.

EDIT: realized I'd made a stupid error
Last edited by Token on Sun Aug 08, 2010 1:19 am UTC, edited 1 time in total.
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thc
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### Re: Eight problems for intuition in basic classical mechanic

Here are my guesses. It's been several years since any sort of mechanics. Also, could you draw a picture for #4?

Spoiler:
1. No
2. Block, hoop, Ball
4. ???
5. 2v
6. The water level is lower (less water displaced).
7. Linear decrease until it reaches zero at the center
8. 1st question: no change. 2nd: orbit velocity would speed up by factor of sqrt(10). Planet would fall into an elliptical orbit.

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

I haven't even taken physics yet (freshman in high school), but here are my guesses
Spoiler:
1. No
2. Block, Hoop, Ball
4.If I understand the question correctly, the sum of the mass of all the components
5.Just a guess, but maybe velocity of light block = (mass of massive block/mass of light block) * v?
6.It is lower after the rock sinks
7.When outside planet, gravity pulls you to center of planet, after that, I am not entirely sure, I would agree with Token, but it seems that the mass of the planet on the opposite side of where you are would have less of an effect than the part of the planet you just passed through. So I will guess it will be slightly less gravity than if you were standing on a planet with a radius of your distance to the center of the planet.
8.A. The planets orbit stays almost exactly the same, possibly decreases in radius by 9/10ths of the original stars size. B. I really don't have a clue...My guess is it falls into the star and burns up, but that's just a guess

Also this is my first post on these forums EDIT: Read question 4 wrong...now I have to agree with jesting rabbit
Last edited by pingiscoolest. on Sun Aug 08, 2010 2:02 am UTC, edited 1 time in total.

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

4.
Spoiler:
I believe it will be twice the light block's weight.
But my physics tends to suck.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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

aleph_one wrote:1. Is there a frictionless track on which a block sent sliding at some initial horizontal velocity on the track will reach a point that is at the starting elevation faster than if the track were level?

Spoiler:
Yes. Consider a ramp down, then level section, then ramp up. Suppose it takes 1 second to descend the ramp, and the block is now moving faster than it would on all level. Let the flat section be long enough that the faster block gets more than 2 seconds ahead.

aleph_one wrote:2. A block, a ball, and a hoop of the same mass are released down a slope. The block slides down frictionlessly, whereas the ball and hoop roll without slipping. In what order do they reach the bottom of the slope? How do their speeds on reaching the bottom of the slope compare?

Spoiler:
Block is fastest, then ball, hoop is slowest. They gained the same energy from delta-h, the difference is how much is kinetic versus rotational.

aleph_one wrote:3. A steady stream of bullets is shot at a target resting on a frictionless surface, accelarating it. The bullets are either foam bullets that stop and fall after hitting the target, lead bullets that embed into the target, or rubber bullets that bounce off the target in the opposite direction. In each case, the same number of bullets are fired, and the bullets the same mass and are fired at the same rate and velocity. How do the final speeds of the target compare in each case?

Spoiler:
Rubber makes it go fastest, then foam, then lead. Think momentum conservation.

aleph_one wrote:4. A pulley consists of a inelastic rope with two unequal weights attached to its ends draped over a frictionless wheel of negligible mass. As the heavier weight falls and lighter weight rises, the pulley is weighted by a spring scale attached to the center of the pulley's wheel. How does the scale's reading compare to the total weight of the pulley's components?

Spoiler:
Twice the weight of the light block. The extra force from the heavier weight goes to accelerate the weights and thus does not contribute to the scale reading.

phlip got this one right further down the thread. I’m editing to include the correct answer:
$\frac{4m_1m_2}{m_1+m_2}$

aleph_one wrote:5. A very massive block sliding at velocity v collides elastically straight-on with a light block that is at rest. With approximately what velocity does the light block travel afterwards?

Spoiler:
2v. Switch to frame of reference where the massive block is stationary and the light block bounces off it.

aleph_one wrote:6. A boat holding a rock floats in a swimming pool. The rock is removed from the boat and sinks to the bottom of the pool. How does the new water level compare to the original water level?

Spoiler:
The water level drops. In the boat the rock displaces a volume of water with mass equal to the rock’s mass, but when it sinks it only displaces its own volume of water. Since the rock sinks it is denser than water, so it displaces less.

aleph_one wrote:7. You travel down a straight tunnel to the center of a planet that has uniform density. How does the gravity you experience change throughout your trip?

Spoiler:
Linearly.

aleph_one wrote:8. A planet orbits a star in a circular orbit. The star collapses to one tenth of its former radius, but keeps the same mass. What happens to the planet's orbit? What if the star were instantaneously replaced by an object ten times as dense?

Spoiler:
Assuming a radially-symmetric spherical star, the shrinking of the star has no effect on the planet’s orbit, assuming the planet’s orbit is outside the body of the star.

If the star suddenly becomes 10 times denser while retaining the same volume, there will be substantial relativistic consequences to the creation of energy. Also, the planet will switch to an elliptical orbit, coming closer to the star and moving faster, then at aphelion getting back to the distance it originally started at the speed it originally started.
Last edited by Qaanol on Tue Aug 10, 2010 11:11 pm UTC, edited 1 time in total.
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pingiscoolest.
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### Re: Eight problems for intuition in basic classical mechanic

Qaanol wrote:
aleph_one wrote:1. Is there a frictionless track on which a block sent sliding at some initial horizontal velocity on the track will reach a point that is at the starting elevation faster than if the track were level?

Spoiler:
Yes. Consider a ramp down, then level section, then ramp up. Suppose it takes 1 second to descend the ramp, and the block is now moving faster than it would on all level. Let the flat section be long enough that the faster block gets more than 2 seconds ahead.

Spoiler:
Right. Read it as "Is there a frictionless track on which a block sent sliding at some initial horizontal velocity on the track will reach a point that is at the starting elevation at a faster velocity than if the track were level?"

Qaanol wrote:
aleph_one wrote:2. A block, a ball, and a hoop of the same mass are released down a slope. The block slides down frictionlessly, whereas the ball and hoop roll without slipping. In what order do they reach the bottom of the slope? How do their speeds on reaching the bottom of the slope compare?

Spoiler:
Block is fastest, then ball, hoop is slowest. They gained the same energy from delta-h, the difference is how much is kinetic versus rotational.

Spoiler:
Does this include friction? It seems a ball would have greater friction due to more contact with the ground (depending of course on the size and material of the ball, and the width of the hoop).

Qaanol wrote:
aleph_one wrote:3. A steady stream of bullets is shot at a target resting on a frictionless surface, accelarating it. The bullets are either foam bullets that stop and fall after hitting the target, lead bullets that embed into the target, or rubber bullets that bounce off the target in the opposite direction. In each case, the same number of bullets are fired, and the bullets the same mass and are fired at the same rate and velocity. How do the final speeds of the target compare in each case?

Spoiler:
Rubber makes it go fastest, then foam, then lead. Think momentum conservation.

Spoiler:
I don't quite understand, all 3 of the materials have the same amount of energy when they strike the target, but the foam and lead transfer all of thier energy to the block, while the rubber doesn't (it bounces off and goes backwards)
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### Re: Eight problems for intuition in basic classical mechanic

pingiscoolest. wrote:
Qaanol wrote:
aleph_one wrote:3. A steady stream of bullets is shot at a target resting on a frictionless surface, accelarating it. The bullets are either foam bullets that stop and fall after hitting the target, lead bullets that embed into the target, or rubber bullets that bounce off the target in the opposite direction. In each case, the same number of bullets are fired, and the bullets the same mass and are fired at the same rate and velocity. How do the final speeds of the target compare in each case?
Spoiler:
Rubber makes it go fastest, then foam, then lead. Think momentum conservation.
Spoiler:
I don't quite understand, all 3 of the materials have the same amount of energy when they strike the target, but the foam and lead transfer all of thier energy to the block, while the rubber doesn't (it bounces off and goes backwards)
pingiscoolest,
Spoiler:
There is no conservation of kinetic energy. Unless a collision is fully elastic, some of the kinetic energy that goes into deforming the objects is not recovered and gets lost to the environment. There is however conservation of momentum, regardless of how elastic the collisions are. The backwards momentum of the bounced rubber bullets can only arise if the target has gained the same amount of forwards momentum.

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

1:
Spoiler:
I'm not sure exactly what this is asking. Is it asking if the block will reach a certain point faster than on a flat track, or have a higher velocity? If the former, then yes. The block could drop to a lower elevation, then rise back up; at the lower elevation the block will be traveling faster. If the latter, then no, because the block has a constant amount of energy that is the sum of its potential and kinetic energy; at the same elevation it will have the same potential, and therefore the same kinetic energy.

2:
Spoiler:
Block, ball, hoop. Since the block is frictionless, there is nothing to slow it down. The ball and hoop, however will require more kinetic energy as some of that energy will go making the object roll. The ball will come before the hoop because its is a sphere where its mass is as close as possible to its center and therefor will require less work to roll.

3:
Spoiler:
The foam bullets will accelerate the target the fastest, because all their kinetic energy is transferred to the target. The rubber bullets will be second, because they keep some of their energy. The lead bullets will be slowest because they will be increasing the mass of the target.

4:
Spoiler:
It will read as twice the weight of the smaller weight.

5:
Spoiler:
v * (mass of heavy block / mass of light block)

6:
Spoiler:
The level will be lower. Before, the rock was displacing water by mass. After, the rock is displacing water by volume. Since the rock sinks, it is clearly more dense than water and will therefor displace less water when it sinks.
Blue, blue, blue

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

jaap wrote:
pingiscoolest. wrote:
Qaanol wrote:
aleph_one wrote:3. A steady stream of bullets is shot at a target resting on a frictionless surface, accelarating it. The bullets are either foam bullets that stop and fall after hitting the target, lead bullets that embed into the target, or rubber bullets that bounce off the target in the opposite direction. In each case, the same number of bullets are fired, and the bullets the same mass and are fired at the same rate and velocity. How do the final speeds of the target compare in each case?
Spoiler:
Rubber makes it go fastest, then foam, then lead. Think momentum conservation.
Spoiler:
I don't quite understand, all 3 of the materials have the same amount of energy when they strike the target, but the foam and lead transfer all of thier energy to the block, while the rubber doesn't (it bounces off and goes backwards)
pingiscoolest,
Spoiler:
There is no conservation of kinetic energy. Unless a collision is fully elastic, some of the kinetic energy that goes into deforming the objects is not recovered and gets lost to the environment. There is however conservation of momentum, regardless of how elastic the collisions are. The backwards momentum of the bounced rubber bullets can only arise if the target has gained the same amount of forwards momentum.
Spoiler:
Ok, makes sense, so more energy is lost as heat in the lead and foam bullet examples? So I supppose then that the target accelerates by the same velocity as the ball bounces backwards (that sounded very confusing...I'm sure theres a better way to write that, what I mean is the velocity of the target increases by the same amount as the velocity of the rubber ball changes when it bounces off the target? Or am I not understanding this correctly
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### Re: Eight problems for intuition in basic classical mechanic

pingiscoolest. wrote:
Spoiler:
what I mean is the velocity of the target increases by the same amount as the velocity of the rubber ball changes when it bounces off the target? Or am I not understanding this correctly

Spoiler:
The momentum of the target increases by the same amount as the momentum of the rubber bullet changes when it bounces off the target. Momentum is defined (in classical mechanics) as mass times velocity.

Say a certain bullet has mass [imath]m_0[/imath] and velocity [imath]v_0[/imath] before striking the target, and mass [imath]m_1[/imath] and velocity [imath]v_1[/imath] after striking the target.

Say the target has mass [imath]M_0[/imath] and velocity [imath]V_0[/imath] before being struck by a certain bullet, and mass [imath]M_1[/imath] and velocity [imath]V_1[/imath] after striking the target.

The conservation of momentum says total momentum before the collision equals momentum after the collisions, so
$m_0v_0+M_0V_0=m_1v_1+M_1V_1$

Rearranging we see the change in momentum of one object is equal and opposite to the change in momentum of the other object(s)
$-(m_1v_1-m_0v_0) = M_1V_1-M_0V_0$

In the case of rubber and foam bullets, neither the mass of the bullet nor the mass of the target changes, so we can leave off the subscripts for the mass and only talk about the change in velocities
$-m\Delta v = M\Delta V$
or
$\Delta V = -\frac{m}{M}\Delta v$

However for the lead bullets, we can consider the bullet to become part of the target so the mass of the target increases by [imath]m_0[/imath], and the bullet as a separate entity disappears so its mass becomes [imath]m_1=0[/imath]. Thus
$m_0v_0=(M_0+m_0)V_1-M_0V_0$

As you can well imagine, this is more difficult to analyze, but it should be obvious that the lead bullet adding mass to the target means the target accelerates more slowly.

Spoiler:
Note that for a heavy target and a long stream of bullets, the final velocities will be very similar to each other, but still in the order rubber-foam-lead.
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aleph_one
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### Re: Eight problems for intuition in basic classical mechanic

1: In "reach a point that is at the starting elevation faster", I mean reach it at an earlier time.

4: A picture of the setup is the first picture here. I'm looking for the weight on the string at the top holding up the contraption.

8: This was meant to be completely classical.
Last edited by aleph_one on Tue Aug 10, 2010 6:19 am UTC, edited 1 time in total.

TheNorm05
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### Re: Eight problems for intuition in basic classical mechanic

pingiscoolest. wrote:
jaap wrote:
pingiscoolest. wrote:
Qaanol wrote:
aleph_one wrote:3. A steady stream of bullets is shot at a target resting on a frictionless surface, accelarating it. The bullets are either foam bullets that stop and fall after hitting the target, lead bullets that embed into the target, or rubber bullets that bounce off the target in the opposite direction. In each case, the same number of bullets are fired, and the bullets the same mass and are fired at the same rate and velocity. How do the final speeds of the target compare in each case?
Spoiler:
Rubber makes it go fastest, then foam, then lead. Think momentum conservation.
Spoiler:
I don't quite understand, all 3 of the materials have the same amount of energy when they strike the target, but the foam and lead transfer all of thier energy to the block, while the rubber doesn't (it bounces off and goes backwards)
pingiscoolest,
Spoiler:
There is no conservation of kinetic energy. Unless a collision is fully elastic, some of the kinetic energy that goes into deforming the objects is not recovered and gets lost to the environment. There is however conservation of momentum, regardless of how elastic the collisions are. The backwards momentum of the bounced rubber bullets can only arise if the target has gained the same amount of forwards momentum.
Spoiler:
Ok, makes sense, so more energy is lost as heat in the lead and foam bullet examples? So I supppose then that the target accelerates by the same velocity as the ball bounces backwards (that sounded very confusing...I'm sure theres a better way to write that, what I mean is the velocity of the target increases by the same amount as the velocity of the rubber ball changes when it bounces off the target? Or am I not understanding this correctly

Spoiler:
I don't believe that the loss in speed from the lead bullet stream would be due to the loss in energy to heat, rather it would dissipate as the lead bullet plunges into the block. The Actual amount lost is due to the kinetic energy equation based on the product of the mass and the velocity squared. The increase in mass of the block reduces the total speed left in the system. This is the reason why cars these days are built to crumble on impact such that the final velocity of the vehicles involved is as low as possible immediately after the collision. The energy is lost as the deformation of the object, which is reflected by the change in mass distribution in the kinetic energy formula.
In the end, the order from greatest velocity to least should go Foam, lead, then rubber. The reason being that in each scenario we assume that each kind of projectile contains the same initial energy. Since the foam bullets stop immediately on impact, we can safely assume that ALL of the energy is transferred into the block. The same holds for the lead bullets however as we have discussed energy is used to deform the block as the mass of the object increases. Lastly the rubber bullets because they retain more of their kinetic energy as they bounce back, providing less to the block's kinetic energy.

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### 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.

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

For those discussing #3:
Spoiler:
It seems as if people misread what the question asked for: the final speed of the target. This depends on the relation of the mass of the target to the mass of the projectiles and the velocity of the projectiles. Specifically, if the target reaches at least the same velocity as the bullets, then the bullets will no longer catch up to it, hence not accelerate it further. Then we must consider how much momentum each bullet adds in a collision.

The lead bullets can never accelerate the target beyond their own velocity, but the velocity of the target will approach the velocity of the bullets as time goes to infinity. This is assuming the target has a non-zero mass.

For rubber bullets: it depends on how well the ball bounces. Assume that, in a reference system travelling at the speed of the target before the impact, the ball bounces back at some fraction f of its incoming velocity, then:

if the target has a mass that is larger than (1+f) times the mass of the projectiles, the target velocity will approach the bullet velocity.
if the target has a mass that is exactly (1+f) times the mass of the projectiles, the target will at the first and only hit acquire the bullet velocity.
if the target has a mass that is less than (1+f) times the mass of the projectiles, the target will at the first and only hit acquire a faster velocity than the bullet.

The foam bullets, however, are different. Since the foam bullets always stop dead in their tracks when they hit the target, they transfer their entire momentum regardless of how fast the target is going (presuming it is going slow enough to be hit by them). Conservation of momentum gives that the target's velocity is increased by the ratio of the mass of the projectile and the target times the projectile velocity(v_p * m_p / m_t). If we call this increase dv, this means that the target will acquire a velocity
v_t_final = v_t_0 + n * dv
such that
v_t_0 + n * dv > v_p > v_t_0 + (n-1) * dv
where n is some positive integer.

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

On 1;
Spoiler:
I believe the fastest track is a parabola, with the exact path defined by a function of gravity, distance and initial velocity of the block. I'm not up to deriving the equation right now, though - I feel as though it should be trivial, but my mechanics has deserted me.

Another way of looking at it to trivially demonstrate that a faster path exists is to set the initial velocity to be zero. In that case, you can clearly see that a block that begins at the top of a frictionless ramp (parabolic or otherwise) will slide down the ramp and retain enough momentum to get back to the same elevation at the end, thus it will have reached the end considerably faster than the block on a flat track, which will never move at all.

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

Spoiler:
I believe the fastest track is a parabola, with the exact path defined by a function of gravity, distance and initial velocity of the block. I'm not up to deriving the equation right now, though - I feel as though it should be trivial, but my mechanics has deserted me.

Another way of looking at it to trivially demonstrate that a faster path exists is to set the initial velocity to be zero. In that case, you can clearly see that a block that begins at the top of a frictionless ramp (parabolic or otherwise) will slide down the ramp and retain enough momentum to get back to the same elevation at the end, thus it will have reached the end considerably faster than the block on a flat track, which will never move at all.

Spoiler:
The easiest explanation for me is to imagine giving the block a ramp at the start, having it travel for a huge amount of time/distance horizontally, then a ramp at the end. Then its reasonable that whatever time it takes to do the ramps can be made up by the huge amount of time that its traveling horizontally fast.
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redrogue
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### Re: Eight problems for intuition in basic classical mechanic

...I recall a video with steel ball bearings on a pair of erector-set tracks that demonstrated the same effect but looked a bit more professional.

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

Spoiler:
This is essentially brachistochrone problem, the solution of which is an upside-down cycloid. It looks somewhat like a parabola at a glance, but it's not the same curve.

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

For #3:
Spoiler:
Wouldn't the rubber bullets rebounding in the oposit direction interfear with future bullets?

pingiscoolest.
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### 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.

Spoiler:
But if we don't assume that no sound or heat is produced, how can we answer this question at all? It seems to me that if we don't assume all of the energy is either maintained by the projectile, or given to the target, we cannot know which bullet will accelerate he target to a higher speed, because we do not know how much energy is lost in each example
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### Re: Eight problems for intuition in basic classical mechanic

Qaanol wrote:
pingiscoolest. wrote:
Spoiler:
what I mean is the velocity of the target increases by the same amount as the velocity of the rubber ball changes when it bounces off the target? Or am I not understanding this correctly

Spoiler:
The momentum of the target increases by the same amount as the momentum of the rubber bullet changes when it bounces off the target. Momentum is defined (in classical mechanics) as mass times velocity.

Say a certain bullet has mass [imath]m_0[/imath] and velocity [imath]v_0[/imath] before striking the target, and mass [imath]m_1[/imath] and velocity [imath]v_1[/imath] after striking the target.

Say the target has mass [imath]M_0[/imath] and velocity [imath]V_0[/imath] before being struck by a certain bullet, and mass [imath]M_1[/imath] and velocity [imath]V_1[/imath] after striking the target.

The conservation of momentum says total momentum before the collision equals momentum after the collisions, so
$m_0v_0+M_0V_0=m_1v_1+M_1V_1$

Rearranging we see the change in momentum of one object is equal and opposite to the change in momentum of the other object(s)
$-(m_1v_1-m_0v_0) = M_1V_1-M_0V_0$

In the case of rubber and foam bullets, neither the mass of the bullet nor the mass of the target changes, so we can leave off the subscripts for the mass and only talk about the change in velocities
$-m\Delta v = M\Delta V$
or
$\Delta V = -\frac{m}{M}\Delta v$

However for the lead bullets, we can consider the bullet to become part of the target so the mass of the target increases by [imath]m_0[/imath], and the bullet as a separate entity disappears so its mass becomes [imath]m_1=0[/imath]. Thus
$m_0v_0=(M_0+m_0)V_1-M_0V_0$

As you can well imagine, this is more difficult to analyze, but it should be obvious that the lead bullet adding mass to the target means the target accelerates more slowly.

Spoiler:
Note that for a heavy target and a long stream of bullets, the final velocities will be very similar to each other, but still in the order rubber-foam-lead.

Spoiler:
I believe I understand what you are saying, but if I understand correctly, then it appears as though the foam bullets should accelerate the target faster. Here's my logic.
If we use your equation $-(m_1v_1-m_0v_0) = M_1V_1-M_0V_0$ Then wouldn't the momentum of the target after the collision ([imath]M_1V_1[/imath]) be greater for reduced momentum values for the bullet after the collision([imath]m_1v_1[/imath]? For example if we set [imath]m_0v_0[/imath] to 2 (just a random value), [imath]m_1v_1[/imath] to 3, [imath]M_0V_0[/imath] to 5, and solved for [imath]M_1V_1[/imath] then we would come up with 4 (assuming my math is correct). And if we set [imath]m_1v_1[/imath] to 0, we would come up with 7. I know these are just arbitrary values, but unless I am missing something, I got a momentum of 7 (insert random unit here) for the foam bullets, and 4 (same unit as before) for the rubber. 7>4 (unless my kindergarten teacher lied), so this would lead me to believe that the foam bullets accelerate the target faster, and therefore to a higher speed (last hit from bullet before bullets can no longer catch up to target gives target more momentum)
--------------------------------------------------------------------------------
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### Re: Eight problems for intuition in basic classical mechanic

pingiscoolest. wrote:
Spoiler:
I believe I understand what you are saying, but if I understand correctly, then it appears as though the foam bullets should accelerate the target faster. Here's my logic.
If we use your equation $-(m_1v_1-m_0v_0) = M_1V_1-M_0V_0$ Then wouldn't the momentum of the target after the collision ([imath]M_1V_1[/imath]) be greater for reduced momentum values for the bullet after the collision([imath]m_1v_1[/imath]? For example if we set [imath]m_0v_0[/imath] to 2 (just a random value), [imath]m_1v_1[/imath] to 3, [imath]M_0V_0[/imath] to 5, and solved for [imath]M_1V_1[/imath] then we would come up with 4 (assuming my math is correct). And if we set [imath]m_1v_1[/imath] to 0, we would come up with 7. I know these are just arbitrary values, but unless I am missing something, I got a momentum of 7 (insert random unit here) for the foam bullets, and 4 (same unit as before) for the rubber. 7>4 (unless my kindergarten teacher lied), so this would lead me to believe that the foam bullets accelerate the target faster, and therefore to a higher speed (last hit from bullet before bullets can no longer catch up to target gives target more momentum)

Spoiler:
Here's what you should be solving:
$\Delta V = -\frac{m}{M}\Delta v$

Conceptually:
The change in a lead bullet's velocity is lowest. It goes from a high speed in one direction to a new speed in the same direction.
The change in a foam bullet's velocity is medium... In fact equal to and opposite of the initial bullet velocity, since it goes from high speed in one direction to zero.
The change in a rubber bullet's velocity is greatest. It goes from a high speed in one direction to a new speed in the opposite direction.

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

pingiscoolest. wrote:
Qaanol wrote:
pingiscoolest. wrote:
Spoiler:
what I mean is the velocity of the target increases by the same amount as the velocity of the rubber ball changes when it bounces off the target? Or am I not understanding this correctly

Spoiler:
The momentum of the target increases by the same amount as the momentum of the rubber bullet changes when it bounces off the target. Momentum is defined (in classical mechanics) as mass times velocity.

Say a certain bullet has mass [imath]m_0[/imath] and velocity [imath]v_0[/imath] before striking the target, and mass [imath]m_1[/imath] and velocity [imath]v_1[/imath] after striking the target.

Say the target has mass [imath]M_0[/imath] and velocity [imath]V_0[/imath] before being struck by a certain bullet, and mass [imath]M_1[/imath] and velocity [imath]V_1[/imath] after striking the target.

The conservation of momentum says total momentum before the collision equals momentum after the collisions, so
$m_0v_0+M_0V_0=m_1v_1+M_1V_1$

Rearranging we see the change in momentum of one object is equal and opposite to the change in momentum of the other object(s)
$-(m_1v_1-m_0v_0) = M_1V_1-M_0V_0$

In the case of rubber and foam bullets, neither the mass of the bullet nor the mass of the target changes, so we can leave off the subscripts for the mass and only talk about the change in velocities
$-m\Delta v = M\Delta V$
or
$\Delta V = -\frac{m}{M}\Delta v$

However for the lead bullets, we can consider the bullet to become part of the target so the mass of the target increases by [imath]m_0[/imath], and the bullet as a separate entity disappears so its mass becomes [imath]m_1=0[/imath]. Thus
$m_0v_0=(M_0+m_0)V_1-M_0V_0$

As you can well imagine, this is more difficult to analyze, but it should be obvious that the lead bullet adding mass to the target means the target accelerates more slowly.

Spoiler:
Note that for a heavy target and a long stream of bullets, the final velocities will be very similar to each other, but still in the order rubber-foam-lead.

Spoiler:
I believe I understand what you are saying, but if I understand correctly, then it appears as though the foam bullets should accelerate the target faster. Here's my logic.
If we use your equation $-(m_1v_1-m_0v_0) = M_1V_1-M_0V_0$ Then wouldn't the momentum of the target after the collision ([imath]M_1V_1[/imath]) be greater for reduced momentum values for the bullet after the collision([imath]m_1v_1[/imath]? For example if we set [imath]m_0v_0[/imath] to 2 (just a random value), [imath]m_1v_1[/imath] to 3, [imath]M_0V_0[/imath] to 5, and solved for [imath]M_1V_1[/imath] then we would come up with 4 (assuming my math is correct). And if we set [imath]m_1v_1[/imath] to 0, we would come up with 7. I know these are just arbitrary values, but unless I am missing something, I got a momentum of 7 (insert random unit here) for the foam bullets, and 4 (same unit as before) for the rubber. 7>4 (unless my kindergarten teacher lied), so this would lead me to believe that the foam bullets accelerate the target faster, and therefore to a higher speed (last hit from bullet before bullets can no longer catch up to target gives target more momentum)

Spoiler:
The thing you're missing is that v is velocity, not speed, meaning it has a direction (sign). The rubber bullet has a negative velocity after it bounces backwards.

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

Malle wrote:For those discussing #3:
Spoiler:
It seems as if people misread what the question asked for: the final speed of the target. This depends on the relation of the mass of the target to the mass of the projectiles and the velocity of the projectiles. Specifically, if the target reaches at least the same velocity as the bullets, then the bullets will no longer catch up to it, hence not accelerate it further. Then we must consider how much momentum each bullet adds in a collision.

The lead bullets can never accelerate the target beyond their own velocity, but the velocity of the target will approach the velocity of the bullets as time goes to infinity. This is assuming the target has a non-zero mass.

For rubber bullets: it depends on how well the ball bounces. Assume that, in a reference system travelling at the speed of the target before the impact, the ball bounces back at some fraction f of its incoming velocity, then:

if the target has a mass that is larger than (1+f) times the mass of the projectiles, the target velocity will approach the bullet velocity.
if the target has a mass that is exactly (1+f) times the mass of the projectiles, the target will at the first and only hit acquire the bullet velocity.
if the target has a mass that is less than (1+f) times the mass of the projectiles, the target will at the first and only hit acquire a faster velocity than the bullet.

The foam bullets, however, are different. Since the foam bullets always stop dead in their tracks when they hit the target, they transfer their entire momentum regardless of how fast the target is going (presuming it is going slow enough to be hit by them). Conservation of momentum gives that the target's velocity is increased by the ratio of the mass of the projectile and the target times the projectile velocity(v_p * m_p / m_t). If we call this increase dv, this means that the target will acquire a velocity
v_t_final = v_t_0 + n * dv
such that
v_t_0 + n * dv > v_p > v_t_0 + (n-1) * dv
where n is some positive integer.

The usual meaning of “stop and fall after hitting the target” is that a foam bullet has final velocity exactly equal to the velocity of the target just before the two collided.
Spoiler:
Thus the foam bullets are just like the rubber bullets with f=0.

jestingrabbit wrote:
Spoiler:
The easiest explanation for me is to imagine giving the block a ramp at the start, having it travel for a huge amount of time/distance horizontally, then a ramp at the end. Then its reasonable that whatever time it takes to do the ramps can be made up by the huge amount of time that its traveling horizontally fast.

Spoiler:
I was thinking that too, originally, but since the block begins with a horizontal velocity, it doesn’t even lose anything by going down a ramp. It just picks up speed, and gets ahead no matter what.

Turtlewing wrote:For #3:
Spoiler:
Wouldn't the rubber bullets rebounding in the oposit direction interfear with future bullets?

This is a highly intelligent statement.
wee free kings

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

For 4, I get:
Spoiler:
Twice the harmonic average (ie the reciprocal of the average of the reciprocals) of the weights of the two objects.

Calculations:
Spoiler:
Say the tension in the rope over the pulley connecting the two masses is T, and say the light mass is m1 and the heavy mass is m2.

Now, the net force on the light mass is T - m1g, upwards, so it'll accelerate upwards at T/m1 - g. Similarly, the heavy mass will accelerate down at g - T/m2. Now, these accelerations have to be equal:
T/m1 - g = g - T/m2
2g = T(1/m1 + 1/m2)
T = 2g/(1/m1 + 1/m2)

So the tension in the rope is the harmonic average of the weight of the two objects... and the force on the pulley is twice the tension in the rope.

My answers for the others, most of which are repeats:

1:
Spoiler:
Yes. IIRC, the best curve is a cycloid, but there are many shapes where the extra time for the extra distance is made up for by a greater speed while it's going. Obviously the object's final speed on arrival will be identical regardless of the path.

2:
Spoiler:
The block will accelerate under gravity without any hinderances. The other two will have the same gravitational force, but will have resistance from rotational inertia, so they'll take off slower. The ring has a higher moment of inertia than the ball, so it'll take off the slowest.

3:
Spoiler:
The foam bullets will impart their full momentum to the target. The lead bullets will impart their full momentum to the target, but also increase the mass of the target (so later bullets will have less effect). The rubber bullets will impart more than their full momentum to the target, as they'll turn around and go back the way they came. Where for all three "full momentum" == the mass of the bullet times the difference in velocity between the bullet and the target (ie the momentum lost by the bullet by coming to rest relative to the target). So rubber is significantly better than foam is marginally better than lead.

5:
Spoiler:
Take a frame-shift to one where the heavy mass is stationary. The light mass will come in at v, hit the heavy mass, and be reflected at v, with negligible change in the heavy object's velocity. So back in the original frame, the light object starts at 0 and ends at 2v.

6:
Spoiler:
The rock initially displaces its weight in water, and later displaces its volume. Since it sinks, the latter must be lower, so the water level will drop. If the former was lower, it would float, and enough of the rock would protrude from the water such that the water level wouldn't change.

7:
Spoiler:
The gravity goes down linearly, as all the earth further from the centre than you forms a spherical shell, which is gravitationally neutral, and the sphere closer to the centre than you has a mass that goes down by r3 (assuming the planet is uniform, which it isn't, but whatever), but gravity goes up by 1/r2, so a total change of just r.

8:
Spoiler:
The only relevant value is the mass of the star, unless the planet's orbit is so close that the planet is actually inside the star at some point (in which case there'd be friction). So changing the size of the star without changing its mass will affect nothing (even in the extreme case of replacing the star with an equally-massive black hole). Changing the star's density without changing its size, though, will increase its mass, which will affect the orbit... if the planet's position and velocity are unchanged, they won't be fast enough to maintain the circular orbit, and it'll fall into an ellipse (with its original position as the apogee).
Last edited by phlip on Tue Aug 10, 2010 7:04 am UTC, edited 1 time in total.

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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Qaanol
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### Re: Eight problems for intuition in basic classical mechanic

phlip wrote:For 4, I get:
Spoiler:
Twice the harmonic average (ie the reciprocal of the average of the reciprocals) of the weights of the two objects.

Calculations:
Spoiler:
Say the tension in the rope over the pulley connecting the two masses is T, and say the light mass is m1 and the heavy mass is m2.

Now, the net force on the light mass is T - m1g, upwards, so it'll accelerate upwards at T/m1 - g. Similarly, the heavy mass will accelerate down at g - T/m2. Now, these accelerations have to be equal:
T/m1 - g = g - T/m2
2g = T(1/m1 + 1/m2)
T = 2g/(1/m1 + 1/m2)

So the tension in the rope is the harmonic average of the weight of the two objects... and the force on the pulley is twice the tension in the rope.

You are right.
wee free kings

Blatm
Posts: 638
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### Re: Eight problems for intuition in basic classical mechanic

Concerning #1:
Spoiler:
It's true that a cycloid has a mass travel from one point to a lower point in the least amount of time when the mass starts at rest. However, this mass starts with a horizontal speed, so I'm not convinced that a cycloid is best, especially if the mass is forced to follow the track. The example I have in mind is a very deep sheer drop, where if the mass is forced to follow the track, will get to the bottom almost instantly, and have a lot of speed when it's down there. I'd expect this to be the fastest if the mass follows the track (and is arbitrarily fast, so I suppose that doesn't mean much). Two interesting problems to consider are where the mass isn't forced to follow the track (so a sheer drop wouldn't do very well, because the mass would have to fall all the way to the bottom, through the air.) and where the length of the track is limited (so a sheer drop might not be best, because it's the longest (convex) track). Any ideas on those?

Concerning #4:
Spoiler:
1) The answer is going to depend on the masses and g, and nothing else. From dimensional analysis, the answer will be linear in both.
2) The answer is symmetric between the two masses.

F = a*g(m1m2/(m1 + m2)) respects these conditions, and looks reasonable. The m1 = m2 case gives that a = 4 without much thinking. Though conceptually the force being proportional to the weight of the smaller mass seems reasonable too, I dislike because it doesn't have a nice mathematical formula like the above possibility. That worries me, because Newtonian problems like these always do, since you can always find them using Free Body Diagrams and Newton's Laws.

What do you think?

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

Blatm wrote:
Spoiler:
[...]The example I have in mind is a very deep sheer drop, where if the mass is forced to follow the track, will get to the bottom almost instantly,[...]

Spoiler:
But where would it get the huge amount of kinetic energy it'd take to get to the bottom of the drop almost instantly?

I guess it depends on exactly how the mass is forced to follow the track, but the best I can imagine is if it's sandwiched between two tracks, so it can't move up or down, and any sharp corners are just sufficiently small circular arcs... so if it enters going horizontally at v, it'll exit going vertically at v, or something like that. And then it'll fall under gravity as normal until it reaches the bottom, and enters another small circular arc at v+gt vertically, and exits at v+gt horizontally. I think that's the best option 'cause it's energy-neutral... regardless of the convolutedness of the track, KE+GPE will be a constant. The other energy-neutral system I can think up is if it's able to leave the track, and then if it falls and lands on the track, it bounces elastically (and the only way for the object to return to just sliding along the track is for the object's trajectory and the track to be tangential where it lands).

In the can't-leave-the-track situation a vertical drop isn't going to be ideal... moving the bottom of the drop epsilon towards the destination would have negligible difference to the length of the drop, but epsilon difference to the length of the horizontal bit after the drop. So it would get there epsilon faster. (Infinitesimals make derivatives fun!)

And in the bouncing situation a vertical drop isn't ideal either, since none of the speed it gains from falling will be translated into horizontal momentum... it'll just bounce up and down a bunch, and arrive at the same time as if the ground was flat. To get more horizontal momentum, it'd need to land on a ground that's at an angle, not just horizontal.

But you might have a point that the object's initial speed could result in the cycloid not being the fastest option... I'm not certain how that fits together. But all the same, the answer to the original question is still "Yes", whatever the optimal curve is.

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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TheSkyMovesSideways
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### Re: Eight problems for intuition in basic classical mechanic

aleph_one wrote:1. Is there a frictionless track on which a block sent sliding at some initial horizontal velocity on the track will reach a point that is at the starting elevation faster than if the track were level?

Here's a video demonstrating the answer to this with beautiful precision: http://www.youtube.com/watch?v=LWZtBGwX1jY
Spoiler:
Those bikes have no brakes and a very high gear ratio (meaning poor acceleration), so conservation of momentum is extremely important during switching, and the switching is necessary because the front guy has to do much more work than the others to fight wind resistance.

phlip wrote:7:
Spoiler:
The gravity goes down linearly, as all the earth further from the centre than you forms a spherical shell, which is gravitationally neutral, and the sphere closer to the centre than you has a mass that goes down by r3 (assuming the planet is uniform, which it isn't, but whatever), but gravity goes up by 1/r2, so a total change of just r.

Not sure (literally) if I'd agree with that...
Spoiler:
Is a hollow, spherical shell really gravitationally neutral? I'd have thought if you were close to one side of the shell, you'd be attracted to that side, since the far side of the shell would be further away, and gravity obeys the inverse square law. Although, I guess the far side of the shell would have greater mass, and the two may be perfectly commensurate. I haven't done the drdθdφ to figure it out.

phlip wrote:8:
Spoiler:
The only relevant value is the mass of the star, unless the planet's orbit is so close that the planet is actually inside the star at some point (in which case there'd be friction). So changing the size of the star without changing its mass will affect nothing (even in the extreme case of replacing the star with an equally-massive black hole). Changing the star's density without changing its size, though, will increase its mass, which will affect the orbit... if the planet's position and velocity are unchanged, they won't be fast enough to maintain the circular orbit, and it'll fall into an ellipse (with its original position as the apogee).

Anyone got a link explaining this one? My (possibly faulty) intuition tells me...
Spoiler:
that an increase in the gravitational pull of the star would cause the satellite's orbit to decay, and it would fall into the star. Do satellites actually go into elliptical orbits when this happens? And does this explain why how orbits can exist, since otherwise they'd be completely unstable?
I had all kinds of plans in case of a zombie attack.
I just figured I'd be on the other side.
~ASW

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

TheSkyMovesSideways wrote:Not sure (literally) if I'd agree with that...
Spoiler:
Is a hollow, spherical shell really gravitationally neutral? I'd have thought if you were close to one side of the shell, you'd be attracted to that side, since the far side of the shell would be further away, and gravity obeys the inverse square law. Although, I guess the far side of the shell would have greater mass, and the two may be perfectly commensurate. I haven't done the drdθdφ to figure it out.

Spoiler:
The usual intuitive explanation (which I learned in some thread in the Science forum) is that in a certain solid angle from your position, the mass of the shell in that direction goes up by the square of the distance, but the gravitational effects goes down by the square of the distance. So the force in every direction is the same, and they balance out.

There's some more details on Wikipedia.

TheSkyMovesSideways wrote:Anyone got a link explaining this one? My (possibly faulty) intuition tells me...
Spoiler:
that an increase in the gravitational pull of the star would cause the satellite's orbit to decay, and it would fall into the star. Do satellites actually go into elliptical orbits when this happens? And does this explain why how orbits can exist, since otherwise they'd be completely unstable?

Spoiler:
Say you have a planet that's moving in a direction perpendicular to a star (the cases where it's moving at an angle towards or away from the star are also solvable, but more complicated and also irrelevant in this case). At a certain speed, the orbit will be circular (that speed depends on the mass of the star and the distance between the star and the planet). At slower speeds the orbit will be elliptical, with the point we're looking at as the apogee. Make it slow enough and the perigee will be close enough to the star that they'll collide (well, inasmuch as something can collide with something that isn't solid). On the other hand, at faster speeds than the one that'd cause a circular orbit, the orbit will also be elliptical, with the point we're looking at being the perigee. Get even faster until you reach the escape velocity... just below the escape velocity it'll still be an ellipse, but the apogee will be absurdly far away; at the escape velocity, it'll be a parabola (which is the limit of an ellipse with the apogee at infinity); at faster speeds, it'll be a hyperbola. At all the points in the orbit other than the apogee and perigee, it's one of those points where it's either moving towards or away from the star that I mentioned we weren't going to worry about.

Now, if the mass of the star goes up, the speed for a circular orbit at the same distance also goes up... so if the planet is still moving the same speed, it'll be moving perpendicular to the star, at a slower speed than the circular orbit speed... so it'll be an ellipse.

Just from orbit mechanics alone, there's nothing that can cause a "decaying" orbit... from every starting position where it's able to make it once around the star, it'll end up back in its same starting position and velocity. That'll keep going unless there's something actively slowing the planet down (like a collision with the star, or friction if the space it's orbiting through isn't empty, or whatever), or the star's mass is continually changing over time (not just a one-time change... like, if the star's mass increased, and the planet fell into the ellipse, and then when the planet was at perigee the mass of the star increased again, the planet could end up in a new circular orbit that's smaller than the first, or a different elliptical orbit), or something's going at relativistic speeds (in which case the whole deal changes).

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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Cosmologicon
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### Re: Eight problems for intuition in basic classical mechanic

phlip wrote:
Blatm wrote:
Spoiler:
[...]The example I have in mind is a very deep sheer drop, where if the mass is forced to follow the track, will get to the bottom almost instantly,[...]

Spoiler:
But where would it get the huge amount of kinetic energy it'd take to get to the bottom of the drop almost instantly?

I guess it depends on exactly how the mass is forced to follow the track, but the best I can imagine is if it's sandwiched between two tracks, so it can't move up or down, and any sharp corners are just sufficiently small circular arcs... so if it enters going horizontally at v, it'll exit going vertically at v, or something like that. And then it'll fall under gravity as normal until it reaches the bottom, and enters another small circular arc at v+gt vertically, and exits at v+gt horizontally. I think that's the best option 'cause it's energy-neutral... regardless of the convolutedness of the track, KE+GPE will be a constant. The other energy-neutral system I can think up is if it's able to leave the track, and then if it falls and lands on the track, it bounces elastically (and the only way for the object to return to just sliding along the track is for the object's trajectory and the track to be tangential where it lands).

In the can't-leave-the-track situation a vertical drop isn't going to be ideal... moving the bottom of the drop epsilon towards the destination would have negligible difference to the length of the drop, but epsilon difference to the length of the horizontal bit after the drop. So it would get there epsilon faster. (Infinitesimals make derivatives fun!)

And in the bouncing situation a vertical drop isn't ideal either, since none of the speed it gains from falling will be translated into horizontal momentum... it'll just bounce up and down a bunch, and arrive at the same time as if the ground was flat. To get more horizontal momentum, it'd need to land on a ground that's at an angle, not just horizontal.

But you might have a point that the object's initial speed could result in the cycloid not being the fastest option... I'm not certain how that fits together. But all the same, the answer to the original question is still "Yes", whatever the optimal curve is.

Nope, none of that stuff really matters. The fastest track is still always
Spoiler:
cycloid. The only assumption behind the brachistochrone problem is that energy is conserved, which means that the speed of the object depends only on its height, via KE + PE = constant. It can be allowed to change direction instantaneously however you want, but its speed is determined. So it doesn't matter whether or how it's confined to the track.

If the object starts with some speed, the brachistochrone is still going to be a cycloid, it'll just be a shallower piece of a cycloid. Consider the original case of the object starting at rest, and label two points A and B at the same height along its ideal track, and consider another object that starts with the same speed as the original object has when it gets to A. If there's a faster way for this new object to get from A to B, there will be a faster way for the original object to get to its destination as well.

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

Cosmologicon wrote:
phlip wrote:
Blatm wrote:
Spoiler:
[...]The example I have in mind is a very deep sheer drop, where if the mass is forced to follow the track, will get to the bottom almost instantly,[...]

Spoiler:
But where would it get the huge amount of kinetic energy it'd take to get to the bottom of the drop almost instantly?

I guess it depends on exactly how the mass is forced to follow the track, but the best I can imagine is if it's sandwiched between two tracks, so it can't move up or down, and any sharp corners are just sufficiently small circular arcs... so if it enters going horizontally at v, it'll exit going vertically at v, or something like that. And then it'll fall under gravity as normal until it reaches the bottom, and enters another small circular arc at v+gt vertically, and exits at v+gt horizontally. I think that's the best option 'cause it's energy-neutral... regardless of the convolutedness of the track, KE+GPE will be a constant. The other energy-neutral system I can think up is if it's able to leave the track, and then if it falls and lands on the track, it bounces elastically (and the only way for the object to return to just sliding along the track is for the object's trajectory and the track to be tangential where it lands).

In the can't-leave-the-track situation a vertical drop isn't going to be ideal... moving the bottom of the drop epsilon towards the destination would have negligible difference to the length of the drop, but epsilon difference to the length of the horizontal bit after the drop. So it would get there epsilon faster. (Infinitesimals make derivatives fun!)

And in the bouncing situation a vertical drop isn't ideal either, since none of the speed it gains from falling will be translated into horizontal momentum... it'll just bounce up and down a bunch, and arrive at the same time as if the ground was flat. To get more horizontal momentum, it'd need to land on a ground that's at an angle, not just horizontal.

But you might have a point that the object's initial speed could result in the cycloid not being the fastest option... I'm not certain how that fits together. But all the same, the answer to the original question is still "Yes", whatever the optimal curve is.

Nope, none of that stuff really matters. The fastest track is still always
Spoiler:
cycloid. The only assumption behind the brachistochrone problem is that energy is conserved, which means that the speed of the object depends only on its height, via KE + PE = constant. It can be allowed to change direction instantaneously however you want, but its speed is determined. So it doesn't matter whether or how it's confined to the track.

If the object starts with some speed, the brachistochrone is still going to be a cycloid, it'll just be a shallower piece of a cycloid. Consider the original case of the object starting at rest, and label two points A and B at the same height along its ideal track, and consider another object that starts with the same speed as the original object has when it gets to A. If there's a faster way for this new object to get from A to B, there will be a faster way for the original object to get to its destination as well.

Spoiler:
The thing is, though, the object doesn't start with a set speed. It starts with a set horizontal speed, so its overall inital speed depends on the slope it begins on. If the object begins by moving at 1 m/s horizontally, then on a flat track its initial speed will be 1 m/s, on a shallow cycloid its initial speed will be somewhat higher, and if it begins on an extremely sheer drop, it could begin by moving arbitrarily fast, allowing it to reach the second point in an arbitrarily short period of time, no matter how far away it is (well, if your ignore relativity).

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

lordatog: It was meant that the block starts out moving horizontally and the initial slope is horizontal.

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

Well, nevermind then. TheSkyMovesSideways
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### Re: Eight problems for intuition in basic classical mechanic

@phlip: Makes sense on both counts. Thanks. I had all kinds of plans in case of a zombie attack.
I just figured I'd be on the other side.
~ASW

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

Vicious Chicken wrote:
Spoiler:
The thing you're missing is that v is velocity, not speed, meaning it has a direction (sign). The rubber bullet has a negative velocity after it bounces backwards.

Spoiler:
Right...thanks --------------------------------------------------------------------------------
All statements to be taken merely as suggestions, as I more often than not am not sure of their validity. juststrange
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### Re: Eight problems for intuition in basic classical mechanic

1
Spoiler:
No. KE+PE == X. Without adding energy from an external source, no.

2.
Spoiler:
Block then ball then hoop. speeds fastest to slowest, same order. Again, KE+PE=X. All have same KE at the bottom. KE = (0.5*m*v^2) + (0.5*I*w^2). 'w' (rate of rotation) is 0 for the block, so v is highest. Between the ball and the ring its the difference in 'I' that makes the difference.

3
Spoiler:
Exact comparison depends on how mass of bullets compares to that of the block, since you are adding mass to the block in the process. As the number of bullets approaches infinity, the final speed of all blocks will approach the speed of the bullets, Obviously it can't exceed that. Conservation of momentum tells me the most energy is transfered with the rubber ball, followed by the foam, followed by the lead. So that'd be the comparison of how they accelerate (aside from the whole change in mass of the block thing!)

4. http://www.physicsforums.com/archive/index.php/t-220168.html Not saying you should look up the answer, I just wanted to share this because it was funny. Shows common misconceptions.

5.
Spoiler:
Fast, very fast. Derivation is in the conservation of kinetic energy and conservation of momentum.

6.
Spoiler:
Slightly lower since the boat rises out of it an displaces less water.

7. I'll have to ponder on this a bit more.

8.
Spoiler:
FIrst situation, no change. Orbit is function of mass, not density. Second situation, the planet would like to orbit farther away. If it was at its original orbit when the star was instantly replaces with a 10X mass change, the planet would be sucked into the star.

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

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.

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

Blatm wrote:All of phlip's answers are correct and well justified.

Sometimes I wish I was the kind of person who'd quote things out of context in his sig...

Code: Select all

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