|
GR8677 #5
|
|
|
|
|
Alternate Solutions |
PorcelainMouse
2008-11-05 15:12:32 | Thanks a19gray2. I completely forgot that. I was working this out and the three step process took me longer than I want. I think there is a quicker way. How about this?...
Recognize that the first and last steps are really the same: conservation of total energy, but where energy moves completely from potential to kinetic and back. For any complete conversion situation:
Right? So, now we know that
 ,
where  is the velocity of ball A and  is it's initial height (where all of the E for this whole situation originates.)
And, we also know, from he same calculation, that
 ,
where  is the final velocity of both (stuck-together) balls and  is their final height.
We want to know the relation ship between and  , so let's try this:
That's pretty close already. (And, you can see where the 1/16 comes from: it's the ratio of the squares of the velocity.) Now, since a19grey2 pointed out we need to use momentum to analyze the impact, we can write:
 v_{AB} )
But, now we only need to solve for  .
Let's square this, to match the ratio of squares we had relating the two heights:
^2 = \left(\frac{m_A + m_B}{m_A}\right)^2 )
Putting this last equation together with the previous ratio of velocities, we can find what we want:
^2 )
That makes 16 times as large as . Okay, it looks harder, but I think it's much faster because the algebra is quicker and easier. |  |
|
|
Comments |
PorcelainMouse
2008-11-05 15:12:32 | Thanks a19gray2. I completely forgot that. I was working this out and the three step process took me longer than I want. I think there is a quicker way. How about this?...
Recognize that the first and last steps are really the same: conservation of total energy, but where energy moves completely from potential to kinetic and back. For any complete conversion situation:
Right? So, now we know that
,
where is the velocity of ball A and is it's initial height (where all of the E for this whole situation originates.)
And, we also know, from he same calculation, that
,
where is the final velocity of both (stuck-together) balls and is their final height.
We want to know the relation ship between and , so let's try this:
That's pretty close already. (And, you can see where the 1/16 comes from: it's the ratio of the squares of the velocity.) Now, since a19grey2 pointed out we need to use momentum to analyze the impact, we can write:
But, now we only need to solve for .
Let's square this, to match the ratio of squares we had relating the two heights:
Putting this last equation together with the previous ratio of velocities, we can find what we want:
That makes 16 times as large as . Okay, it looks harder, but I think it's much faster because the algebra is quicker and easier. | | Blake7
2007-09-11 18:30:50 | The bookkeeping is deceptively simple so get a good night's sleep and commit to the time expenditure in drudge cranking. |  | tercel
2006-12-01 11:18:28 | I'm confused. Why doesn't conservation of energy work to get directly from the initial condition to the solution?
In other words, where does the initial energy,  , go?
VanishingHitchwriter
2006-12-01 14:10:43 |
Not sure if this is what you're asking, but the initial potential energy is converted to kinetic energy. However, when the two objects collide, momentum is transferred, and the velocity is changed (hence kinetic energy changed). The collision is completely inelastic (a la conservation of momentum).
|
a19grey2
2008-11-02 21:07:32 |
To clarify, ANY time that two objects stick together after a collison, energy is lost. Therefore, NEVER use conservation of energy to the before/after parts of a "stick together" collision.
|
|  | jax
2005-12-01 09:01:20 | It seems that only 19% of people got this one right, even though it seems easy. I guess most people probably assumed that the collision was elastic, which is not true. I messed up on that one too... ugh I hope I don't do something dumb like that on the exam! :)
dnvlgm
2007-11-11 19:22:58 |
well, it says the particles stick together, I guess most people read too fast or careless and that's a big problem that I honestly have a lot. Gotta be careful! Good luck with your exam!!! :D
|
| |
|
| Post A Comment! |
You are replying to:
Thanks a19gray2. I completely forgot that. I was working this out and the three step process took me longer than I want. I think there is a quicker way. How about this?...
Recognize that the first and last steps are really the same: conservation of total energy, but where energy moves completely from potential to kinetic and back. For any complete conversion situation:
Right? So, now we know that
,
where is the velocity of ball A and is it's initial height (where all of the E for this whole situation originates.)
And, we also know, from he same calculation, that
,
where is the final velocity of both (stuck-together) balls and is their final height.
We want to know the relation ship between and , so let's try this:
That's pretty close already. (And, you can see where the 1/16 comes from: it's the ratio of the squares of the velocity.) Now, since a19grey2 pointed out we need to use momentum to analyze the impact, we can write:
But, now we only need to solve for .
Let's square this, to match the ratio of squares we had relating the two heights:
Putting this last equation together with the previous ratio of velocities, we can find what we want:
That makes 16 times as large as . Okay, it looks harder, but I think it's much faster because the algebra is quicker and easier. |
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
