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GR8677 #5
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Mechanics$\Rightarrow$}Conservation

This is a three step problem involving conservation of energy in steps one and three and conservation of momentum in step two.

1. $Mgh_0=\frac{1}{2}Mv_{1}^{2}$... Conservation of energy determines the velocity of putty A, $v_1$, (of mass $M$) at the bottom of its trajectory, right before it intersects B.

2. $Mv_1 = (M+3M)v_2=4Mv_2$... Since the putty thingies stick together, conservation of momentum is easy. Solve for $v_2$.

3. $\frac{1}{2}4Mv_{2}^{2}=4Mgh_3$... Conservation of energy, again. Solve for $h_3$.

And voila, plug in the relevant quantities to find that the answer is (A) $\frac{1}{16} h_0$.

Alternate Solutions
 archard2010-08-04 14:05:45 It's super easy if you write the kinetic energy as $\frac{p^2}{2m}$ - since momentum is conserved, you don't even need to worry about whether or not the collision is elastic. Just write $\ M g h 1 = \frac{p^2}{2M}$ $\ 4M g h 2 = \frac{p^2}{8M}$ Divide equations and solve for h2.Reply to this comment PorcelainMouse2008-11-05 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: $m g h = 1/2 m v^2$ $v^2=2gh$ Right? So, now we know that $v_A^2 = 2gh_0$ , where $v_a$ is the velocity of ball A and $h_0$ is it's initial height (where all of the E for this whole situation originates.) And, we also know, from he same calculation, that $v_{AB}^2 = 2gh_{final}$ , where $v_{AB}$ is the final velocity of both (stuck-together) balls and $h_{final}$ is their final height. We want to know the relation ship between $h_0$ and $h_f$, so let's try this: $\frac{v_A^2}{v_A_B^2} =\frac{h_0}{h_f}$ 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: $m_A v_A = ( m_A + m_B ) v_{AB}$ But, now we only need to solve for $v_A / v_{AB}$. $\frac{v_A}{v_{AB}} = \frac{m_A + m_B}{m_A}$ Let's square this, to match the ratio of squares we had relating the two heights: $\left(\frac{v_A}{v_{AB}}\right)^2 = \left(\frac{m_A + m_B}{m_A}\right)^2$ Putting this last equation together with the previous ratio of velocities, we can find what we want: $\frac{v_A^2}{v_A_B^2} =\frac{h_0}{h_f} = \left(\frac{m_A + m_B}{m_A}\right)^2$ That makes $h_0$ 16 times as large as $h_{final}$. Okay, it looks harder, but I think it's much faster because the algebra is quicker and easier.Reply to this comment
archard
2010-08-04 14:05:45
It's super easy if you write the kinetic energy as $\frac{p^2}{2m}$ - since momentum is conserved, you don't even need to worry about whether or not the collision is elastic.

Just write

$\ M g h 1 = \frac{p^2}{2M}$
$\ 4M g h 2 = \frac{p^2}{8M}$

Divide equations and solve for h2.
 flyboy6212010-11-09 20:18:08 Nice!
 timtammy2011-10-11 00:57:49 far and away the best answer
 Shahbaz Ahmed Chughtai2012-06-09 22:45:14 Nice and Comprehensive Answer. :-)
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:

$m g h = 1/2 m v^2$
$v^2=2gh$

Right? So, now we know that
$v_A^2 = 2gh_0$ ,
where $v_a$ is the velocity of ball A and $h_0$ is it's initial height (where all of the E for this whole situation originates.)

And, we also know, from he same calculation, that
$v_{AB}^2 = 2gh_{final}$ ,
where $v_{AB}$ is the final velocity of both (stuck-together) balls and $h_{final}$ is their final height.

We want to know the relation ship between $h_0$ and $h_f$, so let's try this:

$\frac{v_A^2}{v_A_B^2} =\frac{h_0}{h_f}$

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:

$m_A v_A = ( m_A + m_B ) v_{AB}$

But, now we only need to solve for $v_A / v_{AB}$.

$\frac{v_A}{v_{AB}} = \frac{m_A + m_B}{m_A}$

Let's square this, to match the ratio of squares we had relating the two heights:

$\left(\frac{v_A}{v_{AB}}\right)^2 = \left(\frac{m_A + m_B}{m_A}\right)^2$

Putting this last equation together with the previous ratio of velocities, we can find what we want:

$\frac{v_A^2}{v_A_B^2} =\frac{h_0}{h_f} = \left(\frac{m_A + m_B}{m_A}\right)^2$

That makes $h_0$ 16 times as large as $h_{final}$. Okay, it looks harder, but I think it's much faster because the algebra is quicker and easier.
 nakib2010-04-02 11:27:15 Thanks for the neat expression. Memorizing this will save me a minute in the exam for sure [assuming I get a similar kind of problem].
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, $Mgh_0$, go?
 VanishingHitchwriter2006-12-01 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).
 a19grey22008-11-02 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.
 flyboy6212010-11-09 20:17:31 The energy lost in the collision typically goes into thermal energy, i.e. raising the temperature of the masses. Or it could go into vibrations or some other form of energy that does not contribute to translational motion.
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! :)
 dnvlgm2007-11-11 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

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