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Verbatim question for GR8677 #5
Mechanics}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.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
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.
Alternate Solution - Unverified
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.
Alternate Solution - Unverified
Comments
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.
flyboy621
2010-11-09 20:18:08
Nice!
timtammy
2011-10-11 00:57:49
far and away the best answer
Shahbaz Ahmed Chughtai
2012-06-09 22:45:14
Nice and Comprehensive Answer. :-)
Alternate Solution - Unverified
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.
nakib
2010-04-02 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].
Alternate Solution - Unverified
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.NEC
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?
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.
flyboy621
2010-11-09 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.
Answered Question!
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
NEC

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