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February 4th, 2016, 08:09 PM   #1
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please explain this momentum problem

A 40gram bullet is fired with a speed of 300m/s into a ballistic pendulum of mass 5kg suspended from a chord 1m long. Compute the vertical height through which the pendulum rises.

Solution:
m1v1+m2v2 = m1v1'+m2v2'

400(300)+5000(0)=40v1+5000v1

v1=2.38m/s

v2=0

v2^2=v1^2 + 2gS

S=h

0 = 2.38^2 - 2(9.81)(h)

h=0.289m

My Question: this is the solution of the book. I am just wondering why he used S =h instead of S = d? Isn't "S" suppose to be the actual distance that the object travelled?

Please clarify, thanks in advance.

Last edited by skipjack; February 5th, 2016 at 04:42 AM.
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February 4th, 2016, 08:12 PM   #2
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Please consider this diagram
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February 4th, 2016, 08:25 PM   #3
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That should be an equation describing conservation of energy (kinetic energy and potential energy), not the distance calculated from Newton's Third Law.
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February 4th, 2016, 08:28 PM   #4
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In the original post, the book is misquoted.
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February 5th, 2016, 02:13 AM   #5
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So the book is wrong? I knew it...
so what's the right way of solving for h?
I guess it's something like this:

v2^2=v1^2 + 2gS

S=d

0 = 2.38^2 - 2(9.81)(d)

d=0.289m.

After I find the length d, I will solve for h using trigonometry?
Thanks for reply... in advance..

Last edited by skipjack; February 5th, 2016 at 02:38 AM.
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February 5th, 2016, 02:37 AM   #6
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I'm not sure whether the book is wrong, but I doubt that what you posted here is precisely what the book gave, as the arithmetic doesn't work.
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February 5th, 2016, 02:43 AM   #7
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I don't think the book is wrong. They've just decided (slightly oddly) to have the vertical coordinate axis so that $s=0$ at $v=v_2$, the top of the swing.

Then
$\text{(kinetic energy at bottom)} + \text{(potential energy at bottom)} = \text{(kinetic energy at top)}+\text{(potential energy at top)}$
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Last edited by skipjack; February 5th, 2016 at 04:33 AM.
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February 5th, 2016, 04:38 AM   #8
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That doesn't explain why the initial arithmetic is incorrect.
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February 5th, 2016, 05:01 AM   #9
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Oh, yes, that additional zero on the 40 gram bullet. But the value for $v_1$ is correct.

The reuse of variable names is a bit confusing too.
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February 5th, 2016, 05:11 AM   #10
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Using $v_2$ as the speed of the bullet/block combination after the collision ...

$\frac{1}{2}(M+m)v_2^2=(M+m)gh$

$h=\dfrac{v_2^2}{2g} \approx 0.3 \text{ m}$

The book's solution is correct ... horizontal displacement is not considered since there is no change in gravitational potential energy in that dimension.
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Last edited by skipjack; February 6th, 2016 at 12:28 AM.
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