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- anonymous

An air track glider of mass 0.200 kg, moving at 1.0 m/s collides elastically with another glider of mass 0.050 kg equipped with a perfectly elastic spring, which is initially at rest. How much potential energy is stored in the bumper between the two gliders \(during\) the collision? (i.e. Ek loss during the collision at minimum separation of the two gliders).
I just need an explanation on how to do this
Answer is 0.02J

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- anonymous

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- theEric

I really don't know, sorry!
I know that momentum is always conserved.
The total energy is also conserved in an elastic collision.
I think, in this problem, we assume that the first care is moving until it slows down to its post-collision velocity. And the other cart does not move all this time, until the all the energy the first cart lost becomes potential that will then cause the other cart to move.
So we want to know what KE the first cart loses as it gives it all to the potential in the spring.

- theEric

I think I can get it using an equation from Wikipedia.

- theEric

\(\large v_{1} = \dfrac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2}}{m_{1}+m_{2}}\)
Have you seen this before?

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- theEric

- anonymous

sorry , i was afk
i've seen a formula like:
\(V_{f1}=(\frac{m_1-m_2}{m_1+m_2})V_{i1}\) and \( V_{f2}=(\frac{2m_1v_1}{m_1+m_2})\) but not like that..

- anonymous

Is it right if I say
\(E_{ki\ of\ mass\ 1} + E_{ki\ of\ mass\ 2} = E_{kf\ of\ mass\ 1} + E_{kf\ of\ mass\ 2} + E_{potential}\)
\(\frac{1}{2}m_1v_{i1}^2+0=\frac{1}{2}m_1v_{f1}^2+\frac{1}{2}m_2v_{f2}^2 + E_{potential} \)
\( E_{potential}=\frac{1}{2}m_1v_{i1}^2-\frac{1}{2}m_1v_{f1}^2+\frac{1}{2}m_2v_{f2}^2\)
\(E_{potential}=\frac{1}{2}(0.200)(1.0)^2-\frac{1}{2}(0.200)(0.6)^2+\frac{1}{2}(0.05)(1.6)^2\)
oh i forgot to tell you that I already calculated the Final velocities of the gliders
I got \(v_{f1}=0.6 m/s, v_{f2}=1.6m/s\)

- theEric

The \(u\)'s are for before the collision, and the \(v\) is for after.
But this formula didn't work for me.
So if we say the moving cart is cart 1, then we want \(v_1\). When we have the velocity before and after, we can find the change in kinetic energy of cart 1.
\(u_2=0\), which is nice.
Then\[v_{1} = \frac{u_{1}(m_{1}-m_{2})}{m_{1}+m_{2}}\\= \frac{1.0~~(0.200-0.050)}{0.200+0.050}\\= \frac{1.0~~(0.150)}{0.250}\\= \frac{0.150}{0.250}\\= \frac{15}{25}\\=\frac35=v_i\]
\(\Delta(KE)=\frac12m_1v_1^2-\frac12m_1u_1^2\\=\dfrac{m_1}2(v_1^2-u_1^2)\\=\dfrac{0.200}{2}\left((\frac35)^2-1^2\right)\\
=\dfrac{2}{20}\left(\dfrac9{25}-1\right)\\
=\dfrac{1}{10}\left(\dfrac9{25}-\dfrac{25}{25}\right)\\
=\dfrac{1}{10}\left(\dfrac{-16}{25}\right)\\
=\dfrac{-16}{250}\\
=\dfrac{-8}{125}\\=-0.064\neq-0.02\)

- anonymous

for the las line it should be:
\(E_{potential}=\frac{1}{2}(0.200)(1.0)^2-\frac{1}{2}(0.200)(0.6)^2-\frac{1}{2}(0.05)(1.6)^2\)
but still, i got the wrong answer :/

- theEric

When the other cart's velocity is 0, your first equation is the same as Wikipedia's.

- anonymous

oh i solved the same way as yours before and got same results, so i tried a different one..
and really? hmm.. lemme check wikipedia then

- theEric

That's weird! Maybe we're perceiving the problem incorrectly? It might be something to ask your teacher about?

- anonymous

alright , i'll try to ask him tomorrow :)

- anonymous

I got it! I asked my teacher and he said that DURING the collision the velocity of the two objects will be the same and I should only consider the total kinetic energy of the object BEFORE and DURING the collision.. so yeah, that makes sense to me .. I tried solving it and I already got 0.02 J for potential energy ^_^

- theEric

Oh, cool! Thanks :)

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