- #1

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I was figuring it out using hte conservatin of energy principle, but that isnt working:

1/2mu^2 + 1/2mu_2^2 + 1/2kx^2 = 1/2mv^2 + 1/2m_2v_2^2 + 1/2kx_2^2

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- Thread starter UCrazyBeautifulU
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- #1

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I was figuring it out using hte conservatin of energy principle, but that isnt working:

1/2mu^2 + 1/2mu_2^2 + 1/2kx^2 = 1/2mv^2 + 1/2m_2v_2^2 + 1/2kx_2^2

- #2

Doc Al

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- #3

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those are all my numbers....is that right?

- #4

Fermat

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Let both masses have the **same** velocity after impact.

Last edited:

- #5

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i end up with a negative answer.

- #6

Fermat

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Did you use v/2 = 0.78 m/s instead of 1.56 on the rhs of the eqn ?

- #7

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YOu end up with -0.105862 = 706.88(k)

- #8

Fermat

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.5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(0.348)(1.56^2) = 0.4234464

What do you get?

What do you get?

- #9

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.5k(0^2) = 0

so how do I get anything when I can't even solve for k?

so how do I get anything when I can't even solve for k?

- #10

Fermat

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The rhs of your eqn = 0.217232 + 1.7672E-04 * k

Solve for k.

Check your arithmetic.

- #11

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0.423 = 0.529308 + 1.7672 x 10^-4

Which gives a negative answer.

Using what you wrote above I get k = 1166.9 and that answer was not correct.

- #12

Fermat

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Which gives k = 1.2 KN/m

You had 0.529308 on the rhs. You will get that if you used v = 1.56 instead of v = 0.78.

Both masses have the same speed, 0.78 m/s.

- #13

Doc Al

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No. As Fermat has already pointed out, both masses move with the same speed (0.780 m/s) when the spring is fully compressed.UCrazyBeautifulU said:.5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(.348)(1.56^2) + .5(.348)(v/2)^2 + .5k(1.88 x 10^-2)^2

those are all my numbers....is that right?

- #14

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Thank you!

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