# Find the Speed of Carts After Exploding Spring

• robin97531

## Homework Statement

A massless spring of spring constant 20 N/m is placed between two carts. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 3 kg. The carts are pushed toward one another until the spring is compressed a distance 1.3 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?

## Homework Equations

conservation of energy
conservation of momentum

## The Attempt at a Solution

I have worked on this a bit but can't get the right answer...
I have a conservation of energy equation: .5Kx^2 = .5m1v1^2 + .5m2v2^2
and I also have a conservation of momentum one: Pi = Pf, 0 = m1v1 + m2v2
I tried solving for v1 and plugging it into the first eqn to get a velocity, but I can't seem to get it right...any ideas? Thanks!

Your equations are fine and your approach should work. Show exactly what you've done and perhaps we can spot your error.

ok, so I have these two equations

.5Kx^2 = .5m1v1^2 + .5m2v2^2
Pi = Pf, 0 = m1v1 + m2v2

I solve the second one for v2 so I can plug it into the top one and solve for v1
solving the bottom one gets me: -m1v1-/m2 = v2; plug that into the top and you get: .5Kx^2 = .5m1v1^2 + .5m2(-m1v1-/m2)^2

.5(20)1.3 = .5(5)vi^2 + .5(3)(-5*v1/3)^2

I am most likely doing something wrong while totaling up the green section..I'm just multiplying/dividing everything togeter..so I would take the -5 and divide by 3 and square that as well as vi to get 2.778*v1 then multiplying by the .5(3) you get 4.167...is that anywhere near right?

but anyways, at the end my value for v1 is 1.396...and if I plus it back into the equations it seems to work..so not sure what I am doing wrong..

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.5(20)1.3 = .5(5)vi^2 + .5(3)(-5*v1/3)^2
Looks good (except for a few typos: the 1.3 should be squared; vi should be v1)
I am most likely doing something wrong while totaling up the green section..I'm just multiplying/dividing everything togeter..so I would take the -5 and divide by 3 and square that as well as vi to get 2.778*v1 then multiplying by the .5(3) you get 4.167...is that anywhere near right?
Looks good to me. (The v1 should be squared, of course.)

Calculation tip: Before reaching for the calculator, try canceling as much as possible. For example, the .5 drops out from all terms, and you can probably cancel a 5 as well.

Yay

Thanks a lot! I missed that square for the first part - feel dumb now for agonizing over that for so long and not being able to figure it out ^_^
Thanks so much!