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!

 

[Doc Al]

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

 

[robin97531]

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..

 

[Doc Al]

.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.

 

[robin97531]

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!