Air-Track Carts & Spring (Energy)

[sunnnystrong]

Homework Statement

The air-track carts in the figure(Figure 1) are sliding to the right at 2.0 m/s. The spring between them has a spring constant of 140 N/m and is compressed 4.1 cm. The carts slide past a flame that burns through the string holding them together.

What is the final speed of 100-g cart?* I don't actually have a picture of this problem however it is a picture of two carts on an air track separated by a spring.
the left cart has a mass of 100g & the right cart has a mass of 400g.

Homework Equations

Potential Energy of Spring = .5*k*x^2

Conservation of Energy:
.5(m1)(v1i)^2 + .5(m2)(v2i)^2 + .5*k*x^2 = .5(m1)(v1f)^2 + .5(m2)(v2f)^2

Conservation of Momentum:
(m1)(v1i) + (m2)(v2i) = (m1)(v1f) + (m2)(v2f)

The Attempt at a Solution

[/B]
m1 = 0.1kg
v1i = v2i = 2m/s
m2 = 0.4 kg
v2f = ?
v1f = ?
x = 0.041 m

First use the conservation of energy:
.5(m1)(v1i)^2 + .5(m2)(v2i)^2 + .5*k*x^2 = .5(m1)(v1f)^2 + .5(m2)(v2f)^2
1.11767 = .05(v1f)^2 + .2(v2f)^2
Divide by .05
I got this
22.35 = (v1f)^2 + 4(v2f)^2

Next use the conservation of momentum:
(m1)(v1i) + (m2)(v2i) = (m1)(v1f) + (m2)(v2f)
1 = .1(v1f) + .4 (v2f)
Divide by .1
10 = v1f + 4(v2f)
I got this
10-4(v2f) = v1f

Next square v1f to plug into energy equation
(v1f)^2 = 16(v2f)^2 - 80(v2f) + 100

Last step :
22.35 = (v1f)^2 + 4(v2f)^2

22.35 = 16(v2f)^2 - 80(v2f) + 100 + 4(v2f)^2
22.35 = 20(v2f)^2 - 80(v2f) + 100
Divide by 10
2.235 = 2(v2f)^2 - 8(v2f) + 10
0 = 2(v2f)^2 - 8(v2f) + 7.765
Solve for v2f and i got 1.65 m/s.

Plug that into this
10-4(v2f) = v1f
v1f = 3.4 m/s which is incorrect but i don't why?

The answer is 0.628 m/s and I'd like to understand how to do this problem

 

[TomHart]

Maybe I am misunderstanding the problem. Let's say the 2 carts are moving east at 2 m/s. When the string is burned, does the spring push one of the carts toward the north and one toward the south? Or does the spring push one of the carts toward the east and one toward the west?

 

[sunnnystrong]

Maybe I am misunderstanding the problem. Let's say the 2 carts are moving east at 2 m/s. When the string is burned, does the spring push one of the carts toward the north and one toward the south? Or does the spring push one of the carts toward the east and one toward the west?

Oh okay that's probably where my error is... there is a string holding the two carts together where the compressed spring is in the middle. once it passes the flame on the air track the spring will exert a force on the two carts.

 

[sunnnystrong]

Maybe I am misunderstanding the problem. Let's say the 2 carts are moving east at 2 m/s. When the string is burned, does the spring push one of the carts toward the north and one toward the south? Or does the spring push one of the carts toward the east and one toward the west?

But i still am entirely unsure of how i would use that information to solve this problem haha

 

[TomHart]

Both carts are moving in the x direction initially. After the spring is released, are they both still moving only in the x direction, or does the spring cause them to angle off - one towards the north and one towards the south?

Edit: I'm trying to understand where the spring is positioned between the carts. Is it between their sides, or are the cars end-to-end and the spring is touching one car's front bumper and the other car's rear bumper?

 

[sunnnystrong]

Both carts are moving in the x direction initially. After the spring is released, are they both still moving only in the x direction, or does the spring cause them to angle off - one towards the north and one towards the south?

Edit: I'm trying to understand where the spring is positioned between the carts. Is it between their sides, or are the cars end-to-end and the spring is touching one car's front bumper and the other car's rear bumper?

http://tinypic.com/r/124koqc/9

Here is a little sketch

 

[sunnnystrong]

Both carts are moving in the x direction initially. After the spring is released, are they both still moving only in the x direction, or does the spring cause them to angle off - one towards the north and one towards the south?

Edit: I'm trying to understand where the spring is positioned between the carts. Is it between their sides, or are the cars end-to-end and the spring is touching one car's front bumper and the other car's rear bumper?

 

[TomHart]

That's what I figured. Otherwise the answer could not have been 0.628 m/s. Which car is in front the heavy one or light one?

 

[sunnnystrong]

That's what I figured. Otherwise the answer could not have been 0.628 m/s. Which car is in front the heavy one or light one?

The lighter one.

 

[sunnnystrong]

Jk sorry! the heavier one is in the front.

 

[sunnnystrong]

That's what I figured. Otherwise the answer could not have been 0.628 m/s. Which car is in front the heavy one or light one?

sorry ah my brain is fried. the 100 g cart is behind the 400 g cart

 

[TomHart]

Solve for v2f and i got 1.65 m/s.

That's what I got too!
But, when you solve a quadratic there are two solutions. It's the other one that is the right answer.

Edit: It turns out that one of the solutions is for the heavy cart in front and one for the heavy cart in back. At least I think that's the case. I didn't take time to verify it.

 

[sunnnystrong]

That's what I got too!
But, when you solve a quadratic there are two solutions. It's the other one that is the right answer.

Edit: It turns out that one of the solutions is for the heavy cart in front and one for the heavy cart in back. At least I think that's the case. I didn't take time to verify it.

oh interesting so v2f = 2.34m/s?
Than oh yeah you get 0.628 m/s!
cool haha thank you again

How do you know which solution it is? to the quadratic

 

[sunnnystrong]

well that actually makes sense that the velocity would be larger than two as the spring will cause it to accelerate when the string burns.

 

[TomHart]

How do you know which solution it is? to the quadratic

The velocity for the front car HAS TO increase and the velocity of the rear car HAS TO decrease. You defined Vf1 as the 100 g car in the back, so you know VF1 will have end up being less than 2.0 m/s.

 

[sunnnystrong]

The velocity for the front car HAS TO increase and the velocity of the rear car HAS TO decrease. You defined Vf1 as the 100 g car in the back, so you know VF1 will have end up being less than 2.0 m/s.

thanks for your help haha that was a silly mistake

 

[haruspex]

thanks for your help haha that was a silly mistake

It often happens that the equations do not encode all the information, so other solutions may appear. In the present case, the conservation laws produce the same equations regardless of the order of the carts.

 

[sunnnystrong]

It often happens that the equations do not encode all the information, so other solutions may appear. In the present case, the conservation laws produce the same equations regardless of the order of the carts.

interesting thank you haha