# Spring Constant air-track carts

1. Apr 11, 2008

### 1831

Two air-track carts, one on the left with mass 100g and one on the left with mass 300g) are sliding to the right at 1.0 m/s. There is a spring between them that has a spring constant of 100N/m and is compressed 4.2 cm. There is a string that holds the two carts together. The carts slide past a flame that burns through the string holding them together.

(After the flame burns the string)...
Q1: What is the speed of 100g cart?
Q2: What is the direction of the motion of 100g cart? I got to the left
Q3:What is the speed of 300g cart?
Q4:What is the direction of the motion of 300g cart? I got to the right

Here is how I started off for Q1 and Q3:
F=-k$$\Delta$$s
F=(100)(.042) = 4.2 N

@Q1 I drew my force diagram, so the force by the spring on the 100g block points in the negative direction, so I got:
$$\sum$$Fx = -Fspring = ma
-4.2N = (.1kg)(acceleration)
acceleration = -42m/s^2

Then, I used kinematics to solve for final velocity (vf)

Vf^2 = Vi^2 + 2a$$\Delta$$s

since the spring is compressed 4.2cm, then the distance for it to return to equilibrium is .021m on each side...so

Vf^2 = 1^2 +2(-42)(.021)
vf^2=-.764

Now I'm stuck, because I don't want to take the square root of a negative...which makes me think I've done something wrong.

I will follow the same procedure for Q3...once I figure out what I'm doing wrong in Q1.

Thanks!

2. Apr 11, 2008

### Staff: Mentor

That's just the force for the first instant. The force--and resulting acceleration--is not constant.

Hint: Use conservation laws and you can ignore the details of the interaction as the spring pushes the cars.

(Assume that the spring between the cars is not attached to the cars. Or at least not to both cars.)

3. Apr 11, 2008

### 1831

So I guess I should use conservation of momentum, with an initial velocity of 0?

I'm a little confused if I don't need to find the acceleration...

4. Apr 12, 2008

### Staff: Mentor

Yes, but you'll need another conservation law as well.
Why zero? You're given the initial speed.

The acceleration is only relevant while the spring is expanding. Once the carts separate they no longer accelerate. Using conservation laws you can figure out the final speeds without worrying about the (hard to calculate) details of the forces involved as the carts separate.

5. Apr 13, 2008

### 1831

ok...so I've made a little progress, I think.

I am using conservation of momentum and conservation of energy.

m1=.1kg
m2=.3kg
v0=1m/s for both carts

solving for v1f and v2f

using conservation of energy:
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1+m2)(v0)^2 +(1/2)(K)(x)^2 ... (where K is the Force applied, and x is the change in distance ... or amount compressed)
and conservation of momentum:
m1v1+m2v2 = (m1+m2)v0

from the first equation, plugging in numbers i get:

.5(.1)(v1f^2) + (1/2)(.3)(v2f^2) = (1/2)(.1+.3)*1^2+(.5)(100)(.042^2)

simplifying:

.05v1f^2 + .15v2f^2 = (1/2)(.4)*1^2 + .0882
.05v1f^2 + .15v2f^2 = .2882 J
.1v1f + .3v2f = .400, so v1f = 4 - 3v2f

I plugged that into the first equation and got:
v2f = 1.38 and so v1f = -.14m/s...

however, this answer is not right. what am I still doing wrong?

6. Apr 13, 2008

### Staff: Mentor

Looks OK to me. (I get a slightly different answer for v1f, so you might want to recheck your arithmetic in your last step.)

7. Apr 13, 2008

### 1831

I don't get any other answer but -.14 for V1f, can you tell/show me where I went wrong...I don't see how you can get a slightly different answer.

8. Apr 14, 2008

### Staff: Mentor

I don't think you went wrong, just don't round anything off until the end. When I combine those two final equations I get v1f = -.15 m/s.