- #1
sunnnystrong
- 54
- 6
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