# Determining speed and height on a roller coaster

1. Mar 27, 2015

### crushedcorn

1. The problem statement, all variables and given/known data
One car (m=80.0 kg) tracks through the roller coaster in the following diagram. As it passes point A, it has a speed of 50.0 cm/s.

a) Determine the speed at points B, C, and E
b) If the speed at point D is _______, determine it's height (ignore friction)

2. Relevant equations
PE1 + KE1=PE2 + KE2
PE=mgy
KE=1/2mv^2

3. The attempt at a solution
a) I correctly got the velocity for point B as 44.3 m/s (1/2mv^2=mgy, 1/2(80 kg)v^2=(80kg)(9.80m/s^2)(100m). However, when I apply the same formula to points C and E my answer doesn't match that of the answer key.

Point C: (80 kg)(9.80 m/s^2)(38 m)=1/2(80 kg)v^2, v=27.3 m/s. Answer key answer: 34.9 m/s

Point E: (80 kg)(9.80 m/s^s)(20 m)=1/2(80 kg)v^2, v=19.8 m/s. Answer key answer: 39.6 m/s

b) 1/2(80 kg)v^2=(80 kg)(9.80 m/s^2)y. I can simplify this equation but I don't know how to solve this as I would have 2 variables that I need to solve for.

Thank you for your wonderful brains on this!

2. Mar 27, 2015

### jbriggs444

You quoted the equation: PE1 + KE1 = PE2 + KE2
You used the equation: PE = KE

3. Mar 27, 2015

### Matternot

Don't get confused with ΔPE=ΔKE

4. Mar 27, 2015

### crushedcorn

PE1 + KE1=PE2 + KE2 becomes PE=KE because, for example, at point C KE1=0 and PE2=0, so I end up using the PE1=KE2 equation.

How am I confusing ΔPE with ΔKE?

5. Mar 27, 2015

### Matternot

KE1≠0 PE2≠0

You haven't used the fact that it's moving at 0.5ms-1 at A

You aren't confusing ΔKE with ΔPE. From your work it appears you are confusing ΔKE=ΔPE with KE=PE

Last edited: Mar 27, 2015
6. Mar 27, 2015

### Matternot

Total energy conserved:

mgy1+1/2 mv12=mgy2+1/2 mv22

For point B:
1/2(80kg)v2+(80kg)g0=(80kg)(9.80ms-2)(100m)+1/2(80kg)(0.5ms-1)2

The only reason you got the first part "right" is because 1/2(80kg)(0.5ms-1) is really small compared to (80kg)(9.80ms-2)(100m) and so it didn't make a noticable difference to the answer

7. Mar 27, 2015

### Matternot

You want to use:

PEA+KEA=(80kg)(9.80ms-2)(100m)+1/2(80kg)(0.5ms-1)2=PEx+KEx=mgy+1/2mv2

For all but the last parts, y is known

For the last part, v is known

Last edited: Mar 27, 2015
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