How Do You Calculate Roller Coaster Speeds in Physics Problems?

[mickeymouseho]

Homework Statement

Figure 4 shows a partial profile of a roller-coaster ride. Friction and air resistance are negligible. Determine the speed of the coaster at position C if the speed at the top of the highest hill is (a) zero and (b) 5.00m/sFigure 4 (description)

Point A is the highest point of the rollercoaster. It is 37.8m from the ground. Point B is the lowest point of the rollercoaster. It is 12.8m from the ground. Point C is the mid point of the roller-coaster. It is 17.8m from the ground.

*There are no loops
*From A, it curves downwards to B, then upwards to C

I apologize for not having a picture of this.

Homework Equations

E = 0.5mv^2
E = mgh

(sorry, not sure)

The Attempt at a Solution

I have no idea how to start this out...

E = mgh
= m x 9.8m/s^2 x 37.8m *keeping mass as a variable since mass is not given
= 370.44(mass)J

E = 0.5mv^2
370.44(mass)J = 0.5mv^2 *masses cancel out
square root of (370.44J / 0.5) = v
27.22m/s = v

The answer from the textbook: (a) 19.8m/s (b) 20.4m/s

I don't even know how to start out this problem. Help please. Thanks in advance!

EDIT:

for (a) h = 37.8m - 17.8m = 20m

Eg = mgh
= mass x 9.8m/s^2 x 20m
= 196 (mass) J

Ek = 0.5mv^2
196 (mass) J = 0.5mass x v^2 *masses cancel out
Square root: (196J/0.5) = v <- forgot to divide the gravitational potential energy by 0.5
19.8m/s = v

Got the answer for (a) but don't know how to figure out (b)...

 

[mickeymouseho]

bump =)

still trying to figure out (b)

 

[PhanthomJay]

Homework Statement

Figure 4 shows a partial profile of a roller-coaster ride. Friction and air resistance are negligible. Determine the speed of the coaster at position C if the speed at the top of the highest hill is (a) zero and (b) 5.00m/s


Figure 4 (description)

Point A is the highest point of the rollercoaster. It is 37.8m from the ground. Point B is the lowest point of the rollercoaster. It is 12.8m from the ground. Point C is the mid point of the roller-coaster. It is 17.8m from the ground.

*There are no loops
*From A, it curves downwards to B, then upwards to C

I apologize for not having a picture of this.


for (a) h = 37.8m - 17.8m = 20m

Eg = mgh
= mass x 9.8m/s^2 x 20m
= 196 (mass) J

Ek = 0.5mv^2
196 (mass) J = 0.5mass x v^2 *masses cancel out
Square root: (196J/0.5) = v 19.8m/s = v

Got the answer for (a) but don't know how to figure out (b)...

For part a, you set the initial energy (all potential) equal to the final energy (all kinetic). For b, the initial energy is part kinetic and part potential. Calculate it, and set it equal to the final energy to solve for v.

 

[mickeymouseho]

For part a, you set the initial energy (all potential) equal to the final energy (all kinetic). For b, the initial energy is part kinetic and part potential. Calculate it, and set it equal to the final energy to solve for v.

Ah thank you very much good sir =)

(a) Eg = Ek

(b) Eg + Ek = Ek'

 

[happyknife]

Ah thank you very much good sir =)

(a) Eg = Ek

(b) Eg + Ek = Ek'

How did you solve for part B using Eg + Ek = Ek'?