energy question?

Question

A roller-coaster car comes to a near-stop at the top of a hill 40 m in height. It then goes roaring down the hill, and rebounds up a shorter hill 16 m in height. If friction's effects can be ignored, at the top of the 16-m hill its speed will be ___.

I know the answer is about 21 m/s, but I'd appreciate it if someone could explain how to arrive at that answer. Thanks.

Answer 1

Use conservation of energy. At the bottom of the hill, the kinetic energy will be 40 m g; at the top of the small hill, 16 m g of this will have been converted to potential energy. The rest will still be kinetic energy, from which the speed can be calculated.

Answer 2

You apply the concept of energy conservation. The idea is that the total energy of the system never changes -- what exists when the coaster is barely moving at the top of the 40m hill is the same amount the system has when the coaster is moving over the top of the 16m hill.

This problem is nice in that there are only two things that contribute to the total energy; the kinetic energy of the coaster and the gravitational potential energy. At the top of the 40m hill there's negligible kinetic energy and a potential energy of mg*40m, where the first m is the mass of the coaster.

At the top of the 16m hill the energy is part kinetic (1/2mv^2) and part potential (mg(16m)). The two totals must match:

mg(40m) = mg(16m) +1/2mv^2

This equation can be solved for v once you cancel out all the m's (don't get confused by the 'm' that stands for meters).

Answer 3

Assume that the roller coasting is just barely moving at the top of the first hill so it has almost no kinetic energy to speak of. It has a potential energy of

PE1 = mgh1 = E where h1 = height of the first hill and E = total energy (Kinetic plus potential)

Now energy is conserved - there's no friction - so as teh roller coaster goes downward, some of teh potential energy is converted to kinetic energy. At the next hill, the roller coaster has an energy of:

1/2mv^2 + mgh2 where v = speed and h2 = height of hill 2.

But since energy is conserved, the above expression must equal E:

E = mgh1 = 1/2mv^2 + mgh2

g(h1-h2) v^2/2 ---> v =sqrt(2*g*(h1-h2)) = sqrt(2*9.8*(40 -16))

v = 21.7 m/s