1. A roller coaster car starts from rest and rolls down a frictionless track, reaching speed fv at
the bottom.
a. If you want the car to go two times as fast at the bottom, by what factor must you
increase the height of the track?
b. Does your answer to part (a) depend on whether the track is straight or not? Explain.
2. You release a frictionless cart at the top of each of the 2 ramps. On Ramp B, the cart is
released from rest, but on Ramp A the cart is pushed at the top so that it has some initial
velocity. The ramps have the same height as each other at start and finish. One has a small
bump up then down just before the end. The other has a small bump down then up. Which cart has the larger speed at the finish? Why?
1. Twice the speed requires 4 times the kinetic energy. What does that tell you about the height or the top of the track?
2. Forget about the bumps. The car that starts out with kinetic energy (from the push) has more kinetic energy at the bottom.
1. To answer the first question, let's consider the conservation of energy principle. As the roller coaster car rolls down the track without friction, its total mechanical energy remains constant throughout the ride.
a. Let's assume that the initial height of the track is h and its final velocity is fv. According to the conservation of energy, the initial potential energy (mgh) is converted into kinetic energy (0.5mv^2) at the bottom of the track, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, we have:
Potential Energy at the top = Kinetic Energy at the bottom
mgh = 0.5mvf^2
Now, if we want the car to go two times as fast at the bottom, let's call the new final velocity 2fv. Therefore, we have:
mgh = 0.5m(2fv)^2
Simplifying this equation, the mass cancels out:
gh = 2f^2v^2
Next, let's consider the height of the new track, which we'll call h'. By rearranging the equation, we get:
h' = (2f^2v^2)/g
Therefore, if we want the car to go two times as fast at the bottom, we need to increase the height of the track by a factor of (2f^2v^2)/g.
b. Now, the answer to part (a) does not depend on whether the track is straight or not. The factor by which you need to increase the track's height is solely determined by the new desired final velocity (2fv), not the track's shape or curvature.
2. In this case, let's again consider the conservation of energy. The carts are released from the same height and are subject to the same gravitational potential energy.
Now, let's analyze the effect of the small bumps on the carts' speeds at the finish.
For Ramp B, where the cart goes down and then up, the potential energy is converted into kinetic energy as it goes down the ramp, and then again into potential energy as it goes up. Since the energy transformations are in the same order, the final speed at the finish will be the same as the initial velocity.
For Ramp A, where the cart is pushed initially and then goes up and down, the potential energy is first converted into kinetic energy as it goes up and then into potential energy as it goes down. In this case, the initial velocity adds to the potential energy, increasing its total energy at the top. This extra energy contributes to a higher final speed at the finish compared to Ramp B.
Therefore, the cart on Ramp A, which was pushed at the top, will have a larger speed at the finish compared to the cart on Ramp B.
1. a. To make the car go two times as fast at the bottom, you would need to increase the height of the track by a factor of four.
This can be explained using the conservation of energy. At the top of the track, the car has only potential energy (mgh), where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the track. At the bottom of the track, the car has both potential energy and kinetic energy (1/2mv^2), where v is its velocity.
Since the track is frictionless, the total mechanical energy (potential energy + kinetic energy) is conserved. Therefore, we can write:
Potential energy at the top = Potential energy at the bottom + Kinetic energy at the bottom
mgh = 1/2mvf^2
Simplifying the equation, we get:
gh = 1/2vf^2
Now, let's denote the new height of the track as h'. To make the car go two times as fast at the bottom, we want to find h' such that:
1/2vf^2 = 1/2(2vf)^2
Simplifying further, we get:
1/2vf^2 = 1/2(4vf^2)
vf^2 = 4vf^2
Dividing both sides by vf^2, we find that:
1 = 4
This is not true, so the equation cannot be satified. Therefore, there is no way to make the car go two times as fast at the bottom by just increasing the height of the track.
b. The answer to part (a) does not depend on whether the track is straight or not. The factor by which you need to increase the height of the track remains the same regardless of the track's shape.
2. The cart on Ramp A, pushed at the top with an initial velocity, will have a larger speed at the finish compared to the cart on Ramp B.
This can be explained by the conservation of energy again. Both ramps have the same height at the start and finish, so the gravitational potential energy at the start will be the same for both carts. However, since the cart on Ramp A has an initial velocity, it will have some initial kinetic energy.
As the cart moves along the ramp, the initial kinetic energy will be converted into gravitational potential energy, which increases the speed of the cart. On the other hand, the cart on Ramp B starts from rest, so it does not have any initial kinetic energy to convert into gravitational potential energy.
Therefore, the cart on Ramp A, with initial velocity, will have a larger speed at the finish compared to the cart on Ramp B.