Presentation problem. o6: Conservation of Energy Ground rules: 1. One designated

Question

Presentation problem. o6: Conservation of Energy Ground rules: 1. One designated person (the "presenter) within the group will be responsible for writing up (with the group's help) a self-contained, succinct outline of the solution method and the problem's solution 2. Large (I1 17, ledger) sheets will be provided, make sure your lettering/diagrams are large enough and clear enough that others can follow your solution at more than an arm's length away. Be sure your answer includes appropriate units, and that vector quantities include magnitude and direction (or components). A sketch or diagram will almost always be required, sometimes a graph may make things clearer 3. Presentation write-ups are due at the end of the lab session during which they were assigned 4. Each solution write-up will be scored (from zero to ten, using the laboratory grading scale). Each member of the group receives this score (so it is in each contributor's interest to assist the presenter), and the presenter receives twice this score (so it is in the presenter's interest to be as clear and careful as possible). Presenters will be rotated within the assigned lab groups so that each student has the opportunity to serve as presenter at least three times 5. These scores will also be added, and divided by the total number of available presentation points (typically 120), that fraction will determine 20 points of the overall lab grade You are investigating a mishap on a mountain railroad. One railcar was left parked at the top of a hill above another car of the same mass (3300 kg), as shown in the figure. In the morning, both Group E cars are found in the lake. (The tracks go right up to the lake to facilitate loading ore boats.) The supervisor claims that the engineer must not have set the brake on the upper car, it rolled down, hit, and coupled with the lower car, and their momentum carried them into the lake. The engineer disputes this, saying that someone must have released the brake and then pushed the upper car. Can you resolve the dispute? 24 m lake 7 m 50 m

Explanation / Answer

This is a problem where we’ll have to break up the motion into different time intervals. In some of these intervals momentum will be conserved and in some energy will be conserved. The key for momentum conservation will be whether or not the net force on the system is zero. For energy we’ll have to check whether or not there are any non-conservative forces (in other words, mechanical energy converted to other forms).

There are couple ways we can approach this problem (try to answer the question of the engineer’s guilt). The way I’ll go about it in this model is to assume that the first car started at rest (nobody pushed it), and then see if the two cars make it to the top of the second hill. If they do, I’ll assume that they then make it to the lake. If the model shows that the cars do not make it to the top of the second hill, then the engineer is most likely not at fault. In other words, this would show that the only way the two cars could make it to the lake is if they had more energy. This would most likely be due to a push at the top of the first hill.

I would guess that this is going to be a close call- the cars might just make it over the hill. Certainly the two cars won’t go as high as where the first one stared, but it seems possible that they could make it up 7m. (In fact, one might even think that their final height is just 1/2 the original. The 1/2 guess is due to the doubling of the mass.)

For the time between when the first car starts at the first hill, until just before it collides with the second car, let’s assume:

mgyi = 0.5mv^2

with yi = 24m

We can solve for the speed of the car, just before the collision

v = sqrt(2*yi*g)

Now that we know the speeds of the two car before the collision, we can analyze the collision.

Let’s assume:

• The second car is at rest before the collision.

• Momentum is conserved due to a lack of any net external force. This essentially means we’re taking the time of the collision to be small and/ or the ground to be flat. (Also, we are still assuming that the friction is negligible.)

• (Mechanical energy is not conserved as this is a perfectly inelastic collision.)

Equating the momentum before and after the collision:

mv = 2mvf

If we substitute in the speed from the first portion, we can then write the speed after the crash (vf ) in terms of the first hill height.

vf = 0.5*sqrt(2*yi*g)

Now, we return to energy conservation. This last portion of the run is just like the first, just in reverse (converting kinetic into potential energy). We can know answer the question of how high the cars go up the second hill.

0.5*(2m)*vf2 = (2m)gh

h will give us the vertical height that the two cars reach

h = (1/2g)*[0.5*sqrt(2*yi*g)]^2 = (1/2g)*(2*yi*g/4) = yi/4

Since yi = 24m, h= 6m.

It looks like the cars do not make it up the second hill with our model. The only way for the cars to reach the lake would to add energy to the system, most likely with an initial push. So, it appears as though the engineer isn’t guilty. (If we would have added friction, the cars wouldn’t have even reached 6m, so it seems pretty safe to say that the engineer isn’t guilty.)

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