Physics Examkrackers In-Class Exam Lecture 3

[SKaminski]

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Hey,

I have several questions and will attempt to explain each question as thoroughly as possible.

Question 1:
This regards "Puncho" the clown, a clown who rides a unicycle across a tight rope.

51. Why is it easier for Puncho to balance on his unicycle if he carries a long heavy pole centered horizontally at his chest?
A. The pole decreases his rotational inertia.
B. The pole increases his rotational inertia.
C. The weight of the pole increases the frictional force between the unicycle and the tight rope.
D. The weight of the pole increases his momentum when he isn't moving.

Credited Answer: B.
Reasoning: Inertia is your resistance to change in motion.

My answer: C.
My reasoning. Sure, inertia is your resistance to change in motion. But the answer asks specifically about ROTATIONAL inertia. I'm assuming (as i think anybody would before reading the answers) that this relates to the inertia in the tires. The inertia that would keep him balanced would be his HORIZONTAL inertia. Thus, B is invalid. Increased friction will stop him slipping however.

 

[SKaminski]

Q 65
A WWI pilot increases his altitude at an angle of 30 degrees to the horizontal and a velocity of 20 m/s. If he takes 2 kg bomb with him, starting from rest on the ground, how much work has been done on the bomb when the planes reaches an altitude of 200 m? (Note: ignore air resistance)
A. 2000 J
B. 2400 J
C. 4000 J
D. 4400 J

Credited Answer: 4400 J
Reasoning: Don't use vectors to solve this problem. The work done is equal to the change in potential and kinetic energy. W = (change) K. E. + (change) P. E. W = 1/2 mv^2 + mgh = 1/2 X 2 X 400 + 2 X 10 X 200 = 400 + 4000 = 4400

My answer: C: 4000 J.
Reasoning: Well, we all know W = F * D. We also know that even a perfect machine does not change the work, just the force. Because the decreased force is going to counter act the increased distance precicely, we should just be able to calculated this as W = F * d.
W = (2)(10) * 200 = 4000 J.

It should be safe to assume that this is a perfect machine, as we do not have any air resistance. If not, can somebody tell me why its not safe to assume this?

 

[SKaminski]

Q 67.
A rocket is launched from earth to explore our solar system and beyond. As the rocket moves out of the earth's atmosphere and into deep space, the gravitiational constant g decreaes and approaches zero, and the gravitational potential energy of the rocket:
A. also decreases and approaches zero.
B. continually increases.
C. remains constant.
D. increases at first and then decreases and approaches zero.

Credited Answer: B.
Logic: Energy is required to seperate attracting bodies. The rocket is attracted to earth by gravity. Gravity is a conservative force so the added energy goes into potential energy.

My answer: A.
Logic: Examkrackers, dude. If the gravitational constant approaches zero (which it should, since F = (GM1M2)/(r^2), that radius is the most important factor, since its SQUARED, as compared to potentialenergy=mgh. The height their isn't squared.

Of course, bringing those two equations equation F = g. As g decreases (EXPONENTIALLY) it doesn't matter that the h increases (PROPORTIONALLY) because the gravity is becoming significantly less.

I thought this was how we were suppose to think for the mcats? Apply different formula together?

 

[BryanNextStep]

But the answer asks specifically about ROTATIONAL inertia. I'm assuming (as i think anybody would before reading the answers) that this relates to the inertia in the tires. The inertia that would keep him balanced would be his HORIZONTAL inertia. Thus, B is invalid. Increased friction will stop him slipping however.

Heya! I've used the EK books on and off with my tutoring students for years, and there are several of the "Puncho" questions that people get tripped up on all the time.

In this question, they're expecting you to think of the clown, the unicycle, and the bar as all one "system". You don't wanna make it super-complicated for yourself by starting to worry about, say, the friction between Puncho's butt and the seat of the unicycle, or how well his feet are locked into the pedals or whatever.

Just mentally lump the unicycle, the clown, and the bar together as a single system.

So your comment about "inertia in the tires" isn't what the writer was going for here. They were trying to ask about the inertia of the whole system. So if Puncho tips over to the left, then the whole system is rotating to the side. We love dear little Puncho and want his trick to work, so to keep him from tipping over to the side, we give him a big massive pole to increase inertia. That makes it harder for him to tip to the side - any little wobbles won't have as much of an effect since he's got more inertia.

Questions like this can be really frustrating, because near as I can tell the reason you got it wrong is not b/c you don't understand the basic science involved, but just because you mis-read the idea behind the question. Remember the questions on the MCAT are usually testing a pretty simple science concept in a relatively straightforward way. Don't overthink it! 🙂

 

[BryanNextStep]

Q 65
My answer: C: 4000 J.
Reasoning: Well, we all know W = F * D. We also know that even a perfect machine does not change the work, just the force. Because the decreased force is going to counter act the increased distance precicely, we should just be able to calculated this as W = F * d.
W = (2)(10) * 200 = 4000 J.

Looks like you've used the work formula to solve for the PE that the bomb has gained.

The thing is, the problem also says the bomb starts from rest and then is going 200m/s. So you've gotta account for that increase in energy as well.

Usually with stuff like this, before even starting to plug'n'chug equations, I like to just think through conceptually, "Where are the Joules?". Just think about what energy we started with, what energy we ended with, and where any energy came from or was lost. Once the concepts are clear, we can start futzing around with equations to figure stuff out.

In this case, you started from rest and then started moving. So you gained some kinetic joules. And you started on the ground and then ended in the air. So you gained some PE joules. Just add up those two Joules and you've got your answer.

Hope this helps 🙂

 

[BryanNextStep]

Logic: Examkrackers, dude. If the gravitational constant approaches zero (which it should, since F = (GM1M2)/(r^2), that radius is the most important factor, since its SQUARED, as compared to potentialenergy=mgh. The height their isn't squared.

Of course, bringing those two equations equation F = g. As g decreases (EXPONENTIALLY) it doesn't matter that the h increases (PROPORTIONALLY) because the gravity is becoming significantly less.

I thought this was how we were suppose to think for the mcats? Apply different formula together?

Kaminski, dude.

You've gotta realize something's a bit off when your logic ends with "F = g"

The equation is F = mg

😛

Okay, sarcasm aside, here's the mistake:

You're trying to equate mgh to Gmm/r^2

The problem with this is that we basically only ever use "mgh" when we're at or near the surface of the earth, because we're assuming g is a constant 9.8

When you're thinking about gravitational PE between bodies that are really far apart, you've gotta use the equation:

PE = - GMm / r

It's easy to mess this up, since it looks almost exactly like the force equation, except the denominator is just r here. Plus, by convention we put a minus sign out front of the equation.

When something is really, really far away from earth, it doesn't feel much gravitation force at all. That's true. But it does feel some tiny smidge of gravity.

If we were to eliminate all other matter in the universe and just think of the earth and some object really really really far away, eventually that object would fall back to earth. It would take forever, but eventually it would hit the earth. Because it would be falling for a really really long time from really really really far away, it would release a tremendous amount of energy over the course of the fall.

Mathematically, as r approaches infinity in the PE equation, the energy asymptotically approaches a max.

 

[Laureatebarrel]

Hey,

I have several questions and will attempt to explain each question as thoroughly as possible.

Question 1:
This regards "Puncho" the clown, a clown who rides a unicycle across a tight rope.

51. Why is it easier for Puncho to balance on his unicycle if he carries a long heavy pole centered horizontally at his chest?
A. The pole decreases his rotational inertia.
B. The pole increases his rotational inertia.
C. The weight of the pole increases the frictional force between the unicycle and the tight rope.
D. The weight of the pole increases his momentum when he isn't moving.

Credited Answer: B.
Reasoning: Inertia is your resistance to change in motion.

My answer: C.
My reasoning. Sure, inertia is your resistance to change in motion. But the answer asks specifically about ROTATIONAL inertia. I'm assuming (as i think anybody would before reading the answers) that this relates to the inertia in the tires. The inertia that would keep him balanced would be his HORIZONTAL inertia. Thus, B is invalid. Increased friction will stop him slipping however.

There is rotational inertia on the pole, its fulcrum is the hand of the clown. And ad for C it is wrong. Increased friction will not make it easier to balance if not to say harder. Remember the question is asking why a long pole helps him to stay balance. Friction of the wheel has nothing to do with this

 

[Laureatebarrel]

Q 65
A WWI pilot increases his altitude at an angle of 30 degrees to the horizontal and a velocity of 20 m/s. If he takes 2 kg bomb with him, starting from rest on the ground, how much work has been done on the bomb when the planes reaches an altitude of 200 m? (Note: ignore air resistance)
A. 2000 J
B. 2400 J
C. 4000 J
D. 4400 J

Credited Answer: 4400 J
Reasoning: Don't use vectors to solve this problem. The work done is equal to the change in potential and kinetic energy. W = (change) K. E. + (change) P. E. W = 1/2 mv^2 + mgh = 1/2 X 2 X 400 + 2 X 10 X 200 = 400 + 4000 = 4400

My answer: C: 4000 J.
Reasoning: Well, we all know W = F * D. We also know that even a perfect machine does not change the work, just the force. Because the decreased force is going to counter act the increased distance precicely, we should just be able to calculated this as W = F * d.
W = (2)(10) * 200 = 4000 J.

It should be safe to assume that this is a perfect machine, as we do not have any air resistance. If not, can somebody tell me why its not safe to assume this?

This is not perfect machine, but conservation of energy. If you use w = f.d it is very hard to find the force. The total force acting on the bomb is not g. This problem you have to use conservation of energy. Since the bomb starts from rest, it will gain KE = 1/2 mv^2 and it also gains PE = mgh.