Physics question thread

[Shrike]

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All users may post questions about MCAT and OAT physics here. We will answer the questions as soon as we reasonably can. If you would like to know what physics topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm), though be warned, there are subjects listed there that are rarely tested, or that appear in passages only and need not be learned.

Be sure to check the Physics FAQs and Topic Writeups thread if you have a general question; eventually, many of your answers will be located there. Also, a request: to keep this thread at least somewhat neat, when replying to someone else's post please refrain from quoting anything more than what's necessary for clarity.

Acceptable topics:

• general, MCAT-level physics
• particular MCAT-level physics problems, whether your own or from study material
• what you need to know about physics for the MCAT
• how best to approach to MCAT physics passages
• how best to study MCAT physics
• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
• anything you know to be beyond the scope of the MCAT

Side note: anyone who knows how to post subscripts and superscripts in this system, please PM me the method. I also wouldn't mind knowing how to post some obvious symbols, such as Greek letters and an infinty sign. Should be a matter of changing fonts, but I don't know what's available; again, a PM would be appreciated.

If you really know your physics, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current official contributors to the this thread -- a team to which I hope to add more people:

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics, minored in physics, and spent several years accumulating unused school experience (in economics and law).

Also answering questions: Xanthines, a Kaplan instructor. He scored 13 on the PS section of the MCAT and 34 overall.

 

[Monkeymaniac]

In terms of velocity, you don't need to worry about distance traveled. Velocity is defined as the change in position (displacement) over the change in time. So if you have the starting point (x1), the ending point (x2), and the time it took to get from x1 to x2 (delT), your velocity is simply:

v=(x2-x1)/delT

So, to answer your question, yes, you can calculate velocity without knowing distance traveled.

Thanks!

 

[gridiron]

So if I don't know the distance traveled, but know the
starting point and the ending point, then I can calculate
velocity since EP-SP = displacement. Am I right?
And thanks for your reply.

Yes you are. But be careful with the EP-SP notion. For example, if you travel 3 miles north, and then 4 miles east, you would think the displacement is 4-0=4. However, this is not true. You have to take into consideration the net displacement. Drawing this on paper, you get a displacement of 5 and 5. (Draw the triangle out...it helps to see it better). So displacement is final minus intial but it is representative of net quantity.

 

[PBandJ]

Yes you are. But be careful with the EP-SP notion. For example, if you travel 3 miles north, and then 4 miles east, you would think the displacement is 4-0=4. However, this is not true. You have to take into consideration the net displacement. Drawing this on paper, you get a displacement of 5 and 5. (Draw the triangle out...it helps to see it better). So displacement is final minus intial but it is representative of net quantity.

Yes, this is a good point as well. With my explanation above, I assumed that the motion all took place on one line. You have to take into account that distance and velocity are vectors, with both magnitude and direction.

 

[pezzang]

In an ideal fluid through a horizontal pipe, if the radius of the pipe is doubled, shouldn't the pressure be halved by two according to the equation P = F/A? Why is this approach wrong? Thanks for your help in advance. 😀

 

[PBandJ]

In an ideal fluid through a horizontal pipe, if the radius of the pipe is doubled, shouldn't the pressure be halved by two according to the equation P = F/A? Why is this approach wrong? Thanks for your help in advance. 😀

So you have to think about what A is in this equation. It's related to the radius, so your are on the right track in your approach, but you need to know that the area A refers to the cross-sectional area where the fluid is flowing through the pipe. Since the cross-sectional area is a circle in this case, the area A is related to the radius R by the following equation:

A = PI*(R^2)

So, if you plug that back into your P = F/A equation, you get:

P = F/A = F/(PI*(R^2))

From that point, you can see that pressure P is proportional to 1/(R^2). So, as your question asks, if the radius of the pipe is doubled (R' = 2R), then the pressure would in fact be quartered (P' = P/4) because of the R^2 in the denominator.

I hope that makes sense. 🙂

 

[pezzang]

So you have to think about what A is in this equation. It's related to the radius, so your are on the right track in your approach, but you need to know that the area A refers to the cross-sectional area where the fluid is flowing through the pipe. Since the cross-sectional area is a circle in this case, the area A is related to the radius R by the following equation:

A = PI*(R^2)

So, if you plug that back into your P = F/A equation, you get:

P = F/A = F/(PI*(R^2))

From that point, you can see that pressure P is proportional to 1/(R^2). So, as your question asks, if the radius of the pipe is doubled (R' = 2R), then the pressure would in fact be quartered (P' = P/4) because of the R^2 in the denominator.

I hope that makes sense. 🙂

Hi, I completely agree with you, but the answer is "the new pressure cannot be determined without more info" Perhaps I have briefed the question too much. So here is the actual Q:
An ideal fluid with pressure P flows through a horizontal pipe with a radis r. If the radius of the pipe is increased by a factor of 2, which of the following more likely gives the new pressure?
a.P
b.4P
c.16P
d."the new pressure cannot be determined without more info

So what's wrong with the way you described above in the last post about your approach to this question and why is the answer d? Thanks

 

[pezzang]

Here is another physics question that's bothering me. Please help...🙂
Q: Two particles are held in equilibrium by the gravitational and electrostatic forces between them. Particle A has mass m(a) and charge q(a). Particle B has mass m(b) and charge q(b). The distance between the charges is d. Which of the following changes will cause the charges to accelerate toward one another?

A. m(a) is doubled and m(b) is doubled
B. m(a) is doubled and m(b) is halved
C. q(a) is doubled and q(b) is doubled
D. d is doubled

The answer is A, but I think A works only if the heavier particle is located above the lighter one as both their forces will be doubled but the actual magnitude of increase in forces will be proportional to the original mass of the particles as shown in the eq. F = ma.
How is this so? Thanks again!!!!!

 

[gridiron]

Here is another physics question that's bothering me. Please help...🙂
Q: Two particles are held in equilibrium by the gravitational and electrostatic forces between them. Particle A has mass m(a) and charge q(a). Particle B has mass m(b) and charge q(b). The distance between the charges is d. Which of the following changes will cause the charges to accelerate toward one another?

A. m(a) is doubled and m(b) is doubled
B. m(a) is doubled and m(b) is halved
C. q(a) is doubled and q(b) is doubled
D. d is doubled

The answer is A, but I think A works only if the heavier particle is located above the lighter one as both their forces will be doubled but the actual magnitude of increase in forces will be proportional to the original mass of the particles as shown in the eq. F = ma.
How is this so? Thanks again!!!!!

Newton's law of gravitation states that each object exerts a equal but opposite pull for one another (it is a action-reaction pair as Newton's third law states). On the other hand, the electrostatic force between two static charges depends on the magnitude of the charge. Opposite charges attract and like charges repel. In this case, since the charges are in equilibrium, I believe the charges are like in magnitude since the system is in equilibrium and the law of gravitation relies on the pull for one another. That means:
F net = (GM1M2/r^2) = KQ1Q2/r^2. The r in relation cancels. Therefore you can knock out answer choices c and d. Now, since it seems that both charges are of the same sign and magnitude, doubling the magnitude will cause for further repulsion, and F net will have a value. In the question, F net is equal to zero since the system is in equilibrium and the charges are static. Therefore, b can be eliminated and that leaves choice a. Choice a is correct because doubling the mass causes for a four times greater pull of the objects for one another. I hope this helps. Remember, the law of gravitation means the pull of the an object for another object. Good luck!!

 

[pezzang]

Newton's law of gravitation states that each object exerts a equal but opposite pull for one another (it is a action-reaction pair as Newton's third law states). On the other hand, the electrostatic force between two static charges depends on the magnitude of the charge. Opposite charges attract and like charges repel. In this case, since the charges are in equilibrium, I believe the charges are like in magnitude since the system is in equilibrium and the law of gravitation relies on the pull for one another. That means:
F net = (GM1M2/r^2) = KQ1Q2/r^2. The r in relation cancels. Therefore you can knock out answer choices c and d. Now, since it seems that both charges are of the same sign and magnitude, doubling the magnitude will cause for further repulsion, and F net will have a value. In the question, F net is equal to zero since the system is in equilibrium and the charges are static. Therefore, b can be eliminated and that leaves choice a. Choice a is correct because doubling the mass causes for a four times greater pull of the objects for one another. I hope this helps. Remember, the law of gravitation means the pull of the an object for another object. Good luck!!

BioMedEngineer, Thanks so much for your help. But I'm not clear how the signs are the same for the two particles. I would really appreciate if you can explain how you proved that teh charges for the two particles are the same. Thanks!!

 

[gridiron]

BioMedEngineer, Thanks so much for your help. But I'm not clear how the signs are the same for the two particles. I would really appreciate if you can explain how you proved that teh charges for the two particles are the same. Thanks!!

Hey! The law of gravitation states the pull objects exert on one another and is a action-reaction force pair. The electrostatic force is the force two charges feel on one another. The system in the question is in equilibrium meaning the charges are static and fixed in place. If the charges were opposite in magnitude and charge, they would feel a natural pull for one another. But in this case, the law of gravitation balances the electrostatic force. This means that the charges have to be the same in charge and magnitude and. Thus, Fnet is equal to zero and there is no acceleration. However, if the mass is doubled, the gravitational force will slightly overcome the electrostatic force and the objects will pull on one another. Drawing a diagram helps. If the charges are like, than there will be one vector towards each charge representing the gravitational force and one vector away from each charge, they are like so they repel. Since we are dealing with vectors, we assign one as negative and one as positive, depends on the preferred sign convention, and the vectors will cancel out. Thus, the only way to disturb equilibrium is to deal with the gravitational force. I hope this helps. Good luck!!

 

[H and D]

I am having some trouble with harmonics (by trouble I mean I did not get a single question correct in the practice test I took concerning waves and harmonics).
Is it correct to say:
"A higher harmonic will have a higher frequency. Also the frequencies double as the harmonic increases."
For instance if the first harmonic has a frequency of 50hz then the second harmonic will have a frequency of 100 hz, and the third harmonic will have a frequency of 150 hz?
Also any tips or places to look to review harmonics would be appreciated. When I looked on the SDN Physics FAQ thread it only listed a wave topic which focused on longitudinal versus transverse waves (nothing about harmonics).

Thanks in advance

 

[gridiron]

I am having some trouble with harmonics (by trouble I mean I did not get a single question correct in the practice test I took concerning waves and harmonics).
Is it correct to say:
"A higher harmonic will have a higher frequency. Also the frequencies double as the harmonic increases."
For instance if the first harmonic has a frequency of 50hz then the second harmonic will have a frequency of 100 hz, and the third harmonic will have a frequency of 150 hz?
Also any tips or places to look to review harmonics would be appreciated. When I looked on the SDN Physics FAQ thread it only listed a wave topic which focused on longitudinal versus transverse waves (nothing about harmonics).

Thanks in advance

Hey. If you are talking about waves, than the frequency is given by the following relation (velocity of the wave in the medium*harmonic number)/(2*the length of the string). The harmonic frequency is given by the following relationship (f👎=nf1) where n is the harmonic number and f1 is the first frequency---frequency of the first harmonic. So, your analysis is correct--if the first harmonic is 50 than the second harmonic is 100. But beware, the second harmonic can mean that there are 3 nodes on the string. For good review of harmonics, I suggest looking at any physics textbook--or in a review book. If you have any other questions, I am glad to help or I can post more information on harmonics--provided I have the permission first to do so. Good luck

 

[QofQuimica]

Hey. If you are talking about waves, than the frequency is given by the following relation (velocity of the wave in the medium*harmonic number)/(2*the length of the string). The harmonic frequency is given by the following relationship (f👎=nf1) where n is the harmonic number and f1 is the first frequency---frequency of the first harmonic. So, your analysis is correct--if the first harmonic is 50 than the second harmonic is 100. But beware, the second harmonic can mean that there are 3 nodes on the string. For good review of harmonics, I suggest looking at any physics textbook--or in a review book. If you have any other questions, I am glad to help or I can post more information on harmonics--provided I have the permission first to do so. Good luck

Yeah, I think harmonics would be a great topic to write an explanations post for if you have the time and inclination, BME. 👍

 

[SeventhSon]

let's see if i remember this, because I think this showed up on my MCAT...

make sure you understand the difference between harmonic # and overtones in a standing wave.

the harmonic number corresponds to the # of antinodes. The number of nodes in a standing wave is n - 1.

All harmonic frequencies are integer multiples of the fundamental frequency.... i.e. F👎 = n*f1. The fundamental frequency has harmonic # n = 1, but not all harmonic frequencies are necessarily valid for the boundary conditions of a standing wave.

This is where the terminology 'overtone' comes from... the fact that some harmonic #s cannot satisfy the boundary conditions. Overtones depend on the boundary conditions... i.e. for bondary conditions of one node and one antinode, the valid harmonic numbers will be n = 1, 3, 5.... and the three standing waves are considered the fundamental wave, the first overtone, and the second overtone.

Someone correct me if i'm telling lies

 

[Monkeymaniac]

I've been studying centrepetal motion today, and I'm stuck at curved plane question.

https://netfiles.uiuc.edu/dylee1/shared/curvedplane.bmp?uniq=-e3fhxs

Here's the picture of the problem. I understand that there is a gravitational force that acts on the ball vertically downward with the magnitude of mg.
And the component of the force that is perpendicular to the surface is mgcos(phata). Since the ball is not jumping toward the center or moving away from the center, I understand that there should be a force that counters mgcos(phata). In case of noncurved surface, that would be normal force with the same magnitude. But there is just too much complication involved with this. And I read from EK that the components for the normal force for curved surface case is mgcos(phata) and mv^2/r, but did not tell me which directinos of the two vectors are.

So my question is,
1) What is the magnitude of the centrepetal force and how did you calculate it?
2) What is the direction of each vector that makes up the normal force vector in this curved surface problem?
3) If you used normal force as a part of calculating the centrepetal force, is that really valid? I mean centrepetal force that we are calculating here is that of the ball, and the normal force is that of the surface that ball has contact with, so shouldn't we use only the forces acted directly on the ball (i.e. mg) when we calculate the CF of the ball? or is it one of those tricky determining "system" question?

I appreciate in advance.

 

[gridiron]

I've been studying centrepetal motion today, and I'm stuck at curved plane question.

https://netfiles.uiuc.edu/dylee1/shared/curvedplane.bmp?uniq=-e3fhxs

Here's the picture of the problem. I understand that there is a gravitational force that acts on the ball vertically downward with the magnitude of mg.
And the component of the force that is perpendicular to the surface is mgcos(phata). Since the ball is not jumping toward the center or moving away from the center, I understand that there should be a force that counters mgcos(phata). In case of noncurved surface, that would be normal force with the same magnitude. But there is just too much complication involved with this. And I read from EK that the components for the normal force for curved surface case is mgcos(phata) and mv^2/r, but did not tell me which directinos of the two vectors are.

So my question is,
1) What is the magnitude of the centrepetal force and how did you calculate it?
2) What is the direction of each vector that makes up the normal force vector in this curved surface problem?
3) If you used normal force as a part of calculating the centrepetal force, is that really valid? I mean centrepetal force that we are calculating here is that of the ball, and the normal force is that of the surface that ball has contact with, so shouldn't we use only the forces acted directly on the ball (i.e. mg) when we calculate the CF of the ball? or is it one of those tricky determining "system" question?

I appreciate in advance.

Let me take a shot at this bad boy....the ball is in somewhat circular motion. As it is moving down the curved slide per say, it feels forces: The normal force due to the force the curved slide exerts on the ball, and two components of the gravity force: mgsin(theta) and mgcos(theta). As the ball is moving, it will feel a centripetal force inwards. Since these are vectors, you have to choose a convention to associate the vectors with signs--i.e. the force pointing in will be positive and out will be negative. The force balance will then yield: F normal - mgcos(theta) = force centripetal. Depending on the values given in the problem, this should be enough to solve for the normal force, or the velocity of the ball.
You have to use the normal force when solving this problem because the curved surface is exerting a force on the ball the same way the ball is exerting a force on the surface. However, the force exerted by the surface is directing inwards, toward the center of the circle, and the force of the ball is broken down into components.
If there is not enough information provided to solve the problem, you can substitute a value of velocity in terms of the angular velocity and radius. Remember, the ball is rotating, so it exhibits a angular velocity. The formula for this is velocity = angular velocity times the radius of the path. In most physics classes, you will be dealing with angular motion, period, and simple harmonic motion because simple harmonic motion can be represented as circular motion since it is one-dimensional.
The gravitational component, mgsin(theta), is perpendicular to the motion and keeps the object moving in a circle. It's purpose can be clarified as follows: If you tie a string to a rock and then hurl it in a circle over your head (don't try this), with constant speed (does not mean constant velocity), and cut the string at a point, then the rock will fly away parallel to the path of the velocity vector.
I hope this helps and good luck. Someone correct me if I am wrong!!!

 

[Monkeymaniac]

Thanks, that really helped.
We decided that the force pointing inward is positive and outward is negative, we know that the centrepetal force that is pointed inward should be positive. And from the equation Fnormal + (-mg*cos(theta)) = centrepetal force,
we see that Fnormal should be larger than mg*cos(theta). Let's suppose that Fnormal be 5 unit and -mg*cos(theta) be -4 unit, and thus centrepetal force be 1 unit.

If I draw a force diagram, I get something like
https://netfiles.uiuc.edu/dylee1/shared/curvedplane_2.bmp?uniq=lnfsj3
this.

As you can see the sum of forces that are perpendicular to the surface
is not zero (with my example, we have Fnet = 5+1+(-4) = 2 is not 0).
Is it because there must be some non zero force acting on the ball
toward the center for the circular motion to take place?
If that is true, I finally understood the concept. 😀

 

[tik-tik-clock]

Which of the following has the greatest inertia?

A. a 5kg mass at rest
B. a 5KG mass moving at 10 m/s
C. a 5 kg mass acceerating at 10 m/s^2
D. all have the same inertia

Ans:mass is a quantitative measure of linear inertia. I dont understand what this statement means. Does it mean that Inertia depends only on mass regardless of what happens to any of the other factors?

Thanks

 

[gridiron]

Which of the following has the greatest inertia?

A. a 5kg mass at rest
B. a 5KG mass moving at 10 m/s
C. a 5 kg mass acceerating at 10 m/s^2
D. all have the same inertia

Ans:mass is a quantitative measure of linear inertia. I dont understand what this statement means. Does it mean that Inertia depends only on mass regardless of what happens to any of the other factors?

Thanks

I would probably guess D because of Newton's first law of motion which states that a object at rest tends to stay at rest and a object in motion tends to stay in motion. Resistance to a change in motion is inertia. Not too sure...

 

[tik-tik-clock]

I would probably guess D because of Newton's first law of motion which states that a object at rest tends to stay at rest and a object in motion tends to stay in motion. Resistance to a change in motion is inertia. Not too sure...

Their ans is D, But the explanation they give for that is what i wrote above. and i m nore too sure what it means..
Thanks 🙂

 

[gridiron]

Their ans is D, But the explanation they give for that is what i wrote above. and i m nore too sure what it means..
Thanks 🙂

Inertia is a objects ability to resist changes in motion. Newton's first law states that a object at rest will stay at rest or an object in motion will stay in motion unless an external force acts on the system. Thus, all objects will display equal amounts of inertia--all will want to stay in their respective states unless a force acts on it. That is why no object from the list has the greater inertia. Choice will want to stay in rest and B anc C will want to stay in motion. I hope this helps and good luck

I meant to say the objects in this question...not all objects in general...oops!

 

[lorelei]

All objects do not display equal amounts of inertia; it depends on mass. Answer D is correct because the masses are all the same.

Inertia is resistance to acceleration (changes in speed or direction of motion). Look at F=ma; the greater the mass, the more force you need to apply to cause a given acceleration. For example, if I want to move my cat, it's very easy. If I want to move my boyfriend, it's significantly harder. My boyfriend's greater mass means that his body has a greater resistance to changes in motion.

That's what the answer explanation is trying to say.

(For another explanation, see: http://en.wikipedia.org/wiki/Inertia#Mass_and_.27inertia.27 )

EDIT: on looking back at the wikipedia article, it appears that this is considered something of an abuse of the term, and the better term for the concept the question is asking about is "inertial mass." But anyway, if they're asking you to compare inertias, that's what they're talking about.

This might be out of scope for the MCAT, and if it isn't, I'd bet they'd ask the question as "which has the greater resistance to changes in velocity" because the real MCAT does quite well at having correct questions.

 

[liverotcod]

For example, if I want to move my cat, it's very easy. If I want to move my boyfriend, it's significantly harder. My boyfriend's greater mass means that his body has a greater resistance to changes in motion.

Yeah, but I'll bet you have means other than force to produce accelerations in your boyfriend.

 

[tik-tik-clock]

I have one more question:

Planets A and B have the same mass. planet A has a radius half as large as planet B. A 5 kg mass is dropped 10 m above the surgace of the planet A and at the same time a 5 kg mass is dropped 10 m above the surgavce of the planet B. If the mass on the Planet B strikes the ground in 10 sec, the mass on the planet A strikes te ground in:

1. 2.5 s
2. 5 s
C. 10 s
D. 20 s

I am so confused with this question.
I mean I dont understand the explanation of this question either.
Here is what the explanation states:

F= GMm/r^ 2. reducing the radius by a factor of 2, increases the gravitational force by four and the acceleration by 4( F= ma). You should know that the accelration of a projectile is independent of its mass. From x = 1/2 at^2 we see that reducing the acceleration by a factor of 4 increases the time by a factor of 2.

only God knows what this means. Please help me out! ;(

 

[kevster2001]

I have one more question:

Planets A and B have the same mass. planet A has a radius half as large as planet B. A 5 kg mass is dropped 10 m above the surgace of the planet A and at the same time a 5 kg mass is dropped 10 m above the surgavce of the planet B. If the mass on the Planet B strikes the ground in 10 sec, the mass on the planet A strikes te ground in:

1. 2.5 s
2. 5 s
C. 10 s
D. 20 s

I am so confused with this question.
I mean I dont understand the explanation of this question either.
Here is what the explanation states:

F= GMm/r^ 2. reducing the radius by a factor of 2, increases the gravitational force by four and the acceleration by 4( F= ma). You should know that the accelration of a projectile is independent of its mass. From x = 1/2 at^2 we see that reducing the acceleration by a factor of 4 increases the time by a factor of 2.

only God knows what this means. Please help me out! ;(

Maybe I'm reading this question wrong but is the answer C? The R refers to the distance between the two masses, since the mass is 10m above the surface of both planets, the r remains the same. And since the masses of the planets and the mass of the rock is the same in bot hcases, wouldn't the fall time be the same as well?

 

[lorelei]

I have one more question:

Planets A and B have the same mass. planet A has a radius half as large as planet B. A 5 kg mass is dropped 10 m above the surgace of the planet A and at the same time a 5 kg mass is dropped 10 m above the surgavce of the planet B. If the mass on the Planet B strikes the ground in 10 sec, the mass on the planet A strikes te ground in:

1. 2.5 s
2. 5 s
C. 10 s
D. 20 s

I am so confused with this question.
I mean I dont understand the explanation of this question either.
Here is what the explanation states:

F= GMm/r^ 2. reducing the radius by a factor of 2, increases the gravitational force by four and the acceleration by 4( F= ma). You should know that the accelration of a projectile is independent of its mass. From x = 1/2 at^2 we see that reducing the acceleration by a factor of 4 increases the time by a factor of 2.

only God knows what this means. Please help me out! ;(

Where did these questions come from, out of curiosity?

I think the explanation is right, but it's so poorly stated that I can't be sure. Here's my explanation (much longer, but should be easier to follow):

Your goal here is to figure out how the radius of the planet affects the time it takes for a dropped object to fall a given distance. As you know, that time is determined by the value of g for that planet -- where g stands for the acceleration due to gravity. How do we find g?

We start with the formula for universal gravitation:
F = GMm/r^2
where G is the universal gravitational constant, M and m are the masses, and r is the distance between them. So here M will be the planet's mass and m will be the object's mass. r is going to be the radius of the planet. [Note that we care about the distance between the CENTERS of mass, not the distance between the SURFACES.]

We don't care about the force, though; we want the acceleration. Here's where we use F=ma. Substitute that in, so we have:
ma = GMm/r^2
Cancel out m (the object's mass) from both sides, leaving:
a = GM/r^2

That's the derivation of the acceleration due to gravity on any given planet. (This is true near the surface of the planet; if we were near the center or far out in orbit, we'd have to use a different value for r rather than just the planet's radius.)

So now we know that the acceleration due to gravity depends on two things: the mass of the planet, and the radius of the planet. In our case, the masses are the same, but Planet A has a radius one-half that of Planet B. Since we have an inverse-square relationship, that means the acceleration due to gravity on Planet A is going to be FOUR TIMES that on Planet B. (You should be comfortable doing these kinds of ratio calculations; they're very popular on the MCAT.)

How does the acceleration due to gravity affect the time it takes an object to fall? Here's where we use the distance formula:
x=0.5at^2
Increasing a by a factor of four means that t is going to be decreased by a factor of two (aka cut in half) in order to keep x the same.

So the object will take half as long to hit the ground on Planet A as on Planet B, and our answer choice is B, 5 seconds.

 

[luckycharms08]

it'd help if you told people that these were straight out of EK's 1001 chem books. anyway, this is #172 and the answer is B. if you use F= Gmm/r^2, then if the radius is 1/2, F is going to be increased by 4. Therefore a will be increased by 4. using x = x0 + v0t + 1/2at^2, you know that x=1/2at^2. x is constant (10 m) for each case, so you could do 1/2at^2 = 1/2at^2 and if a is increased by 4, then t will have to be 1/2 to keep the equality true.

I have one more question:

Planets A and B have the same mass. planet A has a radius half as large as planet B. A 5 kg mass is dropped 10 m above the surgace of the planet A and at the same time a 5 kg mass is dropped 10 m above the surgavce of the planet B. If the mass on the Planet B strikes the ground in 10 sec, the mass on the planet A strikes te ground in:

1. 2.5 s
2. 5 s
C. 10 s
D. 20 s

I am so confused with this question.
I mean I dont understand the explanation of this question either.
Here is what the explanation states:

F= GMm/r^ 2. reducing the radius by a factor of 2, increases the gravitational force by four and the acceleration by 4( F= ma). You should know that the accelration of a projectile is independent of its mass. From x = 1/2 at^2 we see that reducing the acceleration by a factor of 4 increases the time by a factor of 2.

only God knows what this means. Please help me out! ;(

 

[tik-tik-clock]

Where did these questions come from, out of curiosity?

I think the explanation is right, but it's so poorly stated that I can't be sure. Here's my explanation (much longer, but should be easier to follow):

Your goal here is to figure out how the radius of the planet affects the time it takes for a dropped object to fall a given distance. As you know, that time is determined by the value of g for that planet -- where g stands for the acceleration due to gravity. How do we find g?

We start with the formula for universal gravitation:
F = GMm/r^2
where G is the universal gravitational constant, M and m are the masses, and r is the distance between them. So here M will be the planet's mass and m will be the object's mass. r is going to be the radius of the planet. [Note that we care about the distance between the CENTERS of mass, not the distance between the SURFACES.]

We don't care about the force, though; we want the acceleration. Here's where we use F=ma. Substitute that in, so we have:
ma = GMm/r^2
Cancel out m (the object's mass) from both sides, leaving:
a = GM/r^2

That's the derivation of the acceleration due to gravity on any given planet. (This is true near the surface of the planet; if we were near the center or far out in orbit, we'd have to use a different value for r rather than just the planet's radius.)

So now we know that the acceleration due to gravity depends on two things: the mass of the planet, and the radius of the planet. In our case, the masses are the same, but Planet A has a radius one-half that of Planet B. Since we have an inverse-square relationship, that means the acceleration due to gravity on Planet A is going to be FOUR TIMES that on Planet B. (You should be comfortable doing these kinds of ratio calculations; they're very popular on the MCAT.)

How does the acceleration due to gravity affect the time it takes an object to fall? Here's where we use the distance formula:
x=0.5at^2
Increasing a by a factor of four means that t is going to be decreased by a factor of two (aka cut in half) in order to keep x the same.

So the object will take half as long to hit the ground on Planet A as on Planet B, and our answer choice is B, 5 seconds.

Thanks hun that makes much more sense. The answer is indeed B.

That question is from Physics 1001 book 🙂

 

[tik-tik-clock]

Lorelei,

Just wanted to make sure, The r is that distance between the centers of gravity of 5 kg mass and the planet A and B respectively right? It is not the distance between the centers of gravities of the planets is it?

Thanks 🙂

 

[lorelei]

Lorelei,

Just wanted to make sure, The r is that distance between the centers of gravity of 5 kg mass and the planet A and B respectively right? It is not the distance between the centers of gravities of the planets is it?

Thanks 🙂

You're correct. Whenever using the gravitational formula, r is the distance between the centers of gravity of m and M, the two objects that are getting attracted to each other. So in this case, it's the distance between the center of the 5kg mass and the center of the planet.

 

[tik-tik-clock]

So, when an object is placed on a spring and then the spring is compressed and released, I understand that the gravity acts in the downward direction on the object, but how is the hooke's law in the upward direction?

 

[tik-tik-clock]

SO basically when the compression decreases the hooke's law force decreases right?

Since the force is directly proportional to the change in x.

But I also know that F = - kx,
so what significance does the negative sign hold here? i would assume that we can ignore the negative sign and we will end up with the same logic right?

does the negative sign affect the acceleration in any way? since
m.a=-kx?
if x decreases, the acceleration decreases, right? Don't think that makes sense. Maybe the acceleration is the acceleration due to gravity and it always remains constant?

thanks🙂

 

[gridiron]

SO basically when the compression decreases the hooke's law force decreases right?

Since the force is directly proportional to the change in x.

But I also know that F = - kx,
so what significance does the negative sign hold here? i would assume that we can ignore the negative sign and we will end up with the same logic right?

does the negative sign affect the acceleration in any way? since
m.a=-kx?
if x decreases, the acceleration decreases, right? Don't think that makes sense. Maybe the acceleration is the acceleration due to gravity and it always remains constant?

thanks🙂

Hooke's law is given by F=-kx. The negative sign is more for convention because the force acts opposite of the displaced direction---it is a restoring force. The negative sign does not affect (effect?) the acceleration in any way because a spring system can be described in terms of simple harmonic motion. Set the restoring force equal to ma because it is the force effecting the acceleration of the system. So:
-KX=Ma
In simple harmonic motion, the acceleration is related to the displacement function with the constant of proportionality being - omega squared--the angular speed...so:
-K(x)=-m(w^2)(x)
From here you can derive the frequency and period forms for the system. When it comes to systems such as these, I think it is more appropriate to describe them using simple harmonic motion. Of course, you can find the net force on the system and then use Newtons second law....which would lead to: (g-x)=acceleration...

 

[tik-tik-clock]

Hooke's law is given by F=-kx. The negative sign is more for convention because the force acts opposite of the displaced direction---it is a restoring force. The negative sign does not affect (effect?) the acceleration in any way because a spring system can be described in terms of simple harmonic motion. Set the restoring force equal to ma because it is the force effecting the acceleration of the system. So:
-KX=Ma
In simple harmonic motion, the acceleration is related to the displacement function with the constant of proportionality being - omega squared--the angular speed...so:
-K(x)=-m(w^2)(x)
From here you can derive the frequency and period forms for the system. When it comes to systems such as these, I think it is more appropriate to describe them using simple harmonic motion. Of course, you can find the net force on the system and then use Newtons second law....which would lead to: (g-x)=acceleration...

Thanks BioMedEngg,
Makes sense🙂

One more 😳
Am I right in deriving this conclusion for a mass hanging from a rope?

• deceleration = velocity slowing down in the opposite direction
o acceleration increases in the opposite direction of which the velocity is decreasing.
Then the tension increases in the direction that the acceleration is increasing.

 

[tik-tik-clock]

• the minimum force necessary to pull an initially stationary mass up the plane is the force necessary to move it at constant velocity. How? Why the constant velocity?
I know that a system with constant velocity is at equilibrium. Thus we can equate the forces up the plane to the forces down the plane.

But I dont understand the logic behind the bolded portion.

 

[gridiron]

Thanks BioMedEngg,
Makes sense🙂

One more 😳
Am I right in deriving this conclusion for a mass hanging from a rope?

• deceleration = velocity slowing down in the opposite direction
o acceleration increases in the opposite direction of which the velocity is decreasing.
Then the tension increases in the direction that the acceleration is increasing.

Consider a simple pendulum which consists of a negligible mass hanging from a string. As the string is set into motion, there is a force (from drawing the free body diagram) of mgsintheta that acts to oppose the direction of motion and acts as a restoring force. This force can be expressed in terms of torque by multiplying the action of the force from the point of movement---in this case the length of the string. This in turn can be related to the angular acceleration and moment of inertia---(hang with me on this...I don't believe this will be tested on the MCAT but the derivation helps to understand your question). Therefore you have: (-mgsintheta)*length = Moment of inertia * angular acceleration. Divide by I and assume theta is a small. Angular acceleration is related to the displacement by the angular speed squared. However, as the restoring force, it seems from the equation that the angular acceleration increases. Tension can be related using circular motion formulas and it seems to increase with increased angular acceleration. I hope I didn't confuse you on this....I hope this helps 👍

 

[shantster]

• the minimum force necessary to pull an initially stationary mass up the plane is the force necessary to move it at constant velocity. How? Why the constant velocity?
I know that a system with constant velocity is at equilibrium. Thus we can equate the forces up the plane to the forces down the plane.

But I dont understand the logic behind the bolded portion.

To initially pull a mass, you need to overcome the force of gravity. At that point, the force of gravity equals the force that is being applied by pulling on the object up the incline. Since the two forces equal, then the net force is 0 so the object moves at constant velocity.

Another way to look at it is that if the object is sliding down the incline, it is accelerating at g. If you want it to go up the incline, you start pulling on it, and the downward acceleration starts to decrease to the point that it equals zero, and then will start to increase in the upward direction (or become a greater negative in the downward direction). At the point where the acceleration = 0, the object has turned directions and is starting to go up the incline at a constant speed.

Hope that helps.

 

[wolfram241]

15uC<---------10m------------>-1uC

-5uC charge of 3.5g released from rest 1.5m from negative charge, and moves towards the 15uC charge. What's its velocity at 3m from 15uC?

Help!

Do I use Ki+Ui=Kf+Uf or more classical kinematics?

 

[gridiron]

15uC<---------10m------------>-1uC

-5uC charge of 3.5g released from rest 1.5m from negative charge, and moves towards the 15uC charge. What's its velocity at 3m from 15uC?

Help!

Do I use Ki+Ui=Kf+Uf or more classical kinematics?

Hey. You have to use Newton's second law motion and then apply kinematics. First, you have to find the electrostatic force on the object. You can use (KQ1Q2/r^2), but remember that you have that force over two bodies because there is a force of attraction to the positive charge and repulsion from the negative charge. Once you have solved for the force, you can solve for acceleration of the particle using: F=ma. Then you can apply kinematics by using the equation: (velocity final^2) = (velocity initial^2) +2ad where a is the acceleration you solved for and d is the distance travelled. Try that and good luck!!

 

[tik-tik-clock]

At that point, the force of gravity equals the force that is being applied by pulling on the object up the incline.

Wouldn't the force required to overcome the net force of zero be a little greater to get it in motion? If force is zero, how is the object going to move?

At the point where the acceleration = 0, the object has turned directions and is starting to go up the incline at a constant speed

I dont get this part. Is there a logical reasoning to this, or is it just something we should memorize?


Thanks🙂

 

[gridiron]

Wouldn't the force required to overcome the net force of zero be a little greater to get it in motion? If force is zero, how is the object going to move?

Thanks🙂

Remember, forces are used to determine the effect on the acceleration of a system. If the net force is zero, that means there is zero acceleration. However, zero acceleration does not mean the object isn't moving--it is just moving at constant velocity ( I have heard that they interchange speed and velocity on the MCAT which is bad because speed and velocity are two different entities).

 

[tik-tik-clock]

Remember, forces are used to determine the effect on the acceleration of a system. If the net force is zero, that means there is zero acceleration. However, zero acceleration does not mean the object isn't moving--it is just moving at constant velocity ( I have heard that they interchange speed and velocity on the MCAT which is bad because speed and velocity are two different entities).

Oh yes, totally forgot abt that.
Ty🙂

 

[Dave_D]

Hey. You have to use Newton's second law motion and then apply kinematics. First, you have to find the electrostatic force on the object. You can use (KQ1Q2/r^2), but remember that you have that force over two bodies because there is a force of attraction to the positive charge and repulsion from the negative charge. Once you have solved for the force, you can solve for acceleration of the particle using: F=ma. Then you can apply kinematics by using the equation: (velocity final^2) = (velocity initial^2) +2ad where a is the acceleration you solved for and d is the distance travelled. Try that and good luck!!

Bio, isn't acceleration in this case not a constant? I mean I would think you'd get an acceleration function and then take the integral of it. What I mean is that if the charge travelled a meter going from 4 meters to 3 meters from one of the charges I'd expect a different final speed than if it travelled 1 meter from 5 meters to 4 of that same charge.

 

[I am here to ?]

Ok guys not sure if this is the place but hopefully it can help others...

regarding gases
the EK books (newest ones) say that V real is greater than V ideal (volume) and p real is less than p ideal (pressure)

The tpr books say the same thing except they claim that BOTH v real and p real are less than their ideal counterparts

How is the volume of a gas judged? the volume of the gas molecules or the space that they take up in a container...

and then after researching this subject I found something regarding that the answer depends....positive deviation due to molecular volume and negative deviation due to IM forces

can someone please clarify this with some explanation....its really bugging the heck out of me

 

[xanthomondo]

Ok guys not sure if this is the place but hopefully it can help others...

regarding gases
the EK books (newest ones) say that V real is greater than V ideal (volume) and p real is less than p ideal (pressure)

The tpr books say the same thing except they claim that BOTH v real and p real are less than their ideal counterparts

How is the volume of a gas judged? the volume of the gas molecules or the space that they take up in a container...

and then after researching this subject I found something regarding that the answer depends....positive deviation due to molecular volume and negative deviation due to IM forces

can someone please clarify this with some explanation....its really bugging the heck out of me

The volume is always calculated by the volume of the container (because you have a pressure of it, too)

Yes you're on the right track...there are three scenarios

Attractive forces dominant, Repulsive forces dominant, and no forces (ideal)

If the molecules are of the first type, then the pressure of the system (compared to an ideal gas) will be less... its hard to say that both the volume and pressure will be less, since theyre inversely proportional, but if you hold one constant the other has to be less.

Molecules of the second type have yet another thing to push off of (remember thats what creates the pressure of a balloon - molecules constanty ramming into the walls) and the pressure (or volume) will be increased compared to the ideal world

Theres also one more thing to consider: at extremely high pressures/low volumes, you have to consider the size of the individual gas molecules themselves...this alone will cause a deviation from the ideal PV=nRT

...so do you think this last example will be a positive of negative deviation?

 

[wolfram241]

Hey. You have to use Newton's second law motion and then apply kinematics. First, you have to find the electrostatic force on the object. You can use (KQ1Q2/r^2), but remember that you have that force over two bodies because there is a force of attraction to the positive charge and repulsion from the negative charge. Once you have solved for the force, you can solve for acceleration of the particle using: F=ma. Then you can apply kinematics by using the equation: (velocity final^2) = (velocity initial^2) +2ad where a is the acceleration you solved for and d is the distance travelled. Try that and good luck!!

Where do I find the electric field at, where the charge in motion begins or ends? Or does it not matter?

I have the answer but I can't figure out how to get to it. Very frustrating.

Thanks.

 

[gridiron]

Bio, isn't acceleration in this case not a constant? I mean I would think you'd get an acceleration function and then take the integral of it. What I mean is that if the charge travelled a meter going from 4 meters to 3 meters from one of the charges I'd expect a different final speed than if it travelled 1 meter from 5 meters to 4 of that same charge.

Dave_D thanks so much for catching my error! You are absolutely right...the acceleration does change with distance and you need a integral to solve the problem (integrals are not tested on the MCAT from what I remember). Thanks....I made a silly mistake 😳

 

[gridiron]

Where do I find the electric field at, where the charge in motion begins or ends? Or does it not matter?

I have the answer but I can't figure out how to get to it. Very frustrating.

Thanks.

Apologies...I misinformed you because I didn't read the problem correctly 😳 . I believe that you should use scalar quantities rather than vector quantities to solve this problem. I think, and somebody correct me if I am wrong, you can find the electric potential difference between the two points. When you multiply this value by the charge of the particle, you get the difference in potential energy. In this case, the answer should be negative because the particle is accelerating toward the other charge. The kinetic energy is the opposite the electric potential energy. Once you have that value, you can solve for the velocity. If I am incorrect, someone please correct me because I don't want to make the same mistake again 🙁

 

[wolfram241]

Apologies...I misinformed you because I didn't read the problem correctly 😳 . I believe that you should use scalar quantities rather than vector quantities to solve this problem. I think, and somebody correct me if I am wrong, you can find the electric potential difference between the two points. When you multiply this value by the charge of the particle, you get the difference in potential energy. In this case, the answer should be negative because the particle is accelerating toward the other charge. The kinetic energy is the opposite the electric potential energy. Once you have that value, you can solve for the velocity. If I am incorrect, someone please correct me because I don't want to make the same mistake again 🙁

Can't find electric potential because there are three charges.. However, I think I directly found change in potential energy and set it equal to KE, but the velocity didn't come out to my desired result. Argh. Thanks for trying to help though!

edit: could it have something to do with the electric field between the two outside charges? E=kq/r^2

 

[Mutt]

Question: #310 in EK Physics 1001 Questions

Setup:

Two ropes hold a 10m-long board with a mass of 20kg. Rope 1 (T1) is at 7.5m, Rope 2 (T2) is at 5m (from right side). Board is at rotational equilibrium.

A 10kg mass hangs off of the left side. Find T2.

EK answers:

Choose point of rotation to be T2.

Counterclockwise torque = mg = 5kg(10m/s^2)
Clockwise torque = 2.5T1

My problem:
Setting them equal (b/c they are in rotational equilibrium): T1=20N

Since T1+T2 must equal the mass of the board + weight (20kg + 10kg):
T1+T2 = 30kg x 10m/s^2 = 300N

Substituting 20N into above gives T2 of 280N

The answer is 100N?