CONQUERING PHYSICS GRE PDF
Cambridge Core - General and Classical Physics - Conquering the Physics GRE - by PDF; Export citation 9 - Special Tips and Tricks for the Physics GRE. Editorial Reviews. Review. 'I have used Conquering the Physics GRE as a resource for our annual 'Southern California Physics GRE Bootcamp' program for the. You may register online (My GRE) or by mail – registration deadlines are much help ('Conquering the Physics GRE', by Kahn and Anderson is reasonable).
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Conquering the Physics GRE - Yoni Kahn, Adam Anderson - Ebook download as PDF File .pdf), Text File .txt) or read book online. Conquering the Physics. Why Take the Physics GRE? • You're applying to Focuses on problem solving, calculation, basic physics. • Not the most ETS website: • yazik.info subject/about/content/physics. • Conquering the Physics GRE. View Test Prep - Conquering the Physics GRE 3rd yazik.info from PHYS FISI at Universidad de Los Andes. Conquering the Physics GRE Third Edition.
There are other types of potential energy, but all can be summarized by a definition. Probably the only time you might have to use.
Note the signs: The final sign makes sense because gravita- tional potential decreases that is, becomes more negative as the satellite gets closer to the Earth; in other words, it is attracted towards the Earth.
Probably the most confusing part of this whole business is the signs, which the GRE loves to exploit. Rather than worrying about putting the signs in the right place throughout the whole problem, it may be best to just compute the unsigned quantity, then fill in the sign at the end with physical reasoning.
How fast is it traveling when it reaches the bottom? The quarter-circle shape is irrelevant except for the fact that it gives us the initial height: At the top, the block is stationary, so its velocity is zero and there is no kinetic energy; all the energy is potential. Here the obvious choice is to set the zero of gravitational potential energy at the bottom of the ramp, so that the potential at the top is mgR.
We now apply conservation of energy:. Conveniently enough, the mass cancels out since both the kinetic and potential energies are directly proportional to m. There are a couple things to note here:. The block could have had a nonzero speed at the top, in which case it would have nonzero kinetic energy there. Then in addition to its kinetic energy. If the object rolls without slipping. The rolling-without-slipping condition 1. In this situation friction does no work. Without friction.
The moments of inertia are 12 mr2 for the cylinder and 25 M R2 for the sphere neither of which you should memorize. Dimensional analysis dictates where to put the R so that v comes out with the correct units.
A cylinder of mass m and radius r. Because work is a signed quantity. The alternate form 1. How far down this surface does the block p travel before it stops? But that would actually be too much work no pun intended! The answer is simple: Rolling-without-slipping problems almost always boil down to these kinds of cancellations: We never even had to solve for the velocity at the bottom of the ramp.
A 10 m tall ladder of mass 10 kg is held upright 6 m away from a wall.
Never solve more of a problem than you absolutely must! As a sanity check. When the block has stopped. With what approximate speed is the end of the ladder moving when it strikes the wall? You may assume the mass of the ladder is uniformly distributed. In that case. This is actually a pretty interesting problem. Note that we could not simply apply the formulas for uniform circular motion to determine the normal force.
You can do a similar analysis for R! If the block has initial velocity v. Applying the work-energy theorem. What distance should the spring be compressed? You may assume friction is sufficient that the ball begins rolling without slipping immediately after launch.
You want to launch the pinball a sphere of mass m and radius r so that it just barely reaches the top of the ramp without rolling back. August Some of the problems. This caveat about external forces is sometimes important: In fact. In general. Now suppose the ramp is waxed. The trick with momentum problems is just to be sure that you are counting all types of momenta—linear and angular—and writing down the correct conservation equations. What is the distance the spring should be compressed this time?
Suppose the ball of the mass M initially has speed V. What is the scattering angle of the ball of mass m? Applying conservation of momentum in two dimensions. Consider an example in which a ball of mass M strikes another ball of mass m initially at rest. If things are colliding. For practice. After this. Thus all we actually need is the second of the two equations. Collisional forces can be arbitrarily complicated. Remember that rotational motion is always defined with respect to a reference point or axis.
The ball of mass m goes down. Exhaust all limiting cases and dimensional analysis arguments before doing algebra. This often gets complicated. If you think it will be messy. Since you may not even have time to finish the exam. You just need to set the initial momentum equal to the final momentum and solve for the necessary variables. Solving momentum conservation problems like this one invariably reduces to solving systems of linear equations.
We know M. Extending the metaphor. We wrote angular momentum and torque in their vector form for completeness above. The key point to remember is that the angular momentum vector L is generally parallel 2 to the angular velocity vector!.
For instance. But these correspond to various axes of symmetry. The direction of L is determined by the right hand rule: The parallel axis theorem is a fast and frequently invaluable tool for computing the moment of inertia of systems built out of smaller pieces whose moments of inertia are known.
In principle. Typically this is sufficient. The GRE does provide the formulas for the moments of inertia for rods.
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If we know the moment of inertia I of a system of mass M rotating about an axis through its center of mass. Typically the integral for moment of inertia is solved by changing integration variables to spatial variables.
What is the moment of inertia. July 7. A block explodes into three pieces of equal mass. M 0 M L3 0 L 4 4 to the right of the center of the rod. Piece A has speed v after the explosion.
What is the angle between the direction of the piece A and piece B? What is the moment of inertia about their center of mass of three pennies each of mass M attached at their edges as shown in the diagram above? A disk of mass M and radius R rotates at angular velocity!
The puck rotates at radius R when the tension in the string is T. What is the change in energy of the puck when the radius is decreased? You may assume the puck is a point mass. A small puck of mass M is attached to a massless string that drops through a hole in a platform. Another disk of mass M and radius r is dropped on top of the rotating disk such that their centers coincide.
What is!? Both disks now spin at a new angular velocity!. Lagrangian and Hamiltonian mechanics comprise an elegant way of rewriting the results of classical mechanics. Note also that the Lagrangian is a function of the coordinates q.
Lagrangians and Hamiltonians are fascinating in their own right. This is sort of the idea of this formulation of mechanics: The particle is not allowed to take on any old Cartesian coordinates x. Of course. It is a peculiarity of the Lagrangian formalism that the coordinates and their time derivatives are considered as independent variables. While Lagrangians and Hamiltonians make it easy to write down the equations.
Lagrangian and Hamiltonian questions fall into just two categories: Note the minus sign! The Lagrangian is not the total energy of the system. The power of Lagrangian mechanics lies in being able to choose coordinates to describe only the directions in which the system is allowed to move. U is the potential energy.
We are leaving out many subtleties and special cases which are covered in standard treatments. So the most 24 ushnish gmail. Applying this recipe to our example. Unlike in Newtonian mechanics. Here is where things get tricky.
For the latter. The first is that signs are very easy to mix up. Once you have the Lagrangian. Speaking of derivatives. These equations are known as the Lagrangian equations of motion. In this case. There are several important things to note about the Euler-Lagrange equations. Questions about conserved quantities in Lagrangian mechanics are very common on the GRE.
Writing down the correct Lagrangian is typically the hardest part of the problem. Looking at 1. To reiterate. Whether or not this quantity is conserved. But for complicated examples.
As the name implies. With two slight restrictions. The tricky part about this formalism is once again the kinetic term. As we derived above. For a simple example. To construct H. On the GRE. In the case of angular coordinates. As long as the 27 ushnish gmail.
As with the Lagrangian. Looking at the first equation. But this time. Which of the following is a possible Lagrangian for the system? Which of the following is a possible Hamiltonian for the system?
Which of the following is a conserved quantity for the system? Without exception. This means that the force can be derived from a potential function U r.
Reverting to classical reasoning for a bit. By spherical symmetry. While one can discuss orbits quite straightforwardly using only the language of forces and Newtonian dynamics. As a result. And secondly.
The numerator has to contain the product in order for the whole thing to have dimensions of mass. The most common situation is the limit m2 m1 just noted. In doing so. So in a two-body problem. The orbit E3 is a special case. That is not to say that the orbit must be periodic: Three representative orbit energies are marked. To find the radius of these orbits. To learn more. Without getting into the details of the derivation. An orbit with energy E2 is bound. The first part is trivial: Remember that two bodies orbit about their mutual center of mass — the precise statement is that the sun and planet both undergo elliptical orbits.
The sun only sits still at the focus under the approximation that it is much heavier than any of the planets. If T is the period of a planetary orbit. In any case. Planetary orbits sweep out equal areas in equal times. We can even be a little sloppier. In the drawing below. As in the more general discussion.
The second part is fairly difficult to derive. Planets move in elliptical orbits with one focus at the sun. Recall the definition of l: Which of the following is a possible value of r so that the spring stays at a constant extension r throughout the entire motion of the particle?
A particle of mass m is attached to one end of a spring with spring constant k in a zero-gravity environment. The third law is rather tricky to derive. The spring is stretched to a length r and the particle is given an initial angular momentum l. An asteroid of mass m orbits the sun mass M on a parabolic trajectory. Suppose a new planet were discovered orbiting the sun. Which of the following relates its distance of closest approach d to its orbital velocity v at the point of closest approach?
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You may assume m is negligibly small compared to M. The angular frequency is given by r k! This convenient shorthand will be elaborated in the chapter on waves. The phase is the second constant of integration. Like much of the rest of mechanics. Even this last method should not be too bad for problems on the GRE.
To summarize. One very common GRE question is to ask for the normal modes of a system. Simple as it sounds. Try limiting cases. It turns out all of the solutions. Section 3. Try conservation of energy. Consider the general case of n masses attached to springs. The mathematical setup is as follows. Using our guess ansatz.
Let x1 and x2 denote the displacements of the left and right blocks. The force on the left block is the force due to each of the adjacent springs. The normal modes alluded to above are just the solutions at which the entire system oscillates collectively at a fixed normal frequency. This secular equation defines a polynomial whose solutions give n frequen- cies! With great foresight. For this guess. As above. Taking the determinant of the matrix and setting to zero gives.
This should remind you of the discussion of blocks stuck together in Section 1. We might guess that the other normal mode corresponds to motion of the p blocks in exactly opposite directions.
Depending on answer choices given in the problem, these observations may be sufficient to pick the correct answer. Solving for the normal modes of a system is a very common problem, in both quantitative and qualitative contexts. For a quantitative solution, the recipe is exactly as above. In many cases, this machinery is overkill, especially as most GRE problems are designed to be solved in about one minute. For most problems, you should ask yourself.
Drawing a picture is often helpful. A general rule of thumb is that symmetric motion will have lower frequency than asym- metric motion, with the most symmetric mode corresponding to collective oscillations, as in the p case above of the blocks moving in sync having the lowest frequency. As always in classical mechanics, our strategy is to write down the equations of motion of the system and find solutions.
There are three qualitative types of solutions: It is not critical to remember all of the constants in this expression, but it is valuable to remember that underdamped oscillations are a sinusoid with an expo- nentially decaying envelope. This should make intuitive sense: While the solutions to the underdamped case are sinusoidal functions i.
A harmonic system can also be driven by some external sinusoidal force. In this case, the equation of motion picks up an additional term due to the driving; in the most general case including possible damping. Without going into details, the important point is that unlike the case of the oscillator with no driving force, the amplitude of the steady-state solution that is, after a long time has elapsed is not a free parameter but instead depends on the coefficient A and the frequency!
For a given A, the amplitude is maximized when the driving frequency equals the so-called resonant frequency q! The amplitude D of an undamped oscillator of natural frequency! But this proportionality should hold well near resonance in the weak damping limit. We show a few common examples.
In the limit of small displacements, the equation of motion of a pendulum of length L is described by simple harmonic motion. Correspondence of quantities in the analogy between electrical and mechanical systems.
More precisely, R is the distance between the center of mass of the object and the pivot. The oscillation frequency, easily derived using Lagrangian mechanics, is now r mgR! See below for more details. Pursuing the electrical analogy, we can consider springs attached to a block of mass m in both parallel and series configurations.
When the springs are in series the right figure , the situation is slightly more complicated. Call the displacement of the massless joint between the springs x1 , and call the displacement of the block x2.
To determine the equations of motion, we can use a trick: This enforces the condition that there must be zero force acting on the joint between springs; otherwise, since the joint is massless, the acceleration would be infinite. This leads to a subtle pitfall: If we were to replace each spring k by a capacitor C, the rule of Table 1. To keep things straight, its best to remember the two rules as logically distinct:. A ball of mass m is launched at 45 from the horizontal by a spring of spring constant k that is depressed by a displacement d.
Suppose a motor drives a block on a spring at a frequency!. A block of mass M is attached to two springs. If damping is negligible. Another block of mass m is attached to three springs. What horizontal distance x does the ball travel before returning to its height at launch?
What is the ratio of the oscillation frequency of the block of mass M to the frequency of the block of mass m? The amount of fluid flowing past a point in the first section in a time t must equal the amount of fluid flowing past a point in the second section. To see how this works in context.
Errata for Conquering the Physics GRE - Physics GRE Prep
The cross sectional areas of the two pipes are a2 and b2. In our specific case. The first term is a form of kinetic energy. The second uses the concept of buoyant forces. Note that the density entering into both sides of the above equation is the same because of our assumption that the fluid is incompressible. Nearly all common fluids are incompressible.
Since the first two terms are just the usual kinetic and potential energy terms with the mass divided out. The pipe may go up and down. You can remember this equation by noting its relation to conservation of energy. How much do you weigh underwater? Suppose you weigh 60 kg and your volume is 50 L.
But your weight is N. Problems involving floating and submerged blocks can usually be solved by simply assigning all forces as usual. If the block displaces a volume of V of water when it is floating. The mass of the displaced water is. You are displacing 50 kg of water. Substituting equation 1.
You can think of this as the weight of the displaced water pushing up on the submerged object. If you need to know something about the velocity of the fluid. This is. If you need to know something about pressure. Since the pipe is horizontal except in the region between the two pipe sections.
Obviously the force of gravity pushes down on it. A diver in water picks up a lead cube of side length 10 cm. At the top of the hill. At the bottom of the hill. An aqueduct consisting of a pipe filled completely with water passes up a hill that is 10 m high. What is the frictional force required to hold the stopper in the tube? The ball also has a hollow spherical cavity of radius r concentric with the outer surface.
How much force is needed to lift the cube? A vertical cylindrical tube has radius 1 cm. The tube is plugged with a stopper at the bottom end and filled with water so that the top of the water is 1 m above the bottom stopper. As always.
Dividing by the mass to find the acceleration. The total net force p is then 5 2 N down the ramp. We know that p the applied force contributes a normal force of 5 2 N and a force up the ramp of 5 2 N. As explained in the text. So before considering friction. Fbb is the block-on-block force. March The plan is to solve for Fbb using the second block. The happy fact that the plane is at a 45 angle. So as long as the blocks have nonzero masses. Note that the normal force which provides the friction keeping the mass m stationary is the block-on-block force.
If you have time at the end of the test. Kinematics 1. For the vertical forces to balance. This must equal the acceleration of the first block for the two to move together. The GRE may not always be this kind to you. Substituting the result above. Its initial kinetic energy is zero. If the rod is uniform. In a geosynchronous orbit.
When the ladder hits the wall. Standard Pythagorean triples like this multiple of have a way of showing up on the GRE.
As with the rolling-without-slipping problems. But what we actually care about is the height of the center of mass of the ladder. If F0! Without doing any work. Now there is no change in potential energy immediately after being launched. Choice E is the only one which satisfies all these properties. At the top of the ramp. From the 30 60 90 triangle. Because it rolls without slipping. We now apply the work-energy theorem: The total vertical force must be compensated by the normal force of the surface on the block: Setting the zero of grav- itational potential at the bottom of the ramp.
We may be tempted to set this equal to the potential energy of the spring. It is important to keep the notation straight in this problem. Once again. Momentum p 1. So the spring is compressed by precisely the same distance. Straightforward conservation of angular momentum. This means that the velocity vectors can be arranged as the sides of an isosceles triangle. Since the density depends only on radius. Solving for!.
Angular momentum is conserved because tension acts radially and hence provides no torque. As usual. Solving the rest of the problem is just basic geometry. Call the initial angular velocity! Since the momentum must sum to zero. Work is required to pull the string downwards and change the radius of the rotation. This is conventional for pendulums. Lagrangians and Hamiltonians 1. The potential energy is all gravitational.
Note the minus sign in front of the z coordinate! Laboratory Methods Solutions: Laboratory Methods 8. Particles and Interactions Nuclear Physics: Specialized Topics Solutions: Tips and Tricks Solutions: Undeterred, we headed to our local brick-and-mortar bookstores, where we found a similar situation.
There were practice books for every single GRE subject exam, except physics. Further web searches unearthed , containing every problem and solution from every practice test released up to that point, and , a web forum devoted to discussing problems and strategies for the test.
We discovered these sites had sprung up thanks to other frustrated physicists just like us: We have also included review material for all of the nine content areas on the Physics GRE exam: Exam-style practice problems and worked solutions are included for each review chapter, giving over additional GRE-style practice problems in addition to the from the exams.
The shorter chapters have review problems at the very end, while the longer ones have review problems distributed throughout the chapter. The chapter on quantum mechanics and atomic physics is the longest, for two reasons: Unique to our book is a chapter on special tips and tricks relevant for taking the GRE as a standardized multiple-choice test.
Conquering the Physics GRE - Yoni Kahn, Adam Anderson
Some of the standard test-taking wisdom still applies, but we have found that the structure of the multiple answer choices often provides valuable hints on how to solve a problem: Next, a brief word on what this book is not. This is not a detailed review of undergraduate physics: We believe this will help you succeed on the Physics GRE, but if any of the standard subjects are completely unfamiliar to you, please do not try to teach them to yourself from our book.
There are many excellent texts out there relevant for that purpose, and we have included a list of them in the Resources section following this preface. We strongly encourage you to consult these references, as we have found them useful both in writing this present text and x Preface in our careers as active physics researchers.
We will often refer to them throughout the review chapters. Last, a comment on the structure of this book.
We realize that there are many, many equations to learn that are relevant for GRE-style physics problems. To keep the amount you feel you have to memorize to a minimum, we only assign equation numbers to equations we feel are important to remember — everything else you can safely ignore. This is not to say that you should memorize every single numbered equation — Chapter 9 contains useful advice for what to memorize and what to derive.
A book like this could never have been written without the help and support of other people. We especially thank Yichen Shen for his useful contributions to the condensed matter section of the Specialized Topics review. We thank Jen Sierchio and other members of the physicsgre. We could even put a massless string between the two blocks, and the argument would still hold: since the whole system must fall with acceleration g, there can be no tension in the string. Do the free-body analysis and check this yourself!
When an experimenter holds the 20 kg block stationary, the tension in the string is T1. The experiment is repeated with the 20 kg block hanging under the 5 kg block, and the tension in the string is now T2. This intuition is confirmed by a limiting-cases analysis: if the mass of the lower block is zero, then no matter the mass of the upper block, the string just dangles below the block with no tension, so the tension must be proportional to the mass of the lower block but independent of the mass of the upper one.
In order to treat both cases at once, call the mass of the top block m1 and that of the bottom block m2. The forces on the two blocks are illustrated as follows: F is the force applied by the experimenter. Notice how the string tension acts up on the bottom block but down on the top block, and that the magnitude T is the same for both blocks. For the purposes of the GRE, this is the definition of a massless string: it carries the same tension at every point. In the second setup, you might be asked, given friction between the two blocks, what the minimum force is such that the mass m does not fall down due to gravity, or if m is placed on the surface as well, how the force of one block on another changes depending on whether F is applied to M or m.The above information is conveniently represented in terms of complex numbers: The Physics GRE was very similar to the test, so it pays to ensure you understand and can quickly do all practice questions.
Since the first two terms are just the usual kinetic and potential energy terms with the mass divided out. If its tangential acceleration is zero. Consider two waves. What is the ratio of the oscillation frequency of the block of mass M to the frequency of the block of mass m?
Thus there is no concept of polarization for ordinary sound waves. At the top of the ramp. Optics and Waves 4 78 4.
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