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PHYSICS COMPREHENSIVE

REVIEW

By Larry, Jeff and Jeff



Add thin film optics

Not for circulation outside USNA



CAUTION: This guide covers 35 problems that will not appear on your Fields Test. Also under stand that the guide makes statements that are usually true without adding qualifying remarks or apologies. Advanced or special cases are not treated. In some cases, one or two steps are omitted. Solve each problem as you read through this guide.
STRATEGY: The best approach is to recognize the problem and solve it. If you are unsure, find physical reasons to eliminate as many of the five options as possible. Make an educated guess as you pick from among the remaining options. Mark the problem so that you can return to it later at which time you can attempt a more thorough analysis. As you work through the guide, pay special attention to the examples for which options are found not to be viable.
DO NOT MEMORIZE: Memorizing values and facts from this guide is not recommended. Do think about the concepts and orders of magnitude as you read this guide. Think and read just carefully enough to be left with a hazy recollection of the experience. THINK, think, think while you take the test.
The sample questions appear boxed. Search for “Ex.” to find the next problem.

MECHANICS AND RELATIVITY:

Newton’s Laws are best written for particle mechanics. The laws are written in forms appropriate for inertia reference observes. A particle is a body for which every point in the body has the same velocity. If one considers a rotating billiard ball, there is no one velocity and hence no one acceleration for all the mass points comprising the ball. The center-of-mass theorems are the bridge that allows one to generalize Newton’s laws to forms that apply to more general bodies.


I.) A particle maintains its state of uniform motion unless acted upon by an (external) force.
Uniform motion is motion at a constant velocity .
II.) The net force that acts on a particle causes that particle to accelerate. The net force is the vector sum of all the forces that are acting on the particle, and that net force is equal to the mass of that particle times its acceleration. (The word causes is of cosmic importance in physics.)
III.) If particle A exerts a force on particle B then particle B exerts a force of equal magnitude and opposite direction on particle A.
Newton’s Third Law deals with forces and particles.
General Reciprocity (NOT Newton’s Law, an extension of Newton’s third law): If a physical entity A exerts an influence on entity B, then B exerts a proportionate counter influence on A. For example, an electric charge causes an electric field and an electric field exerts a force on electric charges. A moving charge creates a magnetic field, and a magnetic field exerts forces on moving charges. Note that the entities in these examples were not equivalent so the counter influences were not equivalent. The counter nature is observed when a time-varying magnetic field induces a current in a conductor. The magnetic field caused by the induced current fights (counters) the variation in the magnetic field. The magnetic field in the interior of a perfect conductor must be constant in time as the currents induced in such a conductor are as large as required to completely counter the disturbing influence. Newton was only setting down laws for forces and particles, not for these more complex situations.
Newton’s Laws: Use when asked about forces and acceleration at a single position and time.

Circular Motion :

The centripetal part is always in play. The tangential part applies if the speed is changing.


Friction: Static friction acts with a magnitude from 0 to s N to keep the contacting surfaces from slipping relative to one another. Once the surfaces are slipping, the friction force on each surface has magnitude equal to k N directed to oppose the relative motion. Factoid: k  s.
Kinematics, the description of motion, is always in play. It is used in all mechanics problems.
Use Newton’s laws to solve Here & Now problems (related to force and acceleration).
Work-Energy: Work is W = suggesting that work-energy by used when a change is associated with a change in position.

Mechanical energy: Ei = K1 i + K2 i + … + UA i + UB i + …..

K == ½ McmVcm2 + ½ Icm2 = ½ McmVcm2 + ½ mvrel 2 + ... = ....

U = ½ k (stretch)2 + {mgh; - GMm/r} + {q V; kQq/r}

choose near-earth or Universal form choose appropriate form

Ei + Wnon-conserve = Ef Wnc-frintion = - k N *(distance that surfaces slip relative to one another)

A conservative is any force for which the work integral can be evaluated as the difference in the values of a scalar function at the integration endpoints. (Can do if the work integral around any closed path is zero; curl of the force is zero).

Use Work-Energy to solve Here to There (r) problems.
Momentum and Collisions: Impulse is suggesting that impulse-momentum methods be tried for a change that occurs as time changes – even for the time change from before to after.

Internal forces do not change the total momentum.

Use Impulse-Momentum to solve Now to Then (Before to After, t) problems.
(1) Collisions: Before to after or now to then problems
Rule 1: Conserve momentum vectorially (component by component).

Rule 2: Kinetic energy is unchanged for an elastic collision. Some kinetic energy is converted to other forms if the collision is partially or totally inelastic. As the total momentum is conserved, only the kinetic energy associated with motion relative to the center of mass is available to be lost (converted).



;

M: total mass; V: speed of the center of mass (CM); vrelative, i : speed of mass i relative to CM.

In a totally inelastic collision, the particles stick together so Krelative is lost. KCM remains as required by conservation of momentum.

300

300

m, 2 vo

2 m, vo





Ex. 1) The figure to the left shows two particles with masses and velocities as indicated. The objects are moving on a flat, frictionless surface. When they collide, the objects stick together. Their speed after the collection is most nearly:
(A) 0.67vo

(B) 0.87vo

(C) vo

(D) 1.15 vo

(E) 1.73vo


Type: Before to After  impulse-momentum method  collision sub-type

Rule 1 is to conserve momentum vectorially so a coordinate system is adopted.

300

300

m, 2 vo

2 m, vo



y

x
The momentum of the particles before the collision are and . The total mass is 3 m and the total momentum is .

(inelastic; stick together) D

Was kinetic energy converted to other forms as a result of the collision?

The collision was totally inelastic. Was all the kinetic energy lost?
(2) Motion in a uniform gravitational field:
Ex. 2) Ball 1 is dropped from a height h and ball 2 is dropped form a height ½ h. Which of the following gives the ratio of the speed of ball 1 to that of ball 2 just before they impact? (Assume that air resistance is negligible.)

(A) 2 (B) 2 ½ (C) 1 (D) 2 (E) ½


First, the problem is a constant acceleration problem. Write down the master equations.

x(t) = xo + vo t + ½ a t 2; v(t) = vo + a t ; v 2 = vo2 + 2 a (xxo); vave = ½(v +vo)
The balls fall from rest for a distance H. Using v 2 = vo2 + 2 a (xxo), v 2 = 2 g H or .

= B

Sanity Check. Three answers can be discarded immediately. Which three? Why?
If the answer were (A), ball 1 would have twice the average speed of ball 2. What would be the ratio of the fall times in that case given that the first ball falls twice as far? Would you expect them to hit at the same time if dropped at the same time?
(3) Law of Universal gravitation:
Ex. 3) Two planets of mass m and M respectively have center-to-center spacing R. At what distance from the planet of mass M do the gravitational forces of the planets cancel each other.

(A) (B) (C) (D) (E)

One need only consider points along the line joining the planets. Why? What must be true about the directions of two vectors that sum to zero?
Prepare a sketch: A large well-drawn sketch provides the greatest benefit. Assume M > m. R
d

R - d

M

m
The distance from M is sought. Assign the symbol d to that distance.

Thought 1: Gravitational forces are attractive. Therefore, the ‘zero – force’ point must be between the planets and along the line joining them. Compare/contrast this situation with that in Ex. 10.
Thought 2: The force is an inverse square (with distance) law force.

Think about the result. At the point where their influences balance, the distance from the smaller planet is smaller than that from the large planet. (Include sanity checks as you proceed through your solution.) Assume m < M as you reason through the possibilities,



E

Since is less than one, d > ½ R. The point is farther from the more massive planet. Reread the question. The distance d of the point from the more massive M planet is requested. The unknown label d is assigned to the value requested. Always review the question to ensure that you have answered the question that was asked.

Assume m <<< M, say m = 0.01 M: Answers A, B, C and D correspond to points closer to M or to points not between the planets. Why must the correct point be between the planets and be farther from the more massive planet?

If you expect d to be greater than ½ R, you might revise the figure to reflect this expectation.


METHOD: Try a set of values that simplifies the problem: Midn Jasperson (2011) proposed that one might just look at the case that m = M in which case the equilibrium point should be at the mid-point of the line joining the masses, distance equal ½ R. Only response (E) satisfies that test.
(4) Statics and Archimedes Principle
In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques (about any axis  free to choose it where you wish) acting on the body is zero.
A body is buoyed up by the weight of the fluid that it displaces. FB = fluid Vdisplaced g.
Ex. 4) A metal block is suspended in an empty tank from a scale that indicates a weight of W. The tank is then filled with water until the block is covered. If the density of the metal is three times the density of water, what apparent weight of the block does the scale now read?

(A) ½ W (B) 2/3 W (C) W (D) 3/2 W (E) 3 W


Examine your options. The new reading should be less than the original reading. Answer (B) could be chosen without much more as the one third density means that the mass will be buoyed up by 1/3 of its weight if completely immersed, but one should solve the problem before choosing an answer.
Prepare drawings

mVg



T = Wdry

mVg



T = Wwet

wVg


Wdry = m V g = 3 w V g Wwet­ =m V g -w V g = (m -w) V g

* The ratio of the density of a material to that of water is called the specific gravity of the material.

Wwet­ = Wdry ( -w/m) = Wdry ( - /) B
(5) Simple Harmonic Motion
Equation of Motion: x(t) = A cos[ t + ] = C cos[ t ] + D sin[ t ]

Take derivatives to find expressions for v(t) and a(t).



v(t) = -  A sin[ t + ]; a(t) = - 2 A cos[ t + ]

Energy Conservation: ½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical

With the convention that U = 0 when the particle is at the equilibrium position, the energy is all kinetic as the particle passes through the equilibrium position and all potential at the turning points.

 =

A mass hanging from a linear spring executes simple harmonic motion about its equilibrium position. Potential energies are associated with systems of interacting entities. Potential energies are associated with pairs of things (at least) or the system while a kinetic energy can be owned by a single entity. The potential energy can be shifted by an additive constant. For SHM, it is often set to zero for a particle at the equilibrium position.
Ex. 5) A 1-kilogram particle is attached to a spring and exhibits one-dimensional simple harmonic motion. The particle’s distance (!/* displacement) from the equilibrium position is given by the expression: y(t) = A sin[ t + /2], where A = 1 meter and  = 0.5 rad/s. If the potential energy of the particle (system) at its equilibrium position is null (or zero), which of the following gives the total energy of the particle (system)?

(A) 2 J (B) 1 J (C) ½ J (D) 1/8 J (E) 0 J


v(t) = -  A sin[ t + ]; a(t) = - 2 A cos[ t + ]

vmax =  A; amax = 2 A

½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical

The total energy is kinetic at equilibrium as the potential is set to zero at that point.

½ m (vmax)2 = ½ k A2 = Etotal mechanical

The relation v(t) = -  A sin[ t + ] shows that vmax =  A = 0.5 m/s. Using m = 1 kg,

Etotal = ½ m (vmax)2 = ½ (1 kg) (0.5 m/s)2 = 1/8 J. D

Alternative: Etotal = ½ k A2 = ½ (m 2) A2 = ½ (1 kg) (0.5 rad/s)2 (1 m)2 = 1/8 J.

Write down the equation that is to be used. Substitute numerical values with units for each symbol writing the values in the same geometric pattern that was used for the symbols. Adopt procedures that reduce the chance of making a careless error.
(6) Equilibrium with torques
In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques about any axis (  free to choose axis where you wish) acting on the body is zero.



The displacement from the axis to the point of application of the force is , CCW is positive, and  is the angle that takes you from the direction of to the direction of .


Vector cross product: Extend the fingers of your right hand in along the direction of . Orient your hand so that the fingers can curl toward the direction of . The thumb of the right hand will be in the direction of .

axis




line of action of F



r

500


Ex. 6) The figure shows a uniform rod of mass 4 kilograms that is pivoted at one end and supported by a string at the other end. If the rod is at rest (in equilibrium) the tension in the string is most nearly

(A) 20 N (B) 26 N (C) 31 N (D) 40 N (E) 62 N



Use the figure The rod does not have an assigned length. Assign a length of , 1 m or even 4 m. If the value is not given, the answer must be independent of its value.

Choose an axis: The pivot point is a natural choice for the axis. Picture the situation. If the tension were changed, about what point would the rod rotate?

500

400

400



½

W






The tension force tends to cause a CCW (+) rotation so the associated torque about the pivot is T = + T  sin(400). To compute torque, the weight of the boom can be assumed to be applied at its center of mass.

The weight has a lever arm of ½  and tends to cause a CW rotation. W = - W (½  ).


The sum of the torques must be zero so:



= 30.5 N C

lost calculator  approximate using sine of 450; 40 < 45 so smaller and T larger for same  =T

We expect an answer larger than 28 N so response (C) is chosen. The actual value is about 30.5 N.
(7) Relativity Relative speed v

Lengths are contracted along the direction of relative motion, but lengths running transverse to the motion are unchanged.

A clock in motion relative to the observer runs slowly.

If a time t = 10 s elapses as measured by a moving clock, and it appears to be running slowly as viewed by the primed observer, then the corresponding interval t is longer than 10 s.


Consider a primed observer moving at relative to an unprimed observer with the initial conditions that their respective axes are parallel and the origins coincided at t = t  = 0.

Adopt the notations: .





Ex. 7) A stick of length L lies in the x-y plane as shown. An observer moving at 0.8 c in the x direction measures the length of the stick. Which of the following gives the components of the length as measured by the moving observer?

L

x

y

O








Lx


Ly

(A)



L cos

0.60 L sin

(B)



0.6 L cos

0.60 L sin

(C)



0.6 L cos

L sin

(D)



0.64 L cos

0.64 L sin

(E)



0.78 L cos

0.78 L sin




The component along the direction of motion should be contracted by a factor of  = 0.6 while the transverse length should be unchanged.

Lx = 0.6 (L cos) ; Ly = Ly = L sin

Note that only answer (C) has an unchanged transverse component. C



The following relativity material is under development.

There are four vectors, and they all transform according to the same transformation rules (Assuming the same relative velocity and with .).



For the position vector, x0 = ct, x1 = x, x2 = y and x3 = z. The four-momentum is: (p0, p1, p2, p3) =

(E/c, px, py, pz). Note the manner in which factors of c are used to ensure that all components have the same dimensions.
The transformations ensure that there is an invariant metric product.

Invariant interval: s2 = (c t)2 - (x)2 - (y)2 - (z)2 = (c t)2 - (x)2 - (y)2 - (z)2

This result is useful when one considers the interval between two events. If s2 = (c t)2 - (x)2 - (y)2 - (z)2 is positive, the interval is called time-like and the two events can be causally related. If s2 is negative the events are separated by a distance greater than that which light can travel in a time |t|, and the interval is called space-like and the two events cannot be causally related. Points at

http://en.wikipedia.org/wiki/Light_cone



a time-like interval from the current event are in the future and past light cones of that event and can linked causally to events inside the light cone. Points at a space-like interval form the event are outside the light cone and can not be causally related to the event at the apex of the cones.

Similarly: p2 = (E/c)2 - (px)2 - (py)2 - (pz)2 = (E/c)2 - (px)2 - (py)2 - (pz)2



When one considers the momentum of a single particle, the invariant p2 can be evaluated in the rest frame of the particle with the result that in all frames where m is the mass of the particle. The general expression for the moment of a particle is p = (v mc, v m) where is the standard three velocity of the particle, m is the (rest) mass and .

Using the invariant p2 = (E/c)2 - (px)2 - (py)2 - (pz)2 =m c2, E2 = m2c4 + c2 = Eo2 + c2



E2 = Eo2 + c2 (v m v)2 = Eo2 + c2 (v Eo v/c2)2 = (v Eo)2E = v Eo. The kinetic energy of the particle is the increase over the rest energy. K = E = E Eo = (v – 1) Eo.

Total energy: E = v Eo.

Kinetic energy: E = (v – 1) Eo (small v expansion; binomial)
Sample calculations:

In one frame, event B occurs 4 s after event A and event B is 1 km from A. In a second frame, event B occurs 5 s after event A. a.) Could event A have caused event B? b.) What is the observed distance d between the events in the second frame?

a.)s2 = (c t)2 - (x)2 - (y)2 - (z)2 = (3 x 108 m/s * 4 x 10-6 s)2 – (1000 m)2 = 440000 m2 > 0

The interval is time-like so event A could have caused event B.

b.) The interval is invariant sos2 = 440000 m2 = (3 x 108 m/s * 5 x 10-6 s)2d2 d = 1345 m.
An electron has a rest momentum (0.511 MeV/c, 0, 0, 0). a.) What is the electron’s energy when it is if it is moving at 0.8 c relative to the observer? b.) Assuming that the electron is moving in the positive x direction relative to the observer, what is its velocity if its energy is 1.25 * 0.511 MeV?

Using the invariant p2 = (E/c)2 - (px)2 - (py)2 - (pz)2 =m c2, E2 = m2c4 + c2 = Eo2 + c2



E2 = Eo2 + c2 (v m v)2 = Eo2 + c2 (v Eo v/c2)2 = (v Eo)2E = v Eo

a.) E = v mc = v Eo = [1 – 0.82]= 0.852 MeV.

b.) E = v mc =  Eo  [1 – 2]= 1.25 MeV. 1 – 2 = 1/(1.25)2.  = 0.6 so v = 0.6 c.

With the other assumptions, the particle is traveling at 1.8 x 108 m/s in the positive x direction.


Alternative solution: p = (v mc, v m)  (0.511 MeV/c, 0, 0, 0) in the rest frame. Apply the transformations:

a.) pc =  [0.511 MeV/c -  (0)] =  c-1 Eo E =  Eo ()



Attempt to solve b.) using the transformation method.
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