It's Not Easy being Green...(When Earth Day is Long Gone)

2006 solutions are here.

Just a little bit closer now...

Kinetic Theory/Thermo Solutions:
1 B
2 B
3 A
4 E
5 C
6 C
7 C
8 D
9 E
10 A
11 E
12 A
13 C
14 D
15 A
16 A

'03 solutions, and 1993 solutions

I Know These Constants Are Correct, by Wally Sheep

(Anybody know the book by Wally Lamb? I guess it's a little obscure....)

Here is a link to the table of constants given on the exam. Use it as you do the part II problems.

Abbot's Surprise: A Magnetic Personality!

1. a
2. b
3. c
4. a
5. a
6. c
7. b
8. e
9. b
10. b
11. a
12. a
13. c
14. a
15. e.
16. b
17. d
18. a
19. c
20. e

And, for your viewing and char-broiled rubric-flavored enjoyment, the rubric for the 1994 exam.

What do you Quantum?

1D
2A
3D
4C
5C
6B
7B
8B
9C
10C
11D
12A
13A
14D
15E
16A
17E
18A

1981B1
a) F = Ffric = 20 N
b) Wnet = Fnet*delta_x*(1) = 60 J, so the net force is 15 N. Since the friction force is 20 N, this means the new force F' = 20 N + 15 N = 35 N.
c) The net force acting on the block is 15 N, and the block has a mass of 10 kg. The acceleration is then a = Fnet /m = 1.5 m/s^2.

2.
a) The work done to compress the spring is equal to the final kinetic energy of the block.

KE = 1/2mv^2 = Wspring = 150 J
b) When the blocks are held together, the energy stored in the spring is again 150 J. Thus the kinetic energy when the blocks are moving apart is 150 J.

1/2*M1*v1^2 + 1/2*M2*v2^2 = 150 J

Conservation of momentum also applies:

total initial P = total final P

0 = m1*v1 - m2*v2

The system can be solved for v1 = 15 m/s, v2 = 5 m/s

3.
a) Fe to the right, gravity down, T diagonally to the left at 30 degrees from the vertical.
b) T cos(30) - mg = 0 so T = .058 N
T sin(30) - Felec = 0 and Felec = qE so E = 5.8 x 10³ N/C
c) Since there is a net force in both the downwards direction (due to gravity) and the horizontal direction (due to the electric force), the ball will follow a straight path down and to the right.


4.
total Req = 8 ohm, total emf in the circuit is (60 V - 12 V = 48 V, notice that the second battery is backwards!)

This means that the current in the circuit is 48 V / 8 ohms = 6 Amperes.

a) V across the parallel combination is 6 V, so the current through the 2 ohm resistor is 3 A.

b) P = I^2*R = (6 A)^2 * (3 ohms) = 108 W

c) Because of the direction of current, the battery is being charged. The terminal voltage is the battery voltage plus the voltage added by the resistor going backwards.

V = 12 V + 6A*1 ohm = 18 V

5.
a) Ray diagram:

b) The image is real

c) Use the thin-lens equation equation and solve for di = 9 cm

d)


On this part, you can use the trick suggested in the image, OR you can use the image as the object for lens 2. The important thing is to NOT change how the rays are drawn relative to the primary and secondary foci. The first ray enters the lens parallel and towards the left, and is "brought towards" the primary focus.

The only weird thing is that since the primary focus is on the same side as the "object", the light ray is bent as if it comes from the primary focus. I can show this in class tomorrow.

Oh Say, can you c...(where c is the speed of light)

1 D
2 D
3 E
4 E
5 E
6 E
7 B (n1 = 1.2/.8)
8 C
9 D
10 E
11 D (convex mirror is similar to a diverging lens, in that it always produces a virtual image of an object)
12 D (this is a magnifying glass! remember that converging lenses can form both real and virtual images)
13 D
14 D (be sure to draw BOTH the reflected and refracted ray. The angle of the reflected ray will always be 40 degrees from the normal, which means the angle from the surface is 50 degrees. Depending on the index of refraction of the material, the refracted ray will be at most 40 degrees from the vertical, which is an angle of 50 degrees from the surface. Thus the total angle will be between (40 + 50) to (40 + 90 degrees)
15 C

1995B2.
a) R = V^2/P = 24 ohms
b) Energy used = P*delta_t = 6W * (30 days*24 hrs/day*3600 seconds/hr) = 1.6 x 10⁷ J
c) The bulb should be in series with the resistor, and the two should be in parallel with the battery terminals. The toaster should be connected in parallel with the battery so it gets 120 V across it.
d) The bulb needs 12 V across it, which means the resistor must have 108 V across it to add to 120 V. The power of the bulb is 6 W, which by P = IV, means the current must be 0.5 A. Therefore the resistance must be V/I = 108 V/0.5 A = 216 ohms
e)
i. If the variable resistor is increased, the current through the branch will decrease. Since the resistance of the bulb remains the same, the power P = I^2*R will decrease, and the bulb will get dimmer.

ii. The toaster has the same amount of voltage across it, regardless of the resistor. the power will therefore stay the same.

1995B3
a) T directed up, mg down, so T - mg = 0 in equilibrium, and T = mg = 1 N

b)
T sin(th) = Th = ma = (0.1 kg)(5 m/s^2) = 0.5 N
T cos(theta) - mg = 0 with Tv = T cos(theta) = mg = 1 N

c) Constant speed! in equilibrium.

T - mg = 0, so T = Tv = mg = 1 N.
Th = 0 since there are no components of T in the x direction.

d)
Th = ma_x = (0.1 kg)(5 m/s^2)*sin(30) = 2.5 N
Tv = ma_x...

I will post more later, I just realized I am late for an appointment!

Wave/Interference Review

Waves/Interference Review Solutions:
1 D
2 A
3 C
4 C
5 B
6 A
7 D
8 E
9 C
10 E
11 C
12 D
13 A
14. B (the radical symbol should end before the f)
15. B
16. A
17. B
18. A
19. D

Preview for tomorrow...let's shed some light on MC optics questions....

Working Hard, or hardly working?

Hi everyone,

I want to thank those of you that have contacted me thus far. Those of you that have not...it's cranky time....(Azeezat, Maria, Danielle, Brandon, Sam, Marlon...did you lose my number?)

I've been working hard in Cleveland to get wedding details figured out - running around the city, picking things up, dropping them off, and of course, deciding on cake...here's the evidence of my "hard work" after dinner tonight:




Actually, I'm feeling a bit sick. I never thought I could have too much cake.

Thankfully, you can NEVER have enough physics.

Take care everyone,

EMW

So You Think You Can Physics?

Hi everyone,

I hope you are enjoying your time away from school...I just wanted to give you guys a link to the rubrics for the 1986, 1987, and 2007 exams.

Click on the date and save the file to your computer.

Be SURE to contact me and let me know how things are going!

Is this really the end? I was just getting into it....

Problem 1.
a + b)


Don't worry about the fact that the signs are different from what we discussed in class today - your answers to the other parts are still consistent with the solutions below, and will be correct.
c) Highest energy is the 6 eV transition.
d) lowest is the 1 eV transition.
e) This is the reverse of c and d - longest wavelength will be the 1 eV transition since E = hc/lambda
f) shortest wavelength is with the 6 eV transition, which has the highest energy.
g) The difference in energy between these two energy levels is 5 eV. This translates into a photon of wavelength 248 nanometers.
h) Three are possible.
i) The electron has KE = 7.12 x 10^-19 J = 4.45 eV. This is enough energy to cause an n = 1 to n = 3 transition, but not n = 1 to n = 4.

Problem 2.

1996B5.
a) Z = 102

b) K = 1/2*mv^2 = 8.42 MeV = 1.35 x 10^-12 J
alpha particle mass = (4 u)*(1.66 x 10^-27kg
v = (2*K/m)^1/2 = 2.02 x 10⁷ m/s

c) A correct explanation includes a reference to:
1. a change in mass/mass defect being converted into KE fo the alpha particle

2. conservation of energy - binding energy of the nucleus is converted into KE of the alpha particle during the decay.

3. work-energy theorem - the electric force does work on the alpha particle as it moves away from the nucleus, thereby changing its KE.

Nuke it - Mr. Weinberg has no Gas Stove

Chapter 30:

Q1. Isotopes of a single element have the same number of protons, so they act the same way in a chemical reaction as each other. They have different numbers of neutrons, so the mass of isotopes vary.

Q2 & 3.
a) Uranium-232, 92 protons, 140 neutrons
b) Nitrogen-13, 7 protons, 6 neutrons
c) Hydrogen, 1 proton, 0 neutrons
d) Strontium-82, 38 protons, 44 neutrons
e) Berkelium-247, 97 protons, 150 neutrons

Q10.
Gamma rays are photons or electromagnetic waves, so they have no mass and no charge. Alpha particles have a relatively large mass, a positive charge, and high energy. beta particles have a relatively small mass, a negative charge, and also high energy.

Weinberg Problems:
1.
a) 2.42 m/s
b) 0.642 Hz
c) 3.91 N

2.
a) 225 ohms
b) 75 ohms

3.
a) By the right hand rule, positive charges moving west in the Earth's magnetic field (directed North) will experience a downwards force, so the bottom part of the tube will become positive.
b) 200 m/s (pretty fast!)


4.
a) Green light has a wavelength between 500 and 550 nanometers, so any value in this range is a good estimate.
b) Using c = f*lambda and lambda = 525 nm, f = 5.7 x 10¹⁴ Hz
c) Frequency is the same, wavelength becomes 525 nm/ 1.4 = 375 nm.
d) The path difference is 2t, and the wave inverts twice upon reflection because the index of refraction of the reflecting surfaces is larger in both cases.

Thus 2t = m*lambda, and for the lowest nonzero thickness, t = 187.5 nm.

Are you here from the Wave side, or the Particle side of the family?

1.
a) 2.8 eV
b) 1.7 x 10¹⁸ photons
c) 5.5 x 10⁵ m/s
d) 1.3 x 10^-9 m


2.
a) 6.38 x 10^-10m
b)335 nm
c) The diagram consists of three transitions, 1.2, 3.7, and 4.9 eV.



3.
a) The best fit curve is shaped like a cosine curve.
b) 1.408 x 10^-14 m
c) 4.71 x 10^-20 kg*m/s
d) 8.08 x 10^-14 J

4.
a) 1.33 x 10¹⁵ Hz
b) 276 nm (not visible), 497 nm, 621 nm
c) Because of the energy of the photons in the given range, transitions can occur from the ground state to either the 1st or 2nd excited states.

5.
a)122 nm
b)The energy is the difference between the ionization energy and the energy of the second photon.
13.6eV - 10.2 eV = 3.4 eV
c)8.2 x 10¹⁴Hz
d)
1995B5.

I Emitted the Photon...YOU just didn't absorb it...

1997B Part II.
a)
b) The energy must be -1.9 eV, based on the energy emitted by the 600 nm photon.
c) 1240 nm - outside of the visible spectrum


200...Something, #7:
a) There are actually two (correct) ways to draw the diagram:

b) 1240 nm, again, outside of the visible spectrum.


7.
a) 9.9 x 10^-15 J
b) 3.3 x 10^-23 kg*m/s
c) The photon wavelength increases because some of the energy of the incident photon was lost to KE of the electron.
d) 6.0 x 10^-23 kg*m/s

1993B6
a)1.69 x 10¹⁹ Hz
b)3.73 x 10^-23 kg*m/s
c)3.31 x 10^-16 J
d)2.46 x 10^-23 kg*m/s