I can see clearly now!

Solution to O5 exam:


Almost there guys. Hang tough!

When Life Gives You lemons, make a battery....

I want to thank you all for the work you are putting in - it is making a difference, and I am proud of how far you have all come since the start of our review!

Solutions for yesterday's 2003, and the ninety 8 for the weekend.

Here is the worksheet you can use for predicting your score on the 1998 exam.

Give it a try and let me know what you think - I have tried to work out ALL bugs, but I may have missed one....

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.