Friday, December 19, 2008

Happy Holidays!

Hey, y'all. I haven't yet graded the free response portion of the test, but the multiple choice looks pretty good! I will not get the exams graded before the new year, but I'll be sure to post here when final grades for the semester are up.

You should be proud of all the energy, time, and work you put into this semester. You've learned a ton, and I look forward to building on it next semester. Until then, have fun, be safe, and get as much sleep as you can!

Wednesday, December 17, 2008

Section 3 problems 12 - 21

12. (a) and (b)



(c) 5 m / s ^ 2 = a x


13.
(a)


(b) 4.0 N

(c) 11/3 m/s ^2 = a x

14.




(a) 2.0 m / s ^ 2

(b) T = 16 N

15.
(a) 16 N
(b) no acceleration in z-direction
F net external = 20 N

16.(a) 20 N
(b) 20 N
(c) 140 N

17.



(a) 18 m / s ^ 2
(b) to the right
(c) to the left
(d) down
(e) into the center of the circle or up
(f) into the circle or down
(g) acceleration is to the right and the X-axis is to the right.
for all cases F net = 36 N

(h) 56 N
(i) F or at T = + 16 N or it is down
(j) theta = 29 degrees

18. (a) 100 N
(b) 150 N

19.
the elevator accelerates upward, then moves at a constant velocity and finally decelerates and comes to a stop

20. the sphere of the larger mas has the greater acceleration

21. Take the X-axis in the direction of the acceleration or up the plane and the Y-axis perpendicular to it.

Section 4: Answers

Hey, guys.
Section 4 was all about work and energy -- let me know if you have questions -- Some of these are SUPER hard!



1. Proof by drawing
2.(a) 72 J (b)-48 J.(c) = 24 J.
3. thought experiment.
4. thought experiment
5. (a)48 J(b) 0 (c) 0(d) - 40 J(e) 8 J.
6. work done by gravity 3.35 x 10-2 J.Work by frictional force of air =-3.35 x 10-2 J.
7. 51 x 102 J.
8. (a) 8.0 N (b)120 J.(c) 0. (d) -48 J.(e) -32 J.(f) 40 J.(g) 210 m/s.
9. (a) 19 N (b) -76 J(c) 96 J(d) 0(e) 0(f) 20 J(g)25 m/s.
10. (a) 10 J.(b) -10 J.(c) explain...(d) 0
11. (a)12 J.(b) -10 J.(c) explain(d) work energy theorem.
12. (a) -216 J.(b) 8.0 m/s.(c) 8.3 m
13.(a) Fnet=0 (b) = 29 J(c) 26 J.
14. 2.5 N
15. 21.7 x 10-19 J.
16. (a) 800 N(b) 40 J. (c) 40J.(d) 0.05 m.
17. (a) 1(b) 0 (c) 1(d) 0(e) -2 J
18. 7.7 m/s
19. (repeating #18, now with friction) 6.3 m/s
20. (a)12.0 J. (b)4.0 J.(c) 1.6 kg. (d) 5 m/s.
21.(a) = 5R/2. (b) 3 mg
22. 4 m/s
23. (a) (2gL)1/2 (b)T = 5mg(c) -mgL (d) -2mgL




Section 3: Answers 1-11

Hey, guys.
The big ideas in this section were about Newton's Laws.

1.
(a) 1.0 m/s ^ 2
(b) 0.5 m/s ^ 2.
(c) 4.0 kg.
(d) 2.0 N


2. (a) and (b). For m = 2.0 kg, FN = Fg = 20 N
(a) a= 6.0 m/s ^ 2
(b) a=12.0 m/s ^ 2
(c) For m = 1.0 kg, F N = F g = 10 N. a x = 12 m/s ^ 2.


3.
(a) a x = 4.0 m/s ^ 2
(b) a x = 10 m/s ^ 2


4.
(a) F net external = 2 m/s ^ 2
(b) F A B = 8 N
(c) F B A = 8 N


5.The Newton third law of motion forces are: (1) FRP and FPR, (2) FRH and FHR.



6.
(a) (F net external) x = 2 m/s ^ 2 = a
(b) T = 4 N


7. (a) F = 30 N
(b) F= 39N

8. (a) a = 20 m / s ^ 2
(b) 6.0 N
(c) 0.2 N
(d) 5.4 N

9.
(a) applied force = 1 N
(b) 4 Kg
(c) 5 N
(d) the slope increases, the intercept increases
(e) intercept changes

10.
(a)


(b) 4 N
(c) 11 m / s ^ 2

11.
(a)


(b) 2 N
(c) 11 m / s ^ 2

Section 2: Answers

Hey, guys.
Section two was all about motion in two dimensions. This breaks down into projectile motion and circular motion.

1.


2.


3. C=13 cm. Θ = 26 degrees


4. (a) r = (5.00i + 12.0j) m.
(b) The magnitude of r = 13.0 m. Θ= 67.4 degrees.

5. t = 0.454 s. y = 3.3 ft.

6. (a) 3 s
(b) 45 m.

7. 1.8 m.


8. t, y max, v y (t). and x will all remain the same.


9.(a) v i x = 20 m/s. v i y = 15 m/s.
(b) v x = 20 m/s.
(c) horizontal velocity = 20 m/s
final velocity = 25 m/s at an angle of -37 degrees.
(d) and (e) see figure




10.Divide this problem into two one-dimensional problems:

(a) V o x = 20 m/s
(b) V o y = 15 m/s
(c) y max = 31.25 m/s
(d) t = 4 s
(e) x(4s) = 80 m
(f) v(4s) = 32 m/s
(g) Θ = -51.3 degrees

(a), (b), (f), and (g) are drawn on the figure below.



11. 17.8 m

12. v o = 45 m/s

13. (a) and (b) a = 10 m/s ^ 2
shown in figure



(c) neither v nor a is constant

(d) object is subjected to approximately 1 g

14.

(a) The period = 0.4 π s.
(b) The frequency = 2.5/ π s-1
(c) From (a) v = 2 πr/T.
The centripetal acceleration a = 4 π ^ 2 r f ^ 2 = (d)

MC answers

Hey, guys.
My tutorial just ended, and I'm posting the answers for the multiple choice packet. I'll put up the answers to the numeric packet soon. There are a couple of problems that don't apply to us -- I've marked them as such.

Let me know if you have any questions -- phone is 281-768-0351.


10 D
11 D
12 A
13 E
14 A
15 C
16 A
17 B
18 A
19 B
20 B
21 C
22 D
23 C
24 A
25 E
12 K
13 C
14 J
15 D
16 G
17 E
18 G
19 C
20 G
21 A
22 G
1 A
2 D
3 D
4 B
5 D
8 A
9 D
8 DON'T BOTHER!
9 D
10 K
11 C
12 H
13 B
14 J
15 B
16 F
17 D <3 great problem <3
18 H
19 E
20 K
10 A
11 E
12 DON'T BOTHER
16 E
17 B
18 D
19 A
20 E

Now, take a quick walk around the house. breathe deeply. take a nap if you need to. These will still be here when you get back.

Section 1: Answers

Hey, guys.
Section One is all about motion in a single dimension.

Key terms: acceleration, velocity, displacement.
Key equations = vf=vi


Answers – Section 1

1. (a) 10 m/s.
(b) 161 s


2.(a) If you double x, you double y.
(b) If you halve x, you halve y.
(c) a straight line through the origin.

3.(a) If you double x, y goes up by a factor of 4.
(b) If you halve x, y is reduced by a factor of four.
(c) A plot of y as a function of
(i) x is a parabola;
(ii) the square of x is a straight line through the origin.

4.(a) 30 m.
(b) 60 m.

5.(a) 20 m.
(b) 80 m.
(c) 28 m
(d) 96 m.

6.(a) 20 m/s ^2.
(b) 80 m.

7. 108.

8. 0.454 s.

9.(a) v i = 5 m/s.
(b) 5/3 m/s ^2.
(c) 7.5 m.

10. the result is shown in this graph:




11.(a) 4.0 s.
(b) 16 ft.
(c) 320 ft.
(d) 21.3


12.The magnitude 10 m/s , to the right at t = 1.0 s , to the left at t = 5.0 s

13. (a) From t = 0 to t = 1.0 s, a = 10 m/s ^2.
From t = 1.0 s to 2.0 s, a = 0.
From t = 2.0 s to 3.0 s, a = -15 m/s ^2.
From t = 3.0 s to 4.0 s, a = 0.
From t = 4.0 s to 5.0 s, a = -15 m/s ^2.

(b) From t = 0 to t = 1 s, x(1s) = 10 m.
From t = 1 to t = 2 s. x(2s) = 25 m.
From t = 2 to t = 3 s. x(3s) = 32.5 m.
From t = 3 to t = 4 s. x(4s) = 32.5 m.
From t = 4 to t = 5 s. x(5s) = 25 m.

(c) From t = 0 to t = 1 s, the slope and velocity are positive and increasing.
From t = 1 to t = 2 s, the slope and velocity are constant and positive.
From t = 2 to t = 3 s, the slope and velocity are positive, but decreasing.
From t = 3 to t = 4 s, the slope and velocity equal zero.
From t = 4 to t = 5 s, slope and velocity are negative, but increasing.

14.(a) vo = 30 m/s
(b) v(2.0s) = 10 m/s
(c) y = 45 m

15.(a) v = -40 m/s,
(b) t = 2 s.
(c) t = 6 s.

16.(a) a = -2.5 m/s ^ 2.
(b) t = 4.0 s.
(c) and (d) are shown in the figure below.



17. yo = 20 m. on the tenth floor.


18.(a) 2.0 s.
(b) 4.0 m.
(c) -8.0 m/s.

19.(a) t = 1 s, the velocity is zero and acceleration is positive
(b) t = 1 s, the velocity is zero. For t > 1s, the acceleration is negative
(c) t = 1 s, the velocity is negative and acceleration is positive
(d) t = 1 s, the velocity is negative and acceleration is negative
(e) t = 1 s, the speed is increasing for all cases except (c).

Test post

Hey, everyone. I'll post the answers to both review sheets soon.