PHYSICS 121
Spring 2008
| Homework | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Schedule and Solutions All problems listed are from “College Physics by Serway and Faughn”, Seventh Edition. The notation followed is C.NN where C is the chapter number and NN is the problem number. Hence 5.4 stands for 4th problem from chapter 5.
|
SPECIAL PROBLEM SET:The solution to this problem set will be put up on May 06th. If you understand this problem set and write the solutions out in detail explaining all the important steps, filling the missing steps, it will count towards two homeworks. The lowest three homeworks are going to be discarded. If you turn this homework in, the grade in this homework will replace two other lowest homeworks. If you think you do not need this "extra credit", you need not turn it in.
The final exam problems that cover these topics will be based on these problems, but will not be identical. So it is important that you understand these problems really well. Not all the problems that are listed below are from the homework. Please do not hesitate to ask for help on these problems. The solutions to such problems will be put on the website if it is requested for by the student. For the topics that are not listed here, you should follow the homework problems. You should also look at the problems that were asked in your quizzes. YOU SHOULD LOOK AT ALL examples in the book. some examples below are listed a as : ex-6.5
2.47, 2.59, 2.62, 3.31, 3.17, 3.55, 3.33, 4.50, 4.27, 4.57, 4.49, 4.60, 5.33, 5.55, 5.63, 5.71, 5.73, 6.50, 6.31, 6.11, 6.60, 6.45, ex6.5, ex 7.13, 7.10, 7.39, 8.43, ex-8.12, 8.40, 8.50, 8.52, 9.45, 9.47, 9.80, ex 9.15
Guidelines for homework assignments:
- All homework assignments should be neatly written with answers to questions presented in numerical order. Be sure that your name is clearly written at the top of all pages and that you have stapled all pages together. You are responsible for misplaced or lost pages. The TA will NOT grade any homework that does not meet this criterion and you will get a 0 on that homework.
- Be sure to answer all parts of each question.
Please follow the detailed instructions given in class on how to solve homework problems. Some them are also listed below.
- Make sure that you attempt problems starting at the top of the sheet proceeding downward. If you have the habit of solving problems with sequential steps proceeding horizontally, then you cannot have 2 problems next to each other (horizontally). If you are in the habit of having sequential steps below each other, then you can have more than one problem next to each other ONLY IF there is a clear partition between the two. These instructions are probably not very clear. If you do not understand these instructions, please clarify with the instructor.
- Have enough empty space between one problem and the next.
- Your problems must contain words and explanations for your steps. THIS IS A MUST.
- Any answer must be explained with physical principles or concepts. A SIMPLE YES OR NO WILL NEVER DO.
- If you can draw a diagram or a picture of the situation, then you must draw it.
- All answers must have units.
- Each student is allowed to turn in homework late ONLY 2 times during the semester. But these will be graded for 20 % less credit for every day that they are later.
3.33 Find X and Y after 4s. The straight line distance is found using sqrt(x^2+Y^2)`
3.49 This problem proceeds in two phases. First: Phase AB is uniform acceleration of 30 m/s^2. Use kinematic equations to find where the rocket is (X, and Y), what its velocity is. The second phase BC of motion starts after engines fail. It is projectile motion. You know the initial velocity of this phase, that is the velocity of the rocket when this phase starts. You k3.55now the launch angle. You know X0, Y0 (which are X and Y found from previous part). Find max height reached by rocket. total time is time for phase1 plus time it takes for rocket to land on ground. Find final X for horizontl range. (you probably have included X0 already.
3.55 The projectile has to reach a fence 130 m away. Use the X equation. two unknowns t, and Vo. The projectile has to clear a fence 21 m. Y equation (21 m) has two unknowns. t and Vo. Solve for t from X and substitute in Y equation. Some cancellation? Once you find Vo, everything else is striaghtforward.
3.58 Similar method as above.
Chapter 4.
27. What is responsible to move the 2.0 kg mass? If 1kg mass pushes the 2.0 kg mass, remember, the 2.0 kg mass pushes the 1 kg mass also!
30. m2 moves down. So what is the sign of its acceleration term? Pulley is massless. so essentially it is just one rope.
34. The pulley is massles. So how many tensions do we have? In your equations be mindful of the sign of the acceleration. Both masses are attached by the same string. So accelerations are equal? Which mass moves down? So what is the sign of the ma term?
41 If the force is barely enough to hold the block, which way is it about to slide? Which way does the friction act?
48 CORRECTION! "m1 moves up. so what is the sign of the acceleration term? M2 moves down. which way is friction?"
50. We are looking for minimum value of F. It means that the block is barely in its position. Othewise which way would it move? WHat then is the direction of friction. AN INTERESTING ALTERNATIVE. WHAT Is the maximum value of F?
60. The force pushes the block up against the ceiling. So what is the direction of the normal force from the ceiling?
chapter6
8. use conservation of energy to find v before hitting mattress.
find time using distance and velocity and final velocity.
use impulse momemntum ewn to figure out net force, and then force due to
matress
28. follow example 5, but both bullet and block are moving separately after
collision. Only the block rises.
31. there are two collisions. one between gayle and sled.
find final velocity. use conservation of energy to find how the velocity
changes due to height drop
then there is another collision.
repeat the process.
48. use cons of energy to find velocty of m1 before collision.
use cons of momentum and 1 dimensional elastic collision equation 6.14 to
find final velocities. Use cons of energy to find the height to which m1 rises.
50. problem similar to 28.
CHAPTER 8
8.7. Choose your axis at different places and solve for different unknowns. OR it is instructive to solve for one unknown by choosing the axis appropriatelym and solving for other unknown using newton's second law with equilibrium.
8.15. The distances are not given but you can assume L as the length. Weight acts at L/2. T acts at 2/3 L and so on.
8.28 Use Sum Fx=0 (eq1), SumFy=0 (eq2) , Torque about the point on wall, Mu_s*N= fs. condition 1 and 2 and 4 will allow you to find T in terms of w. that will cancel w in eq 3 and leave x.
8.32. Torque that slows the wheel down is due to friction. that equals I*alpha. you can find alpha from angular velocities. you can find frictional force and then Mu_k since you know N=70.
8.40. The loss in potential energy equals the kinetic energy of translation+kinetic energy of rotation. Use v=r*omega to get the equation in terms of just omega.
8.54 The space station creates a artificial gravity. The force acting on the astronauts towards the center is the normal force which plays the role of centripetal force. But the rotation rate is adjusted so that v^2/r, centripetal acceleration is g= 9.8 m/s^2
CHAPTER 9. .33
the forces downward will be weight of the balloon, and weight of the helium (a spehere of helium whose density you know) and tension balanced by the bouyant force due to air.
9.36. The weight in any medium (oil, or air, or water) is the tension in
the string. which balances mg and the bouyant force.
we did this in class. please refer to the notes for the equation.
9.47. The volume flow rate is given and is equal to A1*V1=A2*V2. allows you to calculate one of the Vs. The pressures at both points can be found. They are due to a column of liquid and you know the formula. using these three quantities, and bernoulli's eqn, you cna find the other V.
9.80 The pressure at a horzontal level in a single connected fluid is the same. For part a, pick a level where the fluid below is the same on both sides. Equate the pressures and find h. The pressure on each side is Patm+pressure due to liquid above. For part b, do the same, and get a relation between pressure of air on left and pressure of air on right. Use bernoulli's equation for a point just above the liquid on left and a similar point on right. The pressure of air just above liquid on left is lower beecause it is influenced by the movement of air. The heights are same. The air is not moving on the right. This gives you a relation between the two pressures again. use these two equations to calculate velocities.
Chapter 3.
2. (a) Approximately 484 km (b) Approximately 18.1° N of W
10. 1.31 km north, 2.81 km east
24. 3.19 s, 36.1 m/s at 60.1° below the horizontal
30. (a) clears the bar by 0.85 m (b) falling, Vy=-13.4 m/s
58. 10.7 m/s
62: is just a derivation.
Chapter 4
6. 7.4 min
16. 77.8 N in each wire
30. 6.53 m/s^2, 32.7 N
32. (a) T>w (b) T=w (c) T<w
(d) 1.85 ×10^4N, Yes (e) 1.47 ×10^4N, Yes (f) 1.25 ×10^4N, Yes
34. (a) 36.8 N (b) 2.45 m/s^2 (c) 1.23 m
48. (a)0.125 m/s^2 (b) 39.7 N (c) 0.235
50. (a) 18.5 N (b) 25.8 N
58. (a) 50.0 N (b) 0.500 (c) 25.0 N
60. 0.814
62. (a) 1.63 m/s^2
(b) 57.2 N tension in string connecting 5-kg and 4-kg, 24.5 N tension in string connecting 4-kg and 3-kg
Chapter 5
2. 30.6 m (4).1.6 kJ (6). (a)900 J (b)0.383
8. 31.9 J (b)0 (c)0 (d)31.9 J
10. 160 ms
12. (a)–168 J (b)–184 J (c)500 J (d)148 J (e)5.64 m/s
14. 90.0 J
16. a) 1.2 J (b)5.0 m/s (c)6.3 J 18. 2.0 m
20. 0.5 m (22) (a)0.768 m (b)1.68*10^5 J
32) 5.11 m/s 44). (h/5)[4*(sin(theta))^2 +1] OR (h/5)[5-4* (cos(theta))^2]
54). A) 7.92 Hp B)14.9 HP
56). (a)7.50 J (b)15.0 J (c)7.50 J (d)30.0 J
64) 3.9 kJ 76) 0.115 62) 1.4 m/s 58) 1.9 m/s
Chapter 6
2. (a)5.40 N·s (b)–27.0 J
8 1.91 * 10^4 N upward.
22. Velocity of thrower=2.48 m/s, Velocity of catcher =0.0225 m/s
28. 5.3 * 10^2 m/s
38. (a)2.5 m/s (b)3.75 * 10^4 J
40 (a)0, 1.50 m/s (b)-1.00 m/s, 1.5 m/s
(c)1.00 m/s 1.50m/s
48. 0.556 m 50. (4M/m) Sqrt(gl)
56. (a)9.90 m/s, -9.90 m/s(b)-16.5 m/s, 3.30 m/s
(c)13.9 m, 0.556 m
58. 0.980 m
60. (a)Vred=0, Vblue=3.00 m/s (b)0.212 m
Chapter 7
2. 2.1 m, 1.2*10^2 m, 7.7*10^2 m
10 50 rev 12 1.02 m
7.28 25 kN b) 12m/s. 7.50 a) 0 b) 1.3kN c) 2.1 kN 7.52 a)2.1 m/s b) 54 degrees c) 4.7 m/s
Chapter 13
13.24) 2.2 Hz
13.26) a.) 0.30 m, 0.24 m b) 0.30 m c) 1/6 Hz d) 6.0 s.
13.28 a) 250 N/m b) T=0.281 s, f=3.56 Hz, omega=22.4 rad/s c) 0.313J d) 5.00 cm e) 1.12 m/s, 25.0 m/^2 f) 0.919 cm
13.54 a) 0.25 m b) 0.47 N/m c) 0.23 m d) 0.12 m/s 13.56) 0.990 m
Chapter 8
20. (b)T = 343 N, H = 171 N, V = 683 N (c)5.14 m
22. (a)392 N (b)H = 339 N (to right), V = 0
28. 2.8 m 30. 149 N·m, 66.0 N·m, 215 N·m
32. 0.30
36. (a)5.35 m/s^2 downward (b)42.8 m (c)8.91 rad/s^2
40. 10.9 rad/s 50. (a)1.9 rad/s (b)KE I =2.5 J, KEf=6.4 J
54. 12.3 m/s^2
78 A smooth (frictionless) surface cannot exert a force parallel to itself. Thus, a smooth vertical wall can exert only horizontal forces, normal to its surface.
(b)Lsin(theta) (c)(L/2)cos(theta)_ (d)2.5 m
Chapter 9
16. 1.2 ×106 Pa 20. (a)65.1 N (b)275 N
26. 0.611 kg 28. 10.7% of the volume is exposed
30. (a)1 017.9 N, 1 029.7 N (b)86.2 N (c)11.8 N for both
36. (a) 8.57 ×10^3 Kg/m^3 (b)714 kg/m^3
22. 10.5 m 42. 11.0 m/s, 2.64*10^4 Pa, 78 1.9 m 80 a) 1.25 cm b) 13.8 m/s
Chapter 10.
10. (a)263°C (b) -262 celsius
12. (a)L= 1.3m – 0.49 mm = 1.29951m (b)fast
34. 3.84 m 40) a Vrms_hydrogen=1.73 km/s, b Vrms_carbondioxide= 0.369 km/s c) Hydrogen escapes but carbon dioxide does not.
50. 2.4 m 52 800 celsius or more.
Chapter 11
11.12 467 pellets 11.32 39 cubic meters per day 11.42 3.77 *10^26 watts.
Chapter 12
12.10 a -12.0 MJ b 12.0 MJ 12.12 6PoVo 12.16 a) 12kJ b -12 kJ 12.32 a) 9.10 kW b) 11.9 kJ
12.54 a) 4PoVob) 4PoVo c) 9.07 kJ