course PHY 201
Dear Prof. Smith, Here is my TEST 2.
Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.
Tanya" "Time and Date Stamps (logged): 13:47:10 12-13-2006 °²³Ά°―°±°²±――΅
General College Physics (Phy 201) Test 2
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Completely document your work and your reasoning.
You will be graded on your documentation, your reasoning, and the correctness of your conclusions.
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Test should be printed using Internet Explorer. If printed from different browser check to be sure test items have not been cut off. If items are cut off then print in Landscape Mode (choose File, Print, click on Properties and check the box next to Landscape, etc.).
Tatyana Simmons
Signed by Attendant, with Current Date and Time: 12/13/2006
If picture ID has been matched with student and name as given above, Attendant please sign here: _________
Instructions:
Test is to be taken without reference to text or outside notes.
Graphing Calculator is allowed, as is blank paper or testing center paper.
No time limit but test is to be taken in one sitting.
Please place completed test in Dave Smith's folder, OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va., 24212-0828 OR email copy of document to dsmith@vhcc.edu, OR fax to 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test..
Directions for Student:
Completely document your work.
Numerical answers should be correct to 3 significant figures. You may round off given numerical information to a precision consistent with this standard.
Undocumented and unjustified answers may be counted wrong, and in the case of two-choice or limited-choice answers (e.g., true-false or yes-no) will be counted wrong. Undocumented and unjustified answers, if wrong, never get partial credit. So show your work and explain your reasoning.
Due to a scanner malfunction and other errors some test items may be hard to read, incomplete or even illegible. If this is judged by the instructor to be the case you will not be penalized for these items, but if you complete them and if they help your grade they will be counted. Therefore it is to your advantage to attempt to complete them, if necessary sensibly filling in any questionable parts.
Please write on one side of paper only, and staple test pages together.
Test Problems:
Problem Number 1
If a mass of 4 kg moving at 7 m/s collides with a mass of 9 kg moving at an unknown velocity, then if after the collision the two masses have common velocity -1.307692 m/s, what is the velocity of the second mass before collision? Is this collision possible for the system consisting of the two masses without the conversion of some internal source of potential energy?
Solution:
1. Using the formula for momentum, we can find v2:
The momentum of the first ball immediately before collision is p1 = 4kg * 7m/sec = 28 kg*m/sec
The momentum of the first ball immediately after collision is p2 = 4kg * (-1.307692 m/s) = -5.230768 kg * m/sec.
The momentum of the second ball immediately after collision is p3 = 9 * (-1.307692 m/s) = -11.769228 kg * m/sec.
The total momentum of the two balls immediately before collision is p4 = 28 + 9 * v2
The total momentum of the two balls immediately after collision is p5 = p2 + p3 = -5.230768 kg * m/sec + -11.769228 kg * m/sec = -16.99999 kg * m/sec
The total momentum immediately before collision is equal to the total momentum immediately after collision. Now, I can solve the equation 28 + 9 * v2 = -16.99999 for v2
V2 = (-16.99999 28 ) / 9 = -4.9999m/sec
-4.9999 m/sec was the speed of the second ball.
Good solution.
2. According to the definition of the potential energy
Potential energy is the energy that is by virtue of the relative positions (configurations) of the objects within a physical system. This form of energy has the potential to change the state of other objects around it, for example, the configuration or motion,
this collision would not be possible without the conversion of some internal source of potential energy.
If the KE of the system after collision exceeds the KE before collision, and if the system is isolated, the extra energy must have come from the conversion of some type of PE.
Have you checked to see whether KE after might be greater than KE before?
Problem Number 2
The energy required for a 440 kg mass to climb from the surface of the Earth to a point 6400 km 'above' the surface--i.e., at two Earth radii from the center--can be approximated by first averaging the field strength of the Earth at this point and at its surface.
Given a surface field strength of 9.8 m/s^2, what is the field strength at two Earth radii from its center? What is the average of these two strengths?
1. g = (Gm1m2/r^2)
Gm1m2 = 9.8 m/sec^2, r = 1 at the surface
2. Field strength at two Earth radii equals 50%
This is not so. Field strength falls off as the inverse square of the distance from the center of the Earth. At two Earty radii the field strength would be 1 / 2^2 = 1/4 as great as at the center.
Average =(1+1/4)/2 = .625g
Using this average to approximate the actual average field strength determine the work necessary to climb from the surface to a point two Earth radii from its center.
. 1. F = 440 kg*(.625g) = 2695 N
W = F * D = 2695* 6.4 * 10^6= 1.7248 * 10^10 Joules
The work would not be based on just the force at the maximum distance. The distance varies from 1 Earth radius to 2 Earth radii, and the force will vary accordingly. At every point but the last it will be greater than the force experienced at 2 Earth radii.
Problem Number 3
A car moving at 19 m/s drives over the top of a hill. The top of the hill forms an arc of a vertical circle 140 meters in diameter.
What is the centripetal force holding the car in the circle?
What therefore is the normal force between the car's tires and the road?
1. The centripetal acceleration from an object moving in a circular arc is expressed as
A = V^2/R =19^2 / 140 = 2.5786 m/sec^2, where V is the speed of the object, a car in this case, and R is the radius of the arc being traversed.
2. The centripetal force is then
Fc = M*Ac = M*V^2/R = M * a = M * 2.5786 = N
where M is the mass of the car.
3. To find the normal force, lets look at the force balance
SFy = 0 = N + M*V^2/R - M*g
Good idea, but you can't sum these forces like this, since they act in different directions.
m v^2 / r is toward the center of the circle, while m g is vertically downward and the normal for N is vertically upward.
You can sum N and - m g; since there is 0 acceleration in the vertical direction you get N + (-mg) = 0, and you can conclude that N = m g.
The net force must be m v^2 / r toward the center of the circle. This is the condition for circular motion at constant amgnular velocity.
where N is the normal force and g is the acceleration due to gravity. Solving for N
N = M*g - M*V^2/R
N = M*(g - V^2/R) = M (9.8 2.5786) = 7.2214*M N, where M is the mass of the car.
Problem Number 4
A uniform rod of negligible mass and length 41 cm is constrained to rotate on an axis about its center. Two masses of .204 kg are attached to the rod at distances of 14.76 cm from the axis of rotation, one on either side of the axis. Two masses of 1.7 kg are attached at the ends of the rods An unknown uniform force is applied to the rod at a position 9.43 cm from the axis of rotation, in the plane of motion of the rod, and at an angle of 43 degrees from the rod. The force is applied as the rod rotates through .12 radians from rest, which requires .6 seconds. The applied force is then removed and, coasting only under the influence of friction, the rod comes to rest after rotating through 3.1 radians, which requires 17 seconds.
Find the net torque for each of the two phases of the motion.
What is the unknown force?
What is the maximum angular momentum of the system? How much of this angular momentum resides in the .204 kg mass?
1. Inertia moment I = 2 * (0.204*(0.1476)^2 + 1.7*(0.41)^2) = 0.58 kg.m^2;
the tangent component of force rotating the system is Ft=F*sin(43°), radial component being unimportant; thus torque T=(F*sin(43)*0.0943 B) N*m, where B is torque braking the system; this torque T results in angular acceleration w in formula T=I*w;
don't use w to represent omega; just write omega. Also torque is reqpresented by tau, not T. The equation would be more correctly written tau = I * omega'. However I know what you mean.
I'm not sure where you encountered omega ' in the 201 course, where Newton's Second Law for rotation is given as tau = I * alpha, where alpha is the angular acceleration. Since acceleration is rate of change of velocity, angular acceleration is certainly equal to the derivaitve of angular velocity with respect to clock time, and tau = I omega ' is a valid expression. I believe you said at one point that you're a mathematics major, so certainly you understand this and may use the notation of calculus if you wish.
the system reaches the given angular position 0.12=0.5*w*(0.6)^2;
thus w = 2/3 rad/s^2 and T = 0.58*2/3 = 0.3866666 N*m;
Other than notation and your lack of units through the calcuations, your solution appears correct.
2. Now torque forward is removed, brake torque is left: B=I*d, where d is angular deceleration in formula: given 3.1=0.5*d*17^2, hence d=0.021453rad/s^2 and B=I*d =0.58*d=0.012443 N*m;
Hence, T+B = 0.386666+0.012443 = 0.39911N.m =F*sin(43)*0.0943
hence, F = 0.39911/sin(43)/0.0943 = 6.206N
3. Max angular speed = w*0.6 = 0.4rad/s; max angular momentum = I*0,4 = 0.232kg.m^2/s;
Angular momentum resides in one of 0,204kg = 0.204*(0.1476)^2 * 0.4 = 0.0018kg.m^2/s
Good
Problem Number 5
If in times 6.5 sec, 6 sec, 9.25 sec and 5.75 sec a ball rolls down a ramp from rest, starting at different positions on the ramp and starting from rest each time, and covers respective distances of 3.121377, 3.542011, 4.306563 and 6.133981 cm, does the data support the contention that the acceleration of the ball is independent of velocity and position on the ramp?
Yes, the data supports this hypothesis.
According the obtained results,
V1 = d/t = 3.121377/6.5 = 0.48022 cm/sec
A1= dv/dt = 0.48022/ 6.5 = 0.073879 cm/sec^2
a = `dv/`dt, not vAve / `dt.
V2 = 0.590335 cm/sec
A2 = 0.098389 cm/sec^2
V3 = 0.465574 cm/sec
A3 = 0.050332 cm/sec^2
V4= 1.066779 cm/sec
A4 = 0.185527 cm/sec^2
The data shows that the values for acceleration which were obtained are different.
You don't appear to have calculated accelerations correctly, using vAve / `dt instead of`dv / `dt. However it is correct to calculate acceleration, and given your results for acceleration your conclusion is correct.
Problem Number 6
If one Calorie of food energy is 1000 calories (note large and small c), and if one calorie is about 4.19 Joules of energy, and if your body is capable of converting about 15% of the food energy you consume into useful work, then if you used up all the energy in a 520-Calorie meal to climb vertically up a magic beanstalk, how high could you climb (you may assume your own mass or, if you prefer, assume a mass of 58 Kg).
520c = 520,000 C (actual calories)
520,000 C x 4.19 J/C = 2,178,800 J (energy)
Potential Energy ( energy of object at a certain height from ground) = MGH
M= mass of object (58 kg)
G= Force of gravity (9.8 m/s^2)
H=Height = ?
We stated the bodies efficiency is 15% so multiply our total energy by .15:
2,178,800 J x .15 = 326,820 J (energy used for actual work)
326,820J = 58Kg x 9.8 m/s^2 x H
H = 575 m
575m is about ~1/3 of a mile.
very good
You would have definitely passed this test with the work you have shown here, most likely with a grade of B. Hopefully you will do even better when you take the test under proctored conditions.