course PHY 201
Hello Prof. Smith,Here is my TEST 1.
General College Physics (Phy 201) Test 1
________________________________________
Completely document your work and your reasoning.
You will be graded on your documentation, your reasoning, and the correctness of your conclusions.
________________________________________
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: 12.08.2005
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
Reason out the quantities v0, vf, v, vAve, a, s and t: If an objects initial velocity is 8 cm/s, and it accelerates uniformly through 47.5 cm in 5 seconds, then what is its acceleration?
1. We can find the acceleration by using this formula: A=(vf-v0)/t
2. However, we dont have vf. We can use this equation v=(s-s0)/t to find the vf.
V=(47.5-0)/5=9.5 cm/sec = 0.095 m/sec
`ds / `dt gives you average velocity, not final velocity.
3. Now, we can find the acceleration:
A=(9.5-8)/5=0.3 cm/sec^2
a = `dv / `dt, and `dv is final velocity - initial velocity. Your calculation shows average velocity - initial velocity.
4. Also, we can find a) vAve, which equals to vAve= (v0+vf)/2 = (8+9.5)/2=8.75 cm/sec
vAve = `ds / `dt; if acceleration is uniform then it is also the case that vAve = (vf + v0) / 2. Since acceleration is uniform it is acceptable to use the latter expression, as you have. However you have not calculated the correct final velocity. You are using average velocity where you should be using final velocity.
The reasoning requested on this problem is as follows:
average velocity is 47.5 cm / (5 sec) = 9.5 cm/s.
Initial velocity is 8 cm/s.
Ave velocity is halfway between init and final vel, since accel is uniform. Since ave vel is 1.5 cm/s greater than init vel, final vel is 1.5 cm/s greater than this, or 11 cm/s.
Change in velocity is therefore 3 cm/s, so acceleration is 3 cm/s / (5 s) = .6 cm/s^2.
b) t =(t1-t0)=(5-0) = 5 sec
Using the equations which govern uniformly accelerated motion determine vf, v0, a, s and t for an object which accelerates at .6 cm/s/s through a distance of 47.5 cm, ending with velocity 8 cm/s.
1.Using this equation a= (vf-v0)/t, we can find the difference between two velocities:
vf-v0 = a / t = 0.6 / 5.9375 = 0.1011 cm/sec
a = `dv / `dt, so `dv = a * `dt, not a / t.
2. where t = s / v = 47.5 / 8 = 5.9375 sec
This would be `ds / v0, which is not generally an important quantity. vAve = `ds / `dt, so `dt = `ds / vAve, not `ds / v0.
3. We obtained the equation: vf-v0 = 0.1011 cm/sec
Now we can find v0: v0 = 8-0.1011 = 7.8989 cm/sec
4. Also, s = (sf-s0) = (47.5 - 0) = 47.5 cm
5. and t = d0/v0 = 0/7.8989 = 0 sec
Problem Number 2
An Atwood machine consists of masses of .9 Kg and .9269999 Kg hanging from opposite sides of a pulley.
As the system accelerates 3 meters from rest, how much work is done by gravity on the system?
1. Lets assume that the m2 is a counterweight and equals to 0.9 kg, then m1 = 0.9269999 kg. Also, g is a constant and equals to 9.8 m/sec^2.
2. Now, we can find for counter weight m1g =0.9269999*9.8 =9.08459902 N and for the first object m2g = 0.9 * 9.8 = 8.82 N
3. Now to find acceleration and Ft, we can apply F=ma to each equation, where we take upward as a positive direction: a2=a and a1= - a
4. Ft- m1g = m1a1= - m1a
Ft m2g = m2a2= m2a
We subtract the first equation from the second one to get:
( m1-m2)g = (m1 + m2)a:
We solve this for a:
A = [(m1-m2) / (m1+m2)] g = [(0.9269999 0.9) / (0.9269999 + 0.9)] 9.8 = (0.0269999 /1.8269999) 9.8 = 0.0147782712 * 9.8 = 0.1448270577 m/sec^2 =0.145 m/sec^2
The object accelerates downward at a= 0.1448270577 m/sec^2
Good, but when you consider the system, it consists of both weights, one accelerating upward and the other downward. So the acceleration of the system cannot be upward or downward. The acceleration of the system will be in the direction of the greater mass; depending on how you draw the system this could be clockwise or counterclockwise.
You have not yet answered the question of how much work is done on the system by gravity.
5. The tension in the cord Ft, can be obtained from one of two equations F = ma, changing a = 0.1448270577 m/sec^2
Ft = m1g m1a = 0.9269999 * (9.8 - 0.1448270577) = 8.950344352 = 8.95 N
Or
Ft = m2g + m2a = 0.9 (9.8 + 0.1448270577 ) = 8.950344352 = 8.95 N
very good
6. We know that W = F d cos a, where force F= mg, displacement d = (df- d0), and cos a = 180 degrees in our case, which equals to -1.
7. Now W = -F (df- d0) =- 0.1448270577 * (3 - 0) = - 0.434481173 J = - 0.43 J
You appear to have multiplied the acceleration by the displacement, not the net force by the displacement.
You are not including units in your calculations; had you included units with these quantities you would have ended up with m/s^2 * m = m^2 / s^2, which is not Joules.
Had you multiplied the acceleration by the total mass of the system you would have obtained the net force exerted by gravity on the system, and multiplying this by the displacement would have given you the work. However you do have to be careful with the signs. Is the displacement in the direction of the acceleration or opposite the direction of the acceleration?
Assuming no friction or other dissipative forces, use the definition of KE to determine the velocity of the system after having moved through the 3 meters, assuming that the system was released from rest.
1. We know that KE = ½ * m * v^2, also, Wnet = KE. So that, we can apply Newtons second law: F net = ma, and use Eq. 2-10c, which we write as v2^2 = v1^2 +2ad.
2. Now, we can solve this equation for (v2^2 v1^2):
(v2^2 v1^2) = a/2d = 0.1448270577 / 2 * 3 = 0.024137843 m/sec
You haven't used the definition of KE to obtain this result, which is in any case not correct. The units of your calculation give you .024 m^2 / s^2, not .024 m/s. And this calculation does not give you velocity, but the change in the squared velocity.
From a correct calculation of work you will know the KE of the system; setting this equal to 1/2 m v^2, you can solve for v.
3. However, we know that the system was released from the rest, so the v1 = 0 m/sec
4. Now, we can determine the v2, which equals to: v2 = sqrt(0.024137843) = 0.1553635831 m/sec or = 15.54 cm/sec
This would be a correct calculation of the final velocity. It should agree with the result you obtain based on the work-energy theorem.
Problem Number 3
When masses of 30, 60 and 90 grams are hung from a certain rubber band its respective lengths are observed to be 47, 58 and 69 cm. What are the x and y components of the tension of a rubber band of length 53.03 cm if the x component of its length if 43.76755 cm?
1. According to the way I completed one of the labs: the tension would equal to 13.2575 N, because 1 N = 4 cm.
From the given information a 53 cm rubber band would support about 56 grams, which would have a weight of about .55 Newtons. You estimate the force by graphing suspended weight vs. length.
2. Using the Pythagorean theorem we can find the y-component : y^2 = 53.03^2 43.76755^2
Y= 29.94298694 cm
3. Now, the tension of x component is 10.9418875 N
And the tension of y component is 7.485746735 N
These components would correspond to the components of a 13.25 N force in the direction of the rubber band; had you used the correct tension you probably would have obtained the correct components.
What vertical force, when added to this force, will result in a total force of magnitude 130 grams (a gram force is the force of gravity on a one gram mass)?
1. 130 grams is 1.3 kg
130 grams is .13 kg.
However it isn't completly clear where the 130 grams comes from.
2. The vertical force will equal to F = mg = 1.30 * 9.8 = 12.74 N.
Problem Number 4
If the slope of a graph of number of paper clips needed to maintain equilibrium vs. ramp slope is 33 clips / unit of ramp slope , and if the slope of a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley is ( 27 cm/s2) / clip, then how many cm/s2 of acceleration should correspond to 1 unit of ramp slope? If we require 29 clips to match the mass of the cart, then if we could apply this force to the cart without the extra mass of all those clips, what would be the acceleration of the cart?
1. If the slope is 33 clips per unit and another slope is 27 cm/sec^2 / clip, than 891 cm/s2 of acceleration should correspond to 1 unit of ramp slope.
2. Since the slope is 27 cm/s^2 / clip, we can say that 29 clips would give the acceleration of 783 cm.s^2.
Problem Number 5
Explain why the slope of a position vs. time graph between two clock times is equal to the average acceleration between those two times.
1. Since the slope equals to rise divided by run, we can say that the division position by two times will give us the average of acceleration.
2. Rise = upper position lower position and run = right two times left two times
The statement 'the slope of a position vs. time graph between two clock times is equal to the average acceleration between those two times' is actually incorrect. The rise of a position vs. clock time graph is the change `ds in position and the run is the change `dt in clock time. So the slope is rise / run = `ds / `dt, which is the definition of average velocity, not average acceleration.
In any case the answer to such a question always starts with 'the rise means ... ' and 'the run means ...', and ends up with 'therefore the slope means ... '. You have the right idea.
Problem Number 6
A ball of mass 8 kg rolls off the edge of a ramp with a horizontal speed of 70 cm/s.
What is its KE as it rolls off the ramp?
1. First of all, we have to concert 70 cm/s into 0.7 m/s
2. Now, KE = ½ * m * v^2=1/2 * 8 *0.7 ^2 = 1.96 J
How much work does gravity do on the ball as it falls 22 cm?
22 cm = 0.22 m
W = Fgrav * d * cos a = mg d cos 180 (in our case) = - 8 * 9.8 * 0.22 = - 17.248 N
Good, but in most projectile situations I recomment using a coordinate system with the x axis in the horizontal direction and the y axis in the upward vertical direction. This would mean that the gravitational force is along the negative y axis, at 270 degrees from the positive x axis as measured counterclockwise from that axis, and you would use m g cos(270 deg) = - m g as the force of gravity.
The displacement is also in the negative y direction, so the work woul be - m g cos(270 deg) * (-.22 m) = ... = 17 kg m^2/s^2 = 17 Joules. The work would be positive, since force and displacement are in the same direction, and the units are Joules, not Newtons.
What will be its kinetic energy after falling 22 cm?
KE = ½ m v1^2 = m g y = 8 * 9.8 * 0.22 = 17.248 J
No energy is dissipated, so the KE will be 17 J greater after falling the .22 m. The KE was originally 2 J, which makes the new KE equal to 19 J.
How much of this KE is accounted for by its horizontal velocity, and how much by its vertical velocity?
KE of the ball is 1.96 J when it is rolling down the ramp and 17.248 J when it is falling down.
What then is its vertical velocity at this point?
Using the formula for KE, we can solve this equation for v:
V= sqrt (2KE/m) = sqrt (2 * 17.248 / 8) = 2.0765 m/s is the vertical velocity of the ball.
This is correct.
Had the question asked for the speed of the ball at this point, you would have had to use the 19 J to find the answer. I mention this because some versions of this problem ask for the speed and not just the vertical velocity.
Problem Number 7
A cart of mass 1.9 kg coasts 100 cm up an incline at 7 degrees with horizontal. Assume that frictional and other nongravitational forces parallel to the incline are negligible.
What is the component of the cart's weight parallel to the incline?
Its displacement
The weight is directly vertically downward. If the x axis is directed along the incline, and if the incline goes up and to the right, this means that the weight is at angle 263 degrees with respect to the positive x axis. The components of the weight would therefore be W_x = W cos(263 deg) and W_y = W sin(263 deg). The x component is the component parallel to the incline and W is m g.
How much work does this force do as the cart rolls up the incline?
1. 100 cm = 1 m
1. W = Fd cos a = mg d cos a = 1.9 * 9.8 * 1 * cos 7 = 18.481209 J
The weight does not make an angle of 7 degrees with the x axis; it makes an angle of 263 degrees if your picture is drawn as described above.
How much work does the net force do as the cart rolls up the incline?
1. Since we already found the force, which is 1.9 N, we can say that this is the only one force which affects the cart. So that the net force equals to 1.9 N.
The net force acting on the cart includes the weight and the normal force exerted by the incline. The net force in the y direction is zero, since the cart moves and accelerates only in the x direction. So, since in this situation only weight and normal force act in the y direction, the normal force is equal and opposite to the y component of the weight. The net force is just the x component of the weight, and it acts as the cart is displaced 1 m in this direction.
2. Now, we can find work done by net force:
W = Fnet d cos a = 1.9 * 1 * cos 7 = 1. 885837688 J
The units of your quantities are kg and m, so your calculation would give you kg * m, not J. Again the cos(7 deg) is not correct here.
Using the definition of kinetic energy determine the velocity of the cart after coasting the 100 cm, assuming its initial velocity to be zero.
1. KE = W net = 1. 885837688 J
"
See my notes, consider them carefully and let me know if you have specific questions.