Class 0913

0913 Quiz

1. A steel ball rolls from rest down a constant incline of length 80 cm, requiring 4 seconds to travel the distance. Answer the following and connect your answers as best you can to the definitions you have memorized: Assume the acceleration to be uniform.

Using the definitions as opposed to the equations of motion find everything you need to construct a graph of v vs. t for the time spent on the incline.

Ave vel. is ave. rate of change of position with respect to clock time.

The ball has an average velocity of vAve = `ds / `dt = 80 cm / (4 s) = 20 cm / s.

The v vs. t graph is a straight line from the t = 0 point (0, 0) to the t = 4 sec point, with average vertical coordinate 20 cm/s.

The final velocity must therefore be 40 cm/s, so that average of initial and final velocities is 20 cm/s.

So the graph is a straight line segment from (0, 0) to (4 s, 40 cm/s).

Using your graph determine the acceleration of the ball on the incline.

The acceleration is rate of change of velocity with respect to clock time, or `dv / `dt. This corresponds to the slope of the graph.

The acceleration is thereforea = `dv / `dt = (40 cm/s) / (4 s) = 10 cm/s^2.

Using your graph determine how far the ball traveled on the incline (yes, I know that was given; you can still find the quantity from your graph and see if it is consistent with the given information).

The area under the graph is ave. ht. * ave. width = 20 cm / s * 4 s = 80 cm.

This is consistent with the given length of the incline.

2. A ball starts from rest on an incline of length 12 cm on which its acceleration is 10 cm / s^2. When the ball reaches the end of this incline it rolls without any immediate change in velocity onto a second incline, on which its acceleration is 20 cm / s^2, and whose length is 18 cm.

Use the equations of uniformly accelerated motion to answer the following:

For the first incline which of the variables v0, vf, `ds, a and `dt do you know?

We know that `ds = 12 cm, v0 = 0 and a = 10 cm/s^2.

Find for the first incline the average velocity, change in velocity and any of the 5 equation variables you don't already know.

Using the fourth equation vf^2 = v0^2 + 2 a `ds we can solve for vf:

vf = +-sqrt(v0^2 + 2 a `ds) =

+-sqrt(0^2 + 2 * 10 cm/s^2 * 12 cm) =

+-14.4 cm/s, approx..

Since `ds is positive and a is positive, with v0 = 0, the velocity is increasing from 0 and we conclude that the +15.4 cm/s is the correct solution.

Now using the first equation `ds = (vf + v0) / 2 * `dt we find that

`dt = 2 `ds / (vf + v0) = 2 * 12 cm / (0 + 15.4 cm/s) = 1.5 sec approx.

vAve = (0 + 15.4 cm/s) / 2 = 7.7 cm/s approx.

`dv = vf - v0 = 15.4 cm/s - 0 = 15.4 cm/s.

How fast will the ball be rolling when it starts on the second incline?

Initial velocity on the second incline will equal the final velocity on the first, so for the second incline v0 = 15.4 cm/s.

For the second incline which of the variables v0, vf, `ds, a and `dt do you know?

v0 = 15.4 cm/s, as mentioned.

`ds = 18 cm and

a = 20 cm/s^2.

Find for the second incline the average velocity, change in velocity and any of the 5 equation variables you don't already know.

We have the same vbls we had before so we use the same strategy:

vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (15.4 cm/s)^2 + 2 * 20 cm/s^2 * 18 cm) = +- sqrt(960 cm^2 / s^2) = +-31 cm/s.

We choose the +31 cm/s since the ball is still accelerating thru a positive displacement in the positive direction and starts with positive velocity.

`dt = 2 * `ds / (vf + v0) = 2 * 18 cm / (15.4 cm/s + 31 cm/s) = .75 sec.

`dv = (31 cm/s - 15.4 cm/s) = 15.6 cm/s.

3. For the situation of #2:

On which of the inclines was acceleration greater?

Accel is greater on the second incline (20 cm/s^2 vs. 10 cm/s^2).

On which of the inclines did velocity change more?

This happened on the second incline, but just barely.

If the second incline had been just a little shorter the velocity would have changed by more on the first incline.

Are your answers to these two questions consistent?

Might almost seem contradictory. Greater accel. on the second, greater displacement on the second, but practically equal velocity changes.

The key is that because of the greater initial velocity on the second incline, the ball spent less time on that incline, and acceleration is the rate of change of velocity with respect to clock time.

3. For the marble on an incline:

What is the direction of the gravitational force on the marble?
What is the direction in which the marble accelerates?
If the ramp wasn't there in what direction does the marble accelerate?
What is the ramp doing to influence the direction of the marble's acceleration?
List all the forces acting on the marble.
Give the direction of each of the forces acting on the marble.

4. For a pearl hanging from a thread, held at a distance of 1 cm from its equilibrium position:

What is the direction of the gravitational force on the pearl?
What is the direction of the force you have to exert to hold it back?
List all the forces acting on the pearl in this position.
If you were to release the force holding the pearl at the 1 cm position what would happen and why?
Would it take more force, the same force, or less force to pull the pearl back to a position 2 cm from equilibrium?
If you were to release the force holding the pearl at the 2 cm position what would happen and why, and how would this compare to what happens if you release from the 1 cm position?

Experiments

Experiment 1:

Make the observations necessary to determine the acceleration of a steel ball rolling down an incline vs. the percent incline.

Sketch a graph of acceleration vs. percent incline and find its slope.

Experiment 2:

Set up two unequal inclines so that the ball will roll down the first incline and onto the second with no significant change in velocity.

Measure the two inclines.

Time the ball on both inclines and see if your results are consistent with the results of your first experiment.

Directions for the VHCC LAN Practice/Quiz Program, accessible through the lab in Room 307 and from 310.

If the Multiple Choice program is on your desktop and you don't have instructions to download a new version, just double-click on that icon and Enter your VHCC username in the box where it is requested.
Enter 201 for your course number.
A list of topics will appear. Enter the number of your choice.
Wait until a message pops up telling you how many questions you will have and OK that box.
The next box allows you to choose whether the questions will have just 1 correct answer or may have multiple correct answers.
Once you have answered this box the screen will tell you to click on the 'next question' button. When you do a question will pop up.
Next you click on the Display Choices button and your choices will pop up.
Once you have decided which choice or choices are true you can click on the Answer Questions button. Enter the letter(s) corresponding to the valid choice(s).
Click on the See Results button, which will give you (often repetitive) feedback on each of your answers.
Click on the Next Question button and repeat the process.

If you are instructed to first download the latest version of the Multiple Choice Generating Program:

Go to Web address www.bb.vccs.edu and log in. If you don't yet have your login name you can log in as vhstudent with password college.
Go to Supervised Study Current Semester (NOT to your Physics page) and Click on Course Documents.
Scroll down that page until you see Multiple Choice Generating Program. Either run the program or copy it to your desktop or some other location on your computer. However don't answer any questions yet.

If you're working at home, once the 201_c.txt file is in the c:\vhmthphy folder and the program itself is on your desktop, you just need to double-click to run the program.

It will be equally easy at VHCC once the network is set up to accomodate the program.