Note on reporting experiments:
Most of the preliminary writeups for this experiment were good for this stage of the course, containing a lot of good and correct work, but
It is always up to the writer to figure out how to present his or her information and analysis in a clear and coherent manner. A table such as the one below can be very helpful for the reader. The table should of course be accompanied by explanations of how the various results were obtained.
ramp length | rise of ramp | run of ramp | slope of ramp | `dt on ramp | vAve on ramp | vf on ramp | accel on ramp |
0915 Quiz
2. A ball starts from rest on an incline of length 18 cm on which its acceleration is 20 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 40 cm / s^2, and whose length is 24 cm.
Use the equations of uniformly accelerated motion to answer the following:
We know that `ds = 18 cm, v0 = 0 and a = 20 cm/s^2.
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 * 20 cm/s^2 * 18 cm) =
+-27 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 +27 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 * 18 cm / (0 + 27 cm/s) = 1.3 sec approx.
vAve = (0 + 27 cm/s) / 2 = 13.5 cm/s approx.
`dv = vf - v0 = 27 cm/s - 0 = 27 cm/s.
v0 = 27 cm/s, as mentioned.
`ds = 24 cm and
a = 40 cm/s^2.
We have the same vbls we had before so we use the same strategy:
vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (27 cm/s)^2 + 2 * 40 cm/s^2 * 24 cm) = +- sqrt(2640 cm^2 / s^2) = +-51 cm/s.
We choose the +51 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 * 24 cm / (27 cm/s + 51 cm/s) = .65 sec.
`dv = (52 cm/s - 27 cm/s) = 25 cm/s.
3. For the situation of #2:
Accel is greater on the second incline (40 cm/s^2 vs. 20 cm/s^2).
Velocity changed by more on the first incline, though the acceleration was less and the distance was also less.
Seems contradictory. Greater accel. on the second, greater displacement on the second, but practically more velocity change on the first.
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. So the lesser change in clock time on the second incline gave us less change in velocity on that incline.
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 inclines with unequal slopes so that the ball will roll down the first incline and onto the second with no significant change in velocity.
Measure the two inclines so you know the rise, run and length of each.
When the ball runs from the start of the first incline to the end of the second, determine how long the ball spends on each incline 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 you are instructed to first download the latest version of the Multiple Choice Generating Program:
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.