#$&*
course phy 201
September 5th at 8:50am
A ball rolls down a 60 cm incline and off the end. As it drops to the floor it travels an additional 16 cm in the horizontal direction, and it is known that the drop to the floor requires 0.4 seconds. From this we conclude that the ball was moving at 40 cm/s at the end of the ramp.`q001. If the ball started from rest, and if the acceleration on the ramp was uniform, then what does the graph of velocity vs. clock time look like?
****
Final velocity= 40cm/sec on y-axis at 3sec on x-axis
Average velocity = 20cm/s on y-axis at 1.5sec on x-axis
#$&*
What new quantity or quantities can we determine, based on our graph, and what is the value of each? (Quantities in which we might be interested include acceleration, average velocity, time interval, change in velocity and possibly others; which ones can be determined directly from the graph?).
****
Intial vel. =0cm/sec ave. vel =40/2=20 acceleration=20/1.5=13.3cm/sec^2
Final vel. =40cm/sec time interval = 3sec at end of ramp +.4 off ramp
#$&*
`q002. From the given information we know that the average velocity of the ball is 20 cm/s. Having found this, how can we determine the time interval for the motion down the ramp?
****
60cm/(20cm/sec)= 3sec for 20cm/sec divide it by half =1.5s
#$&*
`q003. If for an interval on which the graph of v vs. t is a straight line we know two of the three quantities initial velocity, final velocity and average velocity, then we can find the third.
Suppose for such an interval we know that the initial velocity is 10 cm / sec and the final velocity is 30 cm/s. What is the average velocity?
****
((30-10)/2) +10 =20cm/sec= average vel.
#$&*
Suppose we know for a different interval that the initial velocity is 20 cm/s and the average velocity is 30 cm/s. What is the final velocity?
****
((x-20)/2)+20 =30cm/sec=final vel.
#$&*
@&
The final velocity isn't 30 cm/s. Part of your statement is
30cm/sec=final vel.
which is therefore incorrect.
However if you solve the equation
((x-20 cm/sec)/2)+20 cm/s =30cm/sec
for x you get x = 40 cm/s, which is the final velocity.
So your equation
((x-20 cm/sec)/2)+20 cm/s =30cm/sec
is very good and its solution gives you the desired result.
*@
On such an interval where the final velocity is 50 cm/s and the average velocity is 30 cm/s, what is the initial velocity?
****
((50-x)/2)+x=30 x= 10cm/sec=initial vel.
#$&*
Describe in words how you would get the final velocity for an interval on which the initial and average velocities are known.
****
((final vel.-initial vel.)/2) +initial vel. =average vel.
#$&*
`q004. The average rate of change of A with respect to B is defined to be the change in A, divided by the change in B, where A and B represent specific quantities.
If the average velocity of an object on some interval is defined to be the average rate of change of position with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
****
Change in position/change in clock time
#$&*
If the average acceleration of an object on some interval is defined to be the average rate of change of velocity with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
****
(final acc.-initial acc.)/change in clock time
#$&*
@&
This isn't correct. It has the right general form but you have to apply the definition using all the right words.
*@
@&
As one counterexample, if acceleration is constant then your result would tell you that since the initial acceleration would equal the final acceleration, the acceleration must be zero. However constant acceleration doesn't imply zero acceleration. For example the acceleration of gravity is always 9.8 m/s^2, which is nonzero.
*@
What quantity could you find for an interval on which the position of an object changes by 40 cm while the clock time changes by 8 seconds?
****
Average vel.
#$&*
What quantity could you find for an interval on which the velocity of an object changes by 300 cm/sec while the clock time changes by 20 seconds?
****
Average acceleration
#$&*
If you know the average rate of change of A with respect to B, and you also know the change in B, how would you find the change in A?
****
Slope(change in B)= change in A
#$&*
If you know the average rate of change of A with respect to B, and you also know the change in A, how would you find the change in B?
****
Change in A/slope =change in B
#$&*
If you know the average velocity of an object on an interval, and know the change in position, how do you find the change in clock time?
****
Change in position/ average vel. =change in clock time
#$&*
If you know the average acceleration of an object on an interval, and know the change in clock time, what other quantity could you find?
****
Change in vel. =average acceleration(change in clock time)
#$&*
`q005. Let's return to situation we started with, the ball on the ramp. We know its initial velocity to be 0 and its final velocity to be 40 cm/s, and the length of the ramp to be 60 cm.
Summarize everything we can reason out from this data, assuming our v vs. t graph to be a straight line and using the definitions of average velocity and average acceleration.
****
Average vel. =60cm/3sec=20cm/sec
Average acc. =(40cm/sec)/3sec =13.3cm/sec^2
#$&*
`q006. For your first short rubber band, what was the color of this rubber band, and what lengths of this rubber band corresponded to what lengths of the rubber band chain?
****
Red chain- 18.1cm single band- 7.7cm
23.9cm 8.8cm
30cm 10.5cm
34cm 13.2cm
#$&*
Give the same information for your second short rubber band.
****
Yellow chain- 18.5cm single band- 8.5cm
24cm 10.5cm
28cm 12.2cm
32cm 14.2cm
#$&*
Sketch a graph of the length of the first short rubber band vs. the length of the chain. Note the convention that a graph of y vs. x has the y quantity on the vertical axis and the x quantity on the horizontal.
Do your points lie pretty close to a single straight line? Do you think there's a tendency of your data points to curve upward or downward?
****
Do your points lie pretty close to a single straight line? Yes
Do you think there's a tendency of your data points to curve upward or downward? upward
#$&*
Sketch the straight line you think lies closest, on the average, to the points of your graph. What is the slope of this line?
****
Slope =3.5/12
#$&*
Repeat for your second short rubber band.
****
Straight line and tendency to curve upward
Slope= 2/5
#$&*
@&
Good.
You aren't asked at this point to compare these slopes, but if you were you would want to include a numerical value for each, with an appropriate number of significant figures.
Your two slopes are about 0.3 and 0.4. It would be OK to include one more figure with an estimated uncertainty, e.g., 0.29 +- .05 and 0.40 +- .05 if you have reason to think that the uncertainty is +- .05.
*@
If your two short rubber bands were allowed to oppose one another, you would be able to take data for the length of one vs. the length of the other. Based on your graphs, what do you think would be the slope of a graph of the length of your first rubber band vs. the length of the second?
****
Slope =6.2/7.5
#$&*
@&
It would have been a good idea to give at least a brief explanation of where these numbers come from and how they are related to your data.
*@
Based on your two graphs, construct a graph of the length of the first short rubber band vs. the length of the second. Describe your graph.
Positive slope, increasing with an increasing rate
#$&*
`q007. Give your data for the number of dominoes vs. the length of the rubber band chain.
****
Dominos- 0 chain length- 18.9cm
2 19.8cm
4 20.8cm
6 22.4cm
8 24.4cm
10 26.4cm
#$&*
Sketch a graph of the number of dominoes vs. the length of the rubber band chain. Describe your graph.
****
Straight line with downward curve
Positive slope
#$&*
Sketch the straight line you think comes closest, on the average, to your data points. What is its slope?
****
Slope= 4/3
#$&*
Relabel your graph using the following assumptions: Each cm marking on the ruler you used corresponds to an actual metric measurement of 0.006 meters, and each domino has a weight of 0.14 kilogram meters / second^2 (University Physics group) or 0.18 kilogram meters / second^2 (General College Physics group). In terms of your relabeling, what now is the slope of your straight line?
****
.72/.018= 40kg/s^2
@&
Good.
An equivalent way of getting this result:
Your previous slope was 4 / 3 domino / cm, which would be equivalent to
4/3 (0.18 kg m / s^2) / (.006 m) = 40 kg / s^2.
*@
#$&*
`q008. Give your data for the number of cycles in 10 seconds vs. the number of dominoes.
****
Dominos- 0 cycles (10sec) - n/a
2 n/a
4 24(6) =144
6 19(6) =114
8 17(6) =102
10 14(6) =84
#$&*
Sketch a graph of your data. Is it plausible that your graph could be fit by a straight line?
****
yes
#$&*
Figure out the time required for a single cycles, for each number of dominoes. Give your results.
****
4 10/24 =.4167
6 10/19 =.526
8 10/17 =.588
10 10/14 =.714
#$&*
Graph the square of the time required for a single cycle vs. the number of dominoes. Is it plausible that your graph could be well fit by a straight line? If so, give its slope.
****
4 10/24 =.4167^2 =.1736
6 10/19 =.526^2 =.2770
8 10/17 =.588^2 =.3460
10 10/14 =.714^2 =.5102
. Is it plausible that your graph could be well fit by a straight line? If so, give its slope.
Yes, slope =.25/6
#$&*
"
@&
Very nice work. You are doing very well so far in the course.
However be sure to see the notes I've inserted.
*@