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course PHY 241
8/31/2010 5:12 p.m.
Questions 100830.1. What was your count for the pendulum bouncing off the bracket, and how many seconds did this take? What therefore is the time in seconds between collisions with the bracket? What was the length of your pendulum?
Answer: I'm not sure exactly what this first part means but I counted about 10 bounces total before stopping but I did not count the time. I used this to get the rhythm of the bounces then I counted, in time with the pendulum, 4 counts of 8 in 32 seconds, which is what I thought we were supposed to do. All of my counts will then be based on 4 8s in 32 seconds. Based on my count and the number of times the pendulum hit I figure the time between bounces to be approximately 1/4 sec. My pendulum was 13 cm in length.
2. What was the period of your pendulum when it was swinging freely? Give your data and briefly explain how you used it to find the period. How does your result compare with the time between 'hits' in the first question?
Answer: My pendulum consistently had 30 swings, in one direction, in 30 seconds with little distinguishable difference. So we see that it had 15 periods, back and forth swings, in 30 seconds which gives a 2 second period. This is much longer than the bouncing from the first part but I would think they should be related closer numerically than this.
3. Give your data for the ball rolling down the ramp, using the bracket pendulum as your timer. Assuming the ball traveled 30 cm each time, what are the resulting average velocities of the ball for each number of dominoes?
Answer: For one domino I had 5 consecutive counts of 5 (in pendulum counts), which gives a velocity of 30 cm/ 5 counts = 6 cm/count. For two dominos I had 3, 4, 5, 4, 4, 4 counts consecutively for an average of 4 counts so the velocity here would be 30 cm/ 4 counts = 7.5 cm /count. For three dominos I had 5 consecutive counts of 3 and thus the velocity is 30 cm/ 3 counts = 10 cm/count. These can easily be figured in seconds to be: 24 cm/s, 30 cm/s, and 40 cm/s respectively.
4. How did your results change when you allowed the ball to fall to the floor? What do you conclude about the time required for the ball to fall to the floor?
Answer: Here I had 6 counts for 1 domino, 5 counts for two dominos, and 4 counts for 3 dominos. This is 1 additional count per domino over the above results, therefore, we can conclude that it takes one count to reach the floor or 1/4 of one second.
5. Look at the marks made on the paper during the last class, when the ball rolled off the ramp and onto the paper. Assuming that the ball required the same time to reach the floor in each case (which is nearly but not quite the case), did the ball's end-of-ramp speed increase by more as a result of the second added domino, or as a result of the third? Explain.
Answer: There was more of an increase in distance from the table for adding the second domino than the third. This was observed to occur between the third and fourth dominos as well, each addition adds less and less to the distance away from the table. Therefore we can conclude that the addition of each new domino has a diminishing affect on the end-of-ramp speed. This can be asserted because we know that there is no new forward acceleration after the ball leaves the ramp and so the forward motion of the ball is due only to the initial speed the ball has when it enters free fall.
6. A ball rolls from rest down a ramp. Place the following in order: v0, vf, vAve, `dv, v_mid_t and v_mid_x, where the quantities describe various aspects of the velocity of the ball. Specifically:
* v0 is the initial velocity,
* vf the final velocity,
* vAve the average velocity,
* `dv the change in velocity,
* v_mid_t the velocity at the halfway time (the clock time halfway between release and the end of the interval) and
* v_mid_x the velocity when the ball is midway between one end of the ramp and the other.
Explain your reasoning.
Answer: Here we have: v0 < vAve = v_mid_t = v_mid_x < 'dv = vf. It is clear why v0 is least, but the equality in the next three might be unexpected. We know vAve = x/t and can find how long it takes to move x' cm in t' seconds from the equality x/t = x'/t', so for v_mid_t we have t'= t/2, substitution gives (x/t) = (x'/(t/2)), and solving for x' = x/2 this works the other way also giving t' = t/2 and this shows that to travel x/2 cm it takes t/2 seconds and so v_mid_t = v_mid_x. It is easy to see that if it takes (t/2) seconds to travel (x/2) cm we have an average velocity of (x/2)/(t/2)= x/t and so the three are all equivalent. 'dv = vf because 'dv= vf-v0= vf-0=vf.
7. A ball rolls from one ramp to another, then down the second ramp, as demonstrated in class. Place the following in order, assuming that v0 is relatively small: v0, vf, vAve, `dv, v_mid_t and v_mid_x .
Place the same quantities in order assuming that v0 is relatively large.
Which of these quantities will be larger when v0 gets larger? Which will get smaller when v0 gets larger? Which will be unchanged if v0 gets larger?
Answer: The relationships between v0, vAve, v_mid_x, v_mid_t and vf never change regardless of how small or large v0 is. The only quantity that changes its position in the inequalities is 'dv. As v0 becomes larger and larger 'dv becomes smaller with respect to the other quantities. So all the quantities will get larger as v0 gets larger and maintain their order of magnitudes. So we will always have: v0 < vAve = v_mid_t = v_mid_x < vf and 'dv will move down from 'dv= vf at v0=0 to 'dv
8. If a ball requires 1.2 seconds to travel 30 cm down the ramp from rest:
* What is its average velocity?
* What is its final velocity?
* What is the average rate of change of its velocity?
Answer: First: vAve= 30 cm / 1.2 s= 25 cm/s, then, with v0=0, we have: vf=2*vAve= 50 cm/s and last: 'dv = (vf-v0)/t= (50-0)/1.2= 41.67 cm/s/s. The reasoning behind the first two of these is laid out above. The average rate of change of its velocity is the velocity change with respect to time change and is thus 'dv/'dt. It is average because even though we have constant gravitational pull there are other smaller factors that alter the exact rate of acceleration at any given instant.
Be sure to explain your reasoning."
Good thinking throughout. I don't think all of your answers will hold up under closer scrutiny (for example the midpoint clock time occurs before the midpoint of the ramp so an accelerating ball won't have the same velocity at both), but most will, as will the quality of your thinking.
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