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course phy 231
9-12-10 3:00 pm
Class Notes 100830One standard assignment is to come to class with three good questions, ready to turn in.
You should try to do the questions below before class on Wednesday. However some of you got these questions late and won't be able to do so. If that's the case, at least try to think about them, which should better prepare you for some of what we'll be doing in class.
Questions for everyone:
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? 10 bounces in 2 seconds. 3.3 bounces per second. Pendulum length=4.75
10 bounces in 2 seconds would correspond to 5 bounces per second. What is the time between bounces?
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? 1.4 seconds. Made 42 full swings in 30 seconds. 1 bounce= 1 half period swing
more than 30 swings in 30 seconds would mean that each swing took less than a second
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? 1 dominoe= 30 cm/6 bounce=16.7 cm/sec. 2 dom=25 cm/sec, 3 dom= 28 cm/sec, 4 dom=33 cm/sec,
Your data would be the number of bounces for each setup. That should have been reported first. Then your analysis would connect your data to your final results.
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? Each result had 1 bounce added, concluding that the fall to floor was unchanged by how many dominoes there were.
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.
Yes, because the ball was able to go further before hitting the notebook.
I don't think that's an answer to the question. Take another look at that question and see if you agree.
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. v0,v_mid_x,v_mid_t,vAve,'dv,vf. V0 will be 0. so it should be frst. V_mid_x, will be slower than v_mid_t.(first half of ramp slower than second half). Vave should be in the middle, 'dv and vf should be the same and also the highest velocities.
We're still investigating v_mid_x and v_mid_t. We'll see whether your ordering corresponds to experimental data.
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 . the quantities should remain unchanged in order??
Place the same quantities in order assuming that v0 is relatively large. Same???
Which of these quantities will larger when v0 gets larger? Which will get smaller when v0 gets larger? Which will be unchanged if v0 gets larger? None??
8. If the ball requires 1.2 seconds to travel 30 cm down the ramp from rest:
What is its average velocity? 12.5 cm/s
What is its final velocity? 25 cm/s
What is the average rate of change of its velocity? 25 cm/s^2
What is the definition of average rate of change of velocity with respect to clock time?
Apply that definition and see if you don't get something close to 21 cm/s^2.
University Physics questions:
Question 1 below concerns the equations of uniformly accelerated motion.
Questions 2-4 are progressive in difficulty. You should be able to make a good attempt at the first one and probably the second. The third is pretty challenging and you might not be able to do it on your first attempt:
1. If we integrate with respect to clock time t the acceleration function a(t) = a, where a is a constant, we get the velocity function v(t) = v0 + a t, where v0 is the velocity at t = 0. If we integrate this velocity function we get position function x(t) = x0 + v0 t + 1/2 a t^2, where x0 is the position at t = 0.
If we abbreviate v(t) as just v, and x(t) as just x, our equations are
v = v0 + a t
and
x = x0 + v0 t + 1/2 a t^2.
Answer the following:
If we solve the first equation for t, what is the result? t=(v-vo)/a
If we substitute this result for t in the second equation, and solve the resulting equation for v, what do we get? V= sqrt(-2ax0+2ax)
You should show (or at least outline) your steps here. Your answer isn't correct, but without your work I can't see where you might have gone wrong.
If we solve the first equation for a, what is the result? a=(v-vo)/t
If we substitute this expression for a in the second equation, and solve the resulting equation for the quantity (x - x0), what do we get? X-x0= v0t+.5(v/t-vo/t)+.5(v0/t-v/t)t^2
there is no term equal to .5 a, so it's not clear where you got .5 (v/t - v0 / t).
2. A ball rolls down a certain incline, starting from rest at the x position 23 cm and accelerating at 20 cm/s^2.
Use one of the equations from question #1 to find this position. ?????
Use one of the equations from question #1 to find the clock time at which the position will be 50 cm. T=2.24 seconds
The first question appears to be incomplete, not having given you enough information.
I can't tell how you got t = 2.24 seconds.
3. If the ball in #2 starts from the same position and accelerates at the same rate, but starting with velocity 15 cm/s, what will be its position when its velocity reaches 40 cm/s?
36.125 cm.
Again you need to at least outline how you got that, what equation(s) you used, etc.
4. A ball is rolled down a ramp, starting from rest at x position 30 cm. Its acceleration on this ramp is 30 cm/s^2. This ramp is parallel to the ramp in #3, right next to that ramp, and the same meter stick is used to measure positions on both ramps. If the ball is released 1 second after the ball in #3:
At what clock time will its velocity match that of the ball in #3, and how far apart will the two balls be at that instant? T=2 23 cm apart
At what clock time will the second ball pass the first, and at that instant what will be the velocity of the second ball relative to the first? After 4.14 seconds
"
I think you're OK, especially on most of the problems, but I'm not sure. I need to see your work. I can't reliably reverse-engineer an incorrect answer to see how you got it.
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