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course Phy 241
9/04 6 pm
Questions `q001 - `q004 are for everybody.Questions `q005 - `q007 require the use of calculus and are for University Physics students only.
`q001. This series of questions uses the basic analysis of a straight-line v vs. t graph for an object which starts from rest:
Sketch a v vs. t graph representing the motion of a ball that starts from rest and moves for 6 seconds, averaging a velocity of 15 cm / sec.
Describe your graph.
**** The graph is an increasing straight line beginning at (0, 0) and ending at (6, 30).
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What are the initial and final velocities of the ball?
**** The initial velocity is 0 and the final velocity is 30 cm/s.
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What is its velocity at the 3-second mark?
**** The velocity at the 3-second mark is the average velocity, 15 cm/s.
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By how much does its velocity change during the 6-second time interval?
**** The velocity changes by 30 cm/s over the 6-second interval.
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How quickly is its velocity changing during the 6-second interval? Note that the answer to this question is also the slope of your v vs. t graph.
**** The velocity is changing by 5 cm/ s^2.
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`q002. This question asks you to do something closely related to today's class, but somewhat different from anything we actually did.
How far does the ball in the preceding question travel during the 6 second interval?
**** To find this, we take the velocity function for this set of data (v = 5t) and find its integral. That gives us the position function: s = 5/2 * t^2. Since t = 6s, the distance traveled by the ball is 5/2 * 6^2 = 90 cm.
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How far does it travel during the first 3 seconds?
**** We do the same for this question as we did for the last question except that here t = 3s. The function then becomes: s = 5/2 * 3^2 = 45/2 = 22.5 cm
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Would a sketch of its position vs. clock time be a straight line, a rising curve with an increasing slope, a rising curve with a decreasing slope, or some other type of curve?
**** The graph would be a rising curve with an increasing slope.
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`q003. Report your data from today's experiment. Your report should be clear and concise, telling the reader what was measured and how, and specifically what the results were.
**** In today’s experiment, we measured the difference in the heights of dominos. To do this, we stacked dominos into two groups of seven, set them about 20 cm apart, placed a track over them and rolled a ball along the track. We timed how long it took for the ball to either roll to the other end of the track or come to a stop, then timed the ball starting from the other end. After that, we mixed the dominos up again, made two more stacks of seven and repeated the process. By rolling the ball along the track and comparing the times from both directions, we could tell which stack was higher, and by comparing the times from multiple trials, it would be possible to guess the percent difference in the heights of the dominos.
My partner and I only made one copy of the data (a mistake I’ll be sure not to repeat), and she has it. So, I will try to make a chart from memory. The chart indicates the trial number, the amount of time it took the ball to roll across the track (or come to a stop) starting from the left, and the amount of time it took the ball to roll across the track (or roll back) starting from the right.
Cycles from Left Cycles from Right
1 2 3
2 2.8 2.5
3 1.5 2
4 2.4 3
5 2.5 3.3
6 3 2.7
7 1.5 2.5
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`q004. For one of your ramps, indicate the displacement of the ball and the number of cycles of your pendulum corresponding to motion to or from rest in one direction, and the displacement of the ball and the number of cycles to or from rest in the other direction.
**** Since it’s the one I remember the best, I will use ramp 3. The ball rolled halfway along the ramp in 1.5 cycles, stopped, and rolled back to where it started from in 2 cycles. Because it only rolled halfway along the ramp before stopping, the displacement was about 10 cm.
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What is the average velocity of the ball as it travels in each direction, according to your data? Your time will be measured in cycles of the pendulum, rather than in seconds. A cycle is a perfectly valid unit of time, as long as you know the length of the pendulum. So for example the average velocity of the ball will be in centimeters / cycle. We can later convert the result to units involving seconds.
**** The average velocity from left to right was 10 cm/ 1.5 cycles, or about 6.67 cm/cycle. From right to left, the velocity was 10 cm/ 2 cycles, or 5 cm/cycle.
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What are the initial and final velocity of the ball in each direction?
**** The final velocity from left to right was 0, so, assuming constant acceleration, the initial velocity was 13.34 cm/cycle. The initial velocity from right to left was 0, so the final velocity was 10 cm/cycle.
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What is the change in velocity in each direction?
**** The change in velocity from left to right was -13.34 cm/cycle. The change in velocity from right to left was 10 cm/cycle.
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How quickly did velocity change in each direction? Note that an answer to this question is also the slope of an appropriate v vs. t graph.
**** From left to right, the velocity changed by -13.34 cm/cycle/1.5 cycles, or -8.89 cm/cycle^2. From right to left, the velocity changed by 10 cm/cycle/ 2cycles, or 5 cm/cycle^2.
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`q005. University Physics students: If the acceleration of a ball on a certain curved ramp is given by the function
• a(t) = 5 centimeters / second^3 * t + 15 centimeters / second^2
then what function represents its velocity v(t) as a function of clock time t?
**** Since t is measured in seconds, the function can be rewritten as a(t) = 5t +15. The velocity function is the integral of the acceleration function, so v(t) = 5/2 t^2 + 15t + C. Without more information, we cannot determine the value of C.
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If this object has an initial t = 0 velocity of 10 centimeters / second, what is its specific velocity function?
**** v(t) = 5/2 t^2 + 15t +10
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How far would this object travel between t = 0 and t = 8 seconds?
**** To find how far the ball would travel, we first need its position function. This is the integral of the velocity function, so : s(t) = 5/6 t^3 + 15/2 t^2 + 10t + C. We then evaluate this for the interval from 0 to 8. That gives us:
5/6 * 8^3 + 15/2 * 8^2 + 10 * 8 = 986.7 cm approx.
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How fast would it be moving at the end of the 8 seconds?
**** v(8) = 5/2 * 8^2 + 15 * 8 + 10 = 290 cm/s
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Would its v vs. t graph be a straight line?
**** No, it would be a quadratic.
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What would be its average velocity for the 8-second interval?
**** v = Δs/ Δt, so assuming the ball started at s = 0, v = 986.7 cm/ 8 s = 123.3 cm/s
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What would be the average of its initial and final velocities?
**** (10 + 290) / 2 = 150 cm/s
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Is the average of the initial and final velocities equal to its average velocity? Why would we or would we not expect it to be so?
**** No, they are not equal. If the ball were to start at 50 cm/s, in two seconds decrease to 10 cm/s, remain moving at that velocity for 27 seconds, and then accelerate to 30 cm/s in one second, the average velocity of the ball for that 30 seconds would not be 40 cm/s because it did not travel 1200 cm during that time frame. The average velocity is the change in position divided by the change in time.
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`q006. University Physics Students:
For the ball in questions 1 and 2:
How far would the ball travel in the first 4 seconds?
**** The function of the v vs. t graph for the ball in questions 1 and 2 is: v(t) = 5t. To find the displacement of the ball at 4 seconds, we take the integral of the velocity function.
v' (t) = s(t) = 5/2 * t^2
s(t) = 40 cm
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At what time would the ball's displacement be 60 centimeters?
**** s(t) = 5/2 * t^2
60 = 5/2 * t^2
t = sqrt(24) = 4.9 s
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At what time would the ball have traveled through half of its 6-second displacement?
**** At 6 seconds, the ball has travelled 90 cm, so half that distance would be 45 cm.
45 = 5/2 * t^2
t = sqrt(18) = 4.24 s
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`q007. University Physics Students:
The acceleration you obtain for the ball on the ramp is calculated based on its displacement, which we will represent as x, and the time t for it to accelerate through that displacement either to or from rest. A formula for its acceleration is
a = 2 x / t^2.
If t is treated as a constant, what is the derivative da / dx? (i.e., what is the derivative of 2 x / t^2 with respect to x?)
**** The derivative of 2 x / t^2 with respect to x is a’ (x) = 2 / t^2
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If x is treated as a constant, what is the derivative da / dt? (i.e., what is the derivative of 2 x / t^2 with respect to t?)
**** To find the derivative with respect to t, I will first rewrite the expression as 2x * t^(-2). The derivative is a’ (t) = -4x * t^(-3) = -4x / t^3.
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The differential of the expression a = 2 x / t^2 is
da = da/dx * dx + da/dt * dt,
where da/dx and da/dt are the derivatives you calculated above.
dx and dt will be the uncertainties in x and t.
You chose one of your ramps as the basis for your answer to question 4 above.
For that particular set of observations, what do you think was the uncertainty in the distance x traveled by the ball, and what do you think was the uncertainty in your timing? Explain also your reasons for these estimated uncertainties.
**** I am not certain that the ramp was exactly 20 cm and we had no way to measure exactly how far the ball travelled, so I would guess the uncertainty in distance would be about 2 cm. Similarly with time, there was no way to measure exactly how much time had passed, so we had to guess as close as we could with the pendulum. However, since the times fell at the points when the pendulum was highest, we can guess the uncertainty in time is not more than 0.1 cycles.
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What is the resulting uncertainty da = da/dx * dx + da/dt * dt in your calculated acceleration?
**** da = (2/ t^2) * dx + (-4x/ t^3) * dt
From left to right, the calculated uncertainty is:
da = (2/ 1.5^2) * 2 + (-4 * 10 / 1.5^3) * 0.1 = 0.593
From right to left, the calculated uncertainty is:
Da = (2/ 2^2) * 2 + (-4 * 10 / 2^3) * 0.1 = 0.5
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Good, but for future reference:
Those quantities all have units, and you should work out the result with units.
Also you don't want to be subtracting contributions to the uncertainty. For that reason you would probably use dt = -0.1.
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