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Old Dec 15th 2018, 10:20 PM   #1
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Elastic Collision in One Dimention

I was told that my percentages of discrepancy are wrong for this problem. I felt like that was the easiest and most straightforward part of the problem, so I am wondering if I did anything else wrong. I was told that all of my answers were correct (except the predictions, no feedback) besides the percentages which were marked wrong. Everything was graded by a computer, so it's very picky about formatting meaning that I could just be formatting my answers wrong. I've tried a couple of different ways, so I thought I would come here for a second opinion. Please help!

In this lab you will study an elastic collision between a moving glider(m1) and a second one which is initially at rest(m2). (They are on an air track, so ignore friction.)

Given Values(Every other number is calculated.)
m1: 266.8g
m2: 474.9g
L1: 14.1cm
L2: 25.3cm
Δt1i: .122955s
Δt1f: .475583s
Δt2f: .311598s

Derive expressions for v1f and v2f in terms of v1, m1, and m2, based on the scenario above. Use conservation of momentum and conservation of kinetic energy (since we are postulating that the collision is elastic).

V1f=V1i(m1-m2/m1+m2) & V2f=V1i(2m1/m1+m2) (Simplified from conservation of momentum equation.)

Calculate v1i = L1 / Δt1i. Then use the expressions you derived in the Analysis section to predict values for the final velocities v1f and v2f .

(v1i(measured): 114.68cm/s )

Predicted Measurements
v1f (predicted): −32.59939086cm/s
v2f (predicted): 82.08060914cm/s
(I got these numbers using the formulas for velocity above.)

Comparison with experiment: Calculate your experimental values of v1f and v2f using Δt1f and Δt2f. Also calculate the percent discrepancies between the predicted and measured values.(to two significant figures, as a positive number.)

v1f (measured): 29.65cm/s [/U]discrepancy 1.096944377% or 1.1%
v2f (measured): 81.19cm/s discrepancy 9.947355346% or 9.9%
Expect discrepancies in the range of 10%.

(The formula I used to calculate the percentage of discrepancy was {percent error = (| T − E |/T) 100%} where T=accepted value and E=experimental value.)

Thank you for taking the time to help me out!
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Old Dec 16th 2018, 07:35 PM   #2
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Please explain how you calculated the percent discrepancy. For example, consider v2f predicted = 82.08 cm/s, whereas the measured value is 81.19 cm/s; I don't understand how you calculated a discrepancy of 9.9%. The difference between measured and predicted values is equal to 81.19-82.08 = -0.89cm/s. Note the minus sign, meaning the measured value is less than the predicted value. Expressed as a percentage of the predicted value that would be -0.89/82.08 = -1.1%. I should also point out the wording of the problem is rather vague, and I am assuming that the answer is intended to be the discrepancy as a percentage of the predicted value. It's possible the answer they're looking for is the discrepancy as a percentage of the measured value.
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Old Dec 16th 2018, 09:54 PM   #3
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Sorry, I accidentally switched the percentage discrepancies at the bottom when I put them in the post. It should be
v1f (measured): 29.65cm/s discrepancy 9.947355346% or 9.9%
v2f (measured): 81.19cm/s discrepancy 1.096944377% or 1.1%

The formula I used was percent error = (| M− P |/M) 100% where M is the measured value and P is the predicted value. I think that's what you mean when you mentioned the discrepancy as a percentage of the measured value. I also tried switching M and P just in case the wanted the percentage of the discrepancy as based off of the experimental value, but v2f still simplifies to 1.1%, so if that were the case, I should have been marked correct. (I wrote the percentages s positives because the question asked me to format them that way.)
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