Pt1420 Unit 6 Lab Answers

OBSERVATIONS AND ANALYSIS
Trial 1 – Logger Pro graphs

Trial 2 – Logger Pro graphs

Trial 3 – Logger Pro graphs Trial 4 – Logger Pro graphs

Trial 5 – Logger Pro graphs Note: the analysis #1 a) and b) will only be demonstrated on the graphs of trial 1 for this report, trial 2-5 have approximately the same results. a) The region where the ball was being tossed but still remained in hand (x). Examine on the velocity vs. time and acceleration vs. time graphs
b) [A, B] marked as where the ball was in free fall and moving upward
B marked as where the ball has stopped at its highest position
[B, C] marked as where the ball was in free fall and moving downward.
Trial hA (m) hB (m) hC (m) vA (m/s) vB (m/s) vC (m/s) aA(m/s2) aB(m/s2) aC(m/s2)

1 0.256
1 0.256 0.476 0.255 2.01 0.02 -2.01 9.6 -9.9 9.6
2 0.285 0.592 0.290 2.40 0.01 -2.13 9.2 -9.0 9.2
3 0.272 0.530 0.269 2.09 0.01 -2.02 6.3 -9.3 7.2
4 0.251 0.456 0.248 1.83 0.01 -1.86 9.6 -9.5 9.2
5 0.296 0.621 0.258 2.46 0.01 -2.84 4.1 -9.4 19.7
Table 1: DATA TABLE
c) Recorded values of position, velocity and acceleration graphs at points A,B, and
C

Trial 1 – Logger Pro Graphs (analyzed #1 a), b))

Note: the analysis #3, 5, 7 will only be demonstrated on trial 1 for this experiment. Trial 2-5 have approximately the same results.
Trial 1 – Logger Pro graphs (analyzed #3, 5) The value of the coefficient of t^2 was at -4.919±0.03868, compared to ½ g, which is -9.81/2 = -4.905m/s^2 is very close to each other. 6. The value of the slope is -10.04m/s^2, compared to gravity which is -9.81m/s^2 only differ by 0.23 8. The mean value of the acceleration graph in [A,B,C] is -9.435, and the average value between the results in 4, 6, 7 are [(-4.919×2)+(-10.04)+(-9.435)]/3 = -9.771m/s^2. Comparing -9.771 with the accepted value of g -9.81, the value only differ by 0.039.
9. The value of gravity obtained from the data differs slightly with the accepted values of gravity. The reason may be the slight mistakes that are made during the trials under the motion sensor. When the ball was thrown, it must land at the exact position of where it was thrown, however it is impossible for human to precisely pinpoint that location, this caused the slight variation in the values of gravity.
10. Mass of the ball 0.397±5×10-5kg

CALCULATIONS AND RESULTS
PE = mgh, KE = ½ mv2
Sample calculation – Trial 1
〖KE〗_A=1/2×0.397×〖2.01〗^2=0.802J
〖PE〗_A=0.397×9.81×0.256=0.997J
〖TE〗_A=〖PE〗_A+ 〖KE〗_A=0.997J+0.802J= 1.80
〖KE〗_B=1/2×0.397×〖0.02〗^2=7.94×〖10〗^(-5) J
〖PE〗_B=0.397×9.81×0.476= 1.85J
〖TE〗_B=〖PE〗_A+ 〖KE〗_A=7.94×〖10〗^(-5) J+1.85J= 1.85J
〖KE〗_C=1/2×0.397×〖-2.01〗^2=0.802J
〖PE〗_C=0.397×9.81×0.255=0.993J
〖TE〗_C=〖PE〗_A+ 〖KE〗_A=0.993J+0.802J=1.80J
Trial KEA(J) PEA(J) TEA(J) KEB(J) PEB(J) TEB(J) KEC(J) PEC(J) TEC(J)
1 0.802 0.997 1.80 7.94×10-5 1.85 1.85 0.802 0.993 1.80
2 1.14 1.11 2.25 1.99×10-5 2.31 2.31 0.901 1.13 2.03
3 0.867 1.06 1.93 1.99×10-5 2.06 2.06 0.810 1.05 1.86
4 0.665 0.978 1.64 1.99×10-5 1.78 1.78 0.687 0.966 1.65
5 1.20 1.15 2.35 1.99×10-5 2.42 2.42 1.60 1.00 2.60