Gordon-McIntosh- Physics Lab 1
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University of Ottawa *
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Course
1122
Subject
Mathematics
Date
Jan 9, 2024
Type
Pages
3
Uploaded by ProfessorLion1229
Shea Gordon-McIntosh
1) Distance Running
Question 1a: Based on the data shown in Table 1, student 1 is a stronger runner. Student 1 ran
at a consistently quicker pace than student 2. Student 1 takes less time to complete each lap.
Calculation 1a:
Average time per lap = (total time/number of laps)
Student #1: 333s/6 laps Student #2: 393s/6 laps
= 55.5 s/lap = 65.5 s/lap
Calculation 1b:
Student #1: =
Student #2: =
53. 5/6
55. 5/6
= 2.986s = 3.041s
SE= 2.986/
SE= 3.041/
6
6
= 1.219s = 1.241s
Table 1:
Table 1- Individual and average lap times, and standard error for two students
Time 1
(s)
Time 2
(s)
Time 3
(s)
Time 4
(s)
Time 5
(s)
Time 6
(s)
Average
time per
lap (s/lap)
Standard
Error (s)
Student
1
55
57
50
56
55
60
55.5
1.219
Student
2
64
64
64
68
71
62
65.5
1.241
Calculation 1c:
Average speed = (total distance/total time)
Student #1: v= (400m x 6)/ 333s Student #2: v= (400m x 6)/ 393s
= 7.21 m/s = 6.11 m/s
Calculation 1d:
Student #1: v = 1/0.1378 Student #2: v = 1/0.1659
= 7.25 m/s = 6.03
Question 1b:
Student #1: % difference = (7.21-7.25)/(½ (7.21+7.25)) x100
= 0.55%
Student #2: % difference = (6.11-6.03)/(½ (6.11+6.03)) x100
= 1.3%
Shea Gordon-McIntosh
2) Projectile Motion
Calculation 2a:
Average value of range = total range for angle/5
25°: 7.905m/5 35°: 9.71m/5 45°: 10.325m/5 50°: 10.14m/5 60°: 8.935m/5 70°: 6.62m/5
= 1.581m = 1.942m = 2.065m = 2.028m = 1.787m = 1.324m
Calculation 2b:
Range = (v
2
sin(2Θ))/g
g= (4.5)
2
/2.068
= 9.79 m/s
2
Question 2:
% error = | (accepted − experimental)/accepted | × 100%
= ((9.81-9.79)/9.81) x 100%
= 0.20%
Calculation 2c:
i) Theoretical time to reach maximum height = (4.5)(sin(60°))/9.81
= 0.397 s
ii) Theoretical maximum height above launch point = (4.5)
2
sin
2
(60°)/(2)(9.81)
= 0.77 m
iii) Theoretical maximum range = (4.5)
2
(sin(2x60°))/9.81
= 1.79 m
3) Ferris Wheels
Calculation 3a:
Radius (m) = (diameter/2) x 0.3048
Wheel #1: r= (36ft/2) x 0.3048 Wheel #2: r= (28ft/2) x 0.3048
= 5.486m = 4.267m
Calculation 3b:
Average period of rotation = total time/5
Wheel #1: 82.5s/5 Wheel #2: 68.5s/5
= 16.5s = 13.7s
Uncertainty =
Uncertainty = 3.8/5
= 0.4 = 0.8
Calculation 3c:
Average speed = ω x radius
Wheel #1: ω = 2π/T
Wheel #2: ω = 2π/T
Shea Gordon-McIntosh
= 2π/16.5s
= 2π/13.7s
= 0.38 rad/s
= 0.46 rad/s
v = (0.38 rad/s) x 5.486m
v = (0.46 rad/s) x 4.267m
= 2.08 m/s
= 1.96 m/s
Centripetal acceleration = v
2
/radius
Wheel #1: (2.08m/s)
2
/5.486m Wheel #2: (1.96m/s)
2
/4.267m
= 0.789 m/s
2
= 0.900 m/s
2
Question 3: The second wheel operator was correct when he stated that his wheel provided a
bigger thrill for his riders because the second wheel's centripetal acceleration is larger. The
second wheel’s centripetal acceleration is 0.900 m/s
2
, while the first wheel’s centripetal
acceleration is 0.789 m/s
2
.
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