PUM Physics II - Kinematics
Lesson 8 Sample Lab
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8.1 Design an Experiment a) Experimental setups will vary by group: To determine the speed of each car: Stop Watch
Meter Stick
b) Will Vary: Time will be measured with a stop watch or clock, Position of the car at each time interval will be measured by a meter stick. The easiest way to measure the position of the car at each time interval is to drop sugar packets or mark with a pencil the location of the car at each second and then to measure the distances between these tick marks. c) Sample Data: Motion Diagram:
Table: Time (s)
Position (m) 0 1 2 3 4 5 6 7 8 9 10
0 0.09 0.17 0.31 0.4 0.49 0.55 0.68 0.79 0.87 0.98
PUM Physics II - Kinematics
Lesson 8 Sample Lab
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Position vs. Time Graph:
d) The uncertainty in the measurements for clock reading and position are half of the smallest division on the device that we used to measure these physical quantities. Time was measured using a stop watch that measured time to the nearest tenth of a second, therefore the uncertainty in the time measurement is 0.05 seconds, or 5% because we measured time each second. The uncertainty in the measurements for position were much smaller, .001 meters. For this experimental the instrumental uncertainty is the largest. e) To represent the instrumental uncertainty on the graph, we can draw uncertainty lines at each point that represent the 5% instrumental uncertainty (if that is our largest uncertainty in the experiment). This would not change where the trend line is drawn, the trend line will still be drawn through the average of the points on the graph. f) The speed of the mystery machine based on our measurements is the slope of the position versus time graph: 0.098 +/- .005 m/s. 8.2 Design an Experiment a) Answers Will Vary, using sample data: The Mystery Machine moves at a constant speed of .098 +/- .005 m/s and starts at 0 meters. x(t) = .098(m/s)t + 0(m) The Scooby Mobile moves at a constant speed of .065 +/- .005 m/s and starts at +3 meters, moving in the negative direction. x(t) = -.065(m/s)t + 3(m)
PUM Physics II - Kinematics
Lesson 8 Sample Lab
Page 3 of 9
b) Using Sample Data: If we want to find when the cars will meet, we can use the x(t) equations to find at what time their position is the same. If we set the equations equal to each other, we are saying that the position is the same, and we can then solve for t to find at what time that is a true statement. .098(m/s)t + 0(m) = -.065(m/s)t + 3(m) .098(m/s)t – (-.065(m/s)t) = 3m t ( .098(m/s) + .065(m/s) ) = 3m .163 (m/s) t = 3m t = 3m / .163 (m/s) = 18.40 +/- .92 seconds c) We are assuming that the instrumental uncertainty and not the random uncertainty was the largest source of uncertainty, that the cars will move at a constant speed that is equal to our calcuated speed, that the cars move at a constant speed immediately, that we are on a smooth and level surface, etc. d) Will Vary e) Will Vary
Homework 8.3 Write a lab report Will Vary 8.4 Represent and Reason a) x = 3t + 5
Position (m)
Velocity (m/s)
The object moves at a constant velocity of 3 m/s, moving in the positive direction, starting at a location of 5 meters. Assuming units are m and s.
3
5 Time (s)
Time (s)
PUM Physics II - Kinematics
Lesson 8 Sample Lab
Page 4 of 9
b) x=(-3t)+5
Position (m)
Velocity (m/s)
The object moves at a constant velocity of -3 m/s starting from the 5 m mark.
5 Time (s)
Time (s) -3
c) x = (-3t) + (-5)
Position (m)
Velocity (m/s)
The object moves at a constant velocity of -3 m/s starting from the -5 m mark.
Time (s) -5
Time (s) -3
PUM Physics II - Kinematics
Lesson 8 Sample Lab
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d) x = 3t + (-5)
Position (m)
Velocity (m/s)
The object moves at a constant velocity of 3 m/s starting from the -5 m mark.
3 Time (s)
Time (s)
-5
e) x = 3t
Position (m)
Velocity (m/s)
The object moves at a constant velocity of 3 m/s starting from the origin.
3 Time (s)
Time (s)
f) x = -3t
Position (m)
Velocity (m/s)
The object moves at a constant velocity of -3 m/s starting from the origin.
3 Time (s)
Time (s)
PUM Physics II - Kinematics
Lesson 8 Sample Lab
Page 6 of 9
b) The car may or may not be traveling at a constant speed if you change the reference frame. If your reference frame is one where the observer is moving at a non constant speed then the car will also appear to be moving at a non constant speed. If your reference frame is one where the observer is moving at a constant speed that is either greater than or less than the speed of the car then the car will appear to be moving at a constant speed as well. If your reference frame is one where the observer is moving at the same constant speed as the car then the car will not appear to be moving at all! c) Object one starts at a location of -12 m and travels in the positive direction at a speed of 3 m/s. Object two starts at a location of +24 m and travels in the negative direction at a speed of 7 m/s. I am assuming that both objects are moving with respect to the same reference frame. Position vs. Time Graph:
Based on the graph, it appears as though the objects meet at approximately t =3.75s and position = -2m. Using the equations, we can find the time and location more precisely. x = (-12 m) + (3 m/s)t x* = (24 m) + (-7 m/s)t Setting the equations equal to each other we can find the time that the two objects intersect. (-12 m) + (3 m/s)t = (24 m) + (-7 m/s)t (10 m/s)t = 36m t = 3.6 s We can then use either equation to find out where the two objects meet: x = (-12 m) + (3 m/s)(3.6 s) = -1.2 m x* = (24 m) + (-7 m/s)(3.6 s) = -1.2 m
PUM Physics II - Kinematics
Lesson 8 Sample Lab
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The cars meet at -1.2 m after 3.6 s 8.5 Represent and Reason The motion of Object A is the same in both graphs. a) Both graphs represent the motion of two objects. On the y-axis is the position, on the xaxis is the time. On both graphs the objects intersect at 6.0 s b) In graph 2, Object B is traveling in the negative direction and starts at a different location as compared to graph 1. In graph 1, Object B is traveling in the positive direction. c) Object B has the smaller speed. Speed is the magnitude of the velocity. In otherwords it is the value of the steepness, regardless of direction. Object B’s position vs. time line is less steep than Object A’s position vs. time line. d) The two objects pass the same point on the position axis. In otherwords at 6 s, both objects are at the same location (next to each other). e) Object A traveled a greater distance during the first 6.0 s. Object A started at the origin and got to the intersection point 6 s later. Object B started closer to the intersection point. Another way to think about it is that Object A is moving faster than Object B since its line is steeper. Therefore it covered a greater distance before getting to the intersection point. f) Many possibilities, just check to make sure they intersect at 6 s and B2 is steeper than B1. xA1 = (3 m/s)t xB1 = (12 m) + (1 m/s)t xA2 = (3 m/s)t xB2 = (30 m) + (-2m/s)t 8.6 Represent and Reason 35 mi/hour
60 miles North
Position of Secaucus = 60 mi Position of Bay Head = 0 mi South Bound Train: xSB = (-52 mi/hour)t + 60 mi North Bound Train: xNB = (35 mi/hour)t + 0 mi
Bay Head
Secaucus
52 mi/hour
PUM Physics II - Kinematics
Lesson 8 Sample Lab
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Assumptions: Trains move at constant velocity, trains do not stop, trains start and end at stations. Solving for when the trains will meet: (-52 mi/hour)t + 60 mi = (35 mi/hour)t + 0 mi 60 mi = (35 mi/hour)t + (52 mi/hour)t 60 mi = 87(mi/hour)t t = (60 mi)/(87 mi/hr) = 0.7 hr To solve for where the trains will meet, we can plug 0.69 hr into either equation: xNB = (35 mi/hour)t + 0 mi xNB = (35 mi/hour)(0.7 hr) = 25 mi Trains will meet at 25 mi and 0.7 hr 8.7 Represent and Reason a) Dot Diagram:
b) Position vs. Time Graph: We are taking the intersection to be the origin with both students starting out on the positive sides of their cooridors and heading in a negative direction towards the center.
c) xwalking student = (20 ft) + (-5 ft/s)t xrunning student = (32 ft) + (-8 ft/s)t
PUM Physics II - Kinematics
Lesson 8 Sample Lab
Page 9 of 9
d) An infinite number of equations can be written for each student depending on the choice of origin and reference frame. If our reference frame is moving with respect to both students then the speeds of each will be different. If we take a different spot to be the origin then the initial positions of the students will be different and whether that position is positive or negative could be different. For example: xwalking student = (-20 ft) + (5 ft/s)t xrunning student = (-32 ft) + (8 ft/s)t Is just as valid. e) Based on the graph, we can see that the students collide at a position of 0 ft and a time of 4 seconds. This can be shown explicitly by the equations: (20 ft) + (-5 ft/s)t = (32 ft) + (-8 ft/s)t (-5 ft/s)t + (8 ft/s)t = 12 ft t(3 ft/s) = 12 ft t = (12 ft)/(3 ft/s) = 4 s Plugging t = 4 seconds into either equation gives the location: xwalking student = (20 ft) + (-5 ft/s)(4 s) = (20 ft) + (-20 ft) xwalking student = 0 ft 8.8Reason II) 4.4 m/s (17.5/4); Average Speed is the Path Length / Total Trip Time, the total trip time is 4 s not 2 s even though from 2-4 s the object is not moving.
