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Name: Bryan Meza
Lab #9: Conservation during Collisions
Objectives:
Collisions can be classified as elastic, inelastic, or completely inelastic. A collision is most frequently described as super-
elastic when kinetic energy is gained. In this experiment, we will observe most of these collisions and see if momentum and energy are preserved in each case. The energy changes that arise in different kinds of collisions with the carts will be
measured, and the outcomes will be classified as totally inelastic, somewhat inelastic, or elastic.
Equipment:
Station #: 8
Partner: Aiden
, rating: #5/5
Computer
Logger Pro Program
LabQuest Min iINterface
Two Vernier Motion Detectors
Track
Carts with Magnetic Bumpers
Digital Scale
Mass Set
Preliminary Questions:
#1:
#2: The collision is elastic, meaning that momentum and energy are both conserved, because the two balls did not adhere to one another. This is assuming that each ball has the same mass. Momentum and kinetic energy are both conserved in elastic collisions.
Procedure:
For this part of the lab, we setup the track that is level and put up motion detectors at each end of the track, so it can read the movement of the carts. This enabled them to read anything towards it as negative and moving away from it as positive. Once everything was setup, we practiced by creating gentle collisions. We then used magnetic bumper carts.
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This enabled us to get necessary data to determine our average velocity before and after collision for each cart to put in our data table.
Steps #1-14: Collision 1
Collision 2 Collision 3
Collision 4 Collision 5
Collision 6
Data Table A:
Velocity of
cart 1 before
collision
Velocity of
cart 2 before
collision
Velocity of
cart 1 after
collision
Velocity of
cart 2 after
collision
Cart 1
mass
Cart 2
mass
Trial
(m/s)
(m/s)
(m/s)
(m/s)
(kg)
(kg)
Part 1:
bounce
1
0.4386
0
-0.000836
0.4197
.499
.495
2
0.3324
0
-0.0007887
0.2992
.499
.495
Part 2:
stick
3
0.5853
0
0.2671
0.2653
.499
.495
4
0.5821
0
0.2683
0.2687
.499
.495
Part 3:
crash
5
0.6181
0
0.1339
0.5103
.499
.495
6
0.5080
0
0.1151
0.4395
.499
.495
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Analysis:
#1:
The Law of Conservation of Momentum asserts that "the total momentum of the system before collision is equal to the total momentum of the system after collision." Pi = Pf. The system's total momentum before and after collisions is expressed as Pf/Pi. Consequently, the total momentum before and after the contact have a ratio of 1.0.
#2:
In the elastic collision, the kinetic energy of the system is conserved. Kf = Ki as a result. Before and after the impact, the system's total kinetic energy ratio was as follows: As Kf/Ki, 1. As a result, the ratio of the system's total kinetic energy after impact to its total kinetic energy before impact is 1.0.
#3:
Mv = m(v1 + v2)
V = v1 + v2
Data Table B:
Momentum
of cart 1
before
collision
Momentum
of cart 2
before
collision
Momentum
of cart 1
after
collision
Momentum
of cart 2
after
collision
Total
momentum
before
collision
Total
momentum
after
collision Ratio of
total
momentum
after/before
Trial
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
1
0.2189
0
-0.000417
0.2078
0.2189
0.2073
0.95
2
0.1659
0
-0.000394
0.1481
0.1659
0.1477
0.89
3
0.2921
0
0.1333
0.1315
0.2921
0.2648
0.91
4
0.2905
0
0.1339
0.1223
0.2905
0.2562
0.88
5
0.3084
0
0.0668
0.2332
0.3004
0.2322
0.75
6
0.2535
0
0.0574
0.1999
0.2535
0.2574
1.02
#4:
Pf= Pi. Pf/Pi = is the ratio of the system's overall momentum before and after collisions
Data Table C:
KE of
cart 1
before
collision
(J)
KE of
cart 2
before
collision
(J)
KE of
cart 1 after
collision
(J)
KE of
cart 2 after
collision
(J)
Total KE
before
collision
(J)
Total KE
after
collision (J)
Ratio of
total KE after/before
Trial
1
0.04800
0
~0
0.04395
0.04800
0.04395
0.91
2
0.02757
0
~0
0.02234
0.02757
0.02234
0.81
3
0.08547
0
0.01780
0.01601
0.08457
0.03381
0.3956
4
0.08454
0
0.01796
0.01643
0.08454
0.03439
0.41
5
0.09532
0
0.00447
0.03924
0.09532
0.0637
0.67
6
0.06439
0
0.003305
0.04394
0.06439
0.04725
0.7338
#5:
Our momentum is conserved in our collisions, since the ratios are close to 1
#6:
(p1)^2/2m
They are conserved since the ratio of total KE. After and before are close to 1 in the first and second collision. The third is
a little off but not by much
#7:
Elastic collisions, such the way pool balls bounce off one another and continue to travel at their previous speeds, happen
when both momentum and kinetic energy are conserved. When kinetic energy is not preserved and just momentum is, an inelastic collision occurs. When two objects collide, no kinetic energy is transferred or gained—this is known as a totally elastic collision.
Extensions:
#1: Collision 1
#2: Our first trial was not conserved, but the rest were since the ratio was close to 1.
Collision 1
Collision 2a
Collision 2c
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Data Table A:
Velocity of cart
1 before
collision
Velocity of
cart 2 before
collision
Velocity of
cart 1 after
collision
Velocity of
cart 2 after
collision
Cart 1
mass
Cart 2
mass
Ext
(m/s)
(m/s)
(m/s)
(m/s)
(kg)
(kg)
1
0.04800
0
0.2769
0.2749
.499
.455
2a
0.3364
0
0.01967
0.2061
.499
.455
2b
0.4578
0
0.1369
0.1321
.499
.455
2c
0.2537
0
0.1302
0.1338
.499
.455
Data Table B:
Momentum
of cart 1
before
collision
Momentum
of cart 2
before
collision
Momentum
of cart 1
after
collision
Momentum
of cart 2
after
collision
Total
momentum
before
collision
Total
momentum
after
collision Ratio of
total
momentum
after/before
Ext.
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
(kg
•
m/s)
1
0.0240
0
0.138
0.125
0.0240
0.263
10.96
2a
0.168
0
0.00982
0.125
0.168
0.135
0.80
2b
0.228
0
0.0683
0.0601
0.228
0.1284
0.56
2c
0.127
0
0.0650
0.0609
0.127
0.1259
0.99
Data Table C:
KE of
cart 1
before
collision
(J)
KE of
cart 2
before
collision
(J)
KE of
cart 1 after
collision
(J)
KE of
cart 2 after
collision
(J)
Total KE
before
collision
(J)
Total KE
after
collision (J)
Ratio of
total KE after/before
Ext.
1
0.000575
0
0.0191
0.0172
0.000575
0.0363
63.1
2a
0.0282
0
0.0000965
0.00966
0.0282
0.00976
0.34
2b
0.0282
0
0.0191
0.00966
0.0282
0.101
3.58
2c
0.0161
0
0.00423
0.00407
0.0161
0.0083
0.52
Conclusion:
In this lab, we tested the momentum and, in some situations, energy conservation that result from the collision of two carts on a track. It is possible to categorize collisions into three groups: totally elastic, inelastic, and elastic. We saw a collision between two carts in order to test the conservation of momentum in this experiment. The method by which we
would test it would be to compute the momentum and kinetic energy for every kind of collision. We could categorize as one of three collisions after we figured it out. I discovered that an elastic collision occurs when the ratio is almost one and the kinetic energy is conserved. The kinetic energy is wasted in an inelastic collision. And in the event of a collision, the objects remain together—completely inelastic. I become more proficient at reading collisions and energy as a result of doing this experiment. I expected the outcomes we predicted because my initial responses were quite accurate. Naturally, there were mistakes made during the entire experiment. Should I repeat this experiment, I would ensure that the track is leveled as it has the potential to distort our results. Additionally, I would ensure that the sensors are accurately reading the carts as there were instances in which our ratios were far off.
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- PAP Chemistry-2903012-42100P-1 / Gases / Lesson 148 Aguilar All changes saved 25. The kinetic molecular theory says that collisions between particles are elastic, but real-world collisions do not follow this assumption. What is some of the kinetic energy changed into after a collision? O heat O light O potential energy O chemical energy SAVE & EXIT SUBMIT ALL ANSWERS PREVIOUS 25 of 25 96arrow_forwardЗа. Bromine and methanoic acid react in aqueous solution. Br, (aq) + HCO0Н (ад) — 2Br (aq) + 2H" (аq) + Со, (g) The reaction was monitored by measuring the volume of carbon dioxide produced as time progressed. o 5 10 15 20 25 [Source: © International Baccalaureate Organization 2019] 30 20 40 60 80 100 120 140 160 Time / s [Source: © International Baccalaureate Organization 2019] Determine from the graph the rate of reaction at 20 s, in cm³ s1, showing your working. 3b. Outline, with a reason, another property that could be monitored to measure the rate of this reaction. 3c. Describe one systematic error associated with the use of the gas syringe, and how the error affects the calculated rate. 3d. Identify one error associated with the use of an accurate stopwatch. Volume of carbon dioxide / cm T m mm moarrow_forwardAnna was asked to identify and explain if the equation CO2 (g) → CO2 (I) was an endothermic or exothermic reaction. However she got it wrong. What is her error and what should the answer have been? Anna's statement: The equation is an endothermic reaction because the energy is released from gaseous CO2 to the surroundings to become liquidCO, causing the molecules to speed up.arrow_forward
- 1955 Robert Wentorf. Graphite to diamond at 2000°C and 106 atm. Even though graphite is more stable than diamond at 25°C and atmospheric pressure, the conversion is very slow. A process called chemical vapor deposition uses Carbon sources like methane to release carbon atoms in presence of hydrogen gas and deposit it as diamond at low pressure and temperature (600°C) Reference: Chemistry: An atoms first approach, Zumdahl & Zumdahl, Cengage Learning, 2012. Question: Based on the above phase diagram: Comparing graphite and diamond, which one is denser? How many triple points? Explainarrow_forward"71 36 K/s ill.l O 10:01 يستخدم التطبيق المیکروفون Me et من 6 / What are the goals of adding vinegar to milk 2/ What are the results of adding vinegar to milk 3 / What is the benefit of adding vinegar to milk 10:01 زهراء 21.03.17 الساعة ص 10:01arrow_forwardLet's say you blew three baloons to the same size and left one baloon at room temperature, and kept one baloon in an ice bath, and one baloon above a hot water container. You left the laboratory for 30 minutes and when you came back, you saw that the size of the baloon in the ice bath was smaller than before and size of the baloon above hot water container was larger than the original size. As chemistry student, how do you explain the reason for this observation.arrow_forward
- According to Wikipedia, Radiocarbon dating (also referred to as carbon dating or carbon-14 dating) is a method for determining the age of an object containing organic material by using the properties of radiocarbon, a radioactive isotope of carbon. The method was developed in the late 1940s at the University of Chicago by Willard Libby, who received the Nobel Prize in Chemistry for his work in 1960. It is based on the fact that carbon- 14 is constantly being created in the atmosphere by the interaction of cosmic rays with atmospheric nitrogen. The resulting carbon-14 combines with atmospheric oxygen to form radioactive carbon dioxide, which is incorporated into plants by photosynthesis; animals then acquire carbon-14 by eating the plants. When the animal or plant dies, it stops exchanging carbon with its environment, and thereafter the amount of carbon-14 it contains begins to decrease as the carbon-14 undergoes radioactive decay. Measuring the amount of carbon-14 in a sample from a…arrow_forwardConstants: • c = 2.9979 x 108 m/s h = 6.626 x 10-34 J s per one photon • R = 8.314 J/(K mol) = 0.08206 L atm/(K mol) NA = 6.022 x 1023 particles/mol RH = 1.097 x 107 m1 = 2.178 x 10-18 J . . You have an aqueous Glucose solution that is 12.0% Glucose by mass. What is the molality of Glucose in the solution? Molar Mass of Water = 18.015 g/mol Molar Mass of Glucose 180.16 g/mol =arrow_forward2.) It’s a bad day in the lab! Two students are doing experiments. Each is 20 feet away from the professor. At the same time, each of them lets the same amount of a smelly gas into the room. One of them releases ammonia, NH3, and the other releases SO2. NH3 has a pungent odor, and SO2 smells like rotten eggs. The professor has no idea that this has happened, until she smell the first gas. Which chemical will the professor smell first? (NH3 or SO2) . If the professor starts to smell the first gas 42. seconds after the gas is released, how long will it take her to smell the second gas? sec. * Note: It is unsafe practice to work with these chemicals in an open lab.arrow_forward
- Peroxynitric acid (HOONO2) is an unstable molecule that decomposes to nitric acid and oxygen: 2HOONO2(aq) → 2HNO3(aq) + O2(g)When the concentration of peroxynitic acid is graphed against time, the resulting plot is curved, but if the logarithm of this concentration is plotted, we instead get a straight line. Based on this, which statement is true? a) This decay is a second order in peroxynitric acid. b) The slope of the straight-line graph is the rate constant. c) One needs the concentration of peroxynitric acid to calculate its half-life. d) The rate law appears to be of the form -Δ[HOONO2]/Δt = k[HOONO2].arrow_forwardNitric oxide reacts with hydrogen to release large amounts of chemiluminescence with the characteristics of the highly cytotoxic species. The following data were measured for the reaction of nitric oxide with hydrogen: Data number 1 2 3 2 NO(g) + 2 H₂(g) → N₂(g) + 2 H₂O(g) [NO] (M) 0.1 0.1 0.2 [H₂] (M) 0.1 0.2 0.1 Based on the data, (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H₂] = 0.150 M Initial rate (M/S) 1.23 x 10-³ 2.46 x 10-³ 4.92 x 10-³arrow_forwardPentaerythritol tetranitrate (abbreviated PETN, chemical formula C₂H₂N₂O₁₂2) is an explosive material with a structure that is very similar to nitroglycerin. It is one of the components of plastic explosives such as Semtex. The explosion of PETN can be represented by: C5H8N4O12 (s) → 2 CO(g) + 4 H₂O(g) + 3 CO2(g) + 2 N₂(g) A plastic explosive charge is formed into a cylinder about 30 cm long and about 2 cm in diameter. It weighs about 221 grams and is 58.5% PETN by mass. What is the total number of moles of gas produced when this plastic explosive charge explodes? mol gasarrow_forward
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