Impulse and Momentum in Collisions PDF

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Lab work for Physics I for Danylov....

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University of Massachusetts - Lowell

Impulse and Momentum in Collisions Physics 1 (PHYS.1410.818)

Student, Lab Partner Instructor Gabrielle Tubman, Doanna Nguyen

Yousef Izadi

December 3th, 2019

Objective(s): Gain a better understanding of elastic and inelastic collisions and the relationship between the integral of force overtime and impulse.

Introduction: Instead of evaluating force of motion with F=ma, momentum can be used. If the integral of a force over time is equal to momentum. The initial and final velocity, as well as the force of the impact will be measured and recorded into a graph. Those values that were measured will then be testes to see how accurate they are. The values will be tested by calculations and formulas. To test the different types of collisions, there will be two different springs (a heavy and light one), an air bag, and nothing, or a wall. The impulse can be calculated by taking the integral, or by finding the area under the curve. ∫𝐹 =

𝑑𝑝 𝑑𝑡

∫ 𝐹 𝑑𝑡 = ∫ 𝑑𝑝 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖 In an elastic collision, the two objects bounce off of each other, which means there is a change in momentum and the initial and final velocity are essentially the same in magnitude. 𝑚𝑣𝑓 = −𝑚𝑣𝑖 ∫ 𝐹 𝑑𝑡 = −2𝑚𝑣𝑖 In an inelastic collision, the object will have a final velocity of zero due to the object ending at rest at the collision point. ∫ 𝐹 𝑑𝑡 = −𝑚𝑣𝑖 To see how accurate the change in momentum from the calculated and the measured, the error percentage formula can acquire that. |∆𝑃 − 𝐹∆𝑡| ∗ 100 ∆𝑃 Apparatus and Procedures: Materials: − − − − − − −

Force Sensor Position Sensor Weight Heavy Spring Light Spring Rubber Bands Air Bag

− − − −

Collision Car Track Weight Scale Pressure Sensor

Diagram:

Procedure: Gather materials and equipment. Open the impulse program on the computer. Weigh the mass of the cart using a scale. Screw in the heavier spring onto the collision spot. Click the start button on the computer. Then, push the cart into the collision spot. Record the initial and final velocities, area, and max force into the chart from the graph displayed on the computer. 8. After the graph is scaled, print it out. 9. Calculate the percent difference using the change of momentum. The equation is

1. 2. 3. 4. 5. 6. 7.

10. Repeat the steps for the lighter spring. 11. Then repeat the same steps with an air bag. The air bag is just a sandwich bag blown up with a straw. This represents an inelastic collision. 12. Then repeat with nothing on the collision spot.

Results and Analysis: Results: Mass of Cart = 0.62 kg Vi (m/s)

Vf (m/s)

∆p (kg m/s)

F∆t (Ns)

% diff

Max Force (N)

Run 1EW

1.15

0.68

1.135

1.11

2.2

30.18

Run 2EW

0.99

0.56

0.961

0.95

1.1

24.66

Run 3EW

1.04

0.63

1.035

1.03

0.5

28.86

Run 1ES

0.74

0.56

0.806

0.79

2.0

14.10

Run 2ES

1.03

0.62

1.023

1.02

0.3

37.26

Run 1in

0.39

0.07

0.2852

0.27

5.3

05.25

Run 2in

0.60

0.08

0.4216

0.40

5.1

12.48

Run 3in

0.63

0.08

0.4402

0.42

4.6

13.76

Run 1nb

0.71

0.22

0.5766

0.56

2.9

35.49

Run 2nb

0.67

0.20

0.5394

0.53

1.7

33.27

Analysis: There are two types of collisions, elastic and inelastic. The data table shows the results of all four tests and the number of trials for each (In order: Weak Spring, Strong Spring, Airbag, No Nothing). The mass we got for the cart is 0.62kg. The first two tests are elastic collisions because the springs will push the carts back. We noticed that the weak spring had the lowest max force. This made sense because the weak spring can increase the duration of the collision due to its ability to compress more before pushing the cart back. To calculate the ∆p, using the equation... ∆p = m(Vi + Vf) There should be a similarity to the impulse measured and the impulse calcuated with the formula. When we compare the weak spring to the strong spring we can see that the ratio of the max force and initial velocity is higher in the stronger spring. This is because the strong spring

will not compress as much which means that the duration of the collision will be shorter, and force will not be absorbed as much. The final two test are inelastic collisions because there should be no force applied on the cart for it to have negative velocity. This means that the hypothesis we made was correct because the airbag was the best at handling forces in the collision. Finally, when we compare the airbag trials with the no airbag trials, we can see that the max force on the no airbag is significantly higher and, in some trials, more than double the force. This is because the total force in the collision without the airbag is dispersed in only a fraction of a second without anything to dampen the impact. Discussion: We calculated the error percentage using the equation...

The heavier spring’s error percentage was very low. The highest percentage was 2.2 percent. The lighter spring’s error percentage was extremely low as well. The percentage didn’t even go over two percent. The airbag’s collision error was around five percent. That one was hard to measure because the cart kept going off the track. And the collision spot with nothing in the way was also a very low error percentage. The percentage didn’t go over three percent. Our main source of error came from the cart falling off the track constantly. The track also wasn’t attached to the table, so the track would slide on the table slightly. The other source of error would be caused by human error.

Conclusion: The purpose of the lab was to illustrate the difference between elastic and inelastic collisions using data. Also, to notice the change in forces, known as impulse. We can conclude that the area under the graph of the force is the impulse. During the airbag collision, the force was dispensed more efficiently. We were able to find the actually values using the equations for the impulse and momentum. And our percent error between the calculated and the measured were extremely low. Questions: 1. Explain the physics of a car crash and how the crumble zone, safety belt, and air bag work in concert to increase the driver’s survivability. a. Crumble zones are mini rigid structures at the front and back of the vehicle, which are designed to absorb the energy and collapse slowly on impact. Safety belts are designed to stop passengers becoming projectile inside the car by restraining them in their seats. And air bags are designed to inflate to cause an inelastic collision. This absorbs most of the kinetic energy of the impact. 2. Compare the average force experienced by a 60 kg person moving at 34 kg/h (20mph) being stopped in a distance of 2 cm in one case (e.g. a padded dashboard) or 40 cm in the other case (e.g. the effective distance due to the crumble zone and air bag).

a. The force of the 60 kg person going 34 kg/h is 133800 N, while the force of the person being stopped that soon is 6690 N. The ratio is then 20:1. 3. If a pressure sensor were incorporated in the experiment, a similar profile to the force impulse should be recorded. Since in the experiment, no crumble zone is included, the air bag must dissipate all the energy in the collision. If p is the overpressure during the collision and V is the volume of vented air, the dissipated energy should be given by the integral ∫ 𝑝 𝑑𝑉 which would be the area under the pressure vs volume graph. Choose the air bag and the average collision with the highest initial velocity. Assume all the initial kinetic energy is dissipated by the air bag and the average overpressure is 850 Pa (Pa is an SI unit called the pascal). What volume of air (in m3) was pushed out of the bag? Convert your answer to cubic centimeters. a. K = 1/2 m v^2 b. K = 1/2 (60kg) (9.44m/s)^2 c. K = 2675.93 J

d. e. f. g.

E = p dV 2675.93 N*m = (850 N/m^2) (dV) dV =3.148 m^3 Volume of air vented out dV =3.148 m^3

For the experiment in our lab: h. K = 1/2 m v^2 i. K = 1/2 (0.6751Kg) (0.67m/s)^2 j. K = 0.1515J

k. l. m. n.

E = p dV 0.1515N*m = (850 N/m^2) (dV) dV =1.78cm^3 Volume air vented from the zip bag was 1.78cm^3...