Impulse and conservation of linear momentum PDF

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Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

New Jersey City University Department of Physics Experiment 9

Impulse and conservation of linear momentum I. Introduction In physics conservation laws play a fundamental role. Under certain conditions, if the behavior of an object is known at a given time, conservation laws can be used to predict its behavior at any other time. The law of conservation of linear momentum is used to study the motion of interacting bodies when their masses and velocities before or after the interaction are known. Elastic and inelastic collisions, the relationship between impulse and linear momentum, and the law of conservation of linear momentum will be studied in this experiment. In this experiment the hypotheses to be tested are: 1. Impulse equals change in linear momentum. 2. Linear momentum is conserved in perfectly inelastic collisions.

Theory 1. Impulse and linear momentum Newton’s second law of motion is given in mathematical form by,

F = ma = m Δv Δt

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

Meaning F represents the net force or the addition of all external forces acting on a body of mass m moving with acceleration a. This equation can be rearranged to get, FΔt = mΔv = mvf − mvi = pf − pi = Δp (9.1) where FΔt is named the impulse of the force F for a time interval Δt and p = mv is defined as the linear momentum of the object. € Thus, with these definitions, equation (9.1) says that,

 during the interval Δt is equal to the change of momentum Δ  p the impulse of the force F of the object. When there are no external forces, F ∆t = 0 = pf – pi ⇒ pf = pi

The linear momentum is conserved. This is a logical consequence of Newton’s first law of motion because if there is no external force, the object moves with constant velocity. 2. Conservation of linear momentum a. Elastic collisions Consider a body of mass m1 and velocity v1i that collides with another body of mass m2 moving with velocity v2i,as shown in Figure 9.1. After the collision, m1 and m2 move with velocities v1f and v2f, respectively. During the collision, m1 applies a force F21 on m2, and  Newton’s third law says that m2 applies a force F12 = - F21 on m1. If all external forces can be  ignored, the impulse of each mass during the collision is given by,

1v1i F12Δt =m1v1f −m  F21Δt =m2v2f −m2v2i

These equations can be combined to provide F12 = -F21 m1v1f − m1v1i = − (m2v2f − m2v2i)

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

m1v1i + m2v2i =  m1v1f + m2v2f p1i +  p2i = p1f + p2f Pi = Pf



(9.2)

(9.3)

where Pi and Pf are the initial and final linear momentum of the system, respectively, and  equation (9.2) is known as the principle or law of conservation of linear momentum that can be expressed as, when no external forces act on a system consisting of two or more objects, the total linear momentum P  i of the system before the collision is equal to the total linear momentum P  f  of the system after the collision. A collision is said to be elastic if the linear momentum and the kinetic energy of the system are conserved. For two objects moving in one dimension, this is provide by,

m1v1i + m2v2i = m1v1f + m2v2f 2 1 2 m 1(v 1i)

+ 21 m 2(v 2i)

2

(9.4)

= 12 m 1(v 1f) 2 + 12m 2(v 2f )

(9.5)

b. Inelastic collisions Inelastic collisions are those in which only the linear momentum is conserved. A perfectly inelastic collision is that in which the interacting masses are stuck together after the collision. Conservation of momentum for the perfectly inelastic collision of two masses moving in one dimension is given by, m1v1i +m2v2i = (m1 +m 2)vf 

(9.6)

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

Experimental procedure 1. Elastic Collisions A. Computer Setup A. I connected the Universal interface to the computer, turned on the interface, and turned on the computer. B. I connected the Motion Sensor’s phone plugs to Digital Inputs 1 and 2 on the interface. Plug the yellow-banded (pulse) plug into Digital Input 1 and the second plug (echo) into Digital Input 2. C. I connected the DIN plug of the Force Sensor to Analog Input A. D. I opened the document titled: Exp 9 - Impulse.cap. Data recording is set for 500 samples per second (500 Hz) for the Force Sensor and 50 Hz for the Motion Sensor.

B. Sensor Calibration and Equipment Setup Sensor Calibration a. I did not need to calibrate the Motion Sensor.

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

b. In order to calibrate the Force Sensor, I followed the instructions provided by the instructor. Equipment Setup a. I place the Dynamics Track on a horizontal surface. b. I mounted the Economy Force Sensor on the Accessory Bracket. I mounted the Accessory Bracket in the T-slot on the side of the Dynamics Track. c. I used a book or other method to raise the end of the Dynamics Track that is opposite to the end with the Force Sensor about 1.5 cm so the cart will have approximately the same speed for each trial. d. I placed the Motion Sensor at the raised end of the Dynamics Track so I was able to measure the motion of the cart. I released the cart from a point at least 15 cm away from the Motion Sensor. However, I did not write on the track. For all trials released the cart was released from the same place. e. I braced the Force Sensor end of the Dynamics Track against a heavy mass or hold it so the Dynamics Track will not move during the collision. f. I measured the mass of the cart. Record the mass value in your lab notebook. mcart =495.4g

g. I unscrewed the hook attachment from the front of the Force Sensor. In its place, put the magnetic bumper from the Accessory Bracket with Bumpers.

C. Data Recording a. Prior to each data run, I pressed the tare button on the side of the Force Sensor to zero the sensor. b. I placed the cart on the Dynamics Track at the starting mark made during setup. c. I released the cart so that it rolls toward the Force Sensor. I started data recording. d. I ended data recording after the cart had rebounded from the collision with the Force Sensors magnetic bumper.

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

D. Analyzing the Data a. I clicked on the Highlight icon on the top graph. A highlighted area appeared. b. I moved and resized the highlighter to select the region in the graph window that corresponded to the collision.

c. I integrated to find the area under the curve. Click the ‘Area menu button. d. I recorded the value of the area under the curve in my lab notebook. The area was the impulse of the force. Impulse (area under the curve) = 495.4g e. I used the calculator to make a calculation for linear momentum with units of kg m/sec. Next, I clicked the ‘Calculator’ button in the main toolbar. The Calculator window opened. The window showed the calculation for Momentum that was

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

created as part of this activity. If the mass value for your cart is different from the default value, highlight the value and type in the mass of your cart. I clicked the ‘Accept’ button to save my changes.

f. I found the change of momentum. I selected the momentum window and I clicked the ‘Add a coordinate’ tool button. Select “Add Coordinates/Delta Tool”.

g. I moved the Smart Tool to the momentum data point corresponding to the same time I started to measure the force in the Force vs. Time graph in the previous step. h. I did a right click the mouse over the Smart Tool and select “Show Delta Tool”.

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

i. A second Smart Tool appeared. I moved the second Smart Tool to the momentum data point corresponding to the same time I stopped to measure the force in the Force vs. Time graph in the previous step. The change in momentum will be shown in the graph.

j. I recorded the change in momentum and compared it with the impulse. Also, I calculated the percent difference. |pf − pi |= Δp |- 0.15− 0.16| = 0.31 kgm/s

change in momentum = 0.31 kgm/s

difference =

|0.31−0.30| 0.305 (100)= 3.28%

Difference =  3.28%

k. I repeated the experiment replacing the magnetic bumper on the force sensor with the bumper that has the light spring.

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

Impulse (area under the curve) = 0.300N*s Momentum before = 0.16kgm/s Momentum after =-0.15kgm/s |pf − pi |= Δp |- 0.15− 0.16| = 0.31 kgm/s change in momentum = 0.31 kgm/s difference =

|0.31−0.30| (100)= 3.28% 0.305

Difference =  3.28%

l. I repeated the experiment replacing the light spring bumper on the force sensor with the bumper that has the harder spring.

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

Impulse (area under the curve) = 0.300N*s Momentum before = 0.13kgm/s Momentum after = -0.12kgm/s |pf − pi |= Δp |- 0.12− 0.13| = 0.25 kgm/s change in momentum =  0.25 kgm/s difference =

|0.25−0.30| (100)= 18.2% 0.275

Difference = 18.2% Aggregated data

Trial

Impulse

Pbefore

Pafter

Δp

difference

#

[Ns]

[kgm/s]

[kgm/s]

[kgm/s]

%

1

Magnet

0.300N

0.16

0.15

0.31

3.28

2

Hard Spring

0.300N

0.16

0.15

0.31

3.28

3

Soft Spring

0.300N

0.13

0.12

0.25

18.2

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

Questions

1. Why is it desirable to have the same initial speed for each data run? Because the impulse, p, only depends on vi and vf. So if we keep vi same, it will be easier to observe similarities and differences in net force F and time t for the various trials.

2. How will raising the end of the Dynamics Track give the cart the same acceleration each time? The acceleration is defined by change in speed divided by change in time. It gives the same value after collision each time because the horizontal component of the force due to gravity that accelerates the cart down the track always remains unchanged at a constant given angle of track elevation.

3. For your data, how does the change in momentum compare to the impulse? The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.

4. What are possible reasons why the change in momentum is different from the measured impulse? Errors in the measurement of variables involved (mass, velocity, average force, time)

5. Compare the impulse and force for an abrupt (hard) and a cushioned (soft) collision. Use words and also sketch the force vs. time graphs.

Physics I – Laboratory Experiment 9 Nov 23, 2020 Erick Ramirez

An abrupt meaning hard collision would be like a force versus time pike that would be narrow in time and high in net force. So, if the change in momentum is the same, meaning that two different types of collision will have. The area under the force against time curves would be the same. Now for the cushioned meaning soft, the force versus time plot for a softer collision will come up more all over the place on time and it would have a lower net force. 6. Use the above comparison to explain why airbags in cars can help to prevent injuries to the occupants during a frontal collision. Using the above comparison an airbag, would possibly lengthen the time in which the impact will occur for the passenger, which it should reduce the forces on the individuals that are on the car. This would result in reducing the injury of the individuals that are in the car.

7. Was kinetic energy conserved in all experiments involving elastic collisions? Explain by doing the calculations for a trial from each case. By referring to the equations above, it is noticed that for trial 1 and 2 are elastic collisions because if we ignore the percent difference we got on these trials which could be because of the fact that we ignore friction as well as human error. In this part we can state that we got pi=pf which it is the same. Now, for the other one, trial 3 we can conclude that it is not the same because the percent difference is too big meaning that is inelastic collision, This is because the percent difference is not accepted....