116 conservation of momentum PDF

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Physics 116

Section 28334

Lab Number: 7

CONSERVATION OF MOMENTUM Name: Abdonnie R. Holder Instructor: Doctor Sasanthi C. Peiris Partners: Aviva Lehrfield & Candy-Lynn Best Date Performed: Thursday, January 11 , 2018 th

Objective: The objective of this experiment was to verify the conservation of momentum in the collision of two air track gliders. In this experiment the theory of the conservation of momentum is tested by

varying the mass and velocity of two colliding objects moving of a frictionless surface and their momentums determined before and after they collide. The momentum of the object was also measured under coupling conditions, where the two objects were allowed to stick together. Principles/laws tested and used: The linear momentum (P) of an object can be defined as the product of that object's mass (m) and velocity (V). It is a quantity which has both magnitude and direction, which is in the same direction as its velocity. Therefore, momentum can be represented by the equation P = mV, which indicates that if two objects are of equal mass the object with the greater velocity will have the greater momentum. However, if two objects are moving at equal velocity, the object with the greater mass will have the greater momentum. Momentum is a conserved quantity, this means when two objects collide, if there is no external force acting on them, the total momentum before the collision will be equal to the total momentum after the collision.

t-i=fThe total amount of time (t) is found when there is an additional reading that the photogate calculates. This time is subtracted from the initial time (i) that is displayed on the photogate. To get the total time, experimenters must flick the switch downwards to have the total time. When the total time is subtracted from the initial time this gives the final time (f) for that object. If the two objects stick together after the collision, some initial kinetic energy will be transformed to some other form of energy and kinetic energy will not be conserved. However, the momentum will still be conserved, this collision is called inelastic collision. t = the time that glider indicated by photogate before the collision. 1i

1

1

t = the time that glider indicated by photogate before the collision. (In cases where v = 0 there 2i

2

is no t since glider begins at rest.) 2i

2

2

2i

t = the time that glider indicated by photogate after the collision. This can also be found by: t1f

1

1

i=f. t = the time that glider indicated by photogate after the collision. This can also be found by: t2f

2

2

i=f. Apparatus: 

Linear Air-Track



Variable output air supply



Hose



Two master photogate timers



Two gliders



One air-track kit



Electronic balance

Procedure: Students are to first retrieve the items from the front of the class and when this is received, students are to begin prepping the lab area. To begin the experiment first setup the air track as shown by the professor on the board. This setup includes the bumpers and the flags. Measure the masses of m1and m2with the bumpers and flags attached. With this information record it in each of its selective procedure category. The length of each flag is 0.1m which is used to find the velocity by the equation: 1i=0.1T . Before beginning the experiments experimenters, must remember to set the photogate timer to GATE mode and press reset. In addition the memory switch must be turned on, or else the time taken will not be remembered. There will be five experiments being conducted: 1. m = m v ≠ 0 v = 0 1

2

1i

2i

2. m= > m v= ≠ 0 v = 0 1

2

li

2i

3. m < m v ≠ 0 v = 0 1

2

1i

2i

4. m > m v ≠ 0 v ≠ 0 1

2

1i

2i

5. m > m v ≠ 0 v = 0 ( coupled ) 1

2

1i

2i

Experimental Data Procedure 1 P

1

m= m 1

2

v ≠0

v =0

1i

m = 211g 0.211kg

2i

m =210.6g 0.2106kg

1

2

Measure t

1f

t = 0.1596s

t = 0s

1i

2i

V = 0.1m0.1596s=0.627ms 1i

P1=(0.211kg)(0.627ms)+(.2106kg)(0ms) = 0.132 J P

2

m= m 1

m = 211g 0.211kg

2i

2f

t = 0.1745s 2f

V = 0.1m0.1745s=0.573ms 2f

v =0

1i

2i

2

Measure V = 0.1T t = 0s

v ≠0

m =210.6g 0.2106kg

1

2f

2

P2=(0.211 kg)(0ms)+(.2106 kg)(0.573ms) = 0.121 J Procedure 2 P

1

m= > m v= ≠ 0 1

2

li

m = 311.6g 0.3116kg

v =0 2i

m =210.6g 0.2106kg

1

2

Measure t and V = 0.1T 1i

2i

2f

t = 0.1774s

t = 0s

1i

V = 0.1m0.1774s=0.564ms

2i

1i

P1=(0.3116kg)(0.564ms)+(.2106kg)(0ms) = 0.176 J P

2

m= > m v= ≠ 0 1

2

li

m = 311.6g 0.3116kg

v =0 2i

m =210.6g 0.2106kg

1

2

Measure V = 0.1T and t 2f

t = 0s 2i

2f

2f

t = 0.1542s 2f

V = 0.1m0.1542s=0.649ms 2f

P2=(0.3116 kg)(0ms)+(.2106 kg)(0.649ms) = 0.137 J Procedure 3 P

1

m m 1

2

v ≠0

v ≠0 v ≠0 v ≠0

1i

2i

1f

m = 311.6g 0.3116kg

2f

m =210.6g 0.2106kg

1

2

t = 0.5913 - 0.2313s = 0.3601s

t = 0.3984 - 0.1979s = 0.2005s

1f

2f

V = -0.1m0.3601s= -0.2777ms

V = 0.1m0.2005s= 0.4987ms

1f

2i

P2=(0.3116kg)(-0.2777ms)+(.2106kg)( 0.4987ms) = 0.0191 J Procedure 5 (Coupled) Needle on m

1

P

1

m >m 1

m = 309.7g 0.3097kg 1

t = 0.2840s 1i

V = 0.1m0.2840s=0.352ms 1i

P1=(0.3097kg)(0.352ms)+(.2102kg)(0ms) = 0.109 J

2

v ≠0

v =0

1i

2i

m =210.2g 0.2106kg 2

t = 0s 2i

V = 0.1m0s= 0 ms 2i

P

2

m >m 1

2

v ≠0 1i

m = 309.7g 0.3097kg 1

t = 0.4840s 1f

v =0 2i

m =210.2g 0.2106kg 2

t = 0.480s 2f

V = V = 0.1m0.4840= 0.207ms 1f

2f

P2=(0.3097kg)(0.352ms)+(0.2106kg)( 0.352ms) = 0.108 J Discussion and Conclusion: It can be concluded that when objects of a similar mass collide the kinetic energy of the system before the collision is equal to the kinetic energy of the system after collision. We can therefore conclude that the total momentum of the system was conserved as expected for an inelastic collision. Since linear momentum is a conserved quantity, the linear momentum of the system of colliding air gliders is expected to be the same before and after the two air gliders collide. Changing the mass and velocities of the objects is not expected to affect the total momentum of the system under elastic or inelastic conditions. Some errors which may have influenced the results include, systematic errors associated with the instruments (air track and photogate), gross errors in reading the time from the photogate, air resistance. If the air track was not leveled it may have also influenced the results....