Lab 126: ​Conservation of Linear Momentum and Impulse ---

Momentum Theorem

Abdulahad Malik Group B

PHYS 102A-007 Professor Cannon

11/3/20

David Rotilyano

1. Introduction

a. The objective of this experiment is to see if the conservation of momentum works

with fully elastic and inelastic collisions.

2. Theoretical Background

a. Conservation of Momentum states that if a collision occurs between object 1 and

object 2 then the total momentum before and after the collision will be equal to

each other. An example of this would be with a gun and a bullet. Before firing a

gun the bullet and the gun remain at rest and the system’s momentum is 0. When

the gun is fired, the bullet gains momentum and to conserve that momentum the

gun recoils. According to the law of the conservation of mass both system will be

equal to each other after the collision.

3. Experimental Setup and Equipment

J Impulse (kg/m/s)

F External force (N)

m Mass in (kg)

v Velocity is (m/s)

Δt Change in time (s)

K Kinetic energy (J)

4. Procedure

a. The procedure in the lab manual is the same procedure as explained to us.

5. Experimental Data

a. Part 1

Velocity at Velocity at
Time at Time at photogate 1 photogate 2
photogate 1 (s) photogate 2 (s) (m/s) part I trial (m/s) part I trial
part I trial 1 part I trial 1 1 1
0.2127 0.2173 0.4701 0.4602
 Velocity at Velocity at
Time at Time at photogate 1 photogate 2
photogate 1 (s) photogate 2 (s) (m/s) part I (m/s) part I
part I trial#2 part I trial#2 trial#2 trial#2
0.1229 0.1544 0.8137 0.6478
0.6812 0.1468
Velocity at Velocity at
Time at Time at photogate 1 photogate 2
photogate 1 (s) photogate 2 (s) (m/s) part I (m/s) part I
part I trial#3 part I trial#3 trial#3 trial#3
0.1269 0.1913 0.7879 0.5229
0.406 0.2463

b. Part 2
 c. Part 3

i.

6. Calculation

a. The calculations for part 1 were done by finding the M1 and M2 measured in

kilograms which was given. Velocities were given in the table of my raw data

which gave my setup for initial data. The next table then asks for p which is linear

momentum. With the equation p=mv in the first trial the mass of the first glider

was .20643 kg and the mass of the second glider is .20644. The velocities are

done by before and after collisions. Which V1 measured in m/s is .4701 m/s and

V1’ which is after the collision is 0 m/s. And V2 for the second glider is 0 m/s

because the glider was at rest. V2’ is .4602 m/s. Using what was given we can use

the equation p=mv for the before and after collision. For the before collision P1
before is p=.20643*.4701 which gives .09704 kg/m/s which is the linear

momentum before the collision. After the collision it is 0 kg/m/s because the

object at rest is cancelled with the moving object. Making the equation

p=.20643*0 which is 0 kg/m/s. The total P would be adding those two values

which would be .09704 kg/m/s which is before the collision and the same is used

after the collision. Once you find total P before and after you can find the percent
|V 1−V 2|
difference which the equation is (V 1+V 2)/2
which would total 2.12456%. The third

table is kinetic energy which the equation is K=1/2mv^2 and putting the values

K= ½ (.20643)(.4701) which equals .02281 J and the same is used for the before

and after collisions. The percent difference used is the same equation used in the
|V 1−V 2|
linear momentum which is (V 1+V 2)/2
and the percent difference is 70.2318%. Part

2 for totally inelastic collision uses p=mv for linear momentum and kinetic energy

which is k=1/2mv^2. In trial 1 the masses measured in kilograms which is M1

which is .20028 kg and M2 which is .18957 kg. Finding velocity which involves

looking in the graphs above V initial which is .639 m/s and V final which is .2835

m/s. The total p is calculated the same way as before using the equation p=mv

which makes the total .639 kg/m/s and the total P after the collision is .112

kg/m/s. The total kinetic energy is calculated through the equation is K=½ mv^2

which makes .04088 J before the collision and .01586 J after the collision. For

Part 3 impulse must be found by gathering data for mass which is given for Trial

1 as .20646 kg and the velocities before and after which would be -.404 m/s and

after would be .383 m/s. Finding ΔP would involve the equation ΔP=P​f-​P​i. ​To find

p we use the equation p=mv which the equation would be p=.230646*-.404 which
 is -.​08341 kg/m/s and doing the same for the after collision. Then you subtract

both values .08341-.07891=-.16232 kg/m/s. Then finding J which is the area

under the curve which would be .1665 J. FInding the percent difference you
|V 1−V 2|
would use (V 1+V 2)/2
which is 2.6%. Finding Δt would involve subtracting final

time and initial time which could be found on the graph. The final time being

.275s and .255s and the total being .02s. To find Favg= J/Δt and it would give

.1665/.02= 8.35N. And the same is calculated for the rest of the table.

b. p=mv

c. J=F​avg​Δt

d. J=ΔP=P​f-​P​i

e. Δt=t2-t1
|V 1−V 2|
f. (V 1+V 2)/2

i. Results

Table 1.
Elastic
Collision
Trial # M1 [kg] M2 [kg] V1 [m/s] V1' [m/s] V2 [m/s] V2' [m/s]
1 0.20643 kg 0.20644 kg 0.4701 m/s 0 m/s 0 m/s -0.4602 m/s
2 0.20693 kg 0.30663 kg 0.8137 m/s -0.1468 m/s 0 m/s -0.6478 m/s
3 0.20643 kg 0.40673 kg 0.7879 m/s -0.2463 m/s 0 m/s -0.5229 m/s
 %
Difference
in total P
before and
Total P Total P' after
Trial # P1 [kg*m/s] P2 [kg*m/s] [kg*m/s] P1' [kg*m/s] P2' [kg*m/s] [kg*m/s] collision
1 0.9704 0 0.09704 0 0.095 0.095 2.12%
2 0.168 0 0.168 0.303 0.1986 0.2269 30.65%
3 0.1626 0 0.1626 0.5084 0.2125 0.2633 47.29%

%
Difference
in total KE
before and
after
Trial # KE1 [J] KE2 [J] Total KE [J] KE1' [J] KE2' [J] Total KE [J] collision
1 0.2281 0 0.02281 0 0.0475 0.0475 70.23%
2 0.06833 0 0.06833 0.002224 0.06434 0.06656 2.62%
3 0.06407 0 0.06407 0.006261 0.05556 0.06182 3.57%

Table II.
Total
Inelastic
Collision
Total P Total P
before after
collision collision KE before KE after
Trial # M1 [kg] M2 [kg] VI [m/s] VF [m/s] [kg*m/s] [kg*m/s] collision [J] collision [J]
1 0.20028 0.18957 0.639 m/s 0.2853 m/s 0.1278 0.1112 0.04088 0.01586
2 0.20028 0.28974 0.481 m/s 0.191 m/s 0.09633 0.09359 0.02317 0.008938
3 0.20028 0.38938 0.340 m/s 0.114 m/s 0.06809 0.06722 0.01158 0.003832

Table III.
Impulse
 %
difference F
between J (average)
Trial # M [kg] V [m/s] V' [m/s] J [kg*m/s] P [kg*m/s] and P T [s] [n]
-0.0404
1 0.20646 m/s 0.383 m/s 0.1665 0.16247 2.45% 0.155 1.0742
2 0.30661 -0.418 m/s 0.404 m/s 0.2779 0.25203 9.76% 0.22 1.2632
3 0.40671 -0.349 m/s 0.346 m/s 0.288 0.28271 1.85% 0.266 1.0827

8. Analysis and Discussion

a. Concepts that were used were the Conservation of momentum and impulse

momentum theorem.

○ 1.) What could cause the difference in momentum is that when the glider collided

with the other glider it would bounce off causing both negative and positive

velocities. Another cause for the difference could be the amount of force that is

exerted by the glider. For the kinetic energy the cause of difference would be the

force exerted on the object at rest. Another cause would be how much air pressure

was on the air track which would make it either faster or slower at hitting the

object.

○ 2.) KEb-KEa/KEb= ½ mv^2-½ mv^2/1/2mv^2 with p=mv we can put an mv

with p which would be ½ p(mv)-½ p(mv)/1/2p(mv) and moving 1/2p(mv) a down

would make i positive. And it would be 1/2p(mv)/1/2p(mv)+1/2p(mv) which

makes it similar to M2/M1+M2.

9. Conclusion
○ What I learned from this experiment is that for any collision occurring in an

isolated system momentum will be conserved and that the amount of momentum

of the objects in the system is the same before and after the collision there is

minimal difference. The lab could improve by making the procedures easier to

understand. One question I have is if conservation of momentum would apply in

real life situations?