Conservation of Linear Momentum
Objectives
Determine the velocity between two objects after a collision and the momentum of each ball before and after the collision.
Collision types and kinetic energy evaluation.
https://phet.colorado.edu/sims/html/collision-lab/latest/collision-lab_en.html
Theory:
Newton’s second law is written with the formula Fext = (d P / dt), where Fext is the net external force acting on the system during the collision and (P) is the net momentum of the system.
This momentum is given to an object by its mass (m) multiplied by its velocity (v); P = mv, which is a vector quantity.
If Fext = 0, then (dP / dt) = 0, which means (P) = constant. Therefore, when two objects collide on a horizontal surface devoid of friction, the total force acting on the system during the collision is equal to zero which means that (Pi) = (Pf) where (Pi) is the vector sum of the moment of the two objects by collision and (Pf) is Vector sum of the moment of the two objects after collision. And for the calculation of speed:
m1 v1i + m2 v2i = m1 v1f + m2 v2f
There are two types of collisions :
(1) Elastic Collision: In this collision the total kinetic energy before collision equals the total kinetic energy after collision .
(2) Inelastic Collision: In which there is a loss in kinetic energy so the total kinetic energy before collision does not equals the total kinetic energy after collision.
Procedure:
In this experiment we have two parts related to elastic collision (part 1) when m1=m2 (part 2) elastic collision when m1 ≠ m2 .
Part(1);
a)m1 = m2 = 2kg:
1) Open this link: https://phet.colorado.edu/sims/html/collision-lab/latest/collision-lab_en.html
2) Select Intro
3) To make the collision elastic drag the slider of Elasticity to the right 100%.
4) Click on more data.
5) To change any value in the table click on it then type: (See the figure shown above)
6) Choose the following values: a- m1 = m2 = 2kg b- Position 1 (-1.00m) and position 2 (0.00m) c- Velocity of ball 1: make v = vx = 1.00m/s d- Velocity of ball 2: make v = vx = 0 (Ball 2 is at rest)
7) This collision will take place between the blue ball (m1) and the pink ball (m2) of the equal masses (m1 = m2 = 2kg), with (m2) initially at rest (v2i=0). The blue ball (m1) will collide with (m2) and essentially stops , then (m2) will move in the same direction that (m1) was moving before the collision.
8) Click on Play, then after collision Pause.
9) Record the values of v1f and v2f. 10) Calculate the valuesd of Pi , Pf , Pf/Pi , Ki , Kf and Kf/Ki and record them in table 1. 11) Repeat the experiment for two different values of v1i (3 m/s) and (6.00 m/s) and always keep v2i ZERO
V1i
(m/s)
V2i
(m/s)
V1f
(m/s)
V2f
(m/s)
Pi = m1v1i
(kg.m/s)
Pf = m2v2f
(kg.m/s)
Pf/Pi
Ki = 0.5m1v1i2
(J)
Kf = 0.5m2v2f2
(J)
Kf/Ki
1.00
0
3.00
0
6.00
0
Do they all have the same Pf / Pi? explain.
……………………………………………………………………………………..
Do they all have the same Kf / Ki? explain.
………………………………………………………………………………………
Part (2) : Elastic Collision
b) m1 ≠ m2:
1) Use the same link: https://phet.colorado.edu/sims/html/collision-lab/latest/collision-lab_en.html
2) KEEP: Elasticity 100% by dragging the blue triangle (Δ) to the right
3) Select: m1 = 4kg and m2 = 3kg
4) Position 1 (-1.00m) and position 2 (0.00m)
5) Select: v1i=1.00m/s and v2i=0.5m/s
6) Click on Play, then after collision Pause.
7) Record the values of v1f and v2f. 8) Calculate the values of Pi , Pf , Pf/Pi , Ki , Kf and Kf/Ki and record them into tables 2 & 3
Table 2
V1i
(m/s)
V2i
(m/s)
V1f
(m/s)
V2f
(m/s)
Pi = m1v1i + m2v2i
(kg.m/s)
Pf = m1v1f + m2v2f
(kg.m/s)
Pf/Pi
1.00
0.50
1.50
0.75
2.00
1.00
2.50
1.25
Table 3
V1i
(m/s)
V2i
(m/s)
V1f
(m/s)
V2f
(m/s)
Ki = (0.5m1v1i2) + (0.5m2v2i2)
(J)
Kf =
(0.5m1v1f2) + (0.5m2v2f2)
(J)
Kf/Ki
1.00
0.50
1.50
0.75
2.00
1.00
2.50
1.25
Do they all have the same Pf / Pi? explain.
……………………………………………………………………………………..
Do they all have the same Kf / Ki? explain.
……………………………………………………………………………………
Questions:
What is the difference between elastic and inelastic collision?
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
Show the difference when you are the mass of the two balls (m1=m2) and (m1 ≠ m2).
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
What does the quantity (Kf – Ki) represent.
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….
Conclusion: ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………