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Lab #2 - Collision Lab - Conservation of Linear Momentum
PHYS 1604 Engineering Physics I Spring 2022 B2 21448 Showalter
https://bartonline.instructure.com/courses/25117
Conservation of Linear Momentum
Objectives:
To understand the 1-D collision using PHET simulator
To determine the momentum and kinetic energy before and after the
collision between two balls.
To understand the elastic and in-elastic collision
Apparatus:
Two balls
Timer
Meter scale
Theory and Background
This lab deals with the collision of two balls in 1-D space. According to newton law
F=dP/dt, where F is the external force while P is the momentum, dp/dt=0 when
external force becomes zero but the linear momentum is conserved regarding any
type of collision.
The phet simulator mentioned in the lab link contains error therefore for solving
this lab updated version of this collision lab is taken from the phet website, which
provides the maximum value of position of 1.7m instead of 2m.
Part 1: Elastic collision
In this part two balls are selected of equal mass m1=2kg while m2=2kg, now 100%
elasticity is chosen from the slider, and experiment is performed, the results before
and after the collision are recorded. Table 1 shows the results
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Table I
The experiment is performed for multiple value of initial velocities of first ball while
keeping second ball velocity zero. Pi represents the initial momentum before
collision while p2 represents after collision, similarly the initial and final kinetic
energies are calculated as well. Figure below shows the simulation matching results
as well. When ball 1 strikes with ball 2, ball 2 start to move with the velocity of ball
1 while ball 1 velocity will go to zero. The kinetic energies can be seen in the figure
below, which verifies the table calculations.
V1i V2i V1f V2f pi=m1*v1i pf=m2*v2f pf/pi ki=1/2 (m1v1i^2) kf=1/2(m2v2f^2) Kf/ki
(m/s) (m/s) (m/s) (m/s) kg.m/s kg.m/s J J
2 0 0 2 4 4 1 4 4 1
2.5 0 0 2.5 5 5 1 6.25 6.25 1
3 0 0 3 6 6 1 9 9 1
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Figure 1. collision between two equal masses
Comments:
Initial momentum was 4 kg m/s after colliding the ball 2 has the final momentum
of 4 kg m/s. similarly the kinetic energy before and after collision also remains the
same as shown in the table I.
Part 1b. different mass:
Now the same experiment is performed with different masses of two balls, same
above-mentioned work is repeated, and results are taken in table 2 and 3. Mass
m1=3kg while mass m2 =2kg.
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Figure 2: collision with 2 different values of masses
Table II
Table III
V1i V2i V1f V2f pi=m1*v1i +m2*v2i pf=m1*v1f +m2*v2f pf/pi
(m/s) (m/s) (m/s) (m/s) kg.m/s kg.m/s
2.00 0.50 0.80 2.30 6.00 4.60 0.77
2.50 0.75 1.10 2.85 7.50 5.70 0.76
3.00 1.00 1.40 3.40 9.00 6.80 0.76
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Comments:
With different values of masses, the results are shown in the table II and table III.
Kinetic energy obtained in the figure 2 matches with the calculations shown in the
table III, while the momentum before and after collision is verified with the data
calculated in table II.
Part 2: In-elastic collision
In this section the collision is kept inelastic, by changing the slider to 0%, now m1=3kg
while m2=2 kg. Change the velocities as mentioned in the table 4. The velocity,
momentum is calculated and can be compared with the one from the simulation. The
next table comprises of initial and final value of the kinetic energy, figure below shows
the last entry of the table. Where velocities are taken as 3 and 0 respectively thus kinetic
energy is 13.50j before collision while reduces to 8.10 J after the collision, same
calculations can be seen in the last entry of the table V.
Table IV
V1i V2i V1f V2f
ki=1/2 (m1v1i^2)+
1/2(m2v2i^2)
kf=1/2 (m1v1f^2)+
1/2(m2v2f^2)
kf/ki
(m/s) (m/s) (m/s) (m/s) kg.m/s kg.m/s
2.00 0.50 0.80 2.30 6.25 6.25 1.00
2.50 0.75 1.10 2.85 9.94 9.94 1.00
3.00 1.00 1.40 3.40 14.50 14.50 1.00
V1i V2i Vf pi=m1*v1i pf=(m1+m2)*vf pf/pi
(m/s) (m/s) (m/s) kg.m/s kg.m/s
2 0 1.2 6 6 1.00
2.5 0 1.5 7.5 7.5 1.00
3 0 1.8 9 9 1.00
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Table V
V1i V2i Vf ki=0.5m1*vi1^2 kf=0.5(m1+m2)*vf^2 Kf/ki
(m/s) (m/s) (m/s) kg.m/s kg.m/s
2 0 1.2 6.00 3.60 0.60
2.5 0 1.5 9.38 5.63 0.60
3 0 1.8 13.50 8.10 0.60
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Question and Answers:
What is the difference between elastic and inelastic collision?
Major difference is that in elastic collision the kinetic energy remains the same while in inelastic collision
the KE does not remains the same during the process of collision where some of the KE in the form of heat
is dissipated in the environment that makes the KE before and after collision different.
Conclusion:
In this lab two major scenarios are taken. First collision between two balls in elastic and second in inelastic
collision. Furthermore, the elastic collision takes two cases one with equal mass collision and second with
different masses. PHET simulator of collision is used to notice the effect of collision. The momentum and
the kinetic energy along with the ball velocities are calculated and results are noted in the table I-V.

Unformatted Attachment Preview

Lab #2 - Collision Lab - Conservation of Linear Momentum PHYS 1604 Engineering Physics I Spring 2022 B2 21448 Showalter https://bartonline.instructure.com/courses/25117 Conservation of Linear Momentum Objectives: • To understand the 1-D collision using PHET simulator • To determine the momentum and kinetic energy before and after the collision between two balls. • To understand the elastic and in-elastic collision Apparatus: • Two balls • Timer • Meter scale Theory and Background This lab deals with the collision of two balls in 1-D space. According to newton law F=dP/dt, where F is the external force while P is the momentum, dp/dt=0 when external force becomes zero but the linear momentum is conserved regarding any type of collision. The phet simulator mentioned in the lab link contains error therefore for solving this lab updated version of this collision lab is taken from the phet website, which provides the maximum value of position of 1.7m instead of 2m. Part 1: Elastic collision In this part two balls are selected of equal mass m1=2kg while m2=2kg, now 100% elasticity is chosen from the slider, and experiment is performed, the results before and after the collision are recorded. Table 1 shows the results Table I V1i V2i V1f V2f pi=m1*v1i pf=m2*v2f (m/s) (m/s) (m/s) (m/s) kg.m/s kg.m/s pf/pi ki=1/2 (m1v1i^2) kf=1/2(m2v2f^2) J Kf/ki J 2 0 0 2 4 4 1 4 4 1 2.5 0 0 2.5 5 5 1 6.25 6.25 1 3 0 0 3 6 6 1 9 9 1 The experiment is performed for multiple value of initial velocities of first ball while keeping second ball velocity zero. Pi represents the initial momentum before collision while p2 represents after collision, similarly the initial and final kinetic energies are calculated as well. Figure below shows the simulation matching results as well. When ball 1 strikes with ball 2, ball 2 start to move with the velocity of ball 1 while ball 1 velocity will go to zero. The kinetic energies can be seen in the figure below, which verifies the table calculations. Figure 1. collision between two equal masses Comments: Initial momentum was 4 kg m/s after colliding the ball 2 has the final momentum of 4 kg m/s. similarly the kinetic energy before and after collision also remains the same as shown in the table I. Part 1b. different mass: Now the same experiment is performed with different masses of two balls, same above-mentioned work is repeated, and results are taken in table 2 and 3. Mass m1=3kg while mass m2 =2kg. Figure 2: collision with 2 different values of masses Table II V1i V2i V1f V2f pi=m1*v1i +m2*v2i pf=m1*v1f +m2*v2f (m/s) (m/s) (m/s) (m/s) kg.m/s kg.m/s 2.00 0.50 0.80 2.30 6.00 4.60 0.77 2.50 0.75 1.10 2.85 7.50 5.70 0.76 3.00 1.00 1.40 3.40 9.00 6.80 0.76 Table III pf/pi V1i V2i V1f V2f ki=1/2 (m1v1i^2)+ 1/2(m2v2i^2) kf=1/2 (m1v1f^2)+ 1/2(m2v2f^2) (m/s) (m/s) (m/s) (m/s) kg.m/s kg.m/s 2.00 2.50 3.00 0.50 0.75 1.00 0.80 1.10 1.40 2.30 2.85 3.40 6.25 9.94 14.50 6.25 9.94 14.50 kf/ki 1.00 1.00 1.00 Comments: With different values of masses, the results are shown in the table II and table III. Kinetic energy obtained in the figure 2 matches with the calculations shown in the table III, while the momentum before and after collision is verified with the data calculated in table II. Part 2: In-elastic collision In this section the collision is kept inelastic, by changing the slider to 0%, now m1=3kg while m2=2 kg. Change the velocities as mentioned in the table 4. The velocity, momentum is calculated and can be compared with the one from the simulation. The next table comprises of initial and final value of the kinetic energy, figure below shows the last entry of the table. Where velocities are taken as 3 and 0 respectively thus kinetic energy is 13.50j before collision while reduces to 8.10 J after the collision, same calculations can be seen in the last entry of the table V. Table IV V1i V2i Vf pi=m1*v1i pf=(m1+m2)*vf (m/s) (m/s) (m/s) kg.m/s kg.m/s 2 0 1.2 6 6 1.00 2.5 0 1.5 7.5 7.5 1.00 3 0 1.8 9 9 1.00 pf/pi Table V V1i V2i Vf ki=0.5m1*vi1^2 kf=0.5(m1+m2)*vf^2 Kf/ki (m/s) (m/s) (m/s) kg.m/s kg.m/s 2 0 1.2 6.00 3.60 0.60 2.5 0 1.5 9.38 5.63 0.60 3 0 1.8 13.50 8.10 0.60 Question and Answers: What is the difference between elastic and inelastic collision? Major difference is that in elastic collision the kinetic energy remains the same while in inelastic collision the KE does not remains the same during the process of collision where some of the KE in the form of heat is dissipated in the environment that makes the KE before and after collision different. Conclusion: In this lab two major scenarios are taken. First collision between two balls in elastic and second in inelastic collision. Furthermore, the elastic collision takes two cases one with equal mass collision and second with different masses. PHET simulator of collision is used to notice the effect of collision. The momentum and the kinetic energy along with the ball velocities are calculated and results are noted in the table I-V. Name: Description: ...
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