Exp 8 Collision

8/10/2019 Exp 8 Collision

1/30

Collisions in

one dimension


2/30

Goal of this Lab

1. The goal is to investigate the conservation ofmomentum and energy in the one-dimensionalcollision of two bodies through the study ofelastic and inelastic collisions

1. Demonstrate that conservation of momentum is

independent of conservation of energy.


3/30

Background

For a single particle, momentum is defined as theproduct of the mass and the velocity:

p= mv

Momentum, as velocity, is a vector.

For a system of more than one particle, the total linearmomentum is the vector sum of the individualmomenta:

p=p1+p2+ = m1v1+ m2v2+ ...


4/30

Background

The total momentum of any system of particles isconserved (constant), provided that the net external forceon the system is zero.

When two freely moving bodies collide, the initialmomentum of the system is equal to its final momentum.

This Conservation of Momentum Law applies no matterif the collision is elastic or inelastic and no matter howcomplicated the interaction force between the collidingbodies may be.

Elastic Collision: V1i-V2i=-(V1f-V2


5/30

Background

Assume we have two particles with masses m1, m2andspeeds v1iand v2iwhich collide, without any externalforce, resulting in speeds of v1fand v2f after the collision.

Conservation of momentum states that the totalmomentum before the collision (pi) is equal to the totalmomentum after the collision (pf)and since:

pi= m1v1i+ m2v2i and pf= m1v1f+ m2v2f

Therefore: m1v1i+ m2v2i= m1v1f+ m2v2f


6/30

Background

Kinetic energy is the form of energy associatedwith motion, and for a single particle:

EK = mv2.

Kinetic energy is a scalar

For a system of more than one particle the totalkinetic energy is the algebraic sum of theindividual kinetic energies of each particle:

EKT=EK1 + EK2 +

EKT= (m1v12

+ m2v22

+ )


7/30

Part 1 Elastic collision

In the first part of this experiment you will analyzeelastic collisions in one dimension. We will:

a.Determine the linear momentum of a physical body.

a.Apply the principle of conservation of linear

momentum and energy to analyze elastic collisions inone dimension.

Procedure


8/30

car 1

Procedure

2. Level the track (the carts do not accelerate ineither direction).

3. Place two photogates, separated from each other bya distance a little over twice the car length.

car 2

Gate 2 Gate 1

1. Select two carts having the same mass (m1= m2).


9/30

Background

A fundamental law of physics is that in anyinteraction, the total energy of a system is alwaysconserved.

However, within a given system, one form ofenergy may be converted into another.

Therefore, kinetic energy alone is not always

conserved.In inelastic collisions, the kinetic energy is notconserved. The difference is converted into other

forms of energy, such as sound, heat, etc.


10/30

car 1 car 1

Procedure

Car 2

Gate 2 Gate 1

5. Place car 1 out of the fotogates

4. Put a Piquet Fence in each one of the cars

6. Conect the Smart Timer to the gates. This

will give you the speed of the cars.


11/30

car 1car 1

Procedure

Car 2

Gate 2 Gate 1

8. Be sure that car 1 passes completely throughphotogate #1 before colliding with car 2.

9. Obtain the initial and final velocities for each car

7. Keep car 2 at rest and push car 1,allowing it toslide freely .

10. Repeat the procedure with m1>m2and m2>m1,adding masses to the cars.


12/30

Background

A fundamental law of physics is that in any interaction,the total energy of a system is always conserved.

However, within a given system, one form of energy maybe converted into another.

In inelastic collisions, kinetic energy is not conserved.

The difference is converted into other forms of energy,such as sound, heat, potential energy, etc.

Part 2: Inelastic collision


13/30

It turns out that the conservation of momentum is stillvalid, but the final kinetic energy is less than the initial

In the inelastic collision analyzed in this lab the two

bodies stick together after collision and have a commonfinal velocity.

We call that:perfectly inelastic collision.

One way to express the change of kinetic energy is withthe parameter, defined in our case as:

=100 KEf/ Kei

Background


14/30

car 2 car 1

Procedure1. Select two carts having the same mass (m1= m2).2. Level the track (the carts do not accelerate in either

direction).

3. Place two photogates, separated from each other bya distance a little over twice the car length.

4. Place the cars facing the velcro bumpers so that,upon collision, the cars stick together.

Gate 2 Gate 1velcro


15/30

car 1car 2 car 1

Procedure

Gate 2 Gate 1

5. Put a Piquet Fence in each one of the cars6. Conect the Smart Timer to the gates. This will give

you the speed of the carts.

7. Place car 1 out of the fotogates


16/30

car 1car 1

Procedure

car 2

Gate 2 Gate 1

8. Keep car 2 at rest and push car 1,allowing it toslide freely .

9. Be sure that car 1 passes completely through

photogate #1 before colliding with car 2.10. They will stick together and move on the track.

11. The Timer will record the velocity of m2(now

connected with m1

) through photogate #2.


17/30

11. Repeat the procedure with m1>m2and m2>m1,attaching masses to the cars.

12. Use your measurements to determine if linear

momentum and Kinetic Energy are conserved.13. Calculate the parameter.

Procedure


18/30

Report

1. For each experiment, find the initial andfinal linear momenta and kinetic energiesofthe system.

2. Calculate the difference, and the percentdifference, of the linear momenta andkineticenergies for elastic and inelastic colision.

3. Make a table with the percent differences ofthe linear momentum and kinetic energies,

for elastic and inelastic.


19/30

Report

4. Are the percent differences of the linearmomentum and kinetic energiesthe same forelasticthan for inelastic? Comment.

4. Is kinetic energy always conserved?

4. What is the meaning of theparameter?

4. What is the main difference between elasticand inelastic collisions?


20/30

Procedure

Data: Equal Masses and Unequal Masses Case

m1, m2, v1i, v2i, vffor case where m1= m2)m1, m2, v1i, v2i, vffor case where m1> m2)

m1, m2, v1i, v2i, vffor case where m1< m2)

Part 2: Inelastic collision


21/30

Procedure

Results:

If m2 moving If m1 movingv2f= (m2-m1)v2i/ (m1+m2) v1f=(m1-m2)v1i/(m1+m2)

v1f= 2m2v2i/ (m1+m2) v2f=2m1v1i/(m1+m2)

Compare to measured values, calculate deviation

Part 1: Elastic collision


22/30

Data

Data Table 2: Inelastic Collision

Trial m1 m2 v1i v2i vf

m1= m2

m1> m2

m1< m2


23/30

Data

Result Table 2: Inelastic Collision

Trial P1i P2i Pf PiT % diff p

m1= m2

m1> m2

m1< m2


24/30

Results

From your data, find the initial momentum and thefinal momentum of the system, for each one of theexperiments.

Calculate the p(% conserved) parameter formomentum.

Review your results and try to explain any deviationfrom the conservation of linear momentum.


25/30

Results

Calculate and compare the initial and final kineticenergies of the system, for the inelastic experiments.

Is kinetic energy always conserved?

What does the parameter mean?

What is the main difference between elastic andinelastic collisions?


26/30

Results

m1v1i+ m2v2i= (m1+m2) vf

p= pf/ pi* 100%

m1v1i2+ m2v2i

2= (m1+m2) vf2

= KEf/ KEi* 100%


27/30

Procedure

Data: Equal Masses and Unequal Masses Case

m1, m2, v1i, v2i, v1f, v2ffor case where m1= m2)m1, m2, v1i, v2i, v1f, v2ffor case where m1> m2)

m1, m2, v1i, v2i, v1f, v2ffor case where m1< m2)

Part 1: Elastic collision


28/30

Data

Data Table 1: Elastic Collision

Trial m1 m2 v1i v2i v1f v2f

m1= m2

m1> m2

m1< m2


29/30

Data

Result Table 1: Elastic Collision

Trial P1i P2i P1f P2f PiT PfT % diff

m1= m2

m1> m2

m1< m2


30/30

Data

Result Table 1: Elastic Collision

Trial KEi KEf % diff

m1= m2

m1> m2

m1< m2

Documents

Exp 8 Collision