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PHYS 105L: Experiment 12 Momentum and Collisions Report Submitted By Lab Partner (if submitting separate reports) I. Tabulated data and calculated results – 23points

1st graph

Data table (with data for all 3 trials)

II. Calculations – 15 points

Show formulas and calculations written out with units on all values. You are showing the calculations for the first trial. For the other two trials, you mustdo the calculations on your calculator or in Excel and recordthe calculated values in your data table (you must do these calculations but you do not need to show these calculations).

1. Calculation of total momentum of the system before the collision

Momentum (p) = mv

Where M-mass and v-velocity

P(before collision)= m1v1 + m2v2

Substituting m1= 0.5642 kg, m2= 0.4876 and v2=0, v1= 0.193 m/s into the conservation Ke equation we get

m1v1 + m2v2=

0.5642 kg x 0.193 m/s +o= 0.1089 (kg*m/s)

2. Calculation of total momentum of the system after the collision

For completely inelastic collision, the velocity of the two objects in collision is the same after they collide. This is because v1’=v2’=v’

First the resulting body has a mass of m1+m2= 0.5642 kg + 0.4876 kg= 1.0518 kg

P (after collision) = 1.0518 kg x 0.094 m/s=0.09887 (kg*m/s)

3. Calculation of kinetic energy of the system before the collision

Kinetic energy before the collision is given by I/2mV12+1/2mV22

Substituting V1= 0.193, m1= 0.5642 kg, m2= 0.4876 and v2=0

This reduces to I/2mV12= ½x (0.5642)2x0.193=0.0614kgm2/s2

4. Calculation of kinetic energy of the system after the collision

After the collision m=mass of m1+m2= 0.5642 kg + 0.4876 kg= 1.0518 kg

Kinetic energy after the collision is given by 1/2*1.0518*V32 = ½*(1.0518)2* 0.094=0.1039kgm2/s2

5. Calculation of kinetic energy lost during the collision

Kinetic energy lost during the collision= kinetic energy of the system after the collision - kinetic energy of the system before the collision= 0.1039kgm2/s2- 0.0614kgm2/s2= 0.0425kgm2/s2

III. Discussion/Analysis – 12 points

1. What happens to the objects, the momentum and the mechanical energy in an elastic collision? What happens to the objects, the momentum and the mechanical energy in acompletelyinelastic collision?

2. In each trial, what happens to the momentum (compare before and after)? Doeseach of your results confirm the Law of Conservation of Momentum?Explain how you arrived at your conclusions. Discuss 2 significant sources of error in the experiment.

3. In completely inelastic collisions, kinetic energy is not conserved; instead some of that energy will be converted to non-mechanical energy (resulting in total energy being conserved). In last week’s lab, a very small amount of kinetic energy was lost to work done by friction; in today’s lab the contribution of friction with the track is negligible. What are the two significant types of non-mechanical energy that the kinetic energy is converted into in today’s experiment (remember that this is energy lost during the very brief collision,not as the cart moves down the track).

Honesty Statement

My signature below indicates that I (1) made an equal contribution to writing this report, (2) agree that my partner has made an equal contribution to writing this report and (3) read and fully agree with the contents (i.e., results, conclusions, analysis) of this document.

___________________________

___________________________

1st graph

Data table (with data for all 3 trials)

II. Calculations – 15 points

Show formulas and calculations written out with units on all values. You are showing the calculations for the first trial. For the other two trials, you mustdo the calculations on your calculator or in Excel and recordthe calculated values in your data table (you must do these calculations but you do not need to show these calculations).

1. Calculation of total momentum of the system before the collision

Momentum (p) = mv

Where M-mass and v-velocity

P(before collision)= m1v1 + m2v2

Substituting m1= 0.5642 kg, m2= 0.4876 and v2=0, v1= 0.193 m/s into the conservation Ke equation we get

m1v1 + m2v2=

0.5642 kg x 0.193 m/s +o= 0.1089 (kg*m/s)

2. Calculation of total momentum of the system after the collision

For completely inelastic collision, the velocity of the two objects in collision is the same after they collide. This is because v1’=v2’=v’

First the resulting body has a mass of m1+m2= 0.5642 kg + 0.4876 kg= 1.0518 kg

P (after collision) = 1.0518 kg x 0.094 m/s=0.09887 (kg*m/s)

3. Calculation of kinetic energy of the system before the collision

Kinetic energy before the collision is given by I/2mV12+1/2mV22

Substituting V1= 0.193, m1= 0.5642 kg, m2= 0.4876 and v2=0

This reduces to I/2mV12= ½x (0.5642)2x0.193=0.0614kgm2/s2

4. Calculation of kinetic energy of the system after the collision

After the collision m=mass of m1+m2= 0.5642 kg + 0.4876 kg= 1.0518 kg

Kinetic energy after the collision is given by 1/2*1.0518*V32 = ½*(1.0518)2* 0.094=0.1039kgm2/s2

5. Calculation of kinetic energy lost during the collision

Kinetic energy lost during the collision= kinetic energy of the system after the collision - kinetic energy of the system before the collision= 0.1039kgm2/s2- 0.0614kgm2/s2= 0.0425kgm2/s2

III. Discussion/Analysis – 12 points

1. What happens to the objects, the momentum and the mechanical energy in an elastic collision? What happens to the objects, the momentum and the mechanical energy in acompletelyinelastic collision?

2. In each trial, what happens to the momentum (compare before and after)? Doeseach of your results confirm the Law of Conservation of Momentum?Explain how you arrived at your conclusions. Discuss 2 significant sources of error in the experiment.

3. In completely inelastic collisions, kinetic energy is not conserved; instead some of that energy will be converted to non-mechanical energy (resulting in total energy being conserved). In last week’s lab, a very small amount of kinetic energy was lost to work done by friction; in today’s lab the contribution of friction with the track is negligible. What are the two significant types of non-mechanical energy that the kinetic energy is converted into in today’s experiment (remember that this is energy lost during the very brief collision,not as the cart moves down the track).

Honesty Statement

My signature below indicates that I (1) made an equal contribution to writing this report, (2) agree that my partner has made an equal contribution to writing this report and (3) read and fully agree with the contents (i.e., results, conclusions, analysis) of this document.

___________________________

___________________________