Calculating Percent of Kinetic Energy Lost

[Caps1394]

Homework Statement

We are doing a lab in my physics class. We have to calculate the percentage of Kinetic energy lost in an inelastic collision. I believe that I have the right answer but it seems a little high (44%) the one marble is rolling down a flat track and hits a stationary ball at the the end and we let the balls collide and fall off onto carbon paper allowing us to have the distances that it fell.

Homework Equations

Kinematic Equation.
M1Vi1=M1V1+M2V2
Ke=.5MV^2

The Attempt at a Solution

We used the distances from where the balls landed and the end of the track and used kinematics to calculated the initial velocitys. Which in turn would be the Velocities after the collision. Then I used the M1Vi1=M1V1+M2V2 Having the masses cancel each other out (since the marbles are identicle) and finding the Intial Velocity of the First marble. Then I took that velocity and found out how much kinetic energy it had right before the collision. Then I used the Velocities after the collision, plugged them into the kinetic energy equation and added those 2 up. Now that I have the Ke initial and Ke final I subtracted them and found out how much was lost. Taking the amount lost over the intial amount of kinetic energy. Getting what I assume is the % of kinetic energy lost.
Im not sure if this is right or if I am doing a step wrong any help would be appreciated.

 

[LawrenceC]

Your process sounds right. Advise me if I have the situation correctly imagined.

A marble moves on a flat path and strikes another. Both immediately fall off the table but are not joined. The two marbles imprints on the paper are measured so you know the distance to the floor and how far each went horizontally. You measure the horizontal and vertical distances the marbles fell and determine both their velocities as they left the flat track and fell.

Using those velocities you compute the final KE when they left the track to begin their fall. You also compute the first ball's velocity from the momentum equation. With the velocity from the momentum equation you compute the initial KE and compare.

Is this correct?

There is one form of KE that you may be ignoring. It is rotational KE. The first marble is rolling. But it may change so slightly that it does not matter much.

 

[Caps1394]

Your process sounds right. Advise me if I have the situation correctly imagined.

A marble moves on a flat path and strikes another. Both immediately fall off the table but are not joined. The two marbles imprints on the paper are measured so you know the distance to the floor and how far each went horizontally. You measure the horizontal and vertical distances the marbles fell and determine both their velocities as they left the flat track and fell.

Using those velocities you compute the final KE when they left the track to begin their fall. You also compute the first ball's velocity from the momentum equation. With the velocity from the momentum equation you compute the initial KE and compare.

Is this correct?

There is one form of KE that you may be ignoring. It is rotational KE. The first marble is rolling. But it may change so slightly that it does not matter much.

Yes. You have everything correct.

 

[LawrenceC]

Your method sounds correct to me. If you would post the horizontal and vertical distances the marbles went after leaving the table, I'll redo the calculations to see if I get the same result.

 

[Caps1394]

Thank You
Two trial were done both had the same time(.421) and Δy(-.87m)
Note: Only the displacements are as is. All other calculations and number I did not round until I got to the percentage of kinetic energy lost(the very end). Used G=-9.8m/s^2

Trial 1
Ball 1
Δx=.131m
V after collision=.3108m/s
Ball 2
Δx=.378m
V after collision=.897
Calculated: The Ball one velocity before collision to be 1.207m/s and the percent of Ke lost to be≈38.2%

Trial 2
Ball 1
Δx=.114m
V after collision=.2705m/s
Ball 2
Δx=.228m
Velocity after collision=.54109m/s
Calculated: The ball one velocity before collision to be .81164m/s and the percent of Ke lost to be≈44.49%

 

[LawrenceC]

I get your answers. I developed a formula for the percentage of KE lost. It is:

%KElost = 2*d1*d2/(d1+d2)^2

where the d1 and d3 are your delta X's.

 

[Caps1394]

Thank you very much. Also for the equation that will come in handy.