# Velocity vs mass modeling

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1. Aug 22, 2015

### hahame

1. The problem statement, all variables and given/known data
Below is a table for some experiment data:
m(kg) - v(m/s)
0.1 - 3.16
0.2 - 2.24
0.3 - 1.83
0.4 - 1.58
0.5 - 1.41
0.6 - 1.29
0.7 - 1.20
0.8 - 1.12

As the mathematical model for these data, I put v^2 (m^2/s^2) = 1.00 (kgm^2/S^2)/m(kg).

Another set of experiments finds for the exact same velocities as before, but the slope is 100 times larger.
What would be the new mathematical equation for this new experiment? (the dataset for this new experiment given in the problem is just the same as the old data above. But Question: if the slope is 100 larger than before, how is it possible for the old and new experiment both to have the same result data?)

2. Relevant equations

v^2 (m^2/s^2) = 1.00 (kgm^2/S^2)/m(kg) (my guess for the old experiment?)

3. The attempt at a solution
At first I thought it would simply be v^2 (m^2/s^2) = 100.00 (kgm^2/S^2)/m(kg),
however, was confused if it should be v^2 (m^2/s^2) = 10000.00 (kgm^2/S^2)/m(kg), since the left part of the equation is v^2, not v.

However, I got really stuck when I solved for m = 0.3, neither of my guessed answers gave me v = 1.83.

I would greatly appreciate if anyone could help me with figuring out the right mathematical model for the new experiment and making sense of the data table being the same for both old and new experiment.

Thanks! :)

2. Aug 22, 2015

### SteamKing

Staff Emeritus
It's not clear why you are squaring the velocity.

Have you plotted this data, to see what kind of curve might be suitable?

3. Aug 22, 2015

### hahame

Yes! I used excel, and it gave me v = 1/m^-0.5, which is the same as v^2 (m^2/s^2) = 1.00 (kgm^2/S^2)/m(kg).
I squared the equation because that was the only suitable option from the given choices in the original problem.