Acceleration vs. mass (kg) graph

[L²Cc]

If in an acceleration vs. mass (kg) graph (where the x-axis reads (1/m1+m2), the inverse of the sum of masses), the best fit does interesect with the origin because if x = zero, or (1/m1+m2) = o, the final solution is unreal since it is mathematically incorrect to divide one by zero.

Likewise, in an acceleration vs. differences of masses, best interesects with origin since when m1-m2=0, acceleration, as as result, equals to zero.

Finally, would it be correct to stay that the y interest of an acceleration vs. mass graph represents the acceleration of the system when 0 masses are involved, or does it repersent the gravitational mass (resistance)?

 

[mfb]

If you plot the inverse mass then the y intercept correspond to infinite mass, which can't be moved with a finite force. No gravity involved (unless it's providing the force being studied).

 

[kuruman]

It looks like you are describing an Atwood machine experiment in which the acceleration is $$a=\frac{m_2-m_1}{m_2+m_1}g=\frac{\Delta M}{M}g.$$The usual procedure is to keep the total mass $M$ constant and move some mass from one side to the other to vary $\Delta M$. Then a plot of acceleration vs. $\Delta M$ has slope $g/M$ from which one can deduce the acceleration of gravity.

This is the the case when the pulley is massless which never happens. A real pulley with moment of inertia $I$ and radius $R$ has effective mass $m_{eff}=I/R^2$ in which case the measured acceleration is $$a=\frac{\Delta M}{M+m_{eff}}g\approx \frac{\Delta M}{M}\left(1-\frac{m_{eff}}{M}\right)g.$$In the non-ideal pulley case, there is an intercept which can be used to find the effective mass.