Meaning of y-int. on graph of acceleration vs. hanging mass

[vetgirl1990]

Homework Statement

In an experiment, I measured the effect of mass on the acceleration of a cart, rolling down a leveled air track.

I then graphed acceleration vs. mass (m'), where:
m' = m / (M+m); M = mass of the cart, m = hanging mass, which was increased with subsequent trials.

Homework Equations

m' = m / (M+m)

The Attempt at a Solution

The equation of the line that I got when I graphed this relationship, is y=9.855x-0.017. I am trying to figure out what the "meaning" of y-intercept value means. I know that at the y-intercept, the x-intercept (in this case, the mass m') is zero... but this doesn't make sense in the context of this problem. Doesn't this essentially mean that when mass is zero, then acceleration is -0.017m/s/s?

Theoretically, if mass is zero, then acceleration is zero.

 

[gneill]

Can you think of any sources of error, neglected effects, or characteristics of the equipment that might contribute to this "offset" of the line? Do you think that the sign of the intercept might be significant in any way?

Why not itemize all the parts and pieces that comprise the equipment used in the lab and consider each individually for possible influences? How might such influences show up in the equations of motion if they were included in the mathematical model of the system? Pick out likely candidates for discussion.

 

[NascentOxygen]

Homework Statement

In an experiment, I measured the effect of mass on the acceleration of a cart, rolling down a leveled air track.

I then graphed acceleration vs. mass (m'), where:
m' = m / (M+m); M = mass of the cart, m = hanging mass, which was increased with subsequent trials.

Homework Equations

m' = m / (M+m)

The Attempt at a Solution

The equation of the line that I got when I graphed this relationship, is y=9.855x-0.017. I am trying to figure out what the "meaning" of y-intercept value means. I know that at the y-intercept, the x-intercept (in this case, the mass m') is zero... but this doesn't make sense in the context of this problem. Doesn't this essentially mean that when mass is zero, then acceleration is -0.017m/s/s?

Is it nonetheless within experimental error?

Sometimes in a bunch of experimental data there can be just one data point that throws the outcome off-course. Often we don't have the opportunity to repeat the experiment to check the value of that suspect data point, but in the analysis it may be possible to demonstrate that if we omit from our analysis that single out-of-place data point then the outcome of the experiment will meet expectations within experimental error.

 

[Ray Vickson]

Homework Statement

In an experiment, I measured the effect of mass on the acceleration of a cart, rolling down a leveled air track.

I then graphed acceleration vs. mass (m'), where:
m' = m / (M+m); M = mass of the cart, m = hanging mass, which was increased with subsequent trials.

Homework Equations

m' = m / (M+m)

The Attempt at a Solution

The equation of the line that I got when I graphed this relationship, is y=9.855x-0.017. I am trying to figure out what the "meaning" of y-intercept value means. I know that at the y-intercept, the x-intercept (in this case, the mass m') is zero... but this doesn't make sense in the context of this problem. Doesn't this essentially mean that when mass is zero, then acceleration is -0.017m/s/s?

Theoretically, if mass is zero, then acceleration is zero.

Is y = your m'? Is x = your m? If so, the exact equation is
$$y = \frac{x}{x+M}$$
If you attempt to represent this as a simple linear function of the form $y = a + bx$ you will not get very good accuracy, and your "fitting" parameters $a$ and $b$ will not necessarily have any convincing meaning. Not every graph can be well-represented as a straight line.

 

[vetgirl1990]

Can you think of any sources of error, neglected effects, or characteristics of the equipment that might contribute to this "offset" of the line? Do you think that the sign of the intercept might be significant in any way?

Why not itemize all the parts and pieces that comprise the equipment used in the lab and consider each individually for possible influences? How might such influences show up in the equations of motion if they were included in the mathematical model of the system? Pick out likely candidates for discussion.

Thanks for the suggestions! Breaking down the possible sources of error, I think it's most likely systematic error, possibly due to human error of applying force by hand. Given that the y-intercept is supposed to be zero (when m' = 0, acceleration is 0), having an acceleration even when m' is zero indicates that there was another source of acceleration, even without the falling mass.

 

[gneill]

Thanks for the suggestions! Breaking down the possible sources of error, I think it's most likely systematic error, possibly due to human error of applying force by hand. Given that the y-intercept is supposed to be zero (when m' = 0, acceleration is 0), having an acceleration even when m' is zero indicates that there was another source of acceleration, even without the falling mass.

Good, anything else? Maybe some masses or sources of friction that aren't accounted for? How was acceleration measured, by some automatic mechanism or by manual timings? How do you know that air track was perfectly level?

 

[Ray Vickson]

Thanks for the suggestions! Breaking down the possible sources of error, I think it's most likely systematic error, possibly due to human error of applying force by hand. Given that the y-intercept is supposed to be zero (when m' = 0, acceleration is 0), having an acceleration even when m' is zero indicates that there was another source of acceleration, even without the falling mass.

So, what is your answer to the questions I asked you in post #4? The issue is pretty fundamental.

 

[NascentOxygen]

Is y = your m'? Is x = your m? If so, the exact equation is
$$y = \frac{x}{x+M}$$

Ray, I believe the plot is accn vs g·m'
i.e., y = kx

 

[Ray Vickson]

Ray, I believe the plot is accn vs g·m'
i.e., y = kx

I wish the OP had the courtesy to respond.