# Presentation Q's

ACLerok
in a month, i give a presentation about an experiment in which two objects of different mass are dropped at the same time and hit the ground at once. I am asked for a verbal and mathematical model for this experiment. Here goes my verbal model: "While mass does affect an objects weight, it does not play a role in determining the rate of fall of an object rather does the gravitational force on the object by the earth." Is this good enough? What should I add/subtract? I am also asked to find a mathematical model but I don't know which equation corresponds with this experiment. It's a constant acceleration equation right? The physics class I'm taking is a level one course in a college in New Jersey so please no complicated answers! :) Oh yeah, and how do I take data from this experiment? I know I'm supposed to record the objects position vs. time but I amsupposed to gather the data from a Quicktime movie on the Internet. It says its at 30fps but how exactly do I analyze the data? Thanks to all of you!

Ambitwistor
Originally posted by ACLerok in a month, i give a presentation about an experiment in which two objects of different mass are dropped at the same time and hit the ground at once. I am asked for a verbal and mathematical model for this experiment. Here goes my verbal model: "While mass does affect an objects weight, it does not play a role in determining the rate of fall of an object rather does the gravitational force on the object by the earth."

The "weight" of an object is defined to be the gravitational force on the object. So it is not correct to claim that the gravitational force plays a role, but the weight does not.

The correct explanation is that the gravitational force on an object is proportional to the mass of that object. From F = ma, that means that the acceleration is independent of the object's mass, and it is the acceleration which directly governs the motion of the body.

I am also asked to find a mathematical model but I don't know which equation corresponds with this experiment.

You can approximate the Earth's gravitational field as F = mg. Since F = ma, that means that a=g, i.e., the body always has an acceleration g (equal to about 9.81 m/s2 near the surface of the Earth), regardless of what its mass is; the mass cancels out of the equation.

If you want to be more accurate, the gravitational field of a planet of mass M on a body of mass m is equal to F = GMm/r2, where G is Newton's gravitational constant and r is the distance between their centers of mass.

Oh yeah, and how do I take data from this experiment? I know I'm supposed to record the objects position vs. time but I amsupposed to gather the data from a Quicktime movie on the Internet. It says its at 30fps but how exactly do I analyze the data?

Well, assume the first frame is at time t=0. Then the second frame will be at time t=1/30 seconds, the third frame will be at time t=2/30=1/15 seconds, etc. From the movie, you can presumably determine how far the object has fallen in any given frame.

The data should fall roughly along the curve x = 1/2 gt2, where g=9.8 m/s2 (you'll have to convert your measured distances to meters and measure time in seconds to use this value of g). If you have a spreadsheet like Excel, you can do a power fit and determine the acceleration of the body yourself.

ACLerok

Originally posted by Ambitwistor

You can approximate the Earth's gravitational field as F = mg. Since F = ma, that means that a=g, i.e., the body always has an acceleration g (equal to about 9.81 m/s2 near the surface of the Earth), regardless of what its mass is; the mass cancels out of the equation.

If you want to be more accurate, the gravitational field of a planet of mass M on a body of mass m is equal to F = GMm/r2, where G is Newton's gravitational constant and r is the distance between their centers of mass.

i'm not supposed to refer to Newton's Laws or force, just kinematics. is the mathematical model an actual equation or just some sentences or ideas?

Ambitwistor

The kinematical model is that of constant acceleration, as you say.