Acceleration of Free-Falling Masses

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SPH UAcceleration of Free-Falling Corey McCormick

Masses 0/14/0

Purpose What effect does the mass of a free-falling object have on the objects acceleration?

Independent/Dependent Variables The independent variable in this lab is the free-falling objects mass, while the dependent variable is the free-falling objects acceleration.

Hypothesis If the mass of the free-falling object is doubled then the acceleration will be the same because everything falls at the same speed no matter its mass. A free-falling object can be defined as an object whose acceleration depends solely on the ¡§acceleration of gravity¡¨, which is measures at .8 m/s/s [downwards].

Materials

„h Ticker Timer

„h x - Ticker Tape (~0 cm long)

„h x - A piece of scotch tape

„h 100g mass

„h 00g mass

„h School bag

„h A table

Procedure

1. The ticker timer was plugged into an outlet on the wall.

. One of the pieces of ticker tape was inserted into the ticker timer (shinny side up).

. The 100g mass was taped to the end of the ticker tape with one of the pieces of the scotch tape.

4. The ticker timer set up was then placed near the edge of the table.

5. The school bag was placed under table (directly below the edge of the table where the ticker timer set up was placed). (The school bag was used simply to stop the mass from hitting the floor when it was dropped from the table).

6. The ticker timer was turned on and set to 60 Hz.

„³ The set up looked like this

Top view

Side View

7. The 100g mass was dropped from the edge of the table allowing the ticker timer to mark the ticker tape every 1/60th of a second showing the objects position every 1/60th of a second.

8. The ticker tape used was labeled ¡§100g mass¡¨.

. Procedure steps # through #8 were repeated using a 00g mass to replace the 100g mass.

10. Using time intervals measuring 0.017 seconds each, the displacements of the 100g and 00g masses were recorded, up to the 10th time interval (0.15s ¡V 0.170s).

Results

100g Mass

Time Interval

(s)Time Change

(s)Displacement

(m) [down]Average Velocity

(m/s) [down]

0.0 - 0.0170.0170.0010.058

0.017 ¡V 0.040.0170.000.176

0.04 ¡V 0.0510.0170.0040.5

0.051 ¡V 0.0680.0170.0070.41

0.068 ¡V 0.0850.0170.010.706

0.085 ¡V 0.100.0170.0150.88

0.10 ¡V 0.110.0170.0171.000

0.11 ¡V 0.160.0170.011.176

0.16 ¡V 0.150.0170.01.4

0.15 ¡V 0.1700.0170.041.41

00g Mass

Time Interval

(s)Time Change

(s)Displacement

(m) [down]Average Velocity

(m/s) [down]

0.0 - 0.0170.0170.0010.058

0.017 ¡V 0.040.0170.000.176

0.04 ¡V 0.0510.0170.0060.5

0.051 ¡V 0.0680.0170.0100.588

0.068 ¡V 0.0850.0170.010.765

0.085 ¡V 0.100.0170.0160.41

0.10 ¡V 0.110.0170.0181.058

0.11 ¡V 0.160.0170.01.4

0.16 ¡V 0.150.0170.011.5

0.15 ¡V 0.1700.0170.071.58

All of the x and y values on all of the graphs are shown at ¡§# x10e-¡¨ for better clarity.

Analysis (For each mass)

1. Create a position-time graph

. Create an average velocity-time graph. Draw a line of best fit

. Use the average velocity-time graph to calculate the acceleration. Show your work.

100g Mass

Line of best fits coordinates (0, 0) and (0.180, 1.600)

=y ¡Vy1

x ¡Vx1

=1.600 ¡V 0

0.180 ¡V 0

=1.600

0.180

= 8.888

=The acceleration of the free-falling 100g mass was 8.888 m/s/s.

00g Mass

Line of best fits coordinates (0, 0.050) and (0.175, 1.600)

=y ¡Vy1

x ¡Vx1

=1.600 ¡V 0.050

0.175 ¡V 0

=1.550

0.175

= 8.85714

=The acceleration of the free-falling 00g mass was 8.857 m/s/s.

4. Write a conclusion to answer your question.

The mass of a free-falling object has no effect what so ever on a free-falling objects acceleration. Merriam-Websters definition of ¡§free-falling¡¨ is ¡§The condition of unrestrained motion in a gravitational field.¡¨ In Laymens terms, this means that every free-falling object on this planet has the same acceleration due to Earths gravity, .8m/s/s [down]. .8 m/s/s [down] is also called the ¡§acceleration of gravity¡¨.

Evaluation

1. Use the information in your textbook to find the accepted value of the acceleration of a free-falling object. Calculate the percent error.

The accepted value of the acceleration of a free-falling object is .8 m/s/s [down] (The acceleration of gravity).

=100¡V Labs acceleration valuex100

Accepted acceleration value

=100 - 8.888 x 100

.8

= 100 - 0.8058 x 100

=100 - 8.058

=10.4

The percentage error of the value for the acceleration of a free-falling object that was obtained in this lab experiment was 10.4 %.

. Suggest at least three possible sources of error for this investigation.

1. Ticker Timer not Accurate

Using a ticker times is not an accurate way of measuring position or displacement.

The ticker tape tends to slide to the left and the right, while it is running through the ticker timer during the experiment, affecting the line of dots on the ticker tape (they do not form a straight line). Since the dots are crooked , it is much more difficult to measure the displacement between each one of them. Instead, some sort of automatic machine that not only counts the time intervals, but also records the displacements between the time intervals, should have been used. The calculations would have been much more exact.

.Ruler not Accurate

Using a ruler is not an accurate method of measuring displacement for this lab. In this investigation, a setting of 60Hz was used. This means that the time intervals being used were 0.017 seconds. When working with very small numbers, using the naked eye to estimate the measurements of positions and displacements using a ruler is not very accurate at all, especially in cases such as this one, where being 0.001 meters off will affect the final readings. Instead, some sort of automatic machine that not only counts the time intervals, but also records the displacements between the time intervals, should have been used. The calculations would have been much more exact.

.Line of Best Fit not Accurate

Since there are no specific equations or steps to reach an exact location, a scientist can only estimate where to place a line of best fit on a graph. When estimated variables are used in an experiment, the end results will almost never be exact, especially when working with detailed numbers, where if an estimation is off by even a very small amount, the final calculations are skewed. Instead, several instantaneous accelerations should have been calculated to help find the average acceleration rate. At least that way, exact numbers would have been used instead of estimations.

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