25-Feb-2015 Free Fall

Purpose: To demonstrate that when no other forces, besides gravity, are acting on a falling object, it will accelerate by 9.8 m/s^2

Theory: A falling object, with no other forces acting on it, will fall with a constant acceleration of 9.8m/s^2

Procedure: Using a free fall apparatus, a falling mass will make a series of marks on a strip of paper every 1/60th of a second, we then measured the distance of each mark, input the data into excel in order to graph it and see if we can determine the acceleration using the graph and basic kinematics equations.

*First, we set up the free fall apparatus.

* As the mass fell it created a series of marks along a strip of paper, every 1/60th of a second.
* We measured the distance in cm of each mark using a meter stick, for the period of 0.4167 seconds


Results:


 

time s	distance cm	∆x cm
0.0000	0	1.1
0.0167	1.1	0.9
0.0333	2	1.4
0.0500	3.4	1.6
0.0667	5	2
0.0833	7	2.2
0.1000	9.2	2.5
0.1167	11.7	2.7
0.1333	14.4	3
0.1500	17.4	3.2
0.1667	20.6	3.6
0.1833	24.2	3.8
0.2000	28	4
0.2167	32	4.3
0.2333	36.3	4.6
0.2500	40.9	4.9
0.2667	45.8	5.2
0.2833	51	5.5
0.3000	56.5	5.8
0.3167	62.3	5.7
0.3333	68	6.2
0.3500	74.2	6.5
0.3667	80.7	6.8
0.3833	87.5	7
0.4000	94.5	7.1
0.4167	101.6	7.5

Graphed:

*The equation for the chart looks remarkably like the distance formula x=1/2(acceleration)(time^2)+velocity(time), with 482.08x^2 equivalent to 1/2(a)(t^2).

*So if we divide the coefficient (482.08) by 1/2 we should get the acceleration of the fall.

*We get 964.16 cm/s^2 or 9.6 m/s^2

*Using excel we create another chart of the mid-interval of time vs the mid-interval of velocity, during the fall.

*The mid-interval velocity is the same as the average velocity of the mid-interval time.

Results:

mid-interval time Mid-interval speed cm/s 0.0083 66 0.0250 54 0.0417 84 0.0583 96 0.0750 120 0.0917 132 0.1083 150 0.1250 162 0.1417 180 0.1583 192 0.1750 216 0.1917 228 0.2083 240 0.2250 258 0.2417 276 0.2583 294 0.2750 312 0.2917 330 0.3083 348 0.3250 342 0.3417 372 0.3583 390 0.3750 408 0.3917 420 0.4083 426 0.4250 450 Graphed *We once again have an equation that looks like a kinematics formula that includes constant acceleration, Velocity_final=acceleration(time)+Velocity_original *Which means 956.18x is the acceleration 956.18 cm/s^2 multiplied by time. *Converting to meters and rounding back to two significant figures, we once again get 9.6 m/s^2 Questions/Analysis
1. The mid-interval speed was calculated by the change in distance divided by 1/60 of a second (time in between each mark), average velocity is determined by the Velocity_final subtracted by the Velocity_original and the difference divided by 2. The mid-interval speed is more precise in the actual velocity at that point, because it is the actual distance/time which velocity stands for 2. The velocity vs time graph could expressed as an equation of y=956.18x+ 44.596, which is the same formula for calculating Velocity_final from acceleration, time, and adding Velocity_original. This means 956.18 had to be our acceleration (in cm/s^2) and after converting to meters and rounding we got 9.6m/s^2. This is only a 2% difference from the actual acceleration of 9.8, so I'm satisfied with the result. 3. The distance vs time graph could be expressed as the equation y=482.08x^2+43.174x+.0611, which is the same formula for calculating distance with 1/2acceleration, times time squared, added Velocity_original times time, and added original velocity. This meads 482.08 is our acceleration multiplied by 1/2, so after dividing 482.08 by 1/2 and converting to meters and rounding to two significant figures we once again get 9.6m/s^2. This is only a 2% difference from the actual acceleration of 9.8, so I'm satisfied with the result.
Conclusion: *With only a 2% difference between the expected and determined acceleration due to the force of gravity, I feel comfortable with stating the experiment was a valid method of determining acceleration due to gravity. *However, some issues are that the experiment doesn't account for air resistance, and measurements from meter sticks can only ever be accurate to two significant figures.