Ball+on+Incline-+Daniel+Hicks+and+Anish+Sahasrasbudhe

__Title of Lab__: Ball on Incline Lab

__Researchers:__Daniel Hicks and Anish Sahasrabudhe

__Research Question__: How does the time it takes for a tennis ball to roll down an incline relate to the distance that it travels?

__Research__: Variables: t=time a=acceleration x=displacement Constants: vi= initial velocity, Angle of Inclination(Theta).

The kinematics equation relevant to this experiment is x= vit+(1/2)at^(2). Since the object was initially at rest, the Vi is equal to zero. Therefore, the equation can be x=(1/2)at^2. Solving for time would make the equation t^2= 2x/a. In order to solve for acceleration, we used Newton's Second Law of Motion which is ΣF= ma. During this experiment, friction is zero, so mgsin ** Θ= ** ma. The masses on both sides cancel out, leaving gsin ** Θ= ** a.

Plugging in this in the kinematics equation we obtain t 2 = 2x/g sin ** Θ. **

__Data Analysis:__

sin Θ ** = ** ( mg x /mg) mg sin Θ = mg x Since F N =mg y, the vertical forces are in equilibrium with the ball. Therefore, mg x is the important force left, and the ball will accelerate horizontally in the right direction. By Newton's second law of motion,

mg x = ma

mg sin Θ = ma

g sin Θ = a (masses get cancelled) Substituting acceleration from this equation in t^2= 2x/a, we get T^2= 2x/ gsin Θ

__Hypothesis__:T^2= 2x/ gsin Θ Squaring the time will help form a linear relationship in the graph.

__Procedure__: __ Variables __ : Time, displacement __ Constants __ : Angle of inclination, gravity __Materials__: tennis ball, incline board, protractor with string and weight attached, stopwatch, meter stick. 1. Set incline at angle measure at 15 degrees. 2. Let the tennis ball slide down the incline at different distances in 5cm increments and note the time for every distance. 3. Repeat step 2 thrice for every distance between 5cm and 80cm. 4. Record data in table form.

__Data__:
 * Displacement(m) || Time(s) || Time 2 || Time 3 ||
 * .80 || 1.0 || .97 || 1.03 ||
 * .75 || .94 || 1.0 || .91 ||
 * .70 || .93 || 1.0 || .97 ||
 * .65 || .94 || .94 || .94 ||
 * .60 || .91 || .90 || .94 ||
 * .55 || .88 || .90 || .90 ||
 * .50 || .84 || .81 || .88 ||
 * .45 || .78 || .81 || .88 ||
 * .40 || .72 || .75 || .75 ||
 * .35 || .69 || .69 || .68 ||
 * .30 || .65 || .63 || .66 ||
 * .25 || .63 || .62 || .59 ||
 * .20 || .53 || .53 || .56 ||
 * .15 || .47 || .47 || .50 ||
 * .10 || .38 || .35 || .37 ||
 * .05 || .28 || .28 || .28 ||

__Data Analysis__:

This graph has a square root curve because at longer distances, acceleration has more time to increase velocity

The time squared vs displacement graph has a slope of 1.2662 s^2/m, and using the equation g= 2/(slope*sin θ), we find the experimental g is equal to 6.10 m/s^2. According to the line of best fit, the y-intercept is .0376 s^2.



__Conclusion__: As the distance increases, the time it takes for the ball to hit the surface increases with decreasing increments in seconds. Based on our data, our the g value was 6.10 m/s^2 with the actual being 6.6 m/s^2. We had 7.5% percent error which could be reduced by using sensors at both starting at ending points so that the data and graphs will be much more accurate. The y-intercept should be (0,0) because it takes 0 seconds squared to travel 0 centimeters, but our experimental y-intercept was .0376 s^2, representing error in our data. While our experiment yielded error, the linear relationship between time squared and displacement shows that our hypothesis is supported by our data.