Lab-01

Ani Sarajian, Heather Dodson, Nicole Serpa, and Maria Butkov __Lab 17:__ Impulse of a Softball

__Period__: 4

__Objective:__ What is the velocity of a softball?

__Hypothesis:__ The faster velocity of the softball, increases the force of the ball on the box, which directly relates to how far the box will move.

__Date Completed/Date Due:__ 2/1/10 ; 2/8/10

__Graphs:__

__Trial 1__

__Trial 2__ __Trial 3__ __Trial 4__ Trial 5

__Link to excel sheet:__

__Data Table:__



__Sample Calculations:__ (All using trial 1)

__W=f (d)__ W=.224J**
 * W= 2 (.112)

__KE- work = 0__ KE = .224 J**
 * KE - .224 = 0

__KE=1/2mv^2__ .224=.642v^2 .3489=v^2 √.3489 = √ v2 .5906 m/s=v**
 * .224=1/2(1.284)v^2

__Impulse=change in momentum= kg*m/s__ =1.284(.5906) =.75844 kg•m/s**
 * impulse = m*v

__Impulse of ball alone__ =1.284(.5906) =.75844 kg*m/s**
 * =mv

__Initial Vel. of ball__ =.75844/.184 =4.12196 m/s**
 * =impulse of ball alone/ mass of ball

__% Diff = |experimental - theoretical|/ average of the two *100__ =(1.00804/ 4.62598) * 100 = 10.89542986 % difference**
 * =((|4.12196-5.13)|)/(4.12196+5.13)/2)*100

__Conclusion:__ Error in this lab came from measuring the distance the box moved just by looking at the meterstick on the floor; we kind of estimated the distance, there was no way to measure the exact distance. It was also very hard to get a good throw without the ball hitting the back of the box or having the box move sideways. Applications for this lab could be for baseball itself: one would need to know what speed to set the automatic pitching machine to so that is comes at the hitter at a certain speed. Another impulse-momentum relationship would be a hammer to nail; the faster the hammer is going, the greater impulse it will have on the nail, pushing it farther into the wood.**
 * The reasoning for this lab was to figure out the initial velocity of the softball, the speed at which it was moving before it reached it's final velocity upon hitting the packed cardboard box. Yes our hypothesis proves correct because the force was dependent on the velocity of the ball, and all of this put together helped the box move a certain distance. The force was greater due to the speed of the ball, that's why our box moved, thus making it successful because we had assumed that the final velocity was zero. The work done to the box was equal to the kinetic energy therefore using work-energy equations helped us to figure out the other calculations and to successfully finished our experiment. Depending on how you threw the ball had an impact on how far is moved and could change the graph of what the process had looked like.

__Discussion Questions:__

1. What assumptions did you make in order to complete the calculations in this experiment?
 * We made assumptions such as the final velocity had been zero. This was because excluding the paper, we made assumptions as to the balls velocity ending when it hit the back of the box which help us to continue on in our experiment.**

2. How else could you have calculated the initial velocity (without using energy or momentum conservation)?
 * We could have used kinematics to calculate the initial velocity without using energy or momentum conservation in this experiment. There are variables such as the distance traveled, time, mass etc. that we had figured out in the beginning, thus helping us solve for one thing and proceed with another to produce an answer.**

3. Why is the impulse-momentum relationship a useful one?
 * They are directly related; the faster the ball is going (greater momentum), the harder it will hit the paper in the box (greater impulse). You can find out how fast the ball is going just by looking at the impulse, therefore it is useful to calculate Vi of the ball.**

4. Find the magnitude of the impulse delivered momentum. J=F*t J=(1000N)(.05s) J=50N/s**
 * The Magnitude of the Impulse to the object is:

5. Determine the change in the object's momentum. 50=m*change in v 50=50*change in v change in velocity = 1N/s**
 * J=m*change in v

6. If the object's original velocity was -10m/sec, calculate its final velocity. 50(Vf+10)=50 Vf+10=1 Vf=-9m/s**
 * The formula for final velocity is m(Vf+Vi)=impulse.

7. What was the magnitude of the "average force" acting on the object during the time graphed? (0.5)(1000)+(0.10)(1000)/2=100N/s.**
 * The average force acting on the object during the graphing process is assumed to be 100N/s.