Lab-4

Lab: Impulse of a softball, Courtney Aurichio, Jordan Friedman, Nikita Kumar, Period 4, 2/1/10... due 2/8/10.

**//__Objective__//**:- What is the velocity of the Softball?

 * //__Hypothesis__//** – The Velocity of the soft ball is directly related to the velocity of the whole ball-box system. The ball is thrown with a certain amount of force and velocity which is delivered on the box causing its displacement with the same velocity as the ball initially. The greater the force used to throw the ball, greater the velocity and farther the displacement of the box.


 * __Data & Observations:-__**

- Link to Excel Spreadsheet

Mass of the softball (Kg) - 0.184 Mass of the packed box (Kg) - 1.4 Total mass of packed box and softball (Kg) - 1.6

__//Excel Spread Sheet:-//__ __//Velocity Time Graph of Trials//__
 * Trial || Initial velocity of the ball from data studio (m/s) || Distance box Traveled (m) || Friction Force (N) || Work (J) || Kinetic Energy (J) || Velocity of box – ball system (m/s) || Impulse (or change in momentum) of box-ball system. || Impulse (change in momentum of ball only) || Initial Velocity of the ball, v1 || % Difference ||
 * ^  ||^   ||^   ||^   ||^   ||^   ||^   || (kg*m/s) || (kg*m/s) || (m/s) || (%) ||
 * 1 || 10.2 || 0.11 || 6 || 0.66 || 0.66 || 1.48 || 2.37 || 2.37 || 12.72 || 19.81 ||
 * 2 || 9.05 || 0.11 || 6 || 0.66 || 0.66 || 1.48 || 2.37 || 2.37 || 12.72 || 28.85 ||
 * 3 || 8.75 || 0.06 || 6 || 0.36 || 0.36 || 1.1 || 1.76 || 1.76 || 9.47 || 7.6 ||
 * 4 || 9.29 || 0.065 || 6 || 0.39 || 0.39 || 1.14 || 1.82 || 1.82 || 9.81 || 5.3 ||
 * 5 || 9.11 || 0.05 || 6 || 0.3 || 0.3 || 1 || 1.6 || 1.6 || 8.61 || 5.81 ||
 * 5 || 9.11 || 0.05 || 6 || 0.3 || 0.3 || 1 || 1.6 || 1.6 || 8.61 || 5.81 ||

W = f.d W = (6)(0.11) W = 0.66J
 * //__Calculations__//**:
 * 1. Work done by friction brining the box to a stop?**

KE = W KE = 0.66J
 * 2. What was the Kinetic Energy that the box had?**

W = weight. Distance W = (16)(0.11) W = 1.76J W = 1/2mv2
 * 3. Calculate the initial velocity of the box-ball system?**

W = KE 1.76 = ½(1.6)V2 1.76 = 0.8V2 1.76/0.8 = V2 2.2 = V2   1.48m/s = V2

Kg * V2 (1.6)(1.48) = 2.37
 * 4. Momentum (impulse of system) of box-ball system?**

mpulse of ball = Impulse on whole system Impulse of ball = 2.37
 * 5. The impulse the ball had on box?**

mball*vball = (ball mass + box mass)(V2) 0.184V1 = (0.184 + 1.4)(1.48) 0.184V1 = 2.34 V1 = 12.72m/s
 * 6. Initial Velocity of ball?**

__recorded velocity – calculated velocity__ * 100 calculated velocity 10.20 – 12.72/12.72 * 100 = 19.81%
 * 7. % difference between recorded velocity and calculated velocity?**

**__//Discussion Questions://__**
1. We made the assumption that velocity final was 0. The assumption that the ball stopped at the box helped us complete all of our calculations. 2. We could have calculated initial velocity by using a motion sensor and timing how long the ball takes to reach the box using a stop watch. We could have measured the total distance that the ball traveled, we could assume that the final velocity of the ball is 0m/s, and plug in this necessary information to a kinematics equation. 3. The impulse-momentum relationship is a useful one because it shows Newton's third law. If there is an impulse then an objects momentum is directly related to the impulse. Newton's third law clearly states that impulse and momentum have the exact same force. 4. The mass is 50 Kg. The graph shows that the force acting on it is 1000 N and the time is 0.05 s. Therefore we can use the formula F*t to figure out the impulse. (1000)(0.05) = 50 N.s. 5. The change in momentum is equal to the impulse which is 50 N.s. 6. If the objects initial velocity is -10m/s the final velocity can be calculated using the equation:- m(Vf - Vi) = impulse. 50 (Vf +10) = 50 Vf +10 = 1 Vf = -9 m/s 7. The average force acting on the object might be (0.05)(1000) + (0.10)(1000) / 2 = 100N/s

=

 * //__Conclusion:-__ The purpose of this lab was to find with what velocity the softball was thrown initially before it hit the packed box. This purpose was satisfied. By assuming that the final velocity of the ball after it hit the box was 0m/s allowed us to complete our calculations. In-order to figure out the initial velocity it was important to go through few steps first. The work done on the box by the ball is equal to the kinetic energy of the box. Therefore the Work on the ball can be set equal to kinetic energy to find out the velocity of the box-ball system. Impulse can also be defined as the change in momentum. The mass of the whole system multiplied by its velocity is the momentum. And, the momentum of the of the box is the impulse that the ball delivered on the box. Thus the momentum of the ball can be set equal to the momentum of the whole system to find out the initial velocity of the ball making Our Hypothesis true. All our errors occurred f rom 28.85% to 5.3 %. The error could have occur from the distance we stood from the box and how we threw it every trial. To fix this we should set a uniform distance from the box such as 8 meters from the box. Also we would have thrown it underhand or over hand and kept it as a constant throughout the experiment. One real life situation would be the firing of a bullet which has a very small mass but an incredibly high velocity leading it to have a high momentum delivering a huge impact on whatever is in its way. //**=====