Lab_The_Ballistic_Pendulum_(1)

Appendix C: Team Lab Report Rubrics Experiment ______________________________________________ Partners: Danilo Roque Felipe, Jose Velazquez ,Andry Valdes Data (20 %) 1. Measurements organized into a neat table; 2. Values are clearly labeled, correct units; 3. Significant figures of data; 4. Quality/range/ multiple trials (when appropriate); 5. Table of generated values, labeled with units. ___4_/4 ___4_/4 ___4_/4 ___4_/4 __4__/4 Evaluation of Data (40 %) 1. Graphs: ● Variables on appropriate axes (use of units); ● Quality of results. 2. Interpretation of graphs and Mathematical Model: ● Brief written statements of relationships shown on the graphs; ● Equation of the relationship obtained from the graph; ● Correct interpretation of slope and y intercept. 3. Sample of Calculations with Units and Significant Figures 4. Answer to the Analysis questions. 5. Correct units and Calculation of % error. comments: ___5_/5 ___5_/5 ___4_/4 ___3_/4 ___2_/4 __4__/4 _10_/10 ____/4 Conclusion (40% ) Quality of written explanation of relationships. The discussion must include all of the following: 1. New terms and concepts: Definitions according to the textbook; 2. Physical meaning of slope / significance of Y-intercept; 3. Conditions and derivations of general equations; 4. Reasonable explanation for divergent results; 5. Textbook correlation. (4 %) comments: TOTAL: Grade: __6__/6 _6___/6 20__/20 __3__/4

PROCEDURES: Part I Ballistic Pendulum WARNING Ensure the trajectory is clear before cocking the spring and while firing. Safety glasses are available upon request 1. Use a scale to measure the mass of the ball: m = 0.05673 kg . 2. Slide the projectile ball (which has a hole in it) onto the rod of the spring gun. 2. When the ball has been placed on the rod, cock the gun by pushing against the ball until the latch catches. Be very careful not to get your hand caught in the spring gun mechanism!!! 3. Fire the gun several times to see how it operates. Ensure the ball lands in the pendulum arm. If you cannot get the arm to catch properly, ask for help. There are several common problems: ● The height of the pendulum arm may be off. ● The opening of the pendulum arm may be at an angle ● The pendulum arm opening may be loose (which will rotate when the ball hits it, instead of catching it). ● The pivots at the top of the pendulum arm may be too tight or too lose. 4. A sharp curved point on the side of the pendulum (or on some models a dot) marks the center of mass of the pendulum-ball system. Let the pendulum bob hang vertically and measure the distance y 1 , of the point marking the center of mass above the base of the gun. ● Record the value of y 1 = 7,5 cm 5. Place the ball on the rod, push against the call to cock the gun, and fire the ball into the stationary pendulum while it hangs freely at rest. The pendulum will catch the ball, swing up, and then lodge in the notched track. 6. Record the vertical distance from the bottom of the base to the center of mass of the ball-pendulum system in Data Table I. 7. Repeat step 6 four more times for a total of five measurements. 8. Loosen the screws holding the pendulum in its support and remove the pendulum. Use a scale and measure the mass of the pendulum: M = 0.160 kg Data Table 1: Record distances in cm and masses in grams. Trial y 2 (cm) y 2 – y 1 (cm) V (m/s) V xo (m/s) 1 24,5 17 1.83 6.99 2 25,5 18 1.88 7.18

4. Calculate the average of the initial velocity of the ball from the five trials: ( V 0 x ) ave =( 6.99 m / s + 7.18 m / s + 7.18 m / s + 7.18 m / s + 7.18 m / s )/ 5 = 7.14 m / s 5. Use the vertical distance, the value for the gravitational acceleration g = 9.81 m / s 2 to calculate the time it takes for the projectile to reach the floor: t = √ ❑ t = √ ❑ = 1.75s 6. Determine the total horizontal distance traveled by the ball for each of the five trials: x = x 1 + x 2 x=0.3m0+0.703m=1.003m 7. Determine the total horizontal distance traveled by the ball for each of the five trials: V ox = x t V ox 1 =( 1.003 m 1.75 s )= 0.571 m / s V ox 2 =( 1.005 m 1.75 s )= 0.575 m / s V ox 3 =( 0.985 m 1.75 s )= 0.563 m / s V ox 4 =( 0.965 m 1.75 s )= 0.551 m / s

V ox 5 =( 0.945 m 1.75 s )= 0.540 m / s 8. Calculate the average initial velocity of the projectile for the five trials ( V 0 x ) ave = ( 0.571 m / s + 0.575 m / s + 0.563 m / s + 0.551 m / s + 0.540 m / s ) 5 = 0.56 m / s 9. Compare this result by calculating a percentage difference with the one obtained in the projectile motion lab: % difference = | 1.83 m / s − 0.571 m / s | 1 2 ( 1.83 m / s + 0.571 m / s ) × 100% = 26.2% 1. Check that the Law of Conservation of Momentum is held in this experiment: ● Calculate the change in momentum of the Ball: ∆ p = p f − p i = m ( V − V 0 x ) 1→ ∆ p = 0.05673 kg ( 1.83 m / s − 6.99 m / s )=− 0.293 Kg .m / s 2-5→ ∆ p = 0.05673 kg ( 1.88 m / s − 7.18 m / s )=− 0.301 Kg .m / s ● Calculate the change in momentum of the Pendulum: 1→ ∆ P = P f − P i = M ( V − 0 ) =0.16kg(1.83m/s)=0.293kg.m/s 2-5→ ∆ P = P f − P i = M ( V − 0 ) =0.16kg(1.88m/s)=0.301kg.m/s ● Calculate the Total change in momentum of the system, explain its meaning: 1–> ∆ p system = ∆ p + ∆ P = -0.293kg.m/s + 0.293kg.m/s=0 2-5—> ∆ p system = ∆ p + ∆ P = -0.301kg.m/s + 0301kg.m/s=0 2. What is the meaning of this result? Kinetic Energy is conserved 3. Check that the Law of Conservation of Energy holds in this experiment: ● Calculate The total energy of the Ball-Pendulum system right after collision; TE 1 = KE ball − pendulum = 1 2 ( m + M ) V 2 = 1 2 ( 0.05673 kg + 0.160 kg ) ( 1.83 m / s ) 2 = 0.363 J TE 2 − 5 = KE ball − pendulum = 1 2 ( m + M ) V 2 = 1 2 ( 0.05673 kg + 0.160 kg ) ( 1.88 m / s ) 2 = 0.383 J ● Calculate The total energy of the Ball-Pendulum system at the highest vertical position; TE 2 = GPE 2 = ( M + m ) g ( y 2 − y 1 ) ave = ( 0.05673 kg + 0.160 kg ) 9.81 m / s 2 ( 0.18 m )= 0.382 J 4. What is the meaning of this result? Total mechanical energy is conserved 5. What are the different types of collisions, how do you classify them? What collision is taking place in this lab, how do you know? Collisions are two types: Elastic Collision Inelastic Collision This is an elastic collision because kinetic energy is conserved.

IV. Textbook correlation. Kinetic Energy Work Theorem(ch 6.6, pg 177) Gravitational Potential Energy(pg 202) 8.2 Conservation of Momentum Example 8.5 page 243