Lab 4 (1D03)

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D4: Conservation of Energy and Momentum: Elastic and Inelastic Collisions NAME: Amanda Lykos STUDENT No.: 400297082 LAB/TUTORIAL SECTION: T7 LECTURE SECTION: C01 COMPLETE ALL SECTIONS OF THIS TEMPLATE AND SUBMIT AS ONE SINGLE DOCUMENT. SOME INSTRUCTIONS HAVE BEEN INCLUDED HERE. SEE LAB MANUAL FOR FULL DIRECTIONS. The following shows the value of all the questions in this lab: Laboratory 4 Grading Scheme Introduction Introduction Q1 Q2 Q3 Totals Points /2 /2 /2 /6 Procedure Procedure Q4 (photo) Q5 Q6 Tables (2) Totals Points /1 /3 /3 /3 /10 Discussion Discussion Q7 /2 Q8 /2 Totals Points /4 TOTAL /20

Introduction The scenario you will be studying is shown in the schematic below. Two balls of equal mass ( M 1 , M 2 ) collide. The first ball begins with velocity V 1,i and the second ball is stationary. After the collision, the balls have velocities V 1,f and V 2,f . We can assume momentum is conserved. However, we cannot say for certain whether energy is conserved. Question 1 : In what circumstances would energy NOT be conserved? Would this be an elastic or inelastic collision? If energy is not conserved, where does it go? Mechanical energy is an excellent example of energy which is not always conserved. Mechanical energy would always be conserved if there is no frictional force or air resistance (external forces). For example, when one ball hits another, it will eventually stop, this is due to friction, without it, Newton’s first law indicates that it would roll forever. The kinetic energy in the mechanical system is lost due to friction making mechanical energy equal zero. If you consider the entire universe in your system, then it is conserved. The loss of energy is an example of inelastic collision has the energy is transferred into friction and lost to the system. In the second part of this lab you will be building an experiment to investigate conservation of momentum, whether the collision is elastic or inelastic, and how much energy is conserved.

Question 2 : Referring to a system like the one above where everything is in 2-dimensions (like a pool table), explain what physical parameters you would want to measure in order to calculate the momentum of the system. You can assume M 1 = M 2 . In order to the calculate the momentum of a system in 2D, we must know the mass of the balls, the velocity at which they are hit (v initial and v final for both balls). It would also be helpful to find the angle at which the first ball is hit at the second and from a district point of reference in order to calculate the x and y components of the system. Question 3 : Now, explain how you could use the measurements you described in Question 2 to test how elastic or inelastic the collision is. Make sure your answer includes a definition of what you mean by elastic and inelastic. Using the velocity and the mass of ball one and then comparing it to ball two will indicate if there was a loss of kinetic energy (inelastic) or if they have equal momentum (elastic). This should be very obvious as the masses are the same and therefore should theoretically have the same momentum if it is elastic collision. Unfortunately, you will find there are some complications that you will need to overcome or otherwise account for in order to perform an experiment like this at home. The following section will describe how we would like you to perform the experiment for this lab. BEFORE CONTINUING, BE SURE YOU HAVE COMPLETED 3 DISCUSSION QUESTIONS. Procedure ***No rigorous error analysis will be done for this experiment. In your discussion you may want to consider sources of error/approximations you had to make during your experiment/analysis.*** Determining V 1,i One complication you may have is with measuring velocity with only household items. Instead of measuring velocity directly, you will use what you know about projectile motion and free-fall to determine the velocities of the different balls. (Remember, velocity is a vector quantity, so you will have to account for the horizontal component (2D) and the vertical component.) The figure below is a schematic outlining how you can determine V 1,i . Follow the procedure described in VIDEO 1 , then use your results to fill in Table 1.

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Δ h =0.7873m a =9.81m/s^2 t =0.40seconds Note: v = p/m Trial Δ d [m] v = Δ d/t [m/s] KE 0 /m= ½ v 2 [J/kg] 1 0.27m 0.675m/s 0.2278J/kg 2 0.26m 0.65m/s 0.2113J/kg 3 0.26m 0.65m/s 0.2113J/kg Avg 0.263m 0.658m/s 0.2165J/kg You may have noticed that your chart does not include a column for momentum. This is because the mass of your ball is unknown. However, since the mass is constant throughout the experiment, the velocity you calculate will be proportional to the momentum of your ball. Likewise, with kinetic energy. Therefore, you can use velocity in place of momentum and KE/m to compare energy before and after collisions. You can now use this same technique to find the velocity of two balls after they collide.

Question 4: Submit a photograph of your experiment.

Follow directions outlined in Video 2 and Video 3 to fill in the following table. KE 0 /m= 0.2165J/kg t =1.20 Note: KE tot = KE 1 + KE 2 Trial v x [m/s] v y [m/s] v [m/s] KE / m = ½ v 2 [J/kg] KE tot / KE 0 [1] Σ v x , Σ v y [m/s, m/s] x Ball 1 -0.139 0.230 0.192m/s 0.018J/kg KE1+KE2/KE0 0.163 0.019,0.397 Ball 2 0.158 0.167 0.1859m/s 0.0173J/kg 2 Ball 1 -0.210 0.239 0.267m/s 0.0356J/kg 0.371 0.059,0.411 Ball 2 0.269 0.172 0.299 m/s 0.0447J/kg 3 Ball 1 -0.199 0.249 0.261m/s 0.0341J/kg 0.325 0.046,0.405 Ball 2 0.245 0.156 0.269m/s 0.0362 J/kg Avg Ball 1 -0.1826 0.239 0.24m/s 0.0292J/kg 0.286 0.0413,0.4043 Ball 2 0.224 0.165 0.25m/s 0.0327J/kg Question 5 : Use the velocities that you calculated in your chart for both balls to determine if momentum is conserved. Report your final momentum as a percent of your initial momentum. Is this what you expected? Why or why not? P2/P1=(V1F+V2F)/(V1I+V2I) =(0.24+0.25)/( 0.658) = 0.745

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The velocities calculated in my data show that momentum is not conserved. This will always be the case when conducting this experiment as there are external forces such as air resistance, gravity and friction action on both balls. However, there is also a lack of accuracy when measuring the angle at which each ball falls and an approximate reading of cm. One way to improve this lab and receive more accurate answers would be to have cameras monitoring the experiment in order to get the exact position as to where each ball dropped respectively. The final momentum is still 74.5% which means that collision was inelastic. Question 6 : Use the velocities for both balls that you calculated in your chart to determine if energy is conserved. If energy is lost, what percentage is lost? E lost %=100%-(KE tot /KE 0 )x(100%) E lost %=100%-[((1/2)v 1f 2 +(1/2)v 2f 2 )/((1/2)v 1i 2 +(1/2)v 2i 2 )]x(100%) E lost %=100%-[((1/2) 0.24 2 +(1/2) 0.25 2 )/((1/2) 0.658 2 +0))x(100%) E lost %=100%-(0.277)x100% E lost %= 72.3% The velocity of both balls calculates determine that energy is not conserved. This is because when the balls collide on the way down the ramp, there is a loss of energy due to friction against the surface of the ramp. The amount of energy lost was 72.3%. BEFORE CONTINUING, BE SURE YOU HAVE COMPLETED 3 DISCUSSION QUESTIONS (INCLUDING 1 PHOTO) AND 2 TABLES.

Discussion If you were unable to find two identical balls, you would have obtained different results. In particular, will no longer be able to claim M 1 = M 2 . Question 7 : How would your results change if M 1 = 2 M 2 ? Your results would change as it would not give you the same kinetic energy or momentum outcome. It would be more difficult to find if the system is elastic or inelastic collision. Mass and velocity are both directly proportional to momentum; as such, if you double your mass you are also doubling your momentum and therefore would change your kinetic energy by doubling its value. Question 8 : Now, consider what would have happened if M 1 = M 2 but one ball was made of a different material such that twice as much energy was lost. How would this affect your results? The change in kinetic energy (loss) would affect the results of kinetic energy and momentum. They are both reliant on mass and velocity therefore momentum would likely change as well. Since mass is constant during this experiment but kinetic energy is reliant on mass and velocity, there is a loss of kinetic energy therefore decreasing the velocity and momentum. BEFORE CONTINUING, BE SURE YOU HAVE COMPLETED 2 DISCUSSION QUESTIONS.

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