Compute the kinetic energy1 before impact. K_b = 1/2mv^2 = 1/2 times 0.6904 time

Question

Compute the kinetic energy1 before impact. K_b = 1/2mv^2 = 1/2 times 0.6904 times 5.97^2 = 12.30 Compute the kinetic energy after impact. K_a = 1/2(m + M)u^2 = 1/2 times (0.6904 + 2.7062) times 1.18^2 = 2.364 Compute the experimental ratio of kinetic energy before to kinetic energy after the collision. K_b/K_a = 12.30/2.364 = 5.20 Using the average mass of the ball and the mass of the pendulum determine the theoretical ratio of the kinetic energy before collision to that of the kinetic energy after collision. (This ratio is the answer to one of your pre-questions. If you don't believe that you have the proper ratio, ask your lab instructor for it.) Theoretical Ratio = m + M/m = 0.6904 + 2.7062/0.6904 = 4.9 Compare the kinetic energy of the system before impact with the kinetic energy of the system after impact. How much energy was lost? What happened to it? How well does the experimental ratio agree with the theoretical ratio of the kinetic energy before impact to the kinetic energy-after impact?

Explanation / Answer

Loss in kinetic energy = K.E before impact - K.E after impact

                                 = 12.3 - 2.364

                                  = 9.936 J

The loss in kinetic energy is converted into heat energy.

Theoretical ratio of kinetic energy before impact to the kinetic nergy after impact

    = (1/2) mv 2 / [(1/2) (m+M) u 2]

    = mv 2 / [(m+M) u 2]

From law of conservation of momentum , mv = (m+ M) u

                                                               v = [(m+M)u / m]

So, Theoretical ratio of kinetic energy before impact to the kinetic nergy after impact

= mv 2 / [(m+M) u 2]     

= m{(m+M)u / m} 2 / [(m+M) u 2]  

= [(m+M) 2 u 2 / m ]/[(m+M) u 2]  

=(m+M) 2 u 2/m(m+M) u 2

= (m+M) / m

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