½(m_t+m_b) v² = ½ m_tv_t² + ½ m_bv_b²

Linear Regresion: distance = A + Bt + Ct²

First derivative: velocity = B + 2Ct

Time was the point right before and after the collision.

v_{bi =} -3.9006m/s + 2 (-3.9216m/s²)(.135s) = -4.959432m/s

v_{ti =} -3.6323m/s + 2 ( -4.926m/s²)(.135s) = -4.962482m/s

v_{bf =} 6.7297m/s + 2 ( -5.3556m/s²)(.335s)= 3.141448m/s

v_{tf =} 10.691m/s + 2 ( -3.8818m/s²)(.335s)= 8.090194m/s

½ [(.0622kg) + (.5939kg)](24.61m/s)² = ½ (.0622kg)(65.45m/s)² + ½ (.5939kg)(9.87m/s)²

16.146 J = 9.932 J

(9.932 – 16.146) / (9.932 + 16.146) * 100 = -23.82 % difference

Analysis: The initial velocities of the two balls are equal to the nearest hundredth. Also there is significant difference in the velocity of the tennis ball.

v_{bi =} -2.5814m/s + 2 (-4.6557m/s²)(.2683s) = -5.07m/s

v_{ti =} -2.57515m/s + 2 ( -4.6384m/s²)(.2683s) = -5.06m/s

v_{bf =} 8.9655m/s + 2 ( -8.5093m/s²)(.335s)= 3.2642m/s

v_{tf =} 13.706m/s + 2 ( -7.5525m/s²)(.335s)= 8.645825m/s

½[(.0622kg) + (.5939kg)](25.654m/s)² = ½ (.0622kg)(74.75m/s)² + ½(.5939kg)(10.65m/s)²

16.825J = 10.974J

(10.974 – 16.825)/ (10.974 + 16.825) * 100 = -21.04 % difference

Analysis: The results are very similar to those in trial one. The initial velocities of both balls are almost the same as in trial one. The energy also comes out very similar to the values in trial one.