## Chemistry: The Central Science (13th Edition)

We first had to solve for the kinetic energy of the ball when it was going 95 mph (42.5 m/s) in part A. This was calculated to be 131. J. Once we did that, we can then solve for the kinetic energy of the ball when it is going 55 mph. To do this (this uses the same procedure we used to solve for the kinetic energy in part A, except now the velocity will be different): first convert the 55 mph into m/s using unit analysis. We get velocity = 24.6 m/s. Then using the kinetic energy equation, KE = $\frac{1}{2}$$mv^{2}$, we substitute in the mass of the ball in kg (which we obtained from dimensional analysis in part A to be 0.145 kg) and the velocity squared times 1/2. We get our kinetic energy to be 43.8 J. The final step is to calculate the ratio between the kinetic energy from part A and the kinetic energy from part B. To do this, we simply divide 131. J by 43.8 J to get 2.99086757 which is approximately 3. This means that the kinetic energy decreased by a factor of about 3. 