91Ó°ÊÓ

To solve this problem consistently, we need to convert velocities from miles per hour \(\mathrm{mi} / \mathrm{h}\) to meters per second \(\mathrm{m} / \mathrm{s}\). Using the conversion factor, 1 mile = 1609.34 meters and 1 hour = 3600 seconds, we obtain the initial and final velocities in SI units. Initial velocity (\(v_1\)): \(20.0\frac{\mathrm{mi}}{\mathrm{h}}\times\frac{1609.34\mathrm{m}}{1\mathrm{mi}}\times\frac{1\mathrm{h}}{3600\mathrm{s}}\approx8.94\frac{\mathrm{m}}{\mathrm{s}}\) Final velocity (\(v_2\)): \(60.0\frac{\mathrm{mi}}{\mathrm{h}}\times\frac{1609.34\mathrm{m}}{1\mathrm{mi}}\times\frac{1\mathrm{h}}{3600\mathrm{s}}\approx26.82\frac{\mathrm{m}}{\mathrm{s}}\)

First, we find the initial and final momentum magnitudes using the formula \(p = m*v\). Then, we calculate the ratio of final to initial momentum. We will not use actual mass, as it will cancel out in the ratio. Initial momentum magnitude (\(p_1\)): \(m * 8.94\frac{\mathrm{m}}{\mathrm{s}}\) Final momentum magnitude (\(p_2\)): \(m * 26.82\frac{\mathrm{m}}{\mathrm{s}}\) Ratio of final to initial momentum: \(\frac{p_2}{p_1}=\frac{m * 26.82}{m * 8.94}=\frac{26.82}{8.94}\approx3\)

First, we find the initial and final kinetic energy magnitudes using the formula \(KE = \frac{1}{2}mv^2\). Then, we calculate the ratio of final to initial kinetic energy. Again, we will not use actual mass, as it will cancel out in the ratio. Initial kinetic energy magnitude (\(KE_1\)): \(\frac{1}{2}m * (8.94\frac{\mathrm{m}}{\mathrm{s}})^2\) Final kinetic energy magnitude (\(KE_2\)): \(\frac{1}{2}m * (26.82\frac{\mathrm{m}}{\mathrm{s}})^2\) Ratio of final to initial kinetic energy: \(\frac{KE_2}{KE_1}=\frac{\frac{1}{2}m * (26.82\frac{\mathrm{m}}{\mathrm{s}})^2}{\frac{1}{2}m * (8.94\frac{\mathrm{m}}{\mathrm{s}})^2}=\frac{(26.82\frac{\mathrm{m}}{\mathrm{s}})^2}{(8.94\frac{\mathrm{m}}{\mathrm{s}})^2}\approx9\) #Conclusion#: (a) The ratio of the final to the initial magnitude of its momentum : 3 (b) The ratio of the final to the initial kinetic energy: 9