Question

a) What is the ratio of the momentum of a 2.3x10^5 kg airplane traveling at 960 km/h to that of a 1.1 kg ball thrown at 11.3 m/s?
b) What is the ratio of the respective kinetic energies?

Answer 1

Would appreciate an explanation.

Answer 2

a) Well, a ratio is one divided by the other, and their respective momenta are given by
\[ p=mv\]
remember that you need 960 km/h = 266.67 m/s for an accurate comparison

b) Kinetic energies are given by
\[K=\frac{1}{2}mv^2\]

Notice that the velocity term is squared in kinetic energy, so the ratio of the plane's K.E. to the baseball's should be much! larger than the ratio of their momenta

Answer 3

\[ \frac{p_{plane}}{p_{ball}} = \frac{(m_{plane})(v_{plane})}{(m_{ball})(v_{ball})} \]
\[ \frac{K_{plane}}{K_{ball}} = \frac{(m_{plane})(v^2_{plane})}{(m_{ball})(v^2_{ball})} \]