A Ratio in a System of Equations

A ratio in a system of equations

Problem

The kinetic energy, e, of a object of mass m moving at speed v is equal to

e=1/2mv_{2}^{2}.

Two objects with the same mass are moving, one twice as fast as the other. What is the ratio of the kinetic energy of the faster object to the kinetic energy of the slower one?

Solution

Call the faster object’s energy e_{2} and its speed v_{2}; call the slower one’s energy e_{1} and its speed v_{1}. You need to find the ratio of the kinetic energies,  \frac{e_{2}}{e_{1}}=\frac{1/2mv_{2}^{2}}{1/2mv_{1}^{2}}=\frac{v_{2}^{2}}{v_{1}^{2}}. And you know that v_{2}=2v_{1}. Substitute that into the equation for the ratio:

\frac{e_{2}}{e_{1}}=\frac{v^{2}_{2}}{v^{2}_{1}}=\frac{(2v_{1})^{2}}{v^{2}_{1}}=4.

The object moving twice as fast has four times as much energy as the slower object.

Why work the problem in terms of the lower speed rather than the higher one? It’s equally valid either way. But getting the lower speed into your calculation will make things go better. Look what happens if you work in terms of the higher speed:

{\color{DarkBlue} v_{1}=\frac{v_{2}}{2}}

{\color{DarkBlue} \frac{e_2}{e_1}=\frac{1/2mv_2^2}{1/2 m (1/2v_2)^2}=\frac{1}{\frac{1}{4}}=4}

The fractions make the second way just a little messier.

Rule of thumb: When you substitute, things usually go better if you choose your substitution to avoid fractions and negative numbers. You could choose to work with fractions and still get the right answer, but you’d probably have to do more work.

Note that doubling the speed quadruples the energy. That’s because speed term gets squared; energy is increased in proportion to the square of the term by which speed is increased. If you tripled the speed you would multiply the energy by nine.