Frank C. answered • 12/21/19

Professional and Passionate Math Tutor

I think you showed a perfect example of why we cannot solve for this without the initial values. Let's prove it!

Let K_{1} = (1/2)mv_{1}^{2}, and K_{2} = (1/2)mv_{2}^{2} for the respective objects that share the same mass m, and let K = (1/2)mv^{2}, where v = (v_{1} + v_{2})/2 is the average velocity and K is the kinetic energy of the average velocity. We want to prove that the kinetic energy of the average velocity is *not equal* to the average kinetic energy. Or otherwise stated: *K* ≠ (*K*_{1}* + K*_{2})/2

K = (1/2)mv^{2} = (1/2)m[(v_{1} + v_{2})/2]^{2} = (1/2)m(v_{1}^{2} + 2v_{1}v_{2} + v_{2}^{2})/4

(K_{1} + K_{2})/2 = [(1/2)mv_{1}^{2} + (1/2)mv_{2}^{2}]/2 = (1/2)m(v_{1}^{2} + v_{2}^{2})/2

So see how these two values are *not the same*. We do know the value of (v_{1}^{2} + 2v_{1}v_{2} + v_{2}^{2})/4, it's 64. But we're left with two variables in one equation, where you could solve for one velocity and plug it into (v_{1}^{2} + v_{2}^{2})/2 in terms of the other.

Frank C.

Here's an example of similar logic. If you had two objects of same mass traveling at equal speeds in opposite directions, their average velocity would be zero. If you used that to calculate kinetic energy, you'd get zero! But that is certainly not the kinetic energy of either object on its own; their kinetic energies are both non-zero and positive.12/21/19