# How to establish conservation of kinetic energy in elastic collision?

I was doing my Physics 1 one homework about collisions in one dimension using 2 objects. It seemed as though they wanted us to prove using our data that kinetic energy is conserved in a completely elastic collision.

From our data we found that if you add the |v| for the two objects, you will get the same value before and after the collision takes place:

We also noted that the total mass does not change (obviously)

We then concluded this (each variable represents the sum for the system):

I then noticed that this statement is not true because these two values are not the same:

$$\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} \neq \frac{1}{2}\left(m_{1}+m_{2}\right)\left(\left|v_{1}\right|+\left|v_{2}\right|\right)^{2}$$

Now let's suppose that m1 and m2 are the masses of 2 objects, and also that u1 and u2 are those objects' velocities before a collision and v1 and v2 are the objects' velocities before and after that collision.

My question is then, given that the total momentum of the system is shown by the data to be conserved (in other words $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$) and it is also shown that $$\left|u_{1}\right|+\left|u_{2}\right|=\left|v_{1}\right|+\left|v_{2}\right|$$, how would one prove that: $$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}=\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$$

Because all I proved is that:$$\frac{1}{2}\left(m_{1}+m_{2}\right)\left(\left|u_{1}\right|+\left|u_{2}\right|\right)^{2}=\frac{1}{2}\left(m_{1}+m_{2}\right)\left(\left|v_{1}\right|+\left|v_{2}\right|\right)^{2}$$

• Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! Mar 24 at 18:39
• Watch the algebra. $a^2 + b^2$ IS NOT EQUAL TO $(a+b)^2$. Mar 24 at 18:39
• In Desmos, when you copy an entry into your clipboard, it is already in LaTeX. When pasted here, just surround the pasted item by a dollar-sign at the start and at the end (for an inline item) or two-dollar-signs at the start and at the end (for a centered-display item). Mar 24 at 18:47
• Edited, sorry for the bad formatting. Mar 24 at 19:00
• @DavidWhite, I understand that, I hope my edits make that more clear. My problem was more that I didn't realize my faulty algebra within my conclusion until I wrote it out in terms of v1 and v2. Mar 24 at 19:13

You are not deriving the conservation of kinetic energy from an elastic collision. That would be redundant since the definition of an elastic collision is one that conserves kinetic energy. Instead, you use the measurements in the table to show that this collision conserved kinetic energy--and was therefore elastic. Calculate the total kinetic energy before the collision ($$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2$$) and after the collision ($$\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$$) and see if they are the same.