# B Need help with algebra in proving kinetic energy is not conserved

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1. Mar 23, 2016

### Obliv

I'm trying to prove that kinetic energy is not conserved in inelastic collisions using the conservation of momentum. This is the set-up. An object A of momentum ${m_1}{v_1}$ collides inelastically with object B of momentum ${m_2}{v_2}$
using momentum conservation $P_i = P_f$
$${m_1}{v_1} + {m_2}{v_2} = ({m_1}+{m_2}){v_f}$$ solving for $v_f$ we obtain
$$v_f = \frac{({m_1}{v_1}+{m_2}{v_2})}{({m_1}+{m_2})}$$ now the fun part
prove that $$KE_i \ne KE_f$$
$$\frac{{m_1}{{v_1}^2}}{2} + \frac{{m_2}{{v_2}^2}}{2} \ne \frac{1}{2}({m_1} + {m_2})(\frac{{m_1}{v_1} + {m_2}{v_2}}{{m_1}+{m_2}})^2$$

I realize that the two quantities are no longer equal, especially if you try plugging in numbers. I was just wondering if this can be simplified to be seen more clearly.
I multiplied out the numerator and denominator and got some pretty nasty algebra.
$$\frac{{m_1}{{v_1}^2}}{2} + \frac{{m_2}{{v_2}^2}}{2} \ne \frac{1}{2}({m_1}+{m_2})\frac{({m_1}{v_1})^2 + 2{m_1}{v_1}{m_2}{v_2} + ({m_2}{v_2})^2}{{m_1}^2 + 2{m_1}{m_2} + {m_2}^2}$$
I'll continue working on it later but if anyone has any shortcuts to simplifying this I would appreciate it.

2. Mar 23, 2016

### Samy_A

You could look at the expression $$X=\frac{{m_1}{{v_1}^2}}{2} + \frac{{m_2}{{v_2}^2}}{2} - \frac{1}{2}({m_1} + {m_2})(\frac{{m_1}{v_1} + {m_2}{v_2}}{{m_1}+{m_2}})^2$$
You want to see under what condition $X \neq 0$.

Simplify the expression by multiplying with $2(m_1+m_2)$:
You get $$2(m_1+m_2)X=({m_1}{{v_1}^2}+ {m_2}{{v_2}^2})(m_1+m_2) - (m_1v_1+m_2v_2)^2$$

Now work out the RHS.

3. Mar 24, 2016

### Obliv

Okay I continued and got this
$$2({m_1}+{m_2})X = ({m_1}{v_1}^2+{m_2}{v_2}^2)({m_1}+{m_2}) - ({m_1}{v_1} + {m_2}{v_2})^2$$
$$2({m_1}+{m_2})X = ({m_1}{v_1})^2 + {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1}^2 + ({m_2}{v_2})^2 - (({m_1}{v_1})^2 + 2{m_1}{v_1}{m_2}{v_2} + ({m_2}{v_2})^2)$$
Which simplifies to $$2({m_1}+{m_2})X = {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1} - 2{m_1}{v_1}{m_2}{v_2}$$
and if you take out an $({m_1}{m_2})$
$$2({m_1} + {m_2})X = ({m_1}{m_2})({v_2}^2 + {v_1} + 2{v_1}{v_2})$$ and we can conclude now that $X \ne 0$ when neither ${m_2} \ne 0$ nor ${m_1} \ne 0$ or if ${v_2}^2 + {v_1} + 2{v_1}{v_2} \ne 0$

Is this correct?

4. Mar 24, 2016

### Samy_A

No.
The result has to be symmetric in $m_1, m_2$, which it is.
But also in $v_1, v_2$, which your result is not.
Check the term ${m_1}{m_2}{v_1}$ ...
There also is a sign error in your last expression.

I think you can take as given that the masses $m_1, m_2$ are strictly positive numbers.

5. Mar 24, 2016

### Obliv

Yes I forgot a power of 2 $$2({m_1}+{m_2})X = {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1} - 2{m_1}{v_1}{m_2}{v_2}$$ here

$$2({m_1}+{m_2})X = {m_1}{m_2}{v_2}^2 + {m_1}{m_2}{v_1}^2 - 2{m_1}{v_1}{m_2}{v_2}$$

and then $$2({m_1} + {m_2})X = ({m_1}{m_2})({v_2}^2 + {v_1}^2 - 2{v_1}{v_2})$$

If this is correct, $X \ne 0$ when either ${m_1} > 0$ or ${m_2} > 0$ as well as if ${v_2}^2 + {v_1}^2 - 2{v_1}{v_2} \ne 0$

6. Mar 24, 2016

### Samy_A

"either ${m_1} > 0$ or ${m_2} > 0$" is not correct, as the masses are both positive numbers.

For $X \ne 0$, the condition is ${v_2}^2 + {v_1}^2 - 2{v_1}{v_2} \ne 0$.
This last expression can be simplified, using the well known identity $(a-b)²=a²-2ab+b²$.
You then will get a nice condition for $X \ne 0$, which will make sense if we remember that $X$ is the difference in kinetic energy (before and after the collision).

Last edited: Mar 24, 2016
7. Mar 24, 2016

### Obliv

so it can be written as $$2({m_1} + {m_2})X = ({m_1}{m_2})({v_1}^2 - {v_2}^2)$$ and expressed such that $X \ne 0$ when (${m_1} > 0$ and ${m_2} > 0$) and ${v_2} \ne {v_1}$? is this correct or should I go further?

8. Mar 25, 2016

### Samy_A

No, this is not totally correct.
From $2({m_1} + {m_2})X = ({m_1}{m_2})({v_2}^2 + {v_1}^2 - 2{v_1}{v_2})$, you get $2({m_1} + {m_2})X = {m_1}{m_2}(v_2-v_1)²$,
$(v_2-v_1)²$ is not necessarily equal to ${v_1}^2 - {v_2}^2$.

The conclusion is correct. The kinetic energies will be different when $(v_2-v_1)² \ne 0$, that is when $v_1 \ne v_2$.

9. Mar 25, 2016

### Obliv

I was very tired and wrote the tex incorrectly. I meant what you wrote and that's what I had on my scrap paper. Thank you for your guidance :)

10. Mar 25, 2016

### Samy_A

You are welcome.