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

[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.

 

[Samy_A]

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.

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.

 

[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?

 

[Samy_A]

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?

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.

 

[Obliv]

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.

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 ${v_2}^2 + {v_1}^2 - 2{v_1}{v_2} \ne 0$

 

[Samy_A]

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 ${v_2}^2 + {v_1}^2 - 2{v_1}{v_2} \ne 0$

"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).

 

[Obliv]

The masses are positive numbers.
So, 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).

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?

 

[Samy_A]

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?

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 $v_1 \ne v_2$.

 

[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 :)

 

[Samy_A]

You are welcome.