Isolating Variable in Equation for conservation of momentum

[jxj]

Homework Statement

So the problem is trying to isolate mA

in the equation for momentum (only focusing on top formula, not bottom hehe) basically by solving the equation I assume. My teacher said because the vA and vB on the right were prime they could not be combined so I'm having trouble computing it. Please help!

Homework Equations

The Attempt at a Solution

am pretty lost. I believe you have to divide the primes? Im unsure

 

[DrClaude]

Start by putting all terms containing $m_\mathrm{A}$ on the left-hand side, and all other terms on the right-hand side.

 

[haruspex]

My teacher said because the vA and vB on the right were prime they could not be combined

You may be misinterpreting that. Ignore it and see where you get to.

 

[jxj]

Start by putting all terms containing $m_\mathrm{A}$ on the left-hand side, and all other terms on the right-hand side.

when doing this, don't you divide the $m_\mathrm{A}$ plus $v_\mathrm{A}$ prime on the right by its counterpart on the left first? even though it contains a prime?

 

[jxj]

You may be misinterpreting that. Ignore it and see where you get to.

ok thanks

 

[DrClaude]

when doing this, don't you divide the $m_\mathrm{A}$ plus $v_\mathrm{A}$ prime on the right by its counterpart on the left first?

No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like
$$
5x -3 = 2x + 1
$$
the first step is to try and gather all terms in $x$ together, and leave the rest on the other side,
$$
\begin{align*}
5x - 3 -2x +3 &= 2x + 1 - 2x + 3 \\
3x &= 4
\end{align*}
$$

 

[jxj]

No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like
$$
5x -3 = 2x + 1
$$
the first step is to try and gather all terms in $x$ together, and leave the rest on the other side,
$$
\begin{align*}
5x - 3 -2x +3 &= 2x + 1 - 2x + 3 \\
3x &= 4
\end{align*}
$$

ok so right now im a little confused because unlike the example you gave with 5x, the formula is essentially equal on both sides minus the primes. in my mind I would try to cross out the mAvA on the right, is that right?

 

[jxj]

No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like
$$
5x -3 = 2x + 1
$$
the first step is to try and gather all terms in $x$ together, and leave the rest on the other side,
$$
\begin{align*}
5x - 3 -2x +3 &= 2x + 1 - 2x + 3 \\
3x &= 4
\end{align*}
$$

I tried putting all of mA on the left and I got to mAvA - mAvA (prime) = mBvB (prime) - mBvB did I mess up?

 

[verty]

I tried putting all of mA on the left and I got to mAvA - mAvA (prime) = mBvB (prime) - mBvB did I mess up?

No, that's correct. How you want to think of it is doing the same thing to both sides. So to move "m_a v'_a" to the left, subtract it from both sides. Or if it was multiplied by that, you would divide it on both sides.

 

[haruspex]

I tried putting all of mA on the left and I got to mAvA - mAvA (prime) = mBvB (prime) - mBvB did I mess up?

Factorise each side.

 

[jxj]

Factorise each side.

when you say factorise could you explain a little more please

 

[DrClaude]

when you say factorise could you explain a little more please

You want to go from $m_\mathrm{A} v_\mathrm{A} - m_\mathrm{A} v_\mathrm{A}'$ to an expression where $m_\mathrm{A}$ appears only once, using the distributive property of multiplication.

 

[jxj]

You want to go from $m_\mathrm{A} v_\mathrm{A} - m_\mathrm{A} v_\mathrm{A}'$ to an expression where $m_\mathrm{A}$ appears only once, using the distributive property of multiplication.

thanks!

 

[jxj]

You want to go from $m_\mathrm{A} v_\mathrm{A} - m_\mathrm{A} v_\mathrm{A}'$ to an expression where $m_\mathrm{A}$ appears only once, using the distributive property of multiplication.

so right now im at distributing mA by itself into (vA - mAv’A) on the left of the equal sign and vice versa for mB on the right of the equal sign. am I on the right track?

 

[DrClaude]

so right now im at distributing mA by itself into (vA - mAv’A) on the left of the equal sign and vice versa for mB on the right of the equal sign.

Check that part I highlighted in red.

 

[jxj]

Check that part I highlighted in red.

for that part, was I supposed to separate them so I would have mA(vA) -(mA)(v’A) ?

 

[DrClaude]

for that part, was I supposed to separate them so I would have mA(vA) -(mA)(v’A) ?

No, you want a single $m_\mathrm{A}$. You need to do the inverse of distribution:
$$
a (b+c) = ab + ac
$$