- #1
jxj
- 18
- 2
Homework Statement
So the problem is trying to isolate mA
Homework Equations
The Attempt at a Solution
am pretty lost. I believe you have to divide the primes? I am unsure
You may be misinterpreting that. Ignore it and see where you get to.jxj said:My teacher said because the vA and vB on the right were prime they could not be combined
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?DrClaude said:Start by putting all terms containing ##m_\mathrm{A}## on the left-hand side, and all other terms on the right-hand side.
ok thanksharuspex said:You may be misinterpreting that. Ignore it and see where you get to.
No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation likejxj said: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?
ok so right now I am 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?DrClaude said: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?DrClaude said: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 said: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 said:I tried putting all of mA on the left and I got to mAvA - mAvA (prime) = mBvB (prime) - mBvB did I mess up?
when you say factorise could you explain a little more pleaseharuspex said:Factorise each side.
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 said:when you say factorise could you explain a little more please
thanks!DrClaude said: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 I am 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 said: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.
Check that part I highlighted in red.jxj said:so right now I am 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.
for that part, was I supposed to separate them so I would have mA(vA) -(mA)(v’A) ?DrClaude said:Check that part I highlighted in red.
No, you want a single ##m_\mathrm{A}##. You need to do the inverse of distribution:jxj said:for that part, was I supposed to separate them so I would have mA(vA) -(mA)(v’A) ?
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