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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? Im unsure
You may be misinterpreting that. Ignore it and see where you get to.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?Start by putting all terms containing ##m_\mathrm{A}## on the left-hand side, and all other terms on the right-hand side.
ok thanksYou 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 likewhen 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 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?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?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?
Factorise each side.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 pleaseFactorise each side.
thanks!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?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.
for that part, was I supposed to separate them so I would have mA(vA) -(mA)(v’A) ?Check that part I highlighted in red.