Isolating Variable in Equation for conservation of momentum

  • #1
jxj
18
2

Homework Statement


So the problem is trying to isolate mA
equations-of-conservation-of-momentum-energy.png
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



equations-of-conservation-of-momentum-energy.png

The Attempt at a Solution


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

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Answers and Replies

  • #2
DrClaude
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Start by putting all terms containing ##m_\mathrm{A}## on the left-hand side, and all other terms on the right-hand side.
 
  • #3
haruspex
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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.
 
  • #4
jxj
18
2
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?
 
  • #5
jxj
18
2
You may be misinterpreting that. Ignore it and see where you get to.
ok thanks
 
  • #6
DrClaude
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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,
$$
\begin{align*}
5x - 3 -2x +3 &= 2x + 1 - 2x + 3 \\
3x &= 4
\end{align*}
$$
 
  • #7
jxj
18
2
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?
 
  • #8
jxj
18
2
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?
 
  • #9
verty
Homework Helper
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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.
 
  • #10
haruspex
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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.
 
  • #11
jxj
18
2
Factorise each side.
when you say factorise could you explain a little more please
 
  • #12
DrClaude
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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.
 
  • #13
jxj
18
2
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!
 
  • #14
jxj
18
2
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?
 
  • #15
DrClaude
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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.
 
  • #16
jxj
18
2
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) ?
 
  • #17
DrClaude
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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:
$$
a (b+c) = ab + ac
$$
 

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