# Isolating Variable in Equation for conservation of momentum

• jxj
No, you want a single ##m_\mathrm{A}##. You need to do the inverse of distribution:$$a (b+c) = ab +...+ ac #### jxj ## Homework Statement So the problem is trying to isolate mA in the equation for momentum (only focusing on top formula, not bottom hehe) basically by solving the equation I assume. My teacher said because the vA and vB on the right were prime they could not be combined so I'm having trouble computing it. Please help! ## Homework Equations ## The Attempt at a Solution am pretty lost. I believe you have to divide the primes? I am unsure #### Attachments Start by putting all terms containing ##m_\mathrm{A}## on the left-hand side, and all other terms on the right-hand side. • jxj jxj said: My teacher said because the vA and vB on the right were prime they could not be combined You may be misinterpreting that. Ignore it and see where you get to. • jxj DrClaude said: Start by putting all terms containing ##m_\mathrm{A}## on the left-hand side, and all other terms on the right-hand side. when doing this, don't you divide the ##m_\mathrm{A}## plus ##v_\mathrm{A}## prime on the right by its counterpart on the left first? even though it contains a prime? haruspex said: You may be misinterpreting that. Ignore it and see where you get to. ok thanks jxj said: when doing this, don't you divide the ##m_\mathrm{A}## plus ##v_\mathrm{A}## prime on the right by its counterpart on the left first? No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like$$
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*}
$$DrClaude said: No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like$$
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*}
$$ok so right now I am a little confused because unlike the example you gave with 5x, the formula is essentially equal on both sides minus the primes. in my mind I would try to cross out the mAvA on the right, is that right? DrClaude said: No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like$$
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? 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? 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. • jxj 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 haruspex said: Factorise each side. when you say factorise could you explain a little more please jxj said: 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. 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. 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? 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. Check that part I highlighted in red. • jxj DrClaude said: 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) ? jxj said: 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


• jxj