# Conservation of momentum equation, how do I factor out v0Final and v1Final?

I am trying to figure out an equation for conservation of momentum. So,

If combined momentum before and after the collision is the same, and momentum is velocity times mass, then for 2 objects (A,B), it will be:

$(m_0 \times v_0) + (m_1 \times v_1) = (m_0 \times v_0^{Final}) + (m_1 \times v_0^{Final})$

The final velocities for A,B are $v_0^{Final}$ and $v_1^{Final}$. The way to solve an equation that has 2 unknowns is to find another equation with the same 2 unknowns.

So we are using kinetic energy equation, $E_k = 0.5 \times m \times v^2$

If kinetic energy before and after a collision is the same it will be:

$E_{k,0} + E_{k,1} = E_{k,0}^{Final} + E_{k,1}^{Final}$

or

$(0.5 \times m_0 \times v_0^2) + (0.5 \times m_1 \times v_1^2) = (0.5 \times m_0 \times {v_0^{Final}}^2) + (0.5 \times m_1 \times {v_1^{Final}}^2)$

So, given:

$(m_0 \times v_0) + (m_1 \times v_1) = (m_0 \times v_0^{Final}) + (m_1 \times v_1^{Final})$

and

$(0.5 \times m_0 \times v_0^2) + (0.5 \times m_1 \times v_1^2) = (0.5 \times m_0 \times {v_0^{Final}}^2) + (0.5 \times m_1 \times {v_1^{Final}}^2)$

How do I factor out $v_0^{Final}$ and $v_1^{Final}$ and express it in $m_0$, $m_1$, $v_0$ and $v_1$? Can somebody show me the steps?

The answer:

$v_0^{Final} = \frac{(m_0 - m_1) \times v_0 + 2 \times m_1 \times v_1}{m_0 + m_1}$

$v_1^{Final} = \frac{(m_1 - m_0) \times v_1 + 2 \times m_0 \times v_0}{m_0 + m_1}$

Thanks! V.

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## 2 Answers

The easy way to do problems like this is to work in the centre of mass frame. The trick is to add a velocity that makes the total momentum zero, then calculate what happens in the collision, and finally add the centre of mass velocity back on.

Working in the centre of mass frame makes things easy because if the total momentum before the collision is zero the total momentum after the collision must be zero as well. That gives you the useful relation:

$$m_0v_{final 0} + m_1v_{final 1} = 0$$

You can calculate the velocity of the centre of mass frame by requiring that the initial momentum be zero:

$$m_0(v_0 + v_{com}) + m_1(v_1 + v_{com}) = 0$$

which gives you:

$$v_{com} = - \frac {m_0v_0 + m_1v_1}{m_0 + m_1}$$

More info: in the centre of mass frame the collision looks like:

because we are in the centre of mass frame we know the momentum after the collision is zero, so:

$$m_1 v_1 + m_2 v_2 = 0$$

the kinetic energy is:

$$\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = E$$

where E is the initial kinetic energy ($1/2m_1u_1^2 + 1/2m_2u_2^2$) so you can use the first equation to substitute into the second e.g. you can rearrange the first equation to be:

$$v_1 = -\frac{m_2}{m_1} v_2$$

Use this to substitute for $v_1$ in the second equation and after some juggling it gives you:

$$\frac{1}{2} m_2 \left( \frac{m_2}{m_1} + 1 \right) v_2^2 = E$$

So now you have $v_2$ you can substitute back in the first equation to get $v_1$. Now just subtract off the centre of mass velocity to get back to your initial frame.

You could have done the calculation without using the centre of mass frame, but it makes the algebra easier. NB be careful about the signs of the velocities; remember that velocities in opposite directions have opposite signs. It's easy to lose a minus sign and get the wrong answer!

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Emmm... I still don't quite get it. How's that centre of mass frame supposes to help factor out v0Final and v1Final from the 2 equations? (m0 x v0) + (m1 x v1) = (m0 x v0Final) + (m1 x v1Final) and (0.5 x m0 x v0^2) + (0.5 x m1 x v1^2) = (0.5 x m0 x v0Final^2) + (0.5 x m1 x v1Final^2) –  Vennsoh Jun 2 '12 at 1:15

From conservation of energy:

$$m_1 (v_1^2 - v_{1f}^2) + m_2 (v_2^2 - v_{2f}^2) = 0$$

$$m_1 \Delta v_1 (v_1 + v_{1f}) + m_2 \Delta v_2(v_2 + v_{2f}) = 0$$

where $\Delta v_1 = v_1 - v_{1f}$ and $\Delta v_2 = v_2 - v_{2f}$.

But $m_1\Delta v_1 = - m_2\Delta v_2$ since this is the momentum transferred between the bodies $\Delta P$ by conservation of momentum, so:

$$\Delta P ( v_1 + v_1^f - v_2 - v_2^f) = 0$$

and this is the factorization you are looking for. The reason it exists is because you automatically know one solution to the quadratic equation--- the particles pass through each other without interacting ($\Delta P$ is zero). This means that the equation must factor out for this root.

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