Elastic Collisions

$$\vec{p}_{1i}+\vec{p}_{2i} = \vec{p}_{1f} + \vec{p}_{2f}$$

$$m_1v_{1i}+m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$ (4.42)

$$E_{k1i} + E_{k2i} = E_{k1f} + E_{k2f}$$

$$\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$$ (4.43)

are the two relations that follow from momentum and energy conservation (in the one, e.g. - x - direction). Assuming we know $m_1,m_2,v_{1i}$ and $v_{2i}$, can we find $v_{1f}$ and $v_{2f}$?

Yes. There are three ways to proceed. One is to directly solve these two equations simultaneously. This involves solving an annoying quadratic and we will avoid it. The second is to note that we can do the following rearrangement of the energy and momentum equations:

$$m_1 v_{1i}^2 - m_1 v_{1f}^2 = m_2 v_{2f}^2 - m_2 v_{2i}^2$$

$$m_1(v_{1i} - v_{1f})(v_{1i} + v_{1f}) = m_2(v_{2f} - v_{2i})(v_{2i} + v_{2f})$$ (4.44)

(from energy conservation) and

$$m_1(v_{1i} - v_{2f}) = m_2(v_{2f} - v_{2i})$$ (4.45)

(from momentum conservation). When we divide the first of these by the second, we get:

$$(v_{1i} + v_{1f}) = (v_{2i} + v_{2f})$$

$$(v_{2f} - v_{1f}) = -(v_{2i} - v_{1i})$$ (4.46)

or the relative speed of approach before a collision equals the relative speed of recession after a collision.

Although it isn't obvious, this equation is independent from the momentum conservation equation and can be used with it to solve for $v_{1f}$ and $v_{2f}$, e.g. -

$$v_{2f} = v_{1f} - (v_{2i}-v_{1i})$$

$$m_1v_{1i}+m_2v_{2i} = m_1v_{1f} + m_2(v_{1f} - (v_{2i}-v_{1i}))$$

$$(m_1+m_2)v_{1f} = 2m_2v_{2i} + (m_1-m_2)v_{1i}$$

$$v_{1f} = \frac{m_1-m_2}{(m_1+m_2)}v_{1i} + \frac{2m_2}{(m_1+m_2)}v_{1i}$$ (4.47)

(back substitute for $v_{2f}$).

This is satisfying and easy to evaluate, but still hard to understand. To understand the collision, it is easiest to put everything into the CM frame, evaluate the collision, and then put the results back into the lab frame! Let's try this:

In the CM frame,

$$u_{1i} = v_{1i} - v_{\rm cm}$$ (4.48)

$$u_{2i} = v_{2i} - v_{\rm cm}$$ (4.49)

$$v_{\rm cm} = \frac{m_1v_{1i}+m_2v_{2i}}{m_1+m_2}$$ (4.50)

so that the momentum conservation equation becomes:

$$m_1u_{1i}+m_2u_{2i} = m_1u_{1f} + m_2u_{2f}$$ (4.51)

$$p_{1i}' + p_{2i}' = p_{1f}' + p_{2f}' = 0$$ (4.52)

Thus $p_i' = p_{1i}' = -p_{2i}'$ and $p_f = p_{1f}' = p_{2f}'$. The energy conservation equation (in terms of the $p$'s) becomes:

$$\frac{p_{i}'^2}{2m_1} + \frac{p_{i}'^2}{2m_2} = \frac{p_{f}'^2}{2m_1} + \frac{p_{f}'^2}{2m_2}$$ (4.53)

$$\frac{p_{i}'^2}{2}\frac{1}{m_1+m_2} = \frac{p_{f}'^2}{2}\frac{1}{m_1+m_2}$$ (4.54)

$$p_i'^2 = p_f'^2$$ (4.55)

so that we can conclude that

$$p_{1f}' = \pm p_{1i}'$$ (4.56)

The + sign satisfies all the conditions, but means the the particles ``missed''. If they hit, their internal momenta relative to the CM reverses! Thus:

$$m_1u_{1f} = - m_1u_{1i}$$ (4.57)

$$u_{1f} = -u_{1i}$$ (4.58)

$$v_{1f} - v_{\rm cm} = -(v_{1i} - v_{\rm cm})$$ (4.59)

$$v_{1f} = -v_{1i} + 2v_{\rm cm}$$ (4.60)

$$= \frac{-(m_1+m_2)v_{1i} + 2(m_1v_{1i}+m_2v_{2i})}{m_1+m_2}$$ (4.61)

$$= \frac{(m_1-m_2)v_{1i} + 2m_2v_{2i}}{m_1+m_2}$$ (4.62)

which can be seen to equal the result above by inspection. It is, however, a bit easier to understand!