Inelastic Collisions

Figure 10.4: Building blocks of inelastic collisions. In the left panel two particles collide to form a third particle. In the right panel a particle breaks up, forming two particles.

An inelastic collision is one in which the particles coming out of the collision are not the same as the particles going into it. Inelastic collisions conserve both total momentum and energy just as elastic collisions do. However, unlike elastic collisions, inelastic collisions generally do not conserve the total kinetic energy of the particles, as some rest energy is generally created or destroyed.

Figure 10.4 shows the fundamental building blocks of inelastic collisions. We can consider even the most complex inelastic collisions to be made up of composites of only two processes, the creation of one particle from two, and the disintegration of one particle into two.

Let us consider each of these in the center of momentum frame. In both cases the single particle must be stationary in this frame since it carries the total momentum of the system, which has to be zero. By conservation of momentum, if particle 1 in the left panel of figure 10.4 has momentum $p$, then the momentum of particle 2 is $-p$. If the two particles have masses $m_1$ and $m_2$, then their energies are $E_1 = (p^2 c^2 + m_1^2 c^4 )^{1/2}$ and $E_2 = (p^2 c^2 + m_2^2 c^4 )^{1/2}$. The energy of particle 3 is therefore $E_3 = E_1 + E_2$, and since it is at rest, all of its energy is in the form of ``$mc^2$'' or rest energy, and so the mass of this particle is

$$m_3 = (E_1 + E_2 )/c^2 ~~~ \mbox{(center of momentum frame)}$$

$$= (p^2 /c^2 + m_1^2 )^{1/2} + (p^2 /c^2 + m_2^2 )^{1/2}$$

$$= m_1 [1 + p^2 /(m_1^2 c^2 )]^{1/2} + m_2 [1 + p^2 /(m_2^2 c^2 )]^{1/2} .$$ (11.14)

The last line in the above equation shows that $m_3 > m_1 + m_2$ because it is in the form $m_1 A + m_2 B$ where both $A$ and $B$ are greater than one. Thus, rest energy is created in the amount $\Delta E_{rest} = (m_3 - m_1 - m_2 )c^2$.

Actually, it is easy to calculate the mass of particle 3 in the above case from any reference frame as long as the momenta and energies of particles 1 and 2 are known in this frame. By conservation of energy and momentum, $E_3 = E_1 + E_2$ and $\mbox{\bf p}_3 = \mbox{\bf p}_1 + \mbox{\bf p}_2$. Furthermore, $E_3^2 = p_3^2 c^2 + m_3^2 c^4$, so we can solve for $m_3$:

$$m_3 = [(E_1 + E_2 )^2 /c^4 - (\mbox{\bf p}_1 + \mbox{\bf p}... ...p}_1 + \mbox{\bf p}_2 ) /c^2 ]^{1/2} ~~~ \mbox{(any frame)} .$$ (11.15)

The right panel of figure 10.4 shows the process of particle decay. This is just the inverse of the particle creation process, and all of the analysis we have done for creation is valid for particle decay except that rest energy is converted to kinetic energy rather than vice versa.