Totally Inelastic Collisions

In a totally inelastic collision, the two particles stick together after colliding, so that they end up moving with the same final velocity, $v_f= v_{1f}=v_{2f}$. In this case,

$$v_f = \frac{m_1\,v_{1i} + m_2\,v_{2i}}{m_1+m_2}= V.$$ (1.145)

In other words, the common final velocity of the two particles is equal to the center-of-mass velocity of the system. This is hardly a surprising result. We have already seen that in the center-of-mass frame the two particles must diverge with equal and opposite momenta after the collision. However, in a totally inelastic collision these two momenta must also be equal (because the two objects stick together). The only way in which this is possible is if the two particles remain stationary in the center-of-mass frame after the collision. Hence, after the collision, the two particles move with the center-of-mass velocity in the laboratory frame.

Suppose that the second object is initially at rest (i.e., $v_{2i} =0$) in the laboratory frame. In this special case, the common final velocity of the two objects is

$$v_f = \frac{m_1}{m_1+m_2}\,v_{1i}.$$ (1.146)

Note that the first object is slowed down by the collision. The fractional loss in kinetic energy of the system due to the collision is given by

$$f = \frac{m_2}{m_1+m_2}.$$ (1.147)

The loss in kinetic energy is small if the (initially) stationary object is much lighter than the moving object (i.e., if $m_2\ll m_1$), and almost $100\%$ if the moving object is much lighter than the stationary one (i.e., if $m_2\gg m_1$). Of course, the lost kinetic energy of the system is converted into some other form of energy; for instance, heat energy.