Given perfect inelasticity, we go back to first principles.
For inelastic collisions momentum is conserved, but energy is lost. Hence, recall,
m_"A"nu_"A" + m_"B"nu_"B" = m_"A"nu_"A"' + m_"B"nu_"B"'
To demonstrate, I will do a problem.
Two asteroids strike head-on: before the collision, asteroid A (m_"A" = 7.5*10^12"kg") has velocity (3.3"km")/"s" and asteroid B (m_"B" = 1.45*10^13"kg") has velocity (1.4"km")/"s" in the opposite direction. If the asteroids stick together, what is the velocity (magnitude and direction) of the new asteroid after the collision?^1
Given,
m_"A" = 7.5*10^12"kg", and
nu_"A" = (3.3"km")/"s" * (10^3"m")/"km" = (3.3*10^3"m")/"s"
m_"B" = 1.45*10^13"kg", and
nu_"B" = (-1.4"km")/"s" * (10^3"m")/"km" = (-1.4*10^3"m")/"s"
Now, recall what we talked about!
m_"A"nu_"A" + m_"B"nu_"B" = m_"A"nu_"A"' + m_"B"nu_"B"'
nu_"A"' = nu_"B"' = nu'
Hence,
(m_"A"nu_"A" + m_"B"nu_"B")/(m_"A" + m_"B") = nu' = (2.0*10^2"m")/"s"
is the velocity of the compound asteroid after the collision (in the positive direction).
1: Giancoli, D. C. (2014). Physics: Principles with Applications. Boston: Pearson. pg. 195
