# How do you calculate inelastic collisions?

Apr 30, 2018

Given perfect inelasticity, we go back to first principles.

For inelastic collisions momentum is conserved, but energy is lost. Hence, recall,

${m}_{\text{A"nu_"A" + m_"B"nu_"B" = m_"A"nu_"A"' + m_"B"nu_"B}} '$

To demonstrate, I will do a problem.

Two asteroids strike head-on: before the collision, asteroid A (${m}_{\text{A" = 7.5*10^12"kg}}$) has velocity (3.3"km")/"s" and asteroid B (${m}_{\text{B" = 1.45*10^13"kg}}$) 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,

${m}_{\text{A" = 7.5*10^12"kg}}$, and

${\nu}_{\text{A" = (3.3"km")/"s" * (10^3"m")/"km" = (3.3*10^3"m")/"s}}$

${m}_{\text{B" = 1.45*10^13"kg}}$, and

${\nu}_{\text{B" = (-1.4"km")/"s" * (10^3"m")/"km" = (-1.4*10^3"m")/"s}}$

Now, recall what we talked about!

${m}_{\text{A"nu_"A" + m_"B"nu_"B" = m_"A"nu_"A"' + m_"B"nu_"B}} '$

${\nu}_{\text{A"' = nu_"B}} ' = \nu '$

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