Momentum and Kinetic Energy in Collisions — Question 3

Question 3

Reduced mass isolates the energy available in approach

Two particles of masses $m_1$ and $m_2$ move along a line with velocities $u_1$ and $u_2$. Let $M=m_1+m_2$, $V=(m_1u_1+m_2u_2)/M$ and $\mu=m_1m_2/M$.

1. Show that $K=\tfrac12MV^2+\tfrac12\mu(u_1-u_2)^2$.

2. For masses $2$ and $6\,\mathrm{kg}$ at $5$ and $1\,\mathrm{m/s}$, find both contributions.

Question 3 - Solution

Reduced mass isolates the energy available in approach

(a) Derivation and calculation

With $M=m_1+m_2$, $u=u_1-u_2$, and $V_C=(m_1u_1+m_2u_2)/M$, $$u_1=V_C+\frac{m_2}{M}u,\qquad u_2=V_C-\frac{m_1}{M}u.$$ Substitution cancels the cross terms: $$\boxed{K=\frac12MV_C^2+\frac12\mu u^2},\qquad\mu=\frac{m_1m_2}{M}.$$

(b) Result and interpretation

Numerically, $$M=8\ \mathrm{kg},\quad V_C=\frac{2(5)+6(1)}8=2\ \mathrm{m/s},\quad\mu=\frac{12}8=1.5\ \mathrm{kg}.$$ $$\boxed{K=\tfrac12(8)(2)^2+\tfrac12(1.5)(4)^2=16+12=28\ \mathrm J}.$$

Physical check

Direct summation gives $25+3=28$ J.