Momentum and Kinetic Energy in Collisions — Question 2

Question 2

The minimum kinetic energy at fixed momentum

A system consists of particles with positive masses $m_i$, fixed total mass $M$ and total momentum $\vec P$. Define $\vec V=\vec P/M$ and $\vec w_i=\vec v_i-\vec V$.

1. Derive a lower bound on total kinetic energy and its equality condition.

2. Apply it to $M=5\,\mathrm{kg}$, $P=10\,\mathrm{kg\,m/s}$ and initial $K=25\,\mathrm J$. What is the largest possible kinetic-energy reduction at fixed momentum?

Question 2 - Solution

The minimum kinetic energy at fixed momentum

(a) Derivation and calculation

Put $v_i=V_C+w_i$, where $V_C=P/M$ and $\sum m_iw_i=0$: $$K=\tfrac12\sum m_i(V_C+w_i)^2=\tfrac12MV_C^2+V_C\sum m_iw_i+\tfrac12\sum m_iw_i^2.$$ $$\boxed{K=\frac{P^2}{2M}+\frac12\sum m_iw_i^2\geq\frac{P^2}{2M}}.$$ Equality requires every $w_i=0$.

(b) Result and interpretation

For the supplied values, $$K_{\min}=\frac{10^2}{2(5)}=10\ \mathrm J,\qquad V_C=10/5=2\ \mathrm{m/s}.$$ $$\boxed{K_{\mathrm{loss,max}}=25-10=15\ \mathrm J}.$$

Physical check

The minimum retains the kinetic energy of center-of-mass translation.