The Linear Momentum of a System of Particles — Question 1

Question 1

A vector inventory reveals the center velocity

Three particles have masses $1$, $2$, $3\,\mathrm{kg}$ and velocities $(4,0)$, $(-1,2)$, $(0,-1)\,\mathrm{m/s}$.

1. Find total momentum and center velocity.

2. Can replacing the particles by a $6\,\mathrm{kg}$ point at the center reproduce their total kinetic energy?

Question 1 - Solution

A vector inventory reveals the center velocity

(a) Derivation and calculation

Sum the vector momenta, then divide by total mass: $$\boxed{\vec P=1(4,0)+2(-1,2)+3(0,-1)=(2,1)\ \mathrm{kg\,m/s}}.$$ $$M=6\ \mathrm{kg},\qquad\boxed{\vec V_C=\vec P/M=(1/3,1/6)\ \mathrm{m/s}}.$$

(b) Result and interpretation

The actual and center-translation energies differ: $$K=\tfrac12(1)(16)+\tfrac12(2)(5)+\tfrac12(3)(1)=14.5\ \mathrm J.$$ $$K_C=\frac{P^2}{2M}=\frac5{12}\ \mathrm J,\qquad \boxed{K_{\mathrm{relative}}=K-K_C=\frac{169}{12}\ \mathrm J}.$$ A single center point reproduces momentum but omits relative-motion energy.

Physical check

The nonnegative relative energy explains why zero or small center speed need not mean little kinetic energy.