The Linear Momentum of a System of Particles — Question 10

Question 10

The same momentum can conceal different internal motion

Two $1\,\mathrm{kg}$ particles move along a line. In state A their velocities are $1$ and $3\,\mathrm{m/s}$; in state B they are $-2$ and $6\,\mathrm{m/s}$.

1. Compare total momentum and center velocity in the two states.

2. Compare kinetic energies in the lab and in the common center frame. What information does total momentum omit?

Question 10 - Solution

The same momentum can conceal different internal motion

(a) Derivation and calculation

Compute the two momentum sums: $$P_A=1(1)+1(3)=4,\qquad P_B=1(-2)+1(6)=4\quad\mathrm{kg\,m/s}.$$ $$\boxed{P_A=P_B=4\ \mathrm{kg\,m/s}},\qquad\boxed{V_{C,A}=V_{C,B}=2\ \mathrm{m/s}}.$$

(b) Result and interpretation

The laboratory energies are $K_A=(1+9)/2=5$ J and $K_B=(4+36)/2=20$ J. Subtract the common frame speed from each velocity: $$K_A^{\prime}=\tfrac12[(-1)^2+1^2]=\boxed{1\ \mathrm J},\qquad K_B^{\prime}=\tfrac12[(-4)^2+4^2]=\boxed{16\ \mathrm J}.$$ $$K=\tfrac12MV_C^2+K^{\prime}=4\ \mathrm J+K^{\prime}.$$ Total momentum omits the relative-motion energy.

Physical check

The common translational contribution is identical while the center-frame energy differs by 15 J.