The Linear Momentum of a System of Particles — Question 2

Question 2

Stationary center with active constituents

A $2\,\mathrm{kg}$ particle moves at $(3,4)\,\mathrm{m/s}$. A $5\,\mathrm{kg}$ particle has an unknown velocity. The total momentum is zero.

1. Find the second velocity and total kinetic energy.

2. Does zero total momentum imply both particles remain at fixed positions?

Question 2 - Solution

Stationary center with active constituents

(a) Derivation and calculation

Zero total momentum fixes the unknown velocity: $$2(3,4)+5\vec v_2=\vec0\Rightarrow \boxed{\vec v_2=(-1.2,-1.6)\ \mathrm{m/s}}.$$ $$v_1=5,\quad v_2=2,\qquad \boxed{K=\tfrac12(2)(25)+\tfrac12(5)(4)=35\ \mathrm J}.$$

(b) Result and interpretation

The center condition is only $$\boxed{\vec V_C=\vec P/(2+5)=\vec0}.$$ Both particles still move. If external force remains zero, the center remains at rest, while internal or force-free relative motion can continue.

Physical check

The particles have equal and opposite momenta, each of magnitude 10 kg m/s.