The Linear Momentum of a System of Particles — Question 6

Question 6

Symmetric moving masses and a small defect

Four particles lie at the corners of a square centered at the origin, at $(\pm a,\pm a)$. Each moves directly away from the origin at speed $v$. Three have mass $m$; the particle at $(a,a)$ has mass $2m$.

1. Find the total momentum and center velocity.

2. Explain how symmetry simplifies the calculation and identify the direction of the result.

Question 6 - Solution

Symmetric moving masses and a small defect

(a) Derivation and calculation

Decompose the heavier corner into one mass $m$ plus an extra mass $m$. The four equal contributions cancel: $$\vec P_{\mathrm{equal}}=\frac{mv}{\sqrt2}\bigl[(1,1)+(-1,1)+(-1,-1)+(1,-1)\bigr]=\vec0.$$ $$\boxed{\vec P=\frac{mv}{\sqrt2}(1,1)},\qquad \boxed{\vec V_C=\frac{\vec P}{5m}=\frac{v}{5\sqrt2}(1,1)}.$$

(b) Result and interpretation

The remaining magnitude and direction are $$\boxed{|\vec V_C|=v/5},\qquad \hat V_C=(1,1)/\sqrt2.$$ The net motion points toward the heavier corner. Symmetry cancels equal weighted vectors, not unequal masses at otherwise symmetric positions.

Physical check

The position scale $a$ cancels because the specified radial speed is the same at all corners.