The Linear Momentum of a System of Particles — Question 4

Question 4

A missing particle in a momentum record

A three-particle system has total mass $10\,\mathrm{kg}$ and center velocity $(1,-2)\,\mathrm{m/s}$. Two particles have masses $2$ and $3\,\mathrm{kg}$ and velocities $(4,1)$ and $(-1,-3)\,\mathrm{m/s}$.

1. Infer the third mass and velocity.

2. Check the result by recomputing the mass-weighted center velocity.

Question 4 - Solution

A missing particle in a momentum record

(a) Derivation and calculation

First recover the remaining mass and required total momentum: $$\boxed{m_3=10-2-3=5\ \mathrm{kg}},\qquad\vec P=M\vec V_C=(10,-20).$$ $$\vec p_3=\vec P-2(4,1)-3(-1,-3)=(5,-13),\qquad \boxed{\vec v_3=(1,-2.6)\ \mathrm{m/s}}.$$

(b) Result and interpretation

Substitute the result back into the defining average: $$\frac{2(4,1)+3(-1,-3)+5(1,-2.6)}{10} =\frac{(10,-20)}{10}=\boxed{(1,-2)\ \mathrm{m/s}}.$$ Both components independently reproduce the supplied center velocity.

Physical check

The missing mass is positive, and its momentum fills exactly the deficit of the two known particles.