Conservation of Linear Momentum — Question 3

Question 3

Inferring an unobserved breakup fragment

A $6\,\mathrm{kg}$ object moving at $(3,0)\,\mathrm{m/s}$ breaks into three $2\,\mathrm{kg}$ fragments. Two emerge with velocities $(5,2)$ and $(1,-3)\,\mathrm{m/s}$. External impulse during breakup is negligible.

1. Find the third velocity.

2. Compute the kinetic-energy increase and identify its source.

Question 3 - Solution

Inferring an unobserved breakup fragment

(a) Derivation and calculation

Subtract the known fragment momenta from the initial total: $$\vec P_i=6(3,0)=(18,0),\qquad \vec p_3=(18,0)-2(5,2)-2(1,-3)=(6,2).$$ $$\boxed{\vec v_3=\vec p_3/2=(3,1)\ \mathrm{m/s}}.$$

(b) Result and interpretation

The energies before and after breakup are $$K_i=\tfrac12(6)(3)^2=27\ \mathrm J,$$ $$K_f=\tfrac12(2)[(25+4)+(1+9)+(9+1)]=49\ \mathrm J, \qquad\boxed{\Delta K=22\ \mathrm J}.$$ Internal stored energy supplies the increase while total momentum remains unchanged.

Physical check

The vertical fragment momenta $4-6+2$ sum to zero.