Momentum and Kinetic Energy in Collisions — Question 5

Question 5

Energy loss is invariant when momentum is conserved

Two $1\,\mathrm{kg}$ carts with velocities $4$ and $0\,\mathrm{m/s}$ stick and move at $2\,\mathrm{m/s}$. Compare the lab with a frame moving at $U=3\,\mathrm{m/s}$.

1. Compute kinetic-energy loss in each frame.

2. Derive the condition under which the energy change is frame independent.

Question 5 - Solution

Energy loss is invariant when momentum is conserved

(a) Derivation and calculation

In the laboratory and moving frame respectively, $$K_i=8,\quad K_f=\tfrac12(2)(2)^2=4\ \mathrm J;\qquad K_i'=\tfrac12(1^2+(-3)^2)=5,\quad K_f'=1\ \mathrm J.$$ $$\boxed{K_i-K_f=K_i'-K_f'=4\ \mathrm J}.$$

(b) Result and interpretation

For a frame translating at $\vec U$, $$K'=\tfrac12\sum m_i|\vec v_i-\vec U|^2=K-\vec U\cdot\vec P+\tfrac12MU^2.$$ For unchanged total mass, $$\Delta K'=\Delta K-\vec U\cdot\Delta\vec P.$$ Momentum conservation makes $\Delta\vec P=0$, so the kinetic-energy change is frame independent.

Physical check

Individual energies change with frame, while this isolated collision loses the same 4 J.