A New Look at Energy — Question 6

Question 6

An asymmetric inelastic collision separates bulk motion from internal energy

A particle of mass m moving at $0.60c$ strikes an identical stationary particle and sticks to it. The system is isolated and radiates no energy. Use $E^2=p^2c^2+m^2c^4$, $E=\gamma mc^2$, $K=E-mc^2$, and $\mathbf v=c^2\mathbf p/E$.

1. Find the composite velocity and invariant mass from total energy and momentum.

2. Find the composite laboratory kinetic energy and the increase of its rest energy above $2mc^2$.

Question 6 - Solution

An asymmetric inelastic collision separates bulk motion from internal energy

(a) Method and calculation

$$E_i=(1.25+1)mc^2=\tfrac94mc^2,\quad P_i=\tfrac34mc\quad\Rightarrow\quad\boxed{V_f=c/3}.$$$$M^2c^4=E_i^2-P_i^2c^2=\tfrac92m^2c^4\quad\Rightarrow\quad\boxed{M=3m/\sqrt2}.$$

(b) Interpretation and further calculation

$$\boxed{K_f=\left(\frac94-\frac3{\sqrt2}\right)mc^2\simeq0.128680mc^2}.$$$$\boxed{\Delta E_{\mathrm{internal}}=\left(\frac3{\sqrt2}-2\right)mc^2\simeq0.121320mc^2}.$$

Physical check

The initial 0.250mc-squared kinetic energy is divided between final bulk motion and retained internal energy.