Momentum and Kinetic Energy in Collisions — Question 4

Question 4

Restitution describes relative speed rather than energy fraction

Two carts have masses $2$ and $3\,\mathrm{kg}$ and approach with $u_1-u_2=5\,\mathrm{m/s}$. For this head-on collision the coefficient $e=(v_2-v_1)/(u_1-u_2)$ is $0.60$. External impulse is negligible.

1. Find the loss of kinetic energy using the reduced mass.

2. What fraction of the initial center-frame kinetic energy remains? Is that necessarily the laboratory energy fraction?

Question 4 - Solution

Restitution describes relative speed rather than energy fraction

(a) Derivation and calculation

The reduced mass and final relative speed are $$\mu=\frac{2(3)}{2+3}=1.2\ \mathrm{kg},\qquad w_f=e w_i=0.60(5)=3\ \mathrm{m/s}.$$ $$\boxed{K_i-K_f=\tfrac12\mu w_i^2(1-e^2)=\tfrac12(1.2)(25)(0.64)=9.6\ \mathrm J}.$$

(b) Result and interpretation

In the center frame, $$\boxed{K_{{\mathrm{rel}},f}/K_{{\mathrm{rel}},i}=e^2=0.36}.$$ In the laboratory, the ratio is instead $$\frac{K_f}{K_i}=\frac{K_C+e^2K_{{\mathrm{rel}},i}}{K_C+K_{{\mathrm{rel}},i}},$$ which equals 0.36 only if $K_C=0$.

Physical check

Center-of-mass kinetic energy is unchanged by an isolated collision.