Momentum and Kinetic Energy in Collisions — Question 6

Question 6

Stored energy can make separation faster than approach

Two $2\,\mathrm{kg}$ carts approach at $+1$ and $-1\,\mathrm{m/s}$. An internal latch releases $6\,\mathrm J$ into translational kinetic energy during contact. External impulse and other energy transfers are negligible.

1. Find the final speeds if the carts separate.

2. Find the effective restitution coefficient and explain why it can exceed one.

Question 6 - Solution

Stored energy can make separation faster than approach

(a) Derivation and calculation

Initial momentum vanishes and remains zero: $$2v_1+2v_2=0\Rightarrow v_2=-v_1.$$ $$K_i=\tfrac12(2)(1^2+1^2)=2\ \mathrm J,\qquad K_f=2+6=8\ \mathrm J.$$ $$2v_1^2=8\Rightarrow\boxed{v_1=-2,\quad v_2=+2\ \mathrm{m/s}}.$$

(b) Result and interpretation

Choose the separating signs for the left and right objects. The restitution coefficient is $$\boxed{e=\frac{v_2-v_1}{u_1-u_2}=\frac{4}{2}=2}.$$ This is a superelastic collision: stored internal energy increases translational kinetic energy.

Physical check

The final opposite momenta cancel despite the higher speeds.