Inelastic Collisions in One Dimension — Question 2

Question 2

Partial restitution with a moving target

Cart 1 of mass $2\,\mathrm{kg}$ moves right at $5\,\mathrm{m/s}$ and catches cart 2 of mass $3\,\mathrm{kg}$ moving right at $1\,\mathrm{m/s}$. The head-on coefficient of restitution is $e=0.50$. External impulse is negligible.

1. Find both final velocities.

2. Find the kinetic-energy loss and verify that the carts separate.

Question 2 - Solution

Partial restitution with a moving target

(a) Derivation and calculation

Momentum conservation and restitution provide two independent equations: $$2v_1+3v_2=2(5)+3(1)=13,\qquad v_2-v_1=0.5(5-1)=2.$$ $$5v_1+6=13\Rightarrow\boxed{v_1=1.4,\quad v_2=3.4\ \mathrm{m/s}}.$$

(b) Result and interpretation

Compute the kinetic-energy decrease: $$K_i=\tfrac12[2(25)+3(1)]=26.5\ \mathrm J,$$ $$K_f=\tfrac12[2(1.4)^2+3(3.4)^2]=19.3\ \mathrm J.$$ $$\boxed{K_i-K_f=7.2\ \mathrm J}.$$ The outgoing gap grows at $3.4-1.4=2.0$ m/s.

Physical check

The same loss is $\tfrac12(6/5)(4)^2(1-0.5^2)=7.2$ J.