Inelastic Collisions in One Dimension — Question 4

Question 4

Capture followed by rough spring compression

A $1\,\mathrm{kg}$ cart at $6\,\mathrm{m/s}$ sticks to a stationary $2\,\mathrm{kg}$ cart. Immediately afterward the combined cart compresses a relaxed spring of $k=100\,\mathrm{N/m}$ while a constant $5\,\mathrm N$ resistive force opposes motion. Neglect external impulse during the brief capture.

1. Find the postcapture speed and maximum spring compression.

2. Separate the energy lost in capture from the energy dissipated during compression.

Question 4 - Solution

Capture followed by rough spring compression

(a) Derivation and calculation

First isolate the brief capture: $$1(6)=(1+2)V\Rightarrow V=2\ \mathrm{m/s},\qquad K_+=\tfrac12(3)(2)^2=6\ \mathrm J.$$ For the subsequent compression, $$6=\tfrac12(100)x^2+5x\Rightarrow 50x^2+5x-6=0.$$ $$x=\frac{-5\pm\sqrt{25+1200}}{100}\Rightarrow\boxed{x=0.30\ \mathrm m}.$$

(b) Result and interpretation

The negative root $-0.40$ m is outside the positive-compression model. Energy is accounted for by $$K_{\mathrm{capture\ loss}}=18-6=12\ \mathrm J,$$ $$W_{\mathrm{friction\ loss}}=5(0.30)=1.5\ \mathrm J,\qquad U_s=50(0.30)^2=4.5\ \mathrm J.$$ Thus $18=12+1.5+4.5$ J.

Physical check

Friction removes energy after capture and does not belong in the instantaneous capture momentum equation.