Conservation of Linear Momentum — Question 4

Question 4

Momentum conservation along a permitted direction

Two carts of masses $2$ and $3\,\mathrm{kg}$ interact on a straight horizontal rail. Before interaction their horizontal velocities are $4$ and $-1\,\mathrm{m/s}$. Later the first has velocity $-2\,\mathrm{m/s}$. The rail can exert vertical forces but has negligible horizontal friction.

1. Find the second final horizontal velocity.

2. Explain why vertical external forces do not prevent the horizontal calculation.

Question 4 - Solution

Momentum conservation along a permitted direction

(a) Derivation and calculation

Conserve the horizontal component for the two-cart system: $$P_{i,x}=2(4)+3(-1)=5\ \mathrm{kg\,m/s}.$$ $$2(-2)+3v_{2f}=5\Rightarrow\boxed{v_{2f}=3.0\ \mathrm{m/s}}.$$

(b) Result and interpretation

The component impulse equations are independent: $$\Delta P_x=J_{{\mathrm{ext}},x}=0,\qquad\Delta P_y=J_{{\mathrm{ext}},y}.$$ Vertical rail forces do not change horizontal momentum. Conservation of one component requires zero external impulse in that component, not a zero full force vector.

Physical check

The two horizontal momentum changes are $-12$ and $+12$ kg m/s.