Collisions in Two Dimensions — Question 8

Question 8

A guide absorbs one momentum component during capture

Two $1\,\mathrm{kg}$ particles approach a small catcher with velocities $(4,3)$ and $(-2,1)\,\mathrm{m/s}$. They stick together while an ideal horizontal guide constrains their combined final motion to $y=0$. The catcher mass is negligible; the guide supplies no horizontal impulse.

1. Find the final velocity and guide impulse on the two particles.

2. Find the change in translational kinetic energy. Is the full vector momentum of the particles conserved?

Question 8 - Solution

A guide absorbs one momentum component during capture

(a) Derivation and calculation

The guide enforces $V_y=0$ while supplying no horizontal impulse: $$P_{ix}=4-2=2=2V_x\Rightarrow\boxed{\vec V=(1,0)\ \mathrm{m/s}}.$$ $$\boxed{\vec J_{\mathrm{guide}}=\vec P_f-\vec P_i=(2,0)-(2,4)=(0,-4)\ \mathrm{N\,s}}.$$

(b) Result and interpretation

The translational kinetic energies are $$K_i=\tfrac12[(4^2+3^2)+((-2)^2+1^2)]=15\ \mathrm J,\qquad K_f=\tfrac12(2)(1)^2=1\ \mathrm J.$$ $$\boxed{\Delta K=-14\ \mathrm J}.$$ Only horizontal momentum is conserved for the particles. The guide receives the opposite vertical impulse.

Physical check

The external guide impulse removes the initial 4 kg m/s of upward momentum.