Collisions in Two Dimensions — Question 7

Question 7

A right-angle turn in the center frame need not look perpendicular in the lab

A $1\,\mathrm{kg}$ particle at $(8,0)\,\mathrm{m/s}$ collides elastically with a stationary $3\,\mathrm{kg}$ particle. In the center frame, the lighter particle's outgoing velocity points along $+y$.

1. Find the laboratory final velocities.

2. Check energy and determine whether the laboratory departures are perpendicular.

Question 7 - Solution

A right-angle turn in the center frame need not look perpendicular in the lab

(a) Derivation and calculation

The center velocity is $$\vec V_C=\frac{1(8,0)+3(0,0)}4=(2,0)\ \mathrm{m/s}.$$ The light particle has initial center-frame speed 6 m/s, retained in elastic scattering. The specified direction and zero center momentum give $$\vec w_1=(0,6),\qquad\vec w_2=-\vec w_1/3=(0,-2).$$ $$\boxed{\vec v_1=(2,6),\qquad\vec v_2=(2,-2)\ \mathrm{m/s}}.$$

(b) Result and interpretation

Check the energy and angle condition: $$K_f=\tfrac12(1)(2^2+6^2)+\tfrac12(3)(2^2+(-2)^2)=32\ \mathrm J=K_i.$$ $$\boxed{\vec v_1\cdot\vec v_2=4-12=-8\ne0}.$$ The laboratory departure directions are not perpendicular; a frame shift changes their dot product.

Physical check

The final momentum is $(2,6)+3(2,-2)=(8,0)$ kg m/s.