Scattering A Rigit Straw

Set up a coordinate system so that the vertices of the frame are (initiialy) at $(-a,0)$ , $(-a,2a)$ , $(a,2a)$ , $(a,0)$ . Then the centre of mass of the frame has coordinates $(0,\tfrac43a)$ . If each straight section of the frame has mass $m$ ,.then each "side arm" of the frame has moment of inertia $\tfrac13ma^2 + m\big((\tfrac13a)^2 + a^2\big) \; = \; \tfrac{13}{9}ma^2$ about the frame's centre of mass, while the "central arm" of the frame as moment of inertia $\tfrac13ma^2 + m\big(\tfrac23a\big)^2 \; = \; \tfrac79ma^2$ about the frame's centre of mass, and hence the whole frame has moment of inertia $\tfrac{11}{3}ma^2$ about its centre of mass. If the velocity of the centre of mass after the collision is $(0,V)$ , and if the angular velocity after the collision is $\omega$ , then $J \; = \; 3mV \hspace{2cm} \tfrac{11}{3} ma^2\omega \; = \; Ja$ so that $V \; = \; \tfrac{1}{3m}J \hspace{2cm} \omega \; = \; \tfrac{3}{11ma}J$ and so the velocity of $P$ , after a $90^\circ$ turn, is $(0,v_P)$ , where $v_P \; = \; V - \tfrac23a \omega \; = \; \tfrac{5}{33m}J$ and hence $\tfrac{\omega}{v_P} \; = \; \tfrac{9}{5a} \; = \; \boxed{0.2}$