Energy is only conserved in the absence of friction. In this scenario there is both friction from the spear on the track and the drag force of moving through the water. Because of this the energy stored in the band will never equal the kinetic energy of the spear.
Jeff the spring rate of the band is definitely not linear but that is a completely different problem.
I think some of the confusion here might be with the use of KE. In this scenario the conservation of mechanical energy doesnt apply because of the force of friction. Instead what is most applicable are the equations: sum(F)=ma. and p=mv. These are Newtons second law of motion and the equation for momentum (p). Also Force(drag)=1/2(p*(v^2)(C)(A) is critical. What is most important in this formula is the fact that the drag force is proportional to the velocity squared. In your test for penetration, using the same bands shooting different mass spears your acceleration will different. the lighter spear will, like you said have a higher velocity.
From the instant the bands no longer apply a force on the spear the only force acting on them will be the the force of drag. For the the cross sectional area for your two bands differs by about .05". so in your drag equation you have .5(v^2)(.5)(.49)=Fdrag(5/16). For the 9/32" shaft you have .5(v^2)(.5)(.44). These simplify to (.12)(v^2) and (.11)(v^2) respectively.
To summarize, the (9/32) shaft has a smaller mass, a higher drag force and a higher initial velocity.
the (5/16) shaft has more mass, less drag and a lower initial velocity.
Next comes newtons second law. sum(F)=ma=Fdrag
after the bands have released the shaft the only force on the spear is drag. just to simplify this without knowing the actual mass difference of the shafts: mass of(5/16)= 4. Mass of (9/32)=1. NOTE: this is an exaggeration to explain the difference. As a result the initial velocity of the smaller shaft is twice that of the heavier shaft. and experiences 4 times the drag.
Now according to Newtons second law sumF=ma the accel for (9/32)= (.11)(4)=.44 and for the (5/16) = (.12)(1)/4 =.03 creating a ratio of about 15:1. Because of this in very short distances the lighter shaft slows down significantly. This will change at longer ranges because the drag force will decrease as the velocity decrease. There will be a point some distance from the barrel that both will penetrate the same distance.
So even though this is a significant exaggeration the lighter shaft slows down a huge amount in the first few feet and because momentum = mv and the the shaft is lighter but not enough faster to compensate. Now because the heavier shaft has higher momentum it then hits the fish/target with more punch.
I really hope this makes sense and that I didnt loose anyone with the simplifications or the notation. Jeff this should explain your experiment. This also explains why heavier shafts can travel farther. Now I need to go do my actual physics homework.