Projectile splitting in mid-air: why don't the two halves travel the same distance? I was checking a question about projectile and found something quiet interesting.

The question is: 

Calculate $R$ if the ball of mass 1kg was split in mid-air evenly.

Looking at the picture, I can imagine this happening in reality, however, physically I cannot understand why the upper part goes further than the lower one even though they had the same speed when splitting and have the same mass.
Can someone please explain this.  
 A: Whatever force acts on the 2 fragments to make them separate, it is an internal force. It acts equally on both fragments for the same amount of time. So the impulse (=change in linear momentum) is the same for each fragment (ie equal in magnitude, opposite in direction). This is true whether or not the fragments have the same mass.
Because the fragments have the same mass, they also have the same speed (but in opposite directions) relative to the their centre of mass when they split up.  But they do not have the same velocity relative to the ground. If they did have the same velocity relative to the ground, they would continue travelling together without moving apart : the projectile would not split up.
While they are still in flight the centre of mass (CM) of the 2 fragments follows the same trajectory as the complete projectile if it had not spit up. The additional momentum given by the internal force will carry one fragment further than the CM and the other fragment not as far as the CM.
However : note that the midpoint of the 2 fragments on the ground is not necessarily the point at which the complete projectile would have hit the ground. If the fragments hit the ground - and stop falling - at different times, then their CM on the ground is likely to be different from the point at which the complete projectile would have landed.
A: The assumption is that momentum is conserved during the explosion.
Relative to the original projectile one fragment moves faster whilst the other moves slower whilst conserving momentum.
I would consider motion in the horizontal plane as that will determine the range of fragments.
A: When the ball split apart, the two halves had, in the rest frame of the ball, equal nd opposite velocities (imparted by whatever force/impulse drove the half-balls apart).
Since you don't know the direction of the splitting impulse (although if the problem remains planar, you do know the plane it lies in) and you don't know the time of splitting, it is by no means a gimme that the distance $R$ is determined by the problem's conditions.  You can, however, assume the direction of the impulse is purely horizontal (because the split in the ball looks to be purely vertical).  But the dependency on the time of splitting and the velocity of splitting is still troublesome., because the landing spot of one ball only fixes one parameter.
However, if you use the principle of "this is a posed problem, the time of splitting and the velocity must conspire to give one unique answer" then this is an easy solve.  
Without the splitting, the ball would have landed at $10\frac{\sqrt{40}}{g} \approx 20.2$ meters.  If the splitting happens at the last split-second, the halves will end up equi-distant from the splitting point, which is at $10$ meters -- given -- and $30.4$ meters -- the answer.
In point of fact, as long as the splitting impulse is horizontal, the answer will remain $30.4$ meters.  The algebra is not too convoluted.
A: According to Wikipedia, mass is irrelevant so the only discrepancy in where they land will come from differences in the angle of trajectory and initial height caused by how the projectile is split.
