Before explaining the two.solutions, i would like to observe that we can solve this very quickly by making one important simplifying observation:
At the top of their curve, they separate. Their velocity at the top of the curve is horizontal. The new energy of explosion is directed horizontally too. So there is zero change to their vertical velocity due to the explosion. So they will take the same time to reach ground, among other things.
If they separated but no extra energy was added, their separation at landing would obviously be zero. So all we need to do is totally ignore everything except two factors - how long till they land, and the relative horizontal velocity of the fragments due to the separation.
Time of flight after separation: vertical velocity 20.sin(60) = 17.3m/s hence 17.3/g = 1.46 seconds up and therefore also 1.46 seconds down.
Relative velocity due to separation is said to be velocities that result in double the previous KE as at highest point. At the highest point the system has no vertical velocity,its only KE is horizontal velocity: 5kg moving at 20.cos(60) m/s = 10 m/s. KE is mv2/2 = 250 J. So KE after explosion is twice this, or 500J.
This is easy now, we solve for no net momentum change, final KE of 500J and fragments of mass 1kg+4kg:
- m1 = 4 and m2 = 1 (labelling larger fragment m1)
- m1v1 + m2v2 = (m1 + m2).10 (no net change to momentum), therefore substituting and rearranging: v2 = 50 - 4.v1
- m1v12/2 + m2v22/2 = 500 (system KE is doubled from 250J to 500J). Rearranging and substituting we get 4.v12 + (50 - 4.v1)2 = 1000 => 20.v12 - 400.v1 + 1500 = 0. Solving the quadratic, we find v1 = 15 or 5. Since v2 = 50 - 4.v1 we have two pairs of solutions:
- (v1, v2) = (15, -10) or (5, 30).
Checking KE and momentum, we find indeed both solutions have KE = 500J and momentum unchanged (50 kg.m/s). And totally unsurprisingly in both cases the relative velocity of the fragments is identical, as we would expect: | 15 - (-10) | = | 30 - 5 | = 25. The relative velocity after separation is 25 m/s in both cases.
So after separation we have particles in flight for 1.46 seconds separating at 25 m/s.
(The fact they accelerate downwards doesnt matter because the downwards element of the fragments' motion have identical velocity at all times, so vertical motion and gravity wont make any relative change to separation or flight time of either fragment.)
Separation at landing is 1.46 x 25 = 36.5 m
Why two solutions?
Because there are 2 ways to double the systems KE. There are 2 ways to describe whats going on.....
Remember the mass reaches separation with positive velocity, 10 m/s. You could add a small amount of energy, both fragments still end up with positive (horizontal) velocities immediately after separation. Add a bit more, one of them ends up with zero horizontal velocity however, and in this case total system KE is less than doubled. Add even more energy and that fragment is sent into reverse, and has positive KE again. Thats why there are 2 solutions.
Another way to look at it is, one fragment is projected forward and one backward. Because they have different masses and they have a non-zero existing velocity (just before separation), there will be 2 sets of solutions, one for when the heavy mass is sent forwards (and the light mass is very greatly sent backwards, into reverse), the other when the heavy mass is sent backwards (but not enough to send it into reverse) and the light mass is very greatly sent forwards. So again, we get 2 solutions.
These two explanations are flip sides of the same coin. If you want a separation that exactly doubles the system KE, you have two choices: a smaller explosion that separates them with 4kg going backward and 1kg going forward, or a larger explosion sufficient to sent 4kg forward and 1kg backwards.
Anything in between these, ends up with less than doubled system KE. Anything beyond these, more than doubles it.
What if they split apart in a different horizontal direction?
Note that for simplicity I've assumed the fragments fly off in line with their existing flight. The question says they fly off horizontally but not that they fly off in the same direction horizontally. They could both fly off sideways, in opposite directions. In that case you'd have an extra variable Θ (the angle they fly off) and an extra equation ( total momentum conserved = 0 "sideways). Some of these equations would have terms in sin Θ and cos Θ, but otherwise nothing would really change. You'd still solve it the exact same way.
What would we expect and why?
We would expect exactly the same distance.
Remember that all we cared about is their relative velocity - not their absolute velocity. Yes the absolute velocity matters for final KE, but i think this logic is correct, and we can bypass that aspect by changing reference frame in a way that should not change the result. Read on......
If we used fragment 1's horizontal position as our frame of reference (we dont care about any velocity in common, nor the actual KE, only the separation at landing), v1 would have been zero, and we know from above that v2 after impact would have been 25 or -25 (still 2 solutions, but now they are clearer as to meaning, the relative direction of the smaller mass).
So intuitively, that itself has a physical interpretation: if they were launched straight up, and split apart, a relative velocity of 25 m/s would be the correct answer - you'd get two solutions again (+5, -20) and (-5, +20), obtained by subtracting 10 m/s from the above solution. This would still, give conservation of momentum and correct separated KE.
But notice if thats the case, then they could launch straight up, and split apart with relative velocity 25 m/s in any horizontal direction (because then every horizontal direction would be effectively identical in that frame), and that would mean that whatever horizontal direction they split, the relative velocity and time to landing would be unchanged. So it would still be 36.5 m at landing.
Notice the change of reference frame to prove this: from the observers frame they moved 10 m/s at peak in a specific direction, switching to a frame of that speed they are moving vertically only and have no horizontal velocity initially, which means we can pick any positive horizontal axis direction and the relative velocity after separation is unchanged by it.
All these frame changes are to other inertial frames, and in any constant horizontal velocity frame, horizontal momentum and relative horizontal velocity are conserved at separation, so we can see that the outcome is unchanged.