Momentum Problem Physics?

Question

A 75 kg shell is fired with an initial speed of 125 m/s at an angle 55 degrees above horizontal. Air resistance is negligible. At its highest point, the shell explodes into two fragments, one four times more massive than the other. The heavier fragment lands directly below the point of the explosion.
Problem 9.66

If the explosion exerts forces only in the horizontal direction, how far from the launch point does the lighter fragment land?

Answer 1

Vx = 125 m/s * cos(55) = 71.7 m/s

75 * 71.7 = 5377.28 kg m/s = 60 kg * 0 + 15 kg * Vx2

Vx2 = 358.5 m/s

Vy = 125 m/s * sin(55) = 102.4 m/s
@ height gt = Vy...t = Vy/g = 10.44s
if 10.44 s up then 10.44 s down

Xup = 10.44 * Vx
Xdown = 10.44 * 5 * Vx

Total distance traveled = Xup+Xdown = 10.44*6*Vx = 4491.1 m==============================
g = 9.81 m/s/s and 15kg is the smaller of two fragments.
A = 4B, A + B = 75 = 4B + B = 5B, B = 15

Answer 2

MV=M1V1
75*125=(75/4)V1(velocity after impact)
Then use the V1 as the Vi for the projectiles part of the problem.
Using the y-component, V1sin55, to find the highest distance with Vf^2=Vi^2+2aX where Vf=0, Vi=V1sin55 and a=-9.8
Now use the new found X to find t in any equation you chose(like Vt=X) and use this t to find X in the x-component of the projectile
X=Vit+1/2 at^2 where Vi=V1cos55, a=-9.8 and t the same as found for the y component