From conservation of momentum we know that in the center of mass frame of the fragments the momentum of any fragment is equal and opposite to the sum of the momenta of the other two fragments, and since all fragments have the same mass, this must also hold for their velocities. Call their velocities U₁, U₂and U₃, with U₃being fastest. We have

(1) U₃ = −(U₁+ U₂).

From conservation of energy we know that ½m(|U₁|²+ |U₂|² + |U₃|²)= E, where E is the total kinetic energy imparted to the fragments by the explosion and m is the mass of each fragment, so

(2) |U₃|²= 2E/m − (|U₁|²+ |U₂|²).

Writing (|U₁|²+ |U₂|²) as ½(|U₁+ U₂|²+ |U₁− U₂|²) and using (1) we find,

(3) |U₃|²= (4E/m − |U₁− U₂|²)/3.

|U₃| is maximal when |U₁− U₂|is minimal, which happens when U₁= U₂= −½U_3. Thus, in the CM frame the fastest fragment is moving in the direction opposite the other two fragments and twice as fast. Furthermore, U₃ must be in the same direction as the shell's velocity V in the ground frame in order for their sum, the velocity of the fastest fragment in the ground frame, to have maximum magnitude. Thus, in the ground frame the fastest fragment is moving in the same direction that the shell was moving before it exploded. Having observed these two points, the problem reduces to simple algebra: Let the speed of the shell in the ground frame be V = 500 m/s. Let |U₃| = 2U, and |U₁| = |U₂| = U. Noting that the speeds of the fragments in the ground frame are V+2U (fastest fragment) and V−U (other two fragments), we can translate the given kinetic energy increase into,

(4) 3/2 * (1/2)(3m)V^2 = (1/2)(2m)(V−U)^2 + (1/2)m(V+2U)^2,

which simplifies to V = 2U. Hence the speed of the fastest fragment is V + 2U = V + V = 1000 m/s.