Verified answer

in non-elastic crashes; m*V = (m+m/2)*V'

V' = 2V/3

initial energy; (m*V^2)/2

final energy; (3m/2*(2V/3)^2)/2

so lost energy is (m*(V^2))/6 and it becomes heat.

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♠ total momentum of the system of 2 ships before clinch q1=M*v +(M/2)*0;

♣ total momentum of the system of 2 ships after clinch q2=M*v2 +(M/2)*v2;

♦ now 2 fundamental laws in effect:

♪ momentum conservation q1=q2 or Mv=1.5Mv2 or v=1.5v2 or v2=2v/3, and

♫ energy conservation E=E1+E2+E3, where E=0.5*Mv^2 is initial kinetic energy of the system, E1=0.5*M*v2^2 is KE of bully ship, E2=0.5*(M/2)*v2^2 is KE of bullied ship, E3 is heat in question; thus

♥ E3=0.5Mv^2 –{0.5Mv2^2 +0.5(M/2)v2^2}=

=0.5Mv^2 –0.5M*(2v/3)^2 *1.5 =0.5Mv^2 *(1/3) joules, if M in kg and v in m/s;

yes, river is right!