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A football player weighing 75 kg running at 2 m/s toward west tackles a 70 kg player running at 1.5 m/s in the opposite direction. what is the final velocity of the players if they both fall together?

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Now add up the initial and final kinetic energies. 

KE1 = (1/2)*75*2^2 + (1/2)*70*(1.5)^2 
= 150+ 78.75 = 228.75 J 
KE2 = (1/2)*145*(0.31)^2 = 6.97 J
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m1v1_1 + m2v1_2 = vf(m1+m2) ..we're assuming they both have the same final velocity 
Plug in: (75)(-2) + (70)(1.5) = vf(75+70) 
Solve for vf = -0.31 m/s west (since I arbitrarily decided - is west) 
Final answer: 0.3 m/s west 75*2 - 70*1.5 = 145 V 
V = 0.31 m/s (west) 

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75*2 +70*(-1.5) = (75+70)V 

or V = (150-112.5)/145 = 0.259m/s west 

K before = 1/2*75*2^2 + 1/2*70*1.5^2 = 228.75J 

K after = 1/2*145*0.259^2 =4.86J 

So change = 4.86-228.75 = -224J