Now, let's solve this problem in full:
I will assume that you gave me the right information, in particular that the collision was fully ELASTIC.
The equations for two objects, 1 and 2, to be solved are:
m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}
This is conservation of momentum, as you had; b and a refer to initial and final velocities (before and after collision).
Energy conservation:
\frac{1}{2}m_{1}v_{1,a}^{2}+\frac{1}{2}m_{2}v_{2,a}^{2}=\frac{1}{2}m_{1}v_{1,b}^{2}+\frac{1}{2}m_{2}v_{2,b}^{2}
This is also, I believe, what you meant.
Note that the energy equation can be simplified in the following manner:
m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a})+m_{2}(v_{2,b}-v_{2,a})(v_{2,b}+v_{2,a})=0
Rewrite the momentum equation as:
m_{2}(v_{2,b}-v_{2,a})=-m_{1}(v_{1,b}-v_{1,a})
Hence, the energy equation may be written as:
m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a}-(v_{2,b}+v_{2,a}))=0
The root v_{1,b}=v_{1,a} corresponds to no collision, so we have the following system to solve:
v_{1,b}+v_{1,a}=v_{2,b}+v_{2,a}
m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}
Solving for v_{1,a},v_{2,a} yields:
v_{1,a}=\frac{(m_{1}-m_{2})v_{1,b}+2m_{2}v_{2,b}}{m_{1}+m_{2}}
v_{2,a}=\frac{(m_{2}-m_{1})v_{2,b}+2m_{1}v_{1,b}}{m_{1}+m_{2}}
Now, we have simplified the symbolic expressions maximally; it is time to enter in the values!
We set:
m_{1}=149.9,v_{1,b}=1181.8
m_{2}=248.2,v_{2,b}=-1025
The answers should be quite different from the ones you gave
