solution:

Because it is an elastic collision, both kinetic energy and momentum are conserved. We can make two equations, one for the conservation of KE, and one for the conservation of momentum. There are two unknowns in each equation ( tex2html_wrap_inline131 of each block) and we will solve for one and plug it into the other equation.

eqnarray6

I will now solve for tex2html_wrap_inline133 in eqn (4) and plug it into eqn (3) and solve for tex2html_wrap_inline135

eqnarray23

eqnarray63

Now use the Quadratic Equation to solve for tex2html_wrap_inline135 :

eqnarray92

It is clear that it can't be 4m/s because that was its original velocity. tex2html_wrap145

eqnarray105

tex2html_wrap147

Another way to solve the problem in a much easier fashion is to know that the relative velocities before and after the collision must be equal. Use the relation tex2html_wrap_inline143 and the conservation of momentum, and you will end up with the same answer without nearly as much algebra.