How could I solve this problem using the ZMF concept? I understand how this would be done in a 1D problem, so could I apply the same logic, finding a vector that makes momentum in each direction zero, however, impulse is only applied in the i direction, so it would not be as simple as to multiply the velocity of A in the ZMF by e?. Here is what I have so far $$2\begin{bmatrix} 3cos(45) \\ 3sin(45) \end{bmatrix} + \begin{bmatrix} -2cos(60) \\ 2sin(60) \end{bmatrix} = \begin{bmatrix} 3x \\ 3y \end{bmatrix} $$ Hence vector of frame, (x,y) $$ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3\sqrt2-1 \\ 3\sqrt2+\frac{\sqrt3}{2} \end{bmatrix} $$ The next step I think would be to find the velocities of A and B in the frame, but I cannot think what would be done afterwards.

How could I solve this problem using the ZMF concept? I understand how this would be done in a 1D problem, so could I apply the same logic, finding a vector that makes momentum in each direction zero, however, impulse is only applied in the i direction, so it would not be as simple as to multiply the velocity of A in the ZMF by e?. Here is what I have so far $$2\begin{bmatrix} 3\cos(45) \\ 3\sin(45) \end{bmatrix} + \begin{bmatrix} -2\cos(60) \\ 2\sin(60) \end{bmatrix} = \begin{bmatrix} 3x \\ 3y \end{bmatrix} $$ Hence vector of frame, $(x,y)$ $$ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3\sqrt2-1 \\ 3\sqrt2+\frac{\sqrt3}{2} \end{bmatrix} $$ The next step I think would be to find the velocities of A and B in the frame, but I cannot think what would be done afterwards.

How could I solve this problem using the ZMF concept? I understand how this would be done in a 1D problem, so could I apply the same logic, finding a vector that makes momentum in each direction zero, however, impulse is only applied in the i direction, so it would not be as simple as to multiply the velocity of A in the ZMF by e?. Here is what I have so far $$2\begin{bmatrix} 3cos(45) \\ 3sin(45) \end{bmatrix} + \begin{bmatrix} -2cos(60) \\ 2sin(60) \end{bmatrix} = \begin{bmatrix} 3x \\ 3y \end{bmatrix} $$ Hence vector of frame, (x,y) $$ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3\sqrt2-1 \\ 3\sqrt2+\frac{\sqrt3}{2} \end{bmatrix} $$ The next step I think would be to find the velocities of A and B in the frame, but I cannot think what would be done afterwards.

How could I solve this problem using the ZMF concept? I understand how this would be done in a 1D problem, so could I apply the same logic, finding a vector that makes momentum in each direction zero, however, impulse is only applied in the i direction, so it would not be as simple as to multiply the velocity of A in the ZMF by e?. Here is what I have so far $$2\begin{bmatrix} 3cos(45) \\ 3sin(45) \end{bmatrix} + \begin{bmatrix} -2cos(60) \\ 2sin(60) \end{bmatrix} = \begin{bmatrix} 3x \\ 3y \end{bmatrix} $$ Hence vector of frame, (x,y) $$ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3\sqrt2-1 \\ 3\sqrt2+\frac{\sqrt3}{2} \end{bmatrix} $$ The next step I think would be to find the velocities of A and B in the frame, but I cannot think what would be done afterwards.