The final velocites cannot be on the same side of the original axis (or else, there's no way linear momentum could be conserved perpendicularly to that axis) so one angle is positive and the other is negative. Reversing both signs leads to a symmetrical situation in which the final speeds are the same. So, without loss of generality, we may assume angles to be +33° and -46°.

Let's call M the mass of either puck. Before the collision, the cartesian coordinates of the total linear momentum is (with v=5.4 m/s):

[ Mv , 0 ] + [ 0, 0 ] = [ Mv , 0 ]

After the collision, the same momentum is given by a new expression:

[ Mu cos 33° + Mw cos 46° , Mu sin 33° - Mw sin 46° ]

where u and w are the final speeds of the two pucks. Therefore, we have:

u cos 33° + w cos 46° = v

u sin 33° - w sin 46° = 0

Solving for u and v this system of two linear equations (by elimination) we obtain:

u [ cos 33° sin 46° + sin 33° cos 46°] = v sin 46°

w [ sin 33° cos 46° + cos 33° sin 46°] = v sin 33°

Both brackets are equal to the determinant sin (33°+46°) = sin 79°. Therefore:

u = v (sin 46° / sin 79°) = 0.73280 v = 3.957 m/s

w = v (sin 33° / sin 79° ) = 0.55483 v = 2.996 m/s

Rounded to the same precision as the question (2 significant digits)

the speeds of the two pucks are respectively 4.0 m/s and 3.0 m/s <==== Final answer.

COMPLEMENTS:

1) Notice that the collision is not elastic; the ratio of the final kinetic energy to the

initial one is (u^2+w^2)/v^2 = 0.84484 = 84 %

(16% of the energy is lost as heat and/or rotational energy.)

In an elastic spinless collision, that ratio would be 100% and the angle between the outgoing velocities would be 90° instead of 79°.

2) We don't know (nor do we care) which puck was initially moving or initially at rest... Either one could be ejected at the lesser angle of 33°; it all depends on details of the collision which are discarded here and would be ***fundamentally*** obscured if the pucks were truly identical tiny quantum objects with slightly uncertain positions (see last link below, which is otherwise irrelevant to this classical problem ;-)