In Fig. 9-58a, a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.20 kg) and embeds itself in block 2 (mass 1.80 kg). The blocks end up with speeds v1 = 0.630 m/s and v2 = 1.40 m/s (Fig. 9-58b). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) leaves and (b) enters block 1. my attempt: mb = mass of bullet vbi1 = initial v of bullet entering block 1 vbf1 = final v of bullet entering block 1 vbi2 = initial v of bullet leaving block 1 vbf2 = final v of bullet leaving block 1 m1 = mass of block 1 m2 = mass of block 2 a) leaves net p initial = net p final mb x vbi2 = mb x vbf2 + m2v2 b) enters mb x vbi1 = mb x vbf1 + m1v1 the solution part a is a) mb x vbi2 = (mb + m2)v2 part b i did correctly. but how can you not calculate vbf2 (final v of bullet leaving block 1) of the bullet, when you have to calculate vbf1 (final v of bullet leaving block 2) for part b ??? I know the equations I got has two unknowns and cannot be solved but I do not understand the solution at all. I made a figure attached to this question, please take a look. I am very confused if I drew the diagram correctly.