Finding bullet speed before and after passing through block 1

[nchin]

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 + m1v1the 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.

 

[Ibix]

If the bullet ends up stuck in the second block then what is vbf2?

Incidentally, I think you are slightly over-complicating this. There are only three velocities you can calculate - the initial velocity, the velocity after passing through block 1, and the final velocity. If I am understanding your notation correctly (I suspect you have been careless with cut-and-paste in your definitions of the vb values) then vbf1=vbi2, and having both is unnecessary. Your diagram, however, seems to suggest that vbf1 is the velocity at some point inside block 1. Perhaps worth a re-draw?

 

[nchin]

If the bullet ends up stuck in the second block then what is vbf2?

Incidentally, I think you are slightly over-complicating this. There are only three velocities you can calculate - the initial velocity, the velocity after passing through block 1, and the final velocity. If I am understanding your notation correctly (I suspect you have been careless with cut-and-paste in your definitions of the vb values) then vbf1=vbi2, and having both is unnecessary. Your diagram, however, seems to suggest that vbf1 is the velocity at some point inside block 1. Perhaps worth a re-draw?

Please go with my drawing as I think it makes a lot more sense than my notations. But isn't the bullet stuck in block one too?? Since caluclating vbf1 is necessary what isn't vbf2 necessary?? The bullet do get stuck in blocks one and two right?

 

[Ibix]

The bullet passes through block 1

According to your original post, no. The bullet exits block 1 and comes to rest in block 2.