Conservation of momentum problem. Very confused Speed of Bullet?

[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.

 

[Nickie]

I can understand your confusion with this conservation of momentum problem. Let me try to explain it in a simpler way.

In this problem, we have a bullet with a mass of 3.50 g, which is fired horizontally at two blocks on a frictionless table. The first block has a mass of 1.20 kg and the second block has a mass of 1.80 kg. The bullet passes through the first block and embeds itself in the second block. After the collision, the first block has a speed of 0.630 m/s and the second block has a speed of 1.40 m/s.

Now, let's look at the two equations you have written:

a) mb x vbi2 = (mb + m2)v2
b) mb x vbi1 = mb x vbf1 + m1v1

In equation a, we are using the conservation of momentum principle, which states that the total momentum before a collision is equal to the total momentum after the collision. In this case, the initial momentum is the momentum of the bullet before it enters the first block, which is mb x vbi2. The final momentum is the combined momentum of the bullet and the second block after the collision, which is (mb + m2)v2. We can use this equation to find the speed of the bullet as it leaves the second block.

In equation b, we are using the conservation of momentum principle again, but this time for the collision between the bullet and the first block. The initial momentum is the momentum of the bullet before it enters the first block, which is mb x vbi1. The final momentum is the combined momentum of the bullet and the first block after the collision, which is mb x vbf1 + m1v1. We can use this equation to find the speed of the bullet as it enters the first block.

Now, to answer your question about why we cannot calculate vbf2 (final v of bullet leaving block 1) in part a, it is because we do not have enough information to do so. We only know the speed of the first block after the collision, which is 0.630 m/s. To find the speed of the bullet as it leaves the first block, we would need to know the mass of the bullet and the second block, as well as the speed of the first block before the collision. Since we do not have