Speed of a bullet after passing through a block of mass

[ttk3]

Homework Statement

A bullet with a mass of 3.55 g is fired horizontally at two blocks resting on a smooth and frictionless table top as shown in the Figure. The bullet passes through the first 1.00 kg block, and embeds itself in a second 1.95 kg block. Speeds v1 = 1.80 m/s and v2 = 3.00 m/s, are thereby imparted on the blocks. The mass removed from the first block by the bullet can be neglected. Find the speed of the bullet immediately after emerging from the first block.

Homework Equations

I think it should be:
(M1 + M2)Vf = M1 x Vo
M1 = bullet
m2 = mass of block 1
vf = v1

The Attempt at a Solution

(.00355 + 1) 1.8 = .00355Vo

Vo = 508.8 m/s

 

[delecticious]

lol, I think you're in my physics class. For the first part the collision is inelastic, so what you would is you'd have the mass of the bullet (m) times the velocity of the bullet after emerging from block 1 (vm) equal to the sum of block 2 and m (m + m2) times v2. So put that together and you have m*vm=v2(m+m2). For the second part you need to use the velocity you found in the first part, the difference is the collision is elastic not inelastic. So what you have is m (the mass of the bullet) times vm (the velocity of the bullet after leaving block 1) plus the m1 (mass of block 1) times the v1 set equal to m times v0 (the bullets initial velocity. Put that together and you have m *vm + m1*v1 = m*v0 and after that you should have your answer.

 

[learningphysics]

The bullet is embedded in block 2 (the 1.95kg block, not the 1.00kg block)

At the beginning you only have the bullet moving... total initial momentum = 0.00355Vo

At the end you have block 1 (1.00kg) moving at 1.80m/s. And the bullet+block 2 moving at 3.00m/s. what is the total final momentum...

set initial momentum = final momentum.

 

[ttk3]

Thanks, that makes sense.