# Thread: How to calculate the final velocity.

1. ## How to calculate the final velocity.

A bullet of mass 0.205 kg traveling horizontally at a speed of 250 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface.

What is the speed of the block after the bullet embeds itself in the block?

2. Originally Posted by TGS
A bullet of mass 0.205 kg traveling horizontally at a speed of 250 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface.

What is the speed of the block after the bullet embeds itself in the block?
Use conservation of momentum.

3. I've already tried that... I got it wrong.

4. Originally Posted by TGS
I've already tried that... I got it wrong.

5. $m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

$m_1 v_{1i} + 0 = 0 + m_2 v_{2f}$

(0.205 kg) (250m/s) = (3.5kg) $(v_{2f})$

$v_{2f} = 14.643$

6. Originally Posted by TGS
$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

$m_1 v_{1i} + 0 = 0 + m_2 v_{2f}$

(0.205 kg) (250m/s) = (3.5kg) $(v_{2f})$ Mr F says: Your mistake is here.

$v_{2f} = 14.643$
Since the bullet is embedded in the block, the correct equation is (0.205 kg) (250) = (3.5 + 0.205) $(v_{2f})$ ....

7. ohhhhh... thank you.