How to calculate the final velocity.

**TGS** · September 25th 2009, 04:14 PM

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?

**mr fantastic** · September 25th 2009, 04:26 PM

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.

**TGS** · September 25th 2009, 04:57 PM

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

**mr fantastic** · September 25th 2009, 05:12 PM

Originally Posted by TGS

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

Please show your work and I will review it.

**TGS** · September 25th 2009, 05:41 PM

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

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

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

$v_{2f} = 14.643$

**mr fantastic** · September 25th 2009, 05:47 PM

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) $(v_{2f})$ ....

**TGS** · September 25th 2009, 05:54 PM

ohhhhh... thank you.

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