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September 25th, 2009, 05:14 PM
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| |  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? |
September 25th, 2009, 05:26 PM
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Quote:
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.
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September 25th, 2009, 05:57 PM
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I've already tried that... I got it wrong. |
September 25th, 2009, 06:12 PM
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Quote:
Originally Posted by TGS I've already tried that... I got it wrong. | Please show your work and I will review it.
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)
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September 25th, 2009, 06:41 PM
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Last edited by mr fantastic; September 25th, 2009 at 06:44 PM.
Reason: Improved the latex and formatting
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September 25th, 2009, 06:47 PM
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Quote:
Originally Posted by TGS | Since the bullet is embedded in the block, the correct equation is (0.205 kg) (250) = (3.5 + 0.205)  ....
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)
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September 25th, 2009, 06:54 PM
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