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Bullet fired at a Block at rest.

  • Thread starter Trojanof01
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  • #1
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Bullet is fired at a block resting on a table. The velocity of the bullet is 390 m/s and has a mass of .007kg. The mass of the block is .850kg. The bullets penetrates the block and exits it with a velocity of 130 m/s causing the block to slide a distance of .410 m.

What is the coefficient of friction?


So, I thought maybe I could use the equation F= (mu)n -------> F = (mu)mg

and v^2 = (Vo)^2 + 2a(x-xo);

I plugged in the final and initial velocities along with the displacement and solved for a.

Plugged a into the equation F=ma to get the force then used that number and set it equal to (mu)mg to solve for (mu). Ended up with some number that was way too large to be a coeff. of friction.

I'm guessing it was a mistake to assume the velocity of the block at the split second after the bullet exited, was the same as the velocity of the bullet.

Any help would be appreciated...
 

Answers and Replies

  • #2
Dick
Science Advisor
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Think conservation of momentum. When the bullet is exiting the block the sum of its momentum and the momentum of the block has to equal the initial momentum of the bullet.
 
  • #3
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Solved it, had the velocity of the block incorrect and that screwed me up in the final calculation. Thanks.
 

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