1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bullet fired at a Block at rest.

  1. Mar 31, 2007 #1
    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...
  2. jcsd
  3. Mar 31, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    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.
  4. Mar 31, 2007 #3
    Solved it, had the velocity of the block incorrect and that screwed me up in the final calculation. Thanks.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook