Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Inelastic Collision with Friction

  1. Oct 16, 2004 #1
    I forget how to solve one dimensional motion problems with friction, which is holding me up in this problem:

    a 5.00 g bullet is fired horizontally into a 1.2 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between the block and surface is .2. The bullet remains embedded in the block and the block moves 0.23 m before stopping. What is the initial speed of the bullet?



    bullet------->block----.230 M-[ ]

    looking at it as a momentum problem:

    M(bullet)*V(bullet) = M(Bullet+Block)*V(bullet+block)

    you need to solve for v(bullet+block) which can be determined by the following problem:

    A block of mass 1.25 kg on a surface with kinetic friction of .2 is acted on upon a force and move .230 M before coming to rest on a horizontal surface. Find the initial speed of the bullet.


    m = 1.25
    v_0 = x
    v_x = 0
    mu_k = .2
     
  2. jcsd
  3. Oct 16, 2004 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, in the second phase, you have constant deceleration equal to 0.2g
     
  4. Oct 16, 2004 #3
    I understand that.

    What I dont understand is the logic to find the initial velocity.

    I know what this answer is, but I need to apply the exact same logic to a seperate problem. So what I am really asking for is someone to solve this problem and show the steps.

    Gallium
     
  5. Oct 16, 2004 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, you have:
    [tex]v_{f}^{2}-v_{0}^{2}=2*(-0.2g)d[/tex]
    Right?
    ([tex]v_{f}=0[/tex], d is 0.230)
    This enables you to solve for [tex]v_{0}[/tex]
     
  6. Oct 16, 2004 #5
    that doesnt yield the correct answer.
     
  7. Oct 16, 2004 #6

    Pyrrhus

    User Avatar
    Homework Helper

    Because that's the speed for the bullet+box system, not for the bullet, now use conservation of momentum to find the initial speed of the bullet.
     
  8. Oct 16, 2004 #7
    .05 * V_1 = .9212*1.25

    V_1 = 23.03 m/s

    V_1 actually = 229 m/s
     
  9. Oct 16, 2004 #8

    Pyrrhus

    User Avatar
    Homework Helper

    1 Kg is 1000 grams not 100 grams.

    5grams is 0.005 Kg.
     
  10. Oct 16, 2004 #9
    i see, thank you. I found the correct logic.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook