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Inelastic Collision Bullet/Block of Wood

  1. Oct 24, 2013 #1
    Let me preface this by saying that this is a take home quiz and I'd really appreciate it if you didn't actually answer the question, I have outlined what I've done and what I plan to do to get an answer and I'm looking for input on my problem solving process.

    I have a problem that I've been staring at for a while now and am not able to come up with a good starting place. The problem is:
    A 20g bullet is fired into a 1000g block hanging from a 120cm long string. The bullet embeds itself into the block, and the block then swings out at an angle of 25 degrees. What is the speed of the bullet?

    So far I've split it into two different time intervals, the first with the block at rest where Ft=-mg=0, and the second interval where the block is at 25 degrees with a velocity of zero. Then I resolved the forces acting on the block at each point. I'm just wondering where I should go from here. Using the inelastic collision equation I don't really get an answer because I don't know what v' is the instant after collision.

    I guess the next step I would try is to use kinematics in 2 directions to find the instantaneous acceleration during impact, then use that to find the force exerted on the block by the bullet, and finally to find the velocity. Would this method get me where I'm going or is there something I'm missing?
     
  2. jcsd
  3. Oct 24, 2013 #2

    tiny-tim

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    hi rmiller70015! :smile:
    call it v', then, and carry on :wink:

    (and force is irrelevant)
     
  4. Oct 24, 2013 #3
    I don't know if I can call it v' since I'm looking for velocity to begin with, that leaves me with two variables. I'm also wondering if I can call v' the same velocity as what would be in the equation v=rω, but here I don't know my time interval.
     
  5. Oct 24, 2013 #4

    tiny-tim

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    i'm not really following you :confused:

    (and what does ω have to do with it anyway?)

    you've divided the motion into two intervals

    write out the collision equation for the first interval

    what equation do you think you can use for the second interval?​
     
  6. Oct 24, 2013 #5
    I split it into two intervals for no real reason other than maybe I'd find something useful in there, but I don't think I did. As for what you're taking about the inelastic eqn is mAvA+mBvB=(mA+mB)v'. Since the only velocity I know is vB=0, I am left with v_A and v' as unknowns.

    What I'm thinking at this point is to use angular motion, since the block would move along an arc.
     
  7. Oct 24, 2013 #6

    tiny-tim

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    since this is a take home quiz, i can't really tell you what equation to use for the second interval

    (angular motion isn't going to help you at all)

    what tells you how far up the pendulum will swing?​
     
  8. Oct 24, 2013 #7
    Hi rmiller70015,

    Have you considered comparing the energy before and after the collision?
     
  9. Oct 28, 2013 #8
    After perusing through the course material I found that while I can't actually use conservation of energy on an elastic collision, I can use the conservation of mechanical energy for the time after the projectile stops deforming the block of wood, until the point where the wood/bullet system is at v=0, thank you for all your help here.
     
  10. Oct 28, 2013 #9

    tiny-tim

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    hi rmiller70015! :smile:
    (you mean inelastic :wink:)

    that's correct …

    you can always split the motion into two parts, one "during" the collision (in which mechanical energy is not conserved), and one "after" (in which it is) :smile:

    (and of course you have to join them together by using an unknown speed v at the "join")
     
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