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Maximum height after a collision (general solution)

  1. Nov 1, 2016 #1
    Here's the situation:
    A small block of wood of inertia mb is released from rest a distance h above the ground, directly above your head. You decide to shoot it with your pellet gun, which fires a pellet of inertia mp. After the block has fallen a distance d, the pellet hits it and becomes embedded in it, kicking it upward. At the instant of impact, the pellet is moving at speed vp.

    I'm supposed to find the maximum height formula, and this is what I got but it's incorrect:
    http://imgur.com/gallery/QQoFRm7
    QQoFRm7.jpg

    *Relevant equations* :
    x = vt + 1/2at^2
    v = at
    p = mv
    vf^2=vi^2+2ad

    Can someone help me on this? What I did was 1) found the velocity of the block after falling distance d, 2) found the velocity of the block + pellet after collision by viewing it as a completely inelastic collision, 3) plugged it into the vf^2 = vi^2+2ad, 4) added (h-d). Does anyone see where I'm going wrong? I have 2 more tries until I don't get any credit for it. I know this is a long problem so much thanks if anyone can point me in the right direction.
     
    Last edited by a moderator: Nov 1, 2016
  2. jcsd
  3. Nov 1, 2016 #2

    haruspex

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    Right approach, but consider the directions of the velocity vectors before impact.
     
  4. Nov 1, 2016 #3
    The algebra gets kind of tricky in this problem, but it looks like your method is good. I'm not sure why that large term in your equation is negative, but I haven't tried to work through all of your sign definitions. But I think that term should be positive. Also, in the denominator of that large term, inside the parentheses is the sum of the momentums of the pellet and the block just before impact. I would expect that one of those terms would be negative since the velocity of the pellet is upward and the velocity of the block is downward. This next thing is probably just my own preference, but also in that large term, I don't know why you have a 'g' in the denominator inside the radical and a 'g' in the numerator outside the radical. It seems simpler to me to just have a 'g' in the numerator inside the radical. Okay, maybe I'm being too picky about that one. :)
     
  5. Nov 1, 2016 #4
    I think it might be a sign error related to g but I'm not sure. When you have g in an equation is it supposed to just be the magnitude? I viewed g as -9.8, which is why I subtracted the second term from (h-d) iirc (it's 5:30 AM right now and I haven't had the chance to really look it over in depth again)

    And additionally w/the direction of the velocities, I've been doing something similar and "adding a negative". Is this wrong to do that?
     
    Last edited: Nov 1, 2016
  6. Nov 1, 2016 #5
    I think you are right that the problem with your solution not being correct is related to gravity being a negative. But I'm not sure what is the best way to deal with that. Some of the more seasoned guys on here may know of a prescribed method to deal with that, but I don't. I think it might work if you define the up direction as negative, and the down direction as positive. That way your g will be a positive value. But the catch there is that, in the final step, when you use one of the kinematic equations to solve for x (the total distance that the block and pellet move together), the resulting expression will be positive in the down direction. So the increase in height of the block and pellet combination would then be the negative of that expression. . . . I think. :)
     
  7. Nov 1, 2016 #6

    haruspex

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    Your choice, as long as you are consistent.
    My view is that if you have adopted the sign convention that up is positive then the symbol g has a negative value, and change in velocity after time t is gt, not -gt. But almost everybody takes g to have a positive value and writes the change in velocity as -gt.
    With regard to your algebra in this question, you have g inside a square root sign, so presumably you are giving it a positive value. So what is the correct sign on the momentum of the block of wood, ##m_b\sqrt{2gd}##, just before impact?
     
  8. Nov 1, 2016 #7
    Ah I think I see where I went wrong, thank you! It should be -mb√2gd.
     
  9. Nov 1, 2016 #8

    haruspex

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    Yes.
     
  10. Nov 1, 2016 #9
    Just wanted to post an update, I got the right answer: it was (h-d) + (-mb√2gd........ thanks for the help everyone! Pesky sign errors...
     
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