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Collision Problem using Conservation of Momentum and Energy

  1. Oct 10, 2003 #1
    Here's the Problem: A bullet of mass m=19.8 g is shot vertically upward into a block of wood of mass M=119 g that is initially at rest on a thin sheet of paper. The bullet passes through the block, which rises to a height of H=0.71m above its initial position before falling back down. The bullet continues upward to a maximum height of h=4.70m. Using energy conservation, you can express the velocities of the bullet and block immediately after the bullet exits the block in terms of the heights h and H. Use this to somehow determine the initial velocity of the bullet.

    This is How I Tackled Throught the Guidance of BOTH My (HORRIBLE) Physics Book and My (EVEN WORSE) T.A.:

    Using Conservation of Energy I came up with the Eq'n:
    (1/2)(m+M)v^2= MgH + mgh

    I solved for v and got 5.01

    Apparently that's not the answer, so then I took a hint given in the book and Used Conservation of Momentum to get the equation:

    mv(initial) = (M + m)v

    I solved for v initial and got 35.1 m/s

    But apparently, It's STILL Wrong.. I've tried slight variations of the above two equations, and yet, I still continue to get it wrong. I'm not sure what the problem is..Can anyone help?
  2. jcsd
  3. Oct 10, 2003 #2
    Well, looking at your conservation of energy equation I think you're on the right track. You got MgH+mgh which is the final energy for both objects at their apex. And the one-half term is close, I think you simplified when you couldn't, see the velocities should be different so you can't factor out (M+m). There's an old demonstration, Galileo I think, leaning tower of Pisa, if you drop a heavy object and a light object they fall at the same rate and hit the ground at the same time because gravity affects them equally, I think Gene Cernan on Apollo 17 dropped a hammer and a feather on the moon (no air resistance) to show that. Well, this applies on the way up too, if they had the same velocity they would both apex at the same height. Now this gets tricky and I'm not sure you have enough information because when I look at this problem I see a lot of assumptions. First, I'd like to see a separation height, I would think the bullet and block would travel a bit before separation and at that height you would need to account for the potential energy in the equations. When I got close to your equation I had set h-sep=0. Nextly I would like to know where the bullet started. See I first drew the bullet way below the block and labelled that height energy level zero, E=0. But then when they didn't say how far that is there's a problem, see if it was really far from the block the bullet could start off real fast and slow down, but if it was close it wouldn't slow down much, it changes all the energies, and you can't set the block height to zero because the bullet would just be in an energy well (what we call a negative energy position). So I started by assuming the bullet was in a very short barreled gun, right under the block and that the bullet zipped through the block in no time so the separation height was zero. But see, with a different set of assumptions I could make your equation work, your eq has the same velocity which must result in the same final height, well what if they do end at the same height and the 0.71m refers to the block height from the block rest and the 4.7m refers to the bullet height from the bullet rest which was below the block by 4.01m? But I don't think that's what they want.

    Regardless, I would look at it like this, with the bullet starting right under the block and making that level be height=0 and energy=0 then all the energy comes from the bullet being fired. Energy can't just appear or disappear from the system so the bullet, when fired, has the energy E=mgh+(1/2)mv^2 potential and kinetic but we said the initial energy level was zero so no potential and E=(1/2)mv^2 Now with conservation of energy this must equal the final energy of the block and the bullet, the energy the block has is most easily taken from its apex, because it has stopped moving (just before it falls) and you know the height and that energy is E=MgH and for the bullet it is E=mgh so (1/2)mv^2=MgH+mgh and I got v=13.26m/s from that. Now for the first part, the separation velocities, I assume a separation height of zero so that there are no potential energies to worry about (this mathematics actually paints the picture that the bullet is fired while in the block and is gone before the block even begins moving, weird huh?) and so the energy in the block is entirely kinetic...well we know the energy imparted to the block, E=MgH, so the blocks initial energy must equal its final energy so the kinetic E=(1/2)mv^2 must =MgH, boom, solve for v. v=sqrt(2gH) do the same for the bullet and v=sqrt(2gh). And I think this is probably what you're being asked for.
  4. Oct 10, 2003 #3


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    Staff Emeritus
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    Notice that the problem said "you can express the velocities of the bullet and block". Velocities is plural!

    You did the problem twice, once using conservation of energy and the other using conservation of momentum and got two different answers because you assumed that the block and bullet have the same velocity at the point that the bullet exits the block. If they did, the bullet wouldn't exit the block!

    Rewrite your equations for total energy and momentum using, say, v for the velocity of the bullet and u for the velocity of the block. Since you now have two unknowns, you will need to use both conservation laws to get two equations.
  5. Oct 10, 2003 #4
    Maybe you can, but I think you don't need to.
    You know both maximum heights & masses. So you know the total energy in the system. This must clearly come from the bullet's initial motion.
  6. Oct 10, 2003 #5
    Thanks y'all, It turns out I was on the right track initially cuz at first when I was doin the problem I kept the velocities separate and figured out each velocity. But then I got stuck and when I went to recitation my confusing TA threw me (and the rest of the class) WAY OFF..But I got it now, Thanks again
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