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!

A bullet-knot-block system with momentum

  1. Sep 15, 2014 #1

    B3NR4Y

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    You shoot a 0.0050-kg bullet into a 2.0-kg wooden block at rest on a horizontal surface (Figure 1) . After hitting dead center on a hard knot that runs through the block horizontally, the bullet pushes out the knot. It takes the bullet 1.0 ms to travel through the block, and as it does so, it experiences an x component of acceleration of -4.5 × 105 m/s2. After the bullet pushes the knot out, the knot and bullet together have an x component of velocity of +10 m/s. The knot carries 10% of the original inertia of the block.
    Using conservation of momentum, compute the final velocity of the block after the collision.
    2. Relevant equations
    [itex] p=mv [/itex]
    [itex] p_{T1} = p_{T2} [/itex]

    3. The attempt at a solution

    The first part of the problem had me calculate the initial velocity of the bullet, which I did and got a correct answer of 460 m/s. Now I am not sure what to do with the problem because all of the problems we had before had instantaneous collisions, not collisions that happen over time, which I assume has integrals involved, so would I integrate over the time given? I assume the step is a pretty easy and I am just missing what to do. I don't know if I should calculate the momentum the moment to block hits and after the block hits and then find out the velocity of the block, or if since there is a time where the bullet is inside of the block and pushing the "knot" if that causes any significant change in the velocity.
     
  2. jcsd
  3. Sep 15, 2014 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    That looks correct to me. :approve:

    Very nice.

    Conservation of momentum applies whether the collision is instantaneous or not (and of course, assuming there are no external forces or torques). The fact that this collision is not instantaneous doesn't matter. Conservation of momentum still applies.

    Just take the system's total momentum before the collision [Edit: i.e., before the bulled strikes the block], and set that equal to the system's total momentum after the collision [Edit: i.e., after the collision is completely finished]. There is only one unknown variable: the velocity of the remainder of the block. Solve for that. :smile:

    [Edit: btw, no integration is necessary.]
     
    Last edited: Sep 15, 2014
  4. Sep 15, 2014 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi B,

    If it says "Using conservation of momentum, it probably means "Using conservation of momentum", so why not try that ? Write down an equation for conservation of momentum.

    Done? OK, let's continue then:
    For the momentum of the bullet before the collision you have all you need.
    For the knot too.
    For the rest of the block also.
    Personally, I don't understand why you don't show that intermediate result.

    Now comes the tough question: what information is really needed to "compute the final velocity of the block after the collision" ?

    [edit] boy, am I a slow typist...
     
  5. Sep 15, 2014 #4

    B3NR4Y

    User Avatar
    Gold Member

    Okay so since momentum is an extensive quantity, [itex] p_{total, i} = p_{block, i} + p_{bullet, i} [/itex] for the initial momentum, and for the final momentum [itex] p_{total, f} = p_{block, f} + p_{bullet+block \, piece} [/itex] and using the definition of momentum, and knowing that hte piece of the block and bullet's masses are (10% of 4 kg) + 0.0050 kg, or 0.405 kg, and I get the mass of the block to be 3.6 kg, and my equations become:

    [itex] 0.0050 (kg) \, * \, 460 (m/s) = 3.6 (kg) * v_{f} + 4.05 kg*m/s [/itex]

    and solving for vf I get

    [itex] (2.3 kg*m/s \, - \, 4.05 kg*m/s)/3.6(kg) = v_{f} [/itex]

    This throws me off cause it is negative, I might be thinking about the problem wrong, but a negative answer doesn't make sense to me. I got an answer of ~-0.49 [edit: which I entered into the program and is wrong]
     
    Last edited: Sep 15, 2014
  6. Sep 15, 2014 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What was the mass of the block ?

    [edit] kudos for smelling something wrong, though!
     
  7. Sep 15, 2014 #6

    B3NR4Y

    User Avatar
    Gold Member

    I feel incredibly dumb. 2 kilograms, not 4

    [edit: I learned a valuable lesson in reading problems today]
     
  8. Sep 15, 2014 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Happens all the time. You're in good company. Finding your own errors and fixing them makes you a good scientist in the long run. (ahem...)
     
  9. Sep 15, 2014 #8

    B3NR4Y

    User Avatar
    Gold Member

    Yes, thanks for your help! I'm quickly starting to learn in this physics course that if an answer seems wrong, it probably is.
     
  10. Sep 21, 2015 #9
    How do you find initial velocity?
     
  11. Sep 22, 2015 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For the block it is a given, so I suppose you mean for the bullet ...
    Write out some conservation of momentum equations and then work backwards.
     
  12. Sep 22, 2015 #11

    B3NR4Y

    User Avatar
    Gold Member

    Thanks for your help! But this is a year old, this homework is long over lol
     
  13. Sep 23, 2015 #12

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. Molly would be better advised to post her/his own thread.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A bullet-knot-block system with momentum
Loading...