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Simple physics problem

  1. Mar 17, 2006 #1
    Ok, here's the question:
    A 7.0 g bullet is fired into a 1.5 kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 9.3 cm. Find the initial speed of the bullet.
    I'm using the formula:
    m1v1i + 0 = (m1 +m2)vf
    so i substitute in the info I'm given:
    .007v1i + 0 = 1.507kg(.002km/s)
    v1i = 0.43057 km/s, or 430.57 m/s but this is obviously not correct. Any suggestions?
     
  2. jcsd
  3. Mar 17, 2006 #2
    A 7.0 g bullet is fired into a 1.5 kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 9.3 cm. Find the initial speed of the bullet.

    Let me explain you the scenario. ENERGY IS CONSERVED!

    Initial:
    The bullet is moving and has an initial KE, the block isn't moving. We assume that the bullet has no GPE.

    Final:
    The bullet is moving, slower, less KE than initially. The block has a kinetic energy and a GPE.

    So...

    [tex]KE_{bullet} = KE_{bullet} + KE_{block} + GPE_{block}[/tex]

    [tex]1/2 M_{bullet} (V_{Binitial})^2 = 1/2 M_{bullet} (V_{Bfinal})^2 + 1/2 M_{block} (V_{block})^2 + M_{block} g h[/tex]

    Basically, remember SI units. Convert cm into m, g to Kg.

    Just plug in everything and it should work fine. Here is an explanation:

    The total energy of the system is conserved. The energy of the system initally of both the bullet and the block must equal the energy of the bullet and the block finally.
     
    Last edited: Mar 17, 2006
  4. Mar 17, 2006 #3

    Curious3141

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    This is wrong. Total energy is conserved, but there is energy lost to sound and heat, which your analysis does not account for.
     
    Last edited: Mar 17, 2006
  5. Mar 17, 2006 #4

    Curious3141

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    Use conservation of linear momentum for the overall problem.

    Initial conditions : you know the mass of the bullet. You don't know the initial velocity (this is the unknown you must determine).

    Final conditions : you know the final momentum of the bullet (mass is given and final velocity is given). You don't know the final momentum of the block.

    How to find that out ? Here you must use conservation of energy. The block gets a certain kinetic energy imparted to it after impact. Because of it's tether the block moves in a circular arc and rises to a certain height after which it stops momentarily. The gain in Gravitational Potential Energy at this point is equal to the loss in Kinetic Energy. Can you form an equation here to determine what the velocity of the block just after impact was ?

    You cannot use conservation of energy in analysing the impact itself because there is energy that is converted to heat and sound which you cannot account for. Nevertheless you can use it to determine what the conditions just after the impact were.
     
  6. Mar 17, 2006 #5

    Integral

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    I do not see anything wrong with Da-force's help. In this level of problem losses due to things like noise are always neglected, whats unusual about that?

    I think a good question would be wether or not the mass of the bullet ought to contribute the GPE of the block.
     
  7. Mar 17, 2006 #6
    In my physics text, a similar problem, although slightly different as the bullet gets stuck inside the block, the initial energy of the system is 236J, after the impact, the energy is 0.589J. In fact, nearly ALL the energy was lost due to noise and heat. The losses can be significant, which is why I think the problem should state outright if those losses can be safely ignored. By not saying so, we are simply taking a guess that its true.
     
    Last edited: Mar 17, 2006
  8. Mar 17, 2006 #7

    Pengwuino

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    An introductory physics textbook shouldn't have questions about noise and friction when it comes to collision problems.
     
  9. Mar 17, 2006 #8

    Curious3141

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    This is incorrect. Solving the problem correctly (my way), the initial velocity of the bullet is found to be 489 m/s. That gives an initial KE of 0.5*0.007*(489^2) = 837 J

    The final KE = mgh (using GPE of block to simplify) + KE of bullet = 1.5*9.8*0.093 + 0.5*0.007*(200^2) = 141 J

    A whopping (837 - 141) = 696 J, or 83 % of the initial energy has been lost to heat and sound !

    Obviously, Da Force's approach is wrong.

    Why should it ?
     
  10. Mar 17, 2006 #9

    Curious3141

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    Why not ? There are some obvious tacit assumptions like neglecting air resistance, but why must you think that the absence of heat and sound on impact is a necessary one ?
     
  11. Mar 18, 2006 #10
    Ok, this thread is in need of some clarity for everyone, myself included. First and foremost, I will do something that is a no no and post a solution, because I feel that it is VERY IMPORTANT to prove a point here.

    To Da-Force, you have to be VERY careful when it comes to collisions. Your energy equation has one error in it:

    [tex]KE_{bullet} = KE_{bullet} + KE_{block} + GPE_{block}[/tex]

    You have double counted the energy of the block in the RHS of your equation. You included its energy as both kinetic and gravitational, you have 'double counted' the energy.

    Let me solve this problem to clarify the energy losses were talking about here:

    First of all, we know that the bullet goes through the block, and comes out the other end. Lets just use energy and see what kind of numbers we are talking about here:

    [tex] \frac{1}{2} mV_{bullet,initial}^2 = \frac{1}{2}mV_{bullet,final}^2+m_{block}gh [/tex]

    lets plug in numbers,

    [tex] \frac{1}{2}.007V_{bullet,initial}^2 = \frac{1}{2}.007*(200)^2+1.5*9.81*.093 [/tex]

    Lets solve this for the intial speed, v, and see what we come up with. we get a value of v equal to: 200.97511 m/s. Does that make sense? No, it does not.


    Lets do this using the correct method, as Curious3141 mentioned. MOMENTUM.

    We can calculate the intial speed of the block just after impact using energy methods. We know that all of the kinetic energy of the block went into its potential energy, thus:

    [tex] \frac{1}{2}m_{block}V^2 = mgh [/tex]

    Now, plug in values:

    [tex] \frac{1}{2}1.5V^2 = 1.5*9.81*0.093 [/tex]

    This gives us an inital speed of the block as 1.35079 m/s.

    From here, lets conserve momentum.

    [tex] m_{block,1}v_{block,1} +m_{bullet,1}v_{bullet,1} = m_{block,2}v_{block,2} + m_{bullet,2}v_{bullet,2} [/tex]

    Lets plug in values and see what the intial speed of the bullet is this time:

    [tex] 0.007v = .007(200) + 1.5(1.35079) [/tex]

    We get an inital speed of, 489.42 m/s. THIS sounds reasonable.

    And the difference in the energy between the two methods is VERY big.

    If I have made a mistake please let me know.
     
  12. Mar 18, 2006 #11

    Pengwuino

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    With the given information, theres no way to tell how much energy is lost to either form of energy so i don't see how they could be included in such a question.
     
  13. Mar 18, 2006 #12

    Integral

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    I stand corrected. :blushing:
     
  14. Mar 18, 2006 #13

    Pengwuino

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    Hmm.... im assuming im wrong because i haven't taken momentum conservation into account....
     
  15. Mar 18, 2006 #14
    I know because when I took phyics 1 a few years ago I made this mistake on the homework. :blushing:

    If you really want to split hairs here, you should argue that my equation,

    [tex] \frac{1}{2} 1.5 v^2 = 1.5*0.093*9.81 [/tex]

    Should REALLY be 1.5 + .007 because the bullet is embeded in the block for that first instance, however, that makes the new intial velocity only1.3476, basically it does not change the anwser.

    In response to:

    It is a very good question to ask here. Because depending on the dynamics of the impact, the bullet may or may not be able to be considered to pass through the blcok so fast that the block does not have time to rise will the bullet is still inside of it or not. But the mass of the bullet is so small that this should be insignificant.
     
    Last edited: Mar 18, 2006
  16. Mar 18, 2006 #15

    Curious3141

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    That's a fair point. It shouldn't affect the analysis much.
     
  17. Mar 18, 2006 #16

    Curious3141

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    Cool. :cool:
     
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