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Impulse and Momentum of a collision between a block and bullet

  1. Nov 4, 2006 #1
    The question is "A 2 kg block initially hangs at rest at the end of two 1-meter strings of negligible mass. A .003 kg bullet, moving horizontally with a speed of 1000 meters per second, strikes the block and becomes embedded in it. After the collision, the bullet/block combination swings upward, but does not rotate.

    a. Caclulate the speed of the bullet/block combination after the collision.
    I got 1.5 m/s for this. I did M1Vo1 + M2Vo2 = M3Vf and solved for Vf.

    b. Calculate the ratio of the initial kinetic energy of the buillet to the kinetic energy of the bullet/block combination immediately after the collision.
    I got 668:1. I used the equation .5MVf^2 - .5MVo^2 = .5 (.003kg)(1000 m/s)^2 = 1500 and 1/2MVf^2 = .5 (2.003)(1.5)^2 = 2.247, and so 1500:2.247 = 668:1

    c. Calculate the maximum vertical height above the intial rest position reached by the bullet/block combination.
    I'm not really sure how to do this one... would I just do something like .5MVo^2 + .5(9.8)(1) to get it? (kinetic energy + potential energy)

    d. Suppose the bullet went through the block instead of imbedding in it. How would this affect the maximum height? Explain.
    I said that the height would decrease because there would not be as much velocity since the bullet is slowed down by the force of the block.

    I'm not really sure on my answers. I made my best attempt, and I was wondering if anyone could help/guide me through the problem so that I understand it without giving me the answers (until I get the right answer). If not, it's okay, I understand you all must be very busy ^^. Thanks in advance for reading this.
     
    Last edited: Nov 5, 2006
  2. jcsd
  3. Nov 4, 2006 #2
    The title should read "between a block and bullet" sorry.
     
  4. Nov 4, 2006 #3
    Okay for part C I did this
    .5(2.000+.03)(1.5m/s)^2 = (2.003)(9.8)(h)
    h = .5(1.5)^2 / 9.8 = .115 m
     
  5. Nov 4, 2006 #4
    Whoever changed the title I appreciate it immensely. I'm extremely sorry again.
     
  6. Nov 4, 2006 #5
    I'm mostly concerned with parts b and d.. I have a good feeling a and c might be right, but b and d leave me extremely confused. The calculations for b look correct for me, but I dont know if there should be such a great difference in the ratio. And in part d, it seems that the maximum height would decrease because the velocity of the block should be slower.. but I'm not entirely sure why.. =(
     
  7. Nov 5, 2006 #6
    a.
    [tex]m_bv_b=(m+m_b)v_f[/tex]
    b.
    [tex]\frac{KE_b}{KE_f}[/tex]
    c.
    [tex]KE=PE[/tex]
    d.
    Think about it, if mass decreases in the momentum and kinetic energy equations, what happens to final velocity and height?
     
  8. Nov 5, 2006 #7
    I'm going to bed now and will be up in a few hours, but I'm going to state what I think are the answers now based on your equations.

    a) (.003 kg)(1000 m/s) / (.003 kg + 2 kg) = Vf...... Vf = 1.5 m/s
    b) .5(.003 kg)(1000 m/s)^2 / .5(2.003 kg)(1.5 m/s)^2 = 666:1 ratio..
    c) .5mv^2 = mgh .... .5V^2=gh... h=.5V^2/g.. h = .5(1.5)^2/9.8 = .115 m
    d) The final velocity and height should decrease because the mass is decreasing, and there is a smaller product when multiplied by the velocity.

    So my answers are a) 1.5 m/s b) 666:1 ratio c) .115 m d) The maximum height decreases

    Thanks for your help! It's making more sense now :) I appreciate it
     
    Last edited: Nov 5, 2006
  9. Nov 5, 2006 #8
    On part d you are still not understanding. Solve the momentum equation for final velocity. You will find that it is inversely proportional to final mass. If final mass decreases, velocity will increase. If you solve the equation we used for c for final height you will find the same result.
     
  10. Nov 5, 2006 #9
    Oh.. wow. That makes sense. If the bullet goes through the block, it maintains its 1000 m/s and it has such a small mass that the velocity has to increase! And if the velocity is increasing and the mass is decreasing, that means the maximum height has too! I get it now! Thanks :D
     
  11. Nov 5, 2006 #10
    Oops. I made a mistake in my last post. I forgot one thing. Look at this:

    [tex]P_(bullet)=P_(block)+P_(bullet final)[/tex]

    We forgot about the third term. When the bullet passes through the block with its initial momentum it gives the block a certain momentum which is the second term. What I forgot about was the momentum that the bullet still has, which depends on its speed as it leaves through the other side of the block. From here you can do some algebra and find what the height of the block depends upon.

    Sorry about the mistake, it was really early in the morning over here :blushing:
     
  12. Nov 5, 2006 #11
    It's not your fault ^^; I'm sorry for keeping you up so late.

    Well I'm thinking that the bullet has to be slowed down when it passes through the block.. it's speed should decrease, and when its speed decreases the maximum height decreases.. so eventually the maximum height.. decreases?
     
  13. Nov 5, 2006 #12
    Let me think again.. the momentum of the bullet when it starts is mv.. which is .003(1000) = 3 kg m/s... when the bullet hits the block is has a momentum of (2.003)(1.5) which is about 3 kg m/s.. but then you have to add the momentum of the bullet afterwards.. so that means the maximum height would have to increase because no matter what its value is the overall momentum afterwards will increase.. so its height has to increase? I think I'm confusing myself.. x.x
     
  14. Nov 5, 2006 #13
    Let's go through the algebra:

    v_b = velocity of bullet before collision
    v_f = velocity of bullet after collision
    v_c = velocity of block after collision

    [tex]1. m_bv_b=m_cv_c+m_bv_f[/tex]

    [tex]2. v_c=\frac{m_bv_b-m_bv_f}{m_c}[/tex]

    [tex]3. \frac{1}{2}m_cv^2_c=m_cgh[/tex]

    [tex]4. h=\frac{\frac{1}{2}m_cv^2_c}{m_cg}[/tex]

    Simplify:

    [tex]h=\frac{\frac{1}{2}v^2_c}{g}[/tex]

    Whereas when the bullet stuck in the block we had:

    v_f=final velocity of bullet-block system.

    [tex]v_f=\frac{m_bv_b}{m_c+m_b}[/tex]

    [tex]h=\frac{\frac{1}{2}(m_c+m_b)v^2_f}{(m_c+m_b)g}[/tex]

    Simplify:

    [tex]h=\frac{\frac{1}{2}v^2_f}{g}[/tex]

    So, what is your answer?
     
    Last edited: Nov 5, 2006
  15. Nov 5, 2006 #14
    Um...okay so the height of the block is zero before the collision.. and when the bullet gets stuck in the block its height is .115m. Even if the bullet goes through the block, it still hits the block in the beginning so it should be pushed upwards with the same momentum.... so the height doesn't change?
     
  16. Nov 5, 2006 #15

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Your intuition on part d was right in the first place, but the maximum height of the block will depend on what portion of the bullets momentum is transferred to the block. You could do an algebraic calculation that shows the maximum height of the block in terms of the final velocity of the bullet when it leaves the block.
     
  17. Nov 5, 2006 #16
    Okay.. so I was right when I thought the height would decrease because not as much momentum is transferred to the block, and the final velocity of the bullet would decrease .. ?
     
  18. Nov 5, 2006 #17
    Lets go through this:

    First off, we have

    h_1=height block reaches when bullet sticks inside
    h_2=height block reaches when bullet passes through

    [tex]h_1=\frac{\frac{1}{2}v^2_f}{g}[/tex]

    [tex]h_2=\frac{\frac{1}{2}v^2_c}{g}[/tex]

    [tex]g[/tex] and [tex]\frac{1}{2}[/tex] are constant so we only need to worry about [tex]v^2_f[/tex] and [tex]v^2_c[/tex]. The equations we have for them are:

    [tex]v_f=\frac{m_bv_b}{m_c+m_b}[/tex]

    [tex]v_c=\frac{m_bv_b-m_bv_f}{m_c}[/tex]

    Assuming the mass of the bullet [tex]m_b[/tex] is very small compared to the momentum of the bullet after the collision in the case where it passes through, which is the most likely case. We have a case where [tex]v_f > v_c[/tex]. More concisely, we can let [tex]m_b[/tex] be negligible (this isn't completely accurate but it gives a more dramatic demonstration of the result we should expect), which leaves us with:

    [tex]v_f=\frac{m_bv_b}{m_c}[/tex]

    [tex]v_c=\frac{m_bv_b-m_bv_f}{m_c}[/tex]

    Looking back at our initial equations for height, we now know that h_1 is greater than h_2, i.e. the block has a larger height when the bullet sticks inside of it.
     
    Last edited: Nov 5, 2006
  19. Nov 6, 2006 #18
    Okay I did all the arithmatic, substituted all the text for vf and vc in the height equations, and it becomes apparent that the max height decreases because the v_c decreases due to subtracting the m_b and v_f, and as a result the overall momentum and max height decreases when the bullet goes through the block.. thanks so much.
     
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