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A bullet strikes a block like boom boom pow

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.48m long. The block is initially at rest. A bullet with mass 0.0118kg is fired at the block with a horizontal velocity V0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.875m , the tension in the cord is 4.94N .


    2. Relevant equations



    3. The attempt at a solution
    So what I first did was I created the equation for a completely inelastic collition

    MAVA=(MA+MB)vf

    Vf= final velocity MA(bullet)= 0.0118kg
    VA= Initial velocity of bullet MB(Block)= 0.750 kg

    I have to find a way to find the final velocity in order to get the initial speed of the bullet.

    So I used the equations

    U1=K2+U2 which is :
    Mgy1=1/2mv^2+mgy2

    squareroot(2(mgy1-mgy2))/m

    The masses cancel and you are left with

    squareroot (2[gy1-gy2]) = 3.44m/s

    Plugging the final velocity into the completely inelastic equations gives 222.08m/s.


    Not sure if I did the problem right, and its for mastering physics so some feedback would be greatly appreciated (: thank you
     
  2. jcsd
  3. Mar 17, 2012 #2
    The problem did not state that the maximum height the block rose to was 0.875 m. It only stated what the tension in the cord was when it was at that height.

    So knowing the tension, you can determine the speed of the block at that point. Go from there.
     
  4. Mar 17, 2012 #3
    So what you are saying is I use

    T-mg=ma

    Solve for the acceleration than use that to find the final velocity?
     
  5. Mar 17, 2012 #4
    You will have to determine the velocity of the block at the point where it has risen 0.875 m. Because it is still moving, the tension has two components. One is weight in the direction of the cord and the other is what you have when something rotates. You have the equation

    T-mg=ma

    You need a mathematical expression for 'a' in terms of velocity.
     
  6. Mar 17, 2012 #5
    What I tried doing is using t-mg=ma to solve for the acceleration. Then once getting the acceleration use the circular motion equation a=v^2/L Where L is 0.875, but my acceleration turns out to be negative and I cant solve for V.
     
  7. Mar 17, 2012 #6
    Write the equation like this:

    T =x mg + mV^2/L

    You only have a component of mg due to the angle, hence the x.
     
    Last edited: Mar 17, 2012
  8. Mar 17, 2012 #7
    Use trigonometry to find the multiplier x, then solve for V.
     
  9. Mar 17, 2012 #8
    That doesn't make sense to me. I understand that you set the acceleration to equal v^2/l

    But why would you isolate the T and add an unknown value x. My brain keeps wanting to isolate the v which gives me square-root (L(t-mg)/m)=v
     
  10. Mar 17, 2012 #9

    cepheid

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    The tension only picks up the component of the weight that is parallel to the cord.
     
  11. Mar 17, 2012 #10
    I want you to come up with what x is. I don't think I should merely tell you. I could have put a ? there instead. mg needs to be multiplied by something.
     
  12. Mar 17, 2012 #11
    mgxcostheta for one and mgxsintheta for another?

    and we solve for cos(theta) and sin(theta)
     
  13. Mar 18, 2012 #12
    You do use trigonometry. The x is either cos(theta) or sin(theta). Only one is correct and only one is used.

    The problem is solved by determining what the speed of the block and bullet together are after the collision. If you know that, you can determine what the bullet speed is from the momentum balance because there would be only 1 unknown.

    So we need to determine the block speed right after impact. The only data we have is the tension in the cord when the block with embedded bullet is 0.875 m above the point of bullet impact. The tension is due to two forces. One is an mV^2/L force. The other is the component of the weight of the block with bullet embedded which is where the sin(theta) or cos(theta) is used to determine the component. Which one to choose is your decision and only one is correct. You determine theta from geometry.

    Once you correctly arrange that force balance, you only have one unknown and that is the velocity of the block with bullet embedded. Knowing the velocity at a point where the block and bullet have risen 0.875 m, you can determine the velocity when it was first struck by the bullet by using conservation of energy.
     
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