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A 470g firework is traveling straight up at 13 m/s

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data

    a 470 g firework is traveling straight up at 13 m/s when it explodes into two pieces. The smaller piece (150 g ) shoots off horizontally towards the East at 20 m/s. Find the speed and direction of the other piece directly after the explosion.


    2. Relevant equations

    p=mv
    pbefore collision = pafter collision


    3. The attempt at a solution

    I first found the mass of the 2nd piece by using mass of whole firework (470g) and the mass of small piece (150g). mt=ms+m2 = 320g = 0.32kg

    then i did pt= p1 + p2
    mtvt = m1v1 + m2v2---> solve for v2
    0.47kg(13m/s) = 0.15kg(20m/s) + 0.32kg(v2)
    v2= 9.7 m/s

    but now i am not sure if i did this right i feel i am missing a step or 2! please help!!
     
  2. jcsd
  3. Dec 14, 2013 #2

    Curious3141

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    Not right. Remember that velocity is a vector, and has a direction. You need to work with vectors here. Or you can just conserve momentum along the vertical and horizontal axes separately.
     
  4. Dec 15, 2013 #3
    not sure what you mean
     
  5. Dec 15, 2013 #4

    Curious3141

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    I mean that what you did (simple addition/subtraction) is only applicable in problems that involve a single axis of motion. For instance, one ball colliding into another initially stationary one on a smooth table. When more axes (dimensions) are involved, you need to consider them separately.

    The easiest approach is to conserve momentum separately along the two perpendicular axes here: vertically and horizontally. Say "up" is positive (and down is negative) along the vertical axis, and "East" is positive (and West is negative) along the horizontal axis. Now write the conservation statements separately for vertical and horizontal axes. Let the other piece have a final velocity wH horizontally and wV vertically. You need to determine these.

    I'll start you off.

    Initial momentum: Vertical = 0.47*13. Horizontal = 0.

    Final momentum (write it in terms of the two w terms and don't forget the 0.15kg fragment): Vertical = ? Horizontal = ?
     
    Last edited: Dec 15, 2013
  6. Dec 15, 2013 #5
    okay so i would find the horizontal and vertical V of the other piece and then using pythagorean theorem find the final v for the other piece? correct?

    1) horizonal: m1v1=m2v2 = 0.15(20)=0.32v2
    2) vertical = 0
    3) pythagorean theorem

    correct?
     
  7. Dec 15, 2013 #6
    sorry scratch the previous msg so i found the vertical and horizontal V's of the other piece and then using pythagorean theorem i find the overall V correct?

    1) horizontal: pinitial=0=m1v1+m2v2 --> v2 = -9.375
    2) vertical: pinitial = 6.11N = m2v2 (p1=0) ---> v2=19.0
    3) pythagorean theorem: v2 = 21.3m/s
     
  8. Dec 15, 2013 #7
    sweet!! thanks :)
     
  9. Dec 15, 2013 #8

    Curious3141

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    Looks right. Except you need to put in units for the speed and the momentum unit should be Ns (Newton-second).

    How do you work out the direction? Think trig.
     
  10. Jan 20, 2015 #9
    Thanks
     
    Last edited: Jan 20, 2015
  11. Mar 31, 2015 #10
    I'm doing this question but I used a different approach. The answer above of 21.3 m/s doesn't seem to make sense to me. If the momentum is conserved why would the heavier piece go faster than the lighter piece?

    My approach was to start with the momentum as a vector. The initial p going in a +y direction. One piece then goes to the east on the positive x-axis. This forms a right angle triangle with the change of momentum and its direction being the hypotenuse.

    the y component was 6.11 kg*m/s
    the x component was (.150g)*(20 m/s) = 3 kg m/s
    the hypotenuse was then 6.81 kg * m/s

    The other piece is heading west opposite in direction to the eastern piece.
    Using the hypotenuse of 6.81 kg *m/s
    and the vertical component of 6.11 Kg * m/s
    it comes to the 3.0 kg m/s which of course makes sense since it should be equal to the opposite of the other piece..

    using the mass of .320kg for the other piece, v = (3 kg*m/s) / .320g = 9.4 m/s

    This seems to be valid as the heavier object will go slower than the smaller object in opposite directions.
     
  12. Mar 31, 2015 #11

    Dick

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    I think you got basically the same answer as in the original post. Except you only found the horizontal component of the velocity. The original question was asking for speed of the heavier part which combines both the horizontal and vertical components. And, yes, the speed of the heavier part is larger than the speed of the lighter part. This is possible because there was an initial momentum.
     
  13. Mar 31, 2015 #12
    I just realized that I did find the horizontal and vertical components... it's the hypotenuse that I calculated from using pathagorems theory..

    If i take the 6.81 kg * m/s / .320 kg = 21.3 m/s

    i was only using the horizontal component when I had the combined magnitude already.

    Thats what I get for doing physics at midnight.
     
  14. Mar 31, 2015 #13

    Dick

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    Well, you did get it all right except for spelling Pythagoras as Pathagorem. Could be worse.
     
  15. Mar 31, 2015 #14
    D'oh. I blame the essay on aristotelian logic that I just wrote tonight for a scientific thinking class... give a guy a break. thanks though.
     
  16. Mar 31, 2015 #15

    Dick

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    That's actually very acute. Aristotle would have made an assumption like the 'heavier part will always have a smaller speed than the lighter part". He was prone to stuff like that. Assume the essay went well? Just kidding about the spelling part.
     
  17. Mar 31, 2015 #16

    The essay is finished that's well enough for now. No sweat on the spelling, made me chuckle.
     
  18. May 4, 2015 #17
    Ok I was just going over my assignment one more time before submitting this question and the other 19 questions with it. I'm still unsure of something. The question asks what speed and direction the larger piece is moving at given that the smaller piece is moving horizontally at a given velocity of 20 m/s. Aren't we then interested in the horizontal speed and direction for this larger piece as well? Aren't we then looking for the answer to be 9.4 m/s in a westward direction?
     
  19. Oct 7, 2015 #18
    I am somewhat confused about how the total velocity of the second piece is being established.

    I understand the horizontal velocity is found to be 9.4m/s (due to an explosion / additional chemical energy being added into the system). The explosion rips the object into two unequal pieces and they push away from each other with equal amounts of momentum, thus conserving the original horizontal momentum of zero.

    Piece a:
    P=(ma)va
    P=(.150)(20)
    P=3

    Piece b:
    P=(mb)v
    -3=(0.320)v
    vb=-9.4m/s

    However, in my approach to the problem, I was thinking that this should not affect the vertical component.

    Initially:
    P=mv
    P=(0.470)(13) = 6.11

    After:
    P=(ma + mb)v
    6.11=(0.320 + .150)v
    v= 13m/s

    If you still have the same amount of total mass, and you consider the vertical component to be isolated from the explosion/friction/gravity... then shouldn't the vertical momentum and therefore velocity remain conserved? Similar to as if you were in a skating rink moving in one direction with a friend, and you pushed away from them in a perpendicular direction (so adding in new energy)... you and your friend would still remain at the same speed in the initial direction regardless of what was going on in the perpendicular direction, wouldn't you?

    Finally, using Pythagorean theory for the vertical and horizontal velocity; (Root of 13^2 + 9.4^2) I ended up with a total velocity of 16.04m/s up and to the West.

    If I take your number of 21.3m/s and I work it backwards using a 9.4m/s horizontal component... you guys are saying that the upwards velocity of the larger piece has somehow increased to 19.1m/s.
     
    Last edited: Oct 7, 2015
  20. Oct 7, 2015 #19
    The first piece is moving horizontally. It has ho vertical component of velocity or momentum.
    So all the vertical momentum goes to the other piece. Its vertical velocity won't be 13 m/s but more (there is less mass moving up with the same momentum).
     
  21. Oct 7, 2015 #20
    Thanks for the reply, that makes sense. It wasn't immediately clear to me in the original question
     
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