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anonymous

  • one year ago

A 5.0 kg stationary grenade explodes into three pieces. If a 0.5 kg piece moves at 35 m/s [E] and a 1.5 kg piece moves at 25 m/s [S 30° E], determine the velocity of the third piece?

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  1. anonymous
    • one year ago
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    I am using the trigonometry law to account for the final piece, well, as far as conservation of energy holds true .

  2. anonymous
    • one year ago
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    The addition of all of these forces must equal zero

  3. anonymous
    • one year ago
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    0.5kg/s*35m/s=17.5kgm/s 1.5kg/s*25m/s=37.5kgm/s Now use the vector arrows to indicate for the hypotenuse

  4. anonymous
    • one year ago
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    |dw:1440660915883:dw|

  5. anonymous
    • one year ago
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    Use law of conservation of momentum \[\vec p=\vec p_{1}+\vec p_{2}+\vec p_{3}\] momentum before=momentum after \[M\vec u=m_{1}\vec v_{1}+m_{2} \vec v_{2} +m_{3} \vec v_{3}\] \[\vec v_{3}=\frac{1}{m_{3}}.(M \vec u-m_{1} \vec v_{1}-m_{2} \vec v_{2})\]

  6. anonymous
    • one year ago
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    |dw:1440660968779:dw|

  7. anonymous
    • one year ago
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    The triangle is imaginary, hypotenuse is pointing towards upper right.

  8. anonymous
    • one year ago
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    We know that \[\vec u=\vec 0\] Therefore our equation becomes \[\vec v_{3}=-\frac{m_{1} \vec v_{1}+m_{2} \vec v_{2}}{m_{3}}\]

  9. anonymous
    • one year ago
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    Right, and final velocity obtained will be hence (17.5+37.5)N/3.0kg

  10. anonymous
    • one year ago
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    Now I am only left to find the direction which satisfies the vector with cosine

  11. anonymous
    • one year ago
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    or so I think

  12. anonymous
    • one year ago
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    \[v_{3}=|\vec v_{3}|=|-\frac{m_{1}\vec v_{1}+m_{2} \vec v_{2}}{m_{3}}|=\frac{1}{m_{3}}.(m_{1}v_{1}+m_{2}v_{2})\] \[v_{1}=|\vec v_{1}|=\sqrt{v_{1x}^2+v_{1y}^2}, v_{2}=|\vec v_{2}|=\sqrt{v_{2x}^2+v_{2y}^2}\]

  13. anonymous
    • one year ago
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    Now I get it

  14. anonymous
    • one year ago
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    But I think this assumes one dimensional momentum though

  15. anonymous
    • one year ago
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    Two dimensions where two are canceling each other out becomes a question of finding the hypotenuse.

  16. Jhannybean
    • one year ago
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    This is a general physics question dealing with the momentum?

  17. anonymous
    • one year ago
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    Look|dw:1440661555489:dw|

  18. anonymous
    • one year ago
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    |dw:1440661454515:dw|

  19. anonymous
    • one year ago
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    If you mean 30 degrees south of east, just switch the angles

  20. anonymous
    • one year ago
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    |dw:1440662053313:dw|

  21. anonymous
    • one year ago
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    Ops.

  22. anonymous
    • one year ago
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    Now left with the sine law to get the vector version of the velocity.

  23. anonymous
    • one year ago
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    This was a good question

  24. Astrophysics
    • one year ago
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    You want a difficult physics question? :P

  25. Astrophysics
    • one year ago
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    Try this |dw:1440662683161:dw|

  26. anonymous
    • one year ago
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    Angle to the horizontal?

  27. Astrophysics
    • one year ago
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    No numbers, but a hint: Both the incline and box are accelerating

  28. anonymous
    • one year ago
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    I would think that it depends on the mass of each object

  29. anonymous
    • one year ago
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    and gravity being 9.8m/s^2

  30. anonymous
    • one year ago
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    and mass cancels out

  31. Astrophysics
    • one year ago
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    Hmm nope, try setting it up, as you would a regular incline plane question, but remember the incline itself is accelerating.

  32. anonymous
    • one year ago
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    |dw:1440663228557:dw|

  33. Astrophysics
    • one year ago
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    Not bad but remember the incline also has weight and a normal force |dw:1440663465987:dw|

  34. anonymous
    • one year ago
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    Center mass originates from the tip of force due to gravity huh?

  35. anonymous
    • one year ago
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    Man gotta love physics

  36. imqwerty
    • one year ago
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    The force that will generate acceleration in the wedge would be the horizontal component of the normal force by the smaller box :)

  37. anonymous
    • one year ago
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    Yes, but how would you calculate that using gravity?

  38. anonymous
    • one year ago
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    Angle would have to be known as well because normal force decreases as the angle increases.

  39. anonymous
    • one year ago
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    Good question anyhow

  40. imqwerty
    • one year ago
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    u can do this by 2 ways - 1) constraint relation 2)and the one that i told u :)

  41. imqwerty
    • one year ago
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    |dw:1440664770880:dw|

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