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Yahoo!

  • 2 years ago

A projectile is fired upwards at a speed of 100 m/s at an angle of 37 with the horizontal.it explodes into two parts at the highest point in the mass ratio 1:3 and the lighter one comes to rest. The distance from the point of projection where the heavier mass lands is (g=10m/s^2)

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  1. Taufique
    • 2 years ago
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    |dw:1358017831169:dw|

  2. Diwakar
    • 2 years ago
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    At highest point v=100cos37=80m/s If total mass of body is m, then masses of two pieces will be m/4 and 3m/4. The lighter one comes to a rest . By momentum conservation, velocity of 3m/4=4*80/3=320/3 m/s This new projectile will now cover a horizontal distance from its point of formation i.e. the highest point in the trajectory of previous projectile ,s=v*sqrt(2h/g) where v=320/3 as found before. and h is the point of maximum height found by applying kinematics to the projectile of mass m. The total distance from point of projection =s+ half of the range of previous projectile. The range can be calculated just like maximum height i.e. apply kinematics to it. I find it convinient to remember the formulas for range and max height h=(v sintheta)^2/2g Range,r=v^2 sin(2*theta)/g here theta=37 v=100

  3. Taufique
    • 2 years ago
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    |dw:1358018149270:dw|

  4. Taufique
    • 2 years ago
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    R is the range ..(X com) is the centre of mass of the system..

  5. Yahoo!
    • 2 years ago
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    @Diwakar i also used Same Kind of ur Approach But Didnt Get the Result..

  6. Yahoo!
    • 2 years ago
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    and..@Taufique i did nt get ur Approach

  7. Taufique
    • 2 years ago
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    there is no external force so centre of mass will be same at intial and final so ,i have applied this concept..

  8. Yahoo!
    • 2 years ago
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    Centre of mass Will Be near to Heavier Body....

  9. Yahoo!
    • 2 years ago
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    and using the Formula for Range i calculated Pos ( CM) = 960 m..ryt ?

  10. Diwakar
    • 2 years ago
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    The c.m. will finally be at R. It will follow a parabolic trajectory

  11. Yahoo!
    • 2 years ago
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    The answer shuld be1120m

  12. Taufique
    • 2 years ago
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    @Yahoo! ..wait one minute

  13. Taufique
    • 2 years ago
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    @Yahoo! ..what is the answer in your text book for this question

  14. Yahoo!
    • 2 years ago
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    1120m

  15. Taufique
    • 2 years ago
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    sorry yahoo (xcom) without explosion =4m *R xcom after explosion =m*R/2 + k*3m ..then equate both then you will get right answer

  16. Taufique
    • 2 years ago
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    @ yahoo! ..i am sorry for this mistake ...

  17. Taufique
    • 2 years ago
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    @Yahoo! ..here k =1121.47 m answer

  18. Yahoo!
    • 2 years ago
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    No Problem bro

  19. Yahoo!
    • 2 years ago
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    4mR = m*R/2 + k*3m hw R/2 ????????

  20. Taufique
    • 2 years ago
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    here explosion is self explosion i.e no external force is here so we apply this method

  21. Yahoo!
    • 2 years ago
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    Any Reason Behind it ?..

  22. Taufique
    • 2 years ago
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    because small part is in rest and it will drop vertically downward at R/2 distance from origin

  23. Taufique
    • 2 years ago
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    we also solve this problem using momentum conservation ...

  24. Yahoo!
    • 2 years ago
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    If u Dont Mind..Can u Show ?

  25. Yahoo!
    • 2 years ago
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    i appl that method..but didnt get the result

  26. Taufique
    • 2 years ago
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    ok..|dw:1358020235037:dw|

  27. Taufique
    • 2 years ago
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    |dw:1358020365409:dw| |dw:1358020404165:dw| and find k..

  28. Taufique
    • 2 years ago
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    |dw:1358020473006:dw|

  29. Taufique
    • 2 years ago
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    |dw:1358020525435:dw|

  30. Taufique
    • 2 years ago
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    any query, yahoo!

  31. Yahoo!
    • 2 years ago
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    Got it..! THXXX

  32. Taufique
    • 2 years ago
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    now i am giving you a question which is similar to it ..you are interested to solve this

  33. Taufique
    • 2 years ago
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    An isolated particle of mass m is moving in horizontal plane (X-Y),along x-axis,at a certain height above the ground.it suddenly exlodes into two fragment of masses m/4 and 3m/4.An instant later the smaller fragment is at y=15 cm .what will be the position of larger fragment at this instant ???

  34. Taufique
    • 2 years ago
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    @Yahoo! ..solve this ..it will give you more concept on this type question

  35. Yahoo!
    • 2 years ago
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    Thxx...Brother.....i will try to Solve it

  36. Taufique
    • 2 years ago
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    ok... try it.

  37. Yahoo!
    • 2 years ago
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    35 ??

  38. Yahoo!
    • 2 years ago
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    @Taufique wat is the answer

  39. Taufique
    • 2 years ago
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    y=-5cm

  40. Yahoo!
    • 2 years ago
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    hw....?

  41. Taufique
    • 2 years ago
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    HINT: the body is moving on the x axis that is there is no motion along y axis this means the momentum will be zero along y-axis at any time

  42. Yahoo!
    • 2 years ago
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    Huh..Silly....Careless

  43. Yahoo!
    • 2 years ago
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    Didnt See it was on x-axis.......@Taufique if it was y axis..then the answer would be 35 ryt ?

  44. Taufique
    • 2 years ago
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    now find momentum along y-axis for this situation.. after the body will explode ..the momentum will be zero along Y-axis ..am i right?

  45. Yahoo!
    • 2 years ago
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    No..i Dont think

  46. Yahoo!
    • 2 years ago
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    Since it is moving..along Y-axis...it will have an Initial P

  47. Taufique
    • 2 years ago
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    no in the question ,it has been given that the body is moving along horizontal direction i.e X-axis

  48. Yahoo!
    • 2 years ago
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    If it was Along Y-Along

  49. Yahoo!
    • 2 years ago
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    *Axis

  50. Taufique
    • 2 years ago
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    then the momentum will be zero along x -axis at any time..because there is no component of velocity along Xaxis

  51. Taufique
    • 2 years ago
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    where is your confusion???

  52. Yahoo!
    • 2 years ago
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    Cleared...

  53. Taufique
    • 2 years ago
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    |dw:1358022098651:dw|

  54. Taufique
    • 2 years ago
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    y=k (let ) for 3m/4 mass |dw:1358022439653:dw|

  55. Yahoo!
    • 2 years ago
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    Got...It

  56. Taufique
    • 2 years ago
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    ok..

  57. Yahoo!
    • 2 years ago
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    Can u Take a luk at mmy Maths Question

  58. Yahoo!
    • 2 years ago
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    http://openstudy.com/users/yahoo!#/updates/50eee432e4b0d4a537cdad0d

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