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Boushra

The following diagram demonstrates an elastic collision between two balls of equal masses. The black ball is at rest and the final velocity v1 of the white ball = 3 m/s Find: 1. the initial velocity of the white ball, 2. the velocity of the black ball after the collision 3. the angle (theta) 4. the change in the kinetic energy Is there a possibility we can discuss everything here? I'm having a hard time understanding. please help!

  • one year ago
  • one year ago

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  1. Boushra
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    We should use the perfect inelastic collision equation m1u1 + m2u2 = (m1+m2)v

    • one year ago
  2. Vaidehi09
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    can u post the diagram as well?

    • one year ago
  3. Boushra
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    yep just a sec

    • one year ago
  4. meera_yadav
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    hii, is the black bal at rest after the collision? if yes then white ball's initial velocity is zero just before collision

    • one year ago
  5. Boushra
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    |dw:1341315928417:dw|

    • one year ago
  6. Boushra
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    the black ball is in motion after collision

    • one year ago
  7. Vaidehi09
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    then first of all, we can't use the eq u posted abv since the 2 bodies are not sticking together after collision.

    • one year ago
  8. Boushra
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    oh... yeah right @Vaidehi09

    • one year ago
  9. Boushra
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    should we use m1u1+m2u2 = m1v1 + m2v2 ??

    • one year ago
  10. Boushra
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    m1 = m2 = m

    • one year ago
  11. meera_yadav
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    |dw:1341316215450:dw|

    • one year ago
  12. Boushra
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    should we write an equation based on components?

    • one year ago
  13. meera_yadav
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    since this is an elastic collision then we must have momentum before collision = momentum after collision !! we must also have change in kinetic energy energy should be zero as there's no external force acting..

    • one year ago
  14. Boushra
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    yes understood :) then

    • one year ago
  15. meera_yadav
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    yep ! we should write moemtum shold be resolved.. i.e. horizontal momentum before collison = horizontal momentum after collision same goes for vertical momentum

    • one year ago
  16. meera_yadav
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    similar equation we can write it for vertical momentum also !

    • one year ago
  17. Boushra
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    so the first one you drew above is for the horizontal momentum?

    • one year ago
  18. meera_yadav
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    |dw:1341316752653:dw|

    • one year ago
  19. meera_yadav
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    yep ! it's for horizontal... 2nd draw is for vertical momentum..

    • one year ago
  20. Vaidehi09
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    @meera_yadav ur first drawing....shouldn't we have vcos(theta) there...instead of sine?

    • one year ago
  21. Boushra
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    for the first drawing: why did you write down sin theta v, shouldn't it be cos theta v? since we're talking about the horizontal? @meera_yadav

    • one year ago
  22. Vaidehi09
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    the eq for horizontal should be u1 = 3cos45 + vcosA

    • one year ago
  23. Vaidehi09
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    supposing theta = A

    • one year ago
  24. Boushra
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    yes, that's what I got @Vaidehi09

    • one year ago
  25. meera_yadav
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    yeah! sorry ! that's a mistake ... thanks for pointing out :) plz correct it to cos theta

    • one year ago
  26. Boushra
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    great!!! now we can move on to 2nd equation:

    • one year ago
  27. Boushra
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    .5 mu (squared) = .5 m 3(squared) + .5 m v(squared) <--- why did we use this?

    • one year ago
  28. Vaidehi09
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    wait a sec, but we haven't yet found the value of u1.

    • one year ago
  29. Vaidehi09
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    thats the eq for conservation of KE

    • one year ago
  30. Boushra
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    yeah, I know,... should we label it as eq 2?

    • one year ago
  31. Vaidehi09
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    for KE, i got the eq as (u1)^2 = 3^2 + v^2

    • one year ago
  32. Boushra
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    I think the v in it should be v subscript x, correct?

    • one year ago
  33. Vaidehi09
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    i don't think so. for KE, we don't need components since it is a scalar.

    • one year ago
  34. meera_yadav
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    no there's no need for the subscript vaidelhi , because K.E is a scalar quantity...

    • one year ago
  35. Boushra
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    ok, great, I'm following.. so what do we need next.

    • one year ago
  36. meera_yadav
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    I think everythings known to u know !!

    • one year ago
  37. Vaidehi09
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    we have another eq, using vertical components for momentum. and that is, vsin theta = 3sin45

    • one year ago
  38. Boushra
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    ok, hold on a sec

    • one year ago
  39. Boushra
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    shouldn't it be 3 sin 45 = - v sin theta?

    • one year ago
  40. Boushra
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    we have it as: 0 = m(3 sin 45) + m (v sin theta) before right?

    • one year ago
  41. Vaidehi09
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    nope. both the vertical components of velocities are in opp direcs. so we get, 0 = 3sin45 - vcosA

    • one year ago
  42. Boushra
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    u mean v sin A ??

    • one year ago
  43. Vaidehi09
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    ah yes....that should be sine

    • one year ago
  44. Boushra
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    so all I need to do is substitute now? with the 3 eq that I've got?

    • one year ago
  45. Vaidehi09
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    yup...try it out. m also doing the same.

    • one year ago
  46. Boushra
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    ok, I'll put my answers down.. let's do that :)

    • one year ago
  47. Boushra
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    do we substitue eq 3 into 1?

    • one year ago
  48. Boushra
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    we need to solve for u1 first, right?

    • one year ago
  49. Vaidehi09
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    i don't know ur numbering. here our the 3 eqs: 1) u = 3cos45 + vcosA 2) u^2 = 9+ v^2 3) vsinA = 3 sin45

    • one year ago
  50. Vaidehi09
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    our unknowns are u, v, A. start anywhere.

    • one year ago
  51. Boushra
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    let's start with u

    • one year ago
  52. Boushra
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    we should substitute eq 2 into 1?

    • one year ago
  53. Vaidehi09
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    yea, u can do that or vice versa. whichever is convenient to u.

    • one year ago
  54. Boushra
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    and the take the square root of it first?

    • one year ago
  55. Vaidehi09
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    that way u can get rid of u.

    • one year ago
  56. Boushra
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    do we go like this: 3 + v = 3 cos 45 + v cos A ??

    • one year ago
  57. Boushra
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    ?

    • one year ago
  58. Vaidehi09
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    how did u get 3 + v ?

    • one year ago
  59. Boushra
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    u^2 = 9 + v^2

    • one year ago
  60. Boushra
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    I took the square root of that

    • one year ago
  61. Vaidehi09
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    u need sqrt( 9+v^2 )

    • one year ago
  62. Vaidehi09
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    no wait....sqrt( a^2 + b^2) is NOT = a + b

    • one year ago
  63. Boushra
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    oh.

    • one year ago
  64. Boushra
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    can you tell me the way you'd solve for the unknowns from the 3 eq we've got, step by step?

    • one year ago
  65. Vaidehi09
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    before that, let me confirm something. is the angle theta = 45 ?

    • one year ago
  66. Boushra
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    no, there are two angles, one is theta and one is 45. different from one another. you can refer back to the diagram above

    • one year ago
  67. Vaidehi09
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    no i mean, do u have the answers?

    • one year ago
  68. Boushra
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    oh yes, it's correct

    • one year ago
  69. Boushra
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    theta = 45

    • one year ago
  70. Vaidehi09
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    great...so i'm on the right path. ok proceed this way...

    • one year ago
  71. Boushra
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    I just need to know the steps, I've having a hard time substituting..

    • one year ago
  72. Vaidehi09
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    no prob, i'll help.

    • one year ago
  73. Boushra
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    :)

    • one year ago
  74. Boushra
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    can we start by knowing how to substitute and figure out theta?

    • one year ago
  75. Vaidehi09
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    first, swap ur RHS and LHS of eq 3 and add it to eq 1.

    • one year ago
  76. Boushra
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    what is RHS and LHS?

    • one year ago
  77. Vaidehi09
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    right hand side and left hand side of an eq.

    • one year ago
  78. Boushra
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    ok, I'd swap them and add them to the 1st equation

    • one year ago
  79. Vaidehi09
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    so ur eq 3 becomes: 3sin45 = vsinA

    • one year ago
  80. Vaidehi09
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    right. sow rite down what u got here.

    • one year ago
  81. Boushra
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    ok good

    • one year ago
  82. Vaidehi09
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    so u should have now, u + 3sin45 = 3cos45 + vcosA + vsinA

    • one year ago
  83. Boushra
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    yes! i got that too

    • one year ago
  84. Vaidehi09
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    cancel out 3sin45 and 3cos45. so now u have, u = v( cosA + sinA)

    • one year ago
  85. Boushra
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    is there a trig rule for cancelling them out that i should know of?

    • one year ago
  86. Vaidehi09
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    what is sin45?

    • one year ago
  87. Boushra
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    0.707106 which is the same is cos 45

    • one year ago
  88. Boushra
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    so that's how we cancel them out.. ok!

    • one year ago
  89. Vaidehi09
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    right...so mathematically, those 2 terms are exactly the same. and on opp sides of the = sign. so we can cancel them out.

    • one year ago
  90. Vaidehi09
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    ok, moving on...substitute that eq in eq2.

    • one year ago
  91. Boushra
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    perfect, now we're left with u = v ( cos A + sin A)

    • one year ago
  92. Vaidehi09
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    u should get : v^2(sinA+cosA)^2 = 9+ v^2

    • one year ago
  93. Boushra
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    do we get: (v(cos A + sin A) ^2 = 9 + v^2 ?

    • one year ago
  94. Boushra
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    yep, the same.

    • one year ago
  95. Vaidehi09
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    that's right, now expand LHS and show me what u get.

    • one year ago
  96. Boushra
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    i think there' a rule saying cos A + sin A = 1?

    • one year ago
  97. Vaidehi09
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    no...that's when cosA and sinA both are squared terms

    • one year ago
  98. Boushra
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    ok, then hold a sec

    • one year ago
  99. Boushra
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    should we get v^2cos^2A +sin^2A = 9 + v^2 ??

    • one year ago
  100. Vaidehi09
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    do u know the expansion of (a+b)^2 ?

    • one year ago
  101. Boushra
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    yes... oh wait, I made a mistake.. just a sec

    • one year ago
  102. Boushra
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    v^2 sin^2a + 2sinAcosA + cos^2 a = 9 + v^2

    • one year ago
  103. Vaidehi09
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    thats right, except that, the trigo terms should be inside the bracket and v^2 outside it.

    • one year ago
  104. Boushra
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    ok so : v^2 (sin^2a + 2sinAcosA + cos^2 a) = 9 + v^2

    • one year ago
  105. Boushra
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    and then...

    • one year ago
  106. Vaidehi09
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    right, now sin^2a + cos^2a = 1 so we get v^2( 1 + sin2a) = 9 + v^2 v^2 + v^2sin2a = 9 + v^2 v^2sin2a = 9

    • one year ago
  107. Vaidehi09
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    sin2a = 2sinacosa...thats the sub i made.

    • one year ago
  108. Boushra
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    wiat.. shouldn't we habe v^2 (1 + 2sinAcosA) = 9 + v^2 ??

    • one year ago
  109. Vaidehi09
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    look at the second last post. i showed how I substituted.

    • one year ago
  110. Boushra
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    ok hold on

    • one year ago
  111. Vaidehi09
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    there's a formula : sin2A = 2sinAcosA. i used it.

    • one year ago
  112. Boushra
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    aha... ok, got it.. and now what's next?

    • one year ago
  113. Boushra
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    so the final formula here is v^2 sin2 a = 9

    • one year ago
  114. Vaidehi09
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    now divide the eq which we got hen, by eq 3. so \[v ^{2} \sin2A / vsinA = 9/\3sin45\] so we'll get vsinAcosA/sinA = 9 sqrt(2)/3.......since 3sin45 = 3/sqrt(2)

    • one year ago
  115. Vaidehi09
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    in the eq, the RHS is 9/3sin45

    • one year ago
  116. Boushra
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    ok, the RHS is 9/3 sin 45 = 4.2426 ???

    • one year ago
  117. Vaidehi09
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    don't go on to the final values just yet. it'll only complicate things further.

    • one year ago
  118. Vaidehi09
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    continuing the above eq, we'll gte vcosA = 3 sqrt(2)

    • one year ago
  119. Boushra
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    ok

    • one year ago
  120. Vaidehi09
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    gr8! so now divide this eq with ur eq3

    • one year ago
  121. Vaidehi09
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    u should get vcosA/vsinA = 3 rt(2)/ [3/rt(2)]

    • one year ago
  122. Boushra
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    yes that's what i got

    • one year ago
  123. Vaidehi09
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    which would lead us to cotA = 2 or tanA = 1/2

    • one year ago
  124. Vaidehi09
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    so now u can find A = tan^(-1) [1/2]

    • one year ago
  125. Boushra
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    wait cos a/sina = tan a?

    • one year ago
  126. Vaidehi09
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    no, cosa /sina = cota = 1/tana

    • one year ago
  127. Boushra
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    what's cota?

    • one year ago
  128. Vaidehi09
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    cotA

    • one year ago
  129. Boushra
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    yeah, i mean what's cot?

    • one year ago
  130. Vaidehi09
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    u dont know ur trigo ratios?

    • one year ago
  131. Boushra
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    this is the first time I see cot, I will definitely look up all the ratios after this

    • one year ago
  132. Vaidehi09
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    no wait..before solving such questions, u should go and work on ur math. u've reached mechanics, u ought to have a solid base in math for it.

    • one year ago
  133. Boushra
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    so what do i need to look up in regards to trig in mechanics?

    • one year ago
  134. Vaidehi09
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    trigo ratios are elementary in trigo. u need to know the various relations between them. their values for basic angles. formulae etc etc.

    • one year ago
  135. Vaidehi09
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    and then there will come inverse trigo ratios...something that we were going to use here, to find out theta.

    • one year ago
  136. Boushra
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    alright then, nonetheless... i really really really appreciate your help. I will go look those basics up first, and then perhaps come back and solve the question here. Thank you soooo very much :)

    • one year ago
  137. Boushra
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    yeah i know inverse ratios :)

    • one year ago
  138. Vaidehi09
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    cool! and considering how much we've solved here, u should be able to go further without any external help. work hard on math! don't ignore it!

    • one year ago
  139. Vaidehi09
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    @Boushra: I had made a calculation error before. after getting v^2sin2A/vsinA = 9/[3/sqrt(2)] we'll get, 2vcosA = 3 rt(2) therefore, vcosA = 3/rt(2) comparing this with eq3, ie., vsinA = 3/rt(2) we can conclude thata cosA = sinA therefore A = 45

    • one year ago
  140. Vaidehi09
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    so after substituting u'll get v = 3m/s u = 3 rt(2) m/s and A = theta = we already got as 45

    • one year ago
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