Boushra Group Title 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! 2 years ago 2 years ago

1. Boushra Group Title

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

2. Vaidehi09 Group Title

can u post the diagram as well?

3. Boushra Group Title

yep just a sec

hii, is the black bal at rest after the collision? if yes then white ball's initial velocity is zero just before collision

5. Boushra Group Title

|dw:1341315928417:dw|

6. Boushra Group Title

the black ball is in motion after collision

7. Vaidehi09 Group Title

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

8. Boushra Group Title

oh... yeah right @Vaidehi09

9. Boushra Group Title

should we use m1u1+m2u2 = m1v1 + m2v2 ??

10. Boushra Group Title

m1 = m2 = m

|dw:1341316215450:dw|

12. Boushra Group Title

should we write an equation based on components?

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..

14. Boushra Group Title

yes understood :) then

yep ! we should write moemtum shold be resolved.. i.e. horizontal momentum before collison = horizontal momentum after collision same goes for vertical momentum

similar equation we can write it for vertical momentum also !

17. Boushra Group Title

so the first one you drew above is for the horizontal momentum?

|dw:1341316752653:dw|

yep ! it's for horizontal... 2nd draw is for vertical momentum..

20. Vaidehi09 Group Title

21. Boushra Group Title

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

22. Vaidehi09 Group Title

the eq for horizontal should be u1 = 3cos45 + vcosA

23. Vaidehi09 Group Title

supposing theta = A

24. Boushra Group Title

yes, that's what I got @Vaidehi09

yeah! sorry ! that's a mistake ... thanks for pointing out :) plz correct it to cos theta

26. Boushra Group Title

great!!! now we can move on to 2nd equation:

27. Boushra Group Title

.5 mu (squared) = .5 m 3(squared) + .5 m v(squared) <--- why did we use this?

28. Vaidehi09 Group Title

wait a sec, but we haven't yet found the value of u1.

29. Vaidehi09 Group Title

thats the eq for conservation of KE

30. Boushra Group Title

yeah, I know,... should we label it as eq 2?

31. Vaidehi09 Group Title

for KE, i got the eq as (u1)^2 = 3^2 + v^2

32. Boushra Group Title

I think the v in it should be v subscript x, correct?

33. Vaidehi09 Group Title

i don't think so. for KE, we don't need components since it is a scalar.

no there's no need for the subscript vaidelhi , because K.E is a scalar quantity...

35. Boushra Group Title

ok, great, I'm following.. so what do we need next.

I think everythings known to u know !!

37. Vaidehi09 Group Title

we have another eq, using vertical components for momentum. and that is, vsin theta = 3sin45

38. Boushra Group Title

ok, hold on a sec

39. Boushra Group Title

shouldn't it be 3 sin 45 = - v sin theta?

40. Boushra Group Title

we have it as: 0 = m(3 sin 45) + m (v sin theta) before right?

41. Vaidehi09 Group Title

nope. both the vertical components of velocities are in opp direcs. so we get, 0 = 3sin45 - vcosA

42. Boushra Group Title

u mean v sin A ??

43. Vaidehi09 Group Title

ah yes....that should be sine

44. Boushra Group Title

so all I need to do is substitute now? with the 3 eq that I've got?

45. Vaidehi09 Group Title

yup...try it out. m also doing the same.

46. Boushra Group Title

ok, I'll put my answers down.. let's do that :)

47. Boushra Group Title

do we substitue eq 3 into 1?

48. Boushra Group Title

we need to solve for u1 first, right?

49. Vaidehi09 Group Title

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

50. Vaidehi09 Group Title

our unknowns are u, v, A. start anywhere.

51. Boushra Group Title

52. Boushra Group Title

we should substitute eq 2 into 1?

53. Vaidehi09 Group Title

yea, u can do that or vice versa. whichever is convenient to u.

54. Boushra Group Title

and the take the square root of it first?

55. Vaidehi09 Group Title

that way u can get rid of u.

56. Boushra Group Title

do we go like this: 3 + v = 3 cos 45 + v cos A ??

57. Boushra Group Title

?

58. Vaidehi09 Group Title

how did u get 3 + v ?

59. Boushra Group Title

u^2 = 9 + v^2

60. Boushra Group Title

I took the square root of that

61. Vaidehi09 Group Title

u need sqrt( 9+v^2 )

62. Vaidehi09 Group Title

no wait....sqrt( a^2 + b^2) is NOT = a + b

63. Boushra Group Title

oh.

64. Boushra Group Title

can you tell me the way you'd solve for the unknowns from the 3 eq we've got, step by step?

65. Vaidehi09 Group Title

before that, let me confirm something. is the angle theta = 45 ?

66. Boushra Group Title

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

67. Vaidehi09 Group Title

no i mean, do u have the answers?

68. Boushra Group Title

oh yes, it's correct

69. Boushra Group Title

theta = 45

70. Vaidehi09 Group Title

great...so i'm on the right path. ok proceed this way...

71. Boushra Group Title

I just need to know the steps, I've having a hard time substituting..

72. Vaidehi09 Group Title

no prob, i'll help.

73. Boushra Group Title

:)

74. Boushra Group Title

can we start by knowing how to substitute and figure out theta?

75. Vaidehi09 Group Title

first, swap ur RHS and LHS of eq 3 and add it to eq 1.

76. Boushra Group Title

what is RHS and LHS?

77. Vaidehi09 Group Title

right hand side and left hand side of an eq.

78. Boushra Group Title

ok, I'd swap them and add them to the 1st equation

79. Vaidehi09 Group Title

so ur eq 3 becomes: 3sin45 = vsinA

80. Vaidehi09 Group Title

right. sow rite down what u got here.

81. Boushra Group Title

ok good

82. Vaidehi09 Group Title

so u should have now, u + 3sin45 = 3cos45 + vcosA + vsinA

83. Boushra Group Title

yes! i got that too

84. Vaidehi09 Group Title

cancel out 3sin45 and 3cos45. so now u have, u = v( cosA + sinA)

85. Boushra Group Title

is there a trig rule for cancelling them out that i should know of?

86. Vaidehi09 Group Title

what is sin45?

87. Boushra Group Title

0.707106 which is the same is cos 45

88. Boushra Group Title

so that's how we cancel them out.. ok!

89. Vaidehi09 Group Title

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

90. Vaidehi09 Group Title

ok, moving on...substitute that eq in eq2.

91. Boushra Group Title

perfect, now we're left with u = v ( cos A + sin A)

92. Vaidehi09 Group Title

u should get : v^2(sinA+cosA)^2 = 9+ v^2

93. Boushra Group Title

do we get: (v(cos A + sin A) ^2 = 9 + v^2 ?

94. Boushra Group Title

yep, the same.

95. Vaidehi09 Group Title

that's right, now expand LHS and show me what u get.

96. Boushra Group Title

i think there' a rule saying cos A + sin A = 1?

97. Vaidehi09 Group Title

no...that's when cosA and sinA both are squared terms

98. Boushra Group Title

ok, then hold a sec

99. Boushra Group Title

should we get v^2cos^2A +sin^2A = 9 + v^2 ??

100. Vaidehi09 Group Title

do u know the expansion of (a+b)^2 ?

101. Boushra Group Title

yes... oh wait, I made a mistake.. just a sec

102. Boushra Group Title

v^2 sin^2a + 2sinAcosA + cos^2 a = 9 + v^2

103. Vaidehi09 Group Title

thats right, except that, the trigo terms should be inside the bracket and v^2 outside it.

104. Boushra Group Title

ok so : v^2 (sin^2a + 2sinAcosA + cos^2 a) = 9 + v^2

105. Boushra Group Title

and then...

106. Vaidehi09 Group Title

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

107. Vaidehi09 Group Title

sin2a = 2sinacosa...thats the sub i made.

108. Boushra Group Title

wiat.. shouldn't we habe v^2 (1 + 2sinAcosA) = 9 + v^2 ??

109. Vaidehi09 Group Title

look at the second last post. i showed how I substituted.

110. Boushra Group Title

ok hold on

111. Vaidehi09 Group Title

there's a formula : sin2A = 2sinAcosA. i used it.

112. Boushra Group Title

aha... ok, got it.. and now what's next?

113. Boushra Group Title

so the final formula here is v^2 sin2 a = 9

114. Vaidehi09 Group Title

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)

115. Vaidehi09 Group Title

in the eq, the RHS is 9/3sin45

116. Boushra Group Title

ok, the RHS is 9/3 sin 45 = 4.2426 ???

117. Vaidehi09 Group Title

don't go on to the final values just yet. it'll only complicate things further.

118. Vaidehi09 Group Title

continuing the above eq, we'll gte vcosA = 3 sqrt(2)

119. Boushra Group Title

ok

120. Vaidehi09 Group Title

gr8! so now divide this eq with ur eq3

121. Vaidehi09 Group Title

u should get vcosA/vsinA = 3 rt(2)/ [3/rt(2)]

122. Boushra Group Title

yes that's what i got

123. Vaidehi09 Group Title

which would lead us to cotA = 2 or tanA = 1/2

124. Vaidehi09 Group Title

so now u can find A = tan^(-1) [1/2]

125. Boushra Group Title

wait cos a/sina = tan a?

126. Vaidehi09 Group Title

no, cosa /sina = cota = 1/tana

127. Boushra Group Title

what's cota?

128. Vaidehi09 Group Title

cotA

129. Boushra Group Title

yeah, i mean what's cot?

130. Vaidehi09 Group Title

u dont know ur trigo ratios?

131. Boushra Group Title

this is the first time I see cot, I will definitely look up all the ratios after this

132. Vaidehi09 Group Title

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.

133. Boushra Group Title

so what do i need to look up in regards to trig in mechanics?

134. Vaidehi09 Group Title

trigo ratios are elementary in trigo. u need to know the various relations between them. their values for basic angles. formulae etc etc.

135. Vaidehi09 Group Title

and then there will come inverse trigo ratios...something that we were going to use here, to find out theta.

136. Boushra Group Title

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 :)

137. Boushra Group Title

yeah i know inverse ratios :)

138. Vaidehi09 Group Title

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!

139. Vaidehi09 Group Title

@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

140. Vaidehi09 Group Title

so after substituting u'll get v = 3m/s u = 3 rt(2) m/s and A = theta = we already got as 45