## Boushra 3 years ago 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!

1. Boushra

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

2. Vaidehi09

can u post the diagram as well?

3. Boushra

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

|dw:1341315928417:dw|

6. Boushra

the black ball is in motion after collision

7. Vaidehi09

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

oh... yeah right @Vaidehi09

9. Boushra

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

10. Boushra

m1 = m2 = m

|dw:1341316215450:dw|

12. Boushra

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

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

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

21. Boushra

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

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

23. Vaidehi09

supposing theta = A

24. Boushra

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

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

27. Boushra

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

28. Vaidehi09

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

29. Vaidehi09

thats the eq for conservation of KE

30. Boushra

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

31. Vaidehi09

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

32. Boushra

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

33. Vaidehi09

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

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

I think everythings known to u know !!

37. Vaidehi09

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

38. Boushra

ok, hold on a sec

39. Boushra

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

40. Boushra

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

41. Vaidehi09

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

42. Boushra

u mean v sin A ??

43. Vaidehi09

ah yes....that should be sine

44. Boushra

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

45. Vaidehi09

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

46. Boushra

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

47. Boushra

do we substitue eq 3 into 1?

48. Boushra

we need to solve for u1 first, right?

49. Vaidehi09

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

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

51. Boushra

52. Boushra

we should substitute eq 2 into 1?

53. Vaidehi09

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

54. Boushra

and the take the square root of it first?

55. Vaidehi09

that way u can get rid of u.

56. Boushra

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

57. Boushra

?

58. Vaidehi09

how did u get 3 + v ?

59. Boushra

u^2 = 9 + v^2

60. Boushra

I took the square root of that

61. Vaidehi09

u need sqrt( 9+v^2 )

62. Vaidehi09

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

63. Boushra

oh.

64. Boushra

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

65. Vaidehi09

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

66. Boushra

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

no i mean, do u have the answers?

68. Boushra

oh yes, it's correct

69. Boushra

theta = 45

70. Vaidehi09

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

71. Boushra

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

72. Vaidehi09

no prob, i'll help.

73. Boushra

:)

74. Boushra

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

75. Vaidehi09

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

76. Boushra

what is RHS and LHS?

77. Vaidehi09

right hand side and left hand side of an eq.

78. Boushra

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

79. Vaidehi09

so ur eq 3 becomes: 3sin45 = vsinA

80. Vaidehi09

right. sow rite down what u got here.

81. Boushra

ok good

82. Vaidehi09

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

83. Boushra

yes! i got that too

84. Vaidehi09

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

85. Boushra

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

86. Vaidehi09

what is sin45?

87. Boushra

0.707106 which is the same is cos 45

88. Boushra

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

89. Vaidehi09

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

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

91. Boushra

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

92. Vaidehi09

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

93. Boushra

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

94. Boushra

yep, the same.

95. Vaidehi09

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

96. Boushra

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

97. Vaidehi09

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

98. Boushra

ok, then hold a sec

99. Boushra

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

100. Vaidehi09

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

101. Boushra

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

102. Boushra

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

103. Vaidehi09

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

104. Boushra

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

105. Boushra

and then...

106. Vaidehi09

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

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

108. Boushra

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

109. Vaidehi09

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

110. Boushra

ok hold on

111. Vaidehi09

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

112. Boushra

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

113. Boushra

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

114. Vaidehi09

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

in the eq, the RHS is 9/3sin45

116. Boushra

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

117. Vaidehi09

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

118. Vaidehi09

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

119. Boushra

ok

120. Vaidehi09

gr8! so now divide this eq with ur eq3

121. Vaidehi09

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

122. Boushra

yes that's what i got

123. Vaidehi09

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

124. Vaidehi09

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

125. Boushra

wait cos a/sina = tan a?

126. Vaidehi09

no, cosa /sina = cota = 1/tana

127. Boushra

what's cota?

128. Vaidehi09

cotA

129. Boushra

yeah, i mean what's cot?

130. Vaidehi09

u dont know ur trigo ratios?

131. Boushra

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

132. Vaidehi09

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

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

134. Vaidehi09

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

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

136. Boushra

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

yeah i know inverse ratios :)

138. Vaidehi09

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

@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

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