## anonymous 4 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. anonymous

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

2. anonymous

can u post the diagram as well?

3. anonymous

yep just a sec

4. anonymous

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

5. anonymous

|dw:1341315928417:dw|

6. anonymous

the black ball is in motion after collision

7. anonymous

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

8. anonymous

oh... yeah right @Vaidehi09

9. anonymous

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

10. anonymous

m1 = m2 = m

11. anonymous

|dw:1341316215450:dw|

12. anonymous

should we write an equation based on components?

13. anonymous

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. anonymous

yes understood :) then

15. anonymous

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

16. anonymous

similar equation we can write it for vertical momentum also !

17. anonymous

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

18. anonymous

|dw:1341316752653:dw|

19. anonymous

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

20. anonymous

21. anonymous

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. anonymous

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

23. anonymous

supposing theta = A

24. anonymous

yes, that's what I got @Vaidehi09

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

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

29. anonymous

thats the eq for conservation of KE

30. anonymous

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

31. anonymous

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

32. anonymous

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

33. anonymous

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

34. anonymous

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

35. anonymous

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

36. anonymous

I think everythings known to u know !!

37. anonymous

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

38. anonymous

ok, hold on a sec

39. anonymous

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

40. anonymous

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

41. anonymous

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

42. anonymous

u mean v sin A ??

43. anonymous

ah yes....that should be sine

44. anonymous

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

45. anonymous

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

46. anonymous

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

47. anonymous

do we substitue eq 3 into 1?

48. anonymous

we need to solve for u1 first, right?

49. anonymous

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. anonymous

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

51. anonymous

52. anonymous

we should substitute eq 2 into 1?

53. anonymous

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

54. anonymous

and the take the square root of it first?

55. anonymous

that way u can get rid of u.

56. anonymous

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

57. anonymous

?

58. anonymous

how did u get 3 + v ?

59. anonymous

u^2 = 9 + v^2

60. anonymous

I took the square root of that

61. anonymous

u need sqrt( 9+v^2 )

62. anonymous

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

63. anonymous

oh.

64. anonymous

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

65. anonymous

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

66. anonymous

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. anonymous

no i mean, do u have the answers?

68. anonymous

oh yes, it's correct

69. anonymous

theta = 45

70. anonymous

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

71. anonymous

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

72. anonymous

no prob, i'll help.

73. anonymous

:)

74. anonymous

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

75. anonymous

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

76. anonymous

what is RHS and LHS?

77. anonymous

right hand side and left hand side of an eq.

78. anonymous

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

79. anonymous

so ur eq 3 becomes: 3sin45 = vsinA

80. anonymous

right. sow rite down what u got here.

81. anonymous

ok good

82. anonymous

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

83. anonymous

yes! i got that too

84. anonymous

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

85. anonymous

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

86. anonymous

what is sin45?

87. anonymous

0.707106 which is the same is cos 45

88. anonymous

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

89. anonymous

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

90. anonymous

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

91. anonymous

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

92. anonymous

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

93. anonymous

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

94. anonymous

yep, the same.

95. anonymous

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

96. anonymous

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

97. anonymous

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

98. anonymous

ok, then hold a sec

99. anonymous

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

100. anonymous

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

101. anonymous

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

102. anonymous

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

103. anonymous

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

104. anonymous

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

105. anonymous

and then...

106. anonymous

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. anonymous

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

108. anonymous

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

109. anonymous

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

110. anonymous

ok hold on

111. anonymous

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

112. anonymous

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

113. anonymous

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

114. anonymous

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. anonymous

in the eq, the RHS is 9/3sin45

116. anonymous

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

117. anonymous

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

118. anonymous

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

119. anonymous

ok

120. anonymous

gr8! so now divide this eq with ur eq3

121. anonymous

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

122. anonymous

yes that's what i got

123. anonymous

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

124. anonymous

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

125. anonymous

wait cos a/sina = tan a?

126. anonymous

no, cosa /sina = cota = 1/tana

127. anonymous

what's cota?

128. anonymous

cotA

129. anonymous

yeah, i mean what's cot?

130. anonymous

u dont know ur trigo ratios?

131. anonymous

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

132. anonymous

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. anonymous

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

134. anonymous

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

135. anonymous

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

136. anonymous

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. anonymous

yeah i know inverse ratios :)

138. anonymous

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. anonymous

@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. anonymous

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