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!
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Boushra
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We should use the perfect inelastic collision equation
m1u1 + m2u2 = (m1+m2)v
Vaidehi09
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can u post the diagram as well?
Boushra
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yep just a sec
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
Boushra
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|dw:1341315928417:dw|
Boushra
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the black ball is in motion after collision
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.
Boushra
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oh... yeah right @Vaidehi09
Boushra
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should we use m1u1+m2u2 = m1v1 + m2v2 ??
Boushra
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m1 = m2 = m
meera_yadav
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|dw:1341316215450:dw|
Boushra
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should we write an equation based on components?
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..
Boushra
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yes understood :) then
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
meera_yadav
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similar equation we can write it for vertical momentum also !
Boushra
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so the first one you drew above is for the horizontal momentum?
meera_yadav
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|dw:1341316752653:dw|
meera_yadav
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yep ! it's for horizontal...
2nd draw is for vertical momentum..
Vaidehi09
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@meera_yadav ur first drawing....shouldn't we have vcos(theta) there...instead of sine?
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
Vaidehi09
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the eq for horizontal should be
u1 = 3cos45 + vcosA
Vaidehi09
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supposing theta = A
Boushra
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yes, that's what I got @Vaidehi09
meera_yadav
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yeah! sorry ! that's a mistake ...
thanks for pointing out :)
plz correct it to cos theta
Boushra
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great!!! now we can move on to 2nd equation:
Boushra
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.5 mu (squared) = .5 m 3(squared) + .5 m v(squared) <--- why did we use this?
Vaidehi09
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wait a sec, but we haven't yet found the value of u1.
Vaidehi09
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thats the eq for conservation of KE
Boushra
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yeah, I know,... should we label it as eq 2?
Vaidehi09
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for KE, i got the eq as (u1)^2 = 3^2 + v^2
Boushra
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I think the v in it should be v subscript x, correct?
Vaidehi09
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i don't think so. for KE, we don't need components since it is a scalar.
meera_yadav
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no there's no need for the subscript vaidelhi , because K.E is a scalar quantity...
Boushra
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ok, great, I'm following.. so what do we need next.
meera_yadav
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I think everythings known to u know !!
Vaidehi09
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we have another eq, using vertical components for momentum.
and that is, vsin theta = 3sin45
Boushra
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ok, hold on a sec
Boushra
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shouldn't it be 3 sin 45 = - v sin theta?
Boushra
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we have it as: 0 = m(3 sin 45) + m (v sin theta) before right?
Vaidehi09
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nope. both the vertical components of velocities are in opp direcs. so we get,
0 = 3sin45 - vcosA
Boushra
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u mean v sin A ??
Vaidehi09
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ah yes....that should be sine
Boushra
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so all I need to do is substitute now? with the 3 eq that I've got?
Vaidehi09
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yup...try it out. m also doing the same.
Boushra
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ok, I'll put my answers down.. let's do that :)
Boushra
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do we substitue eq 3 into 1?
Boushra
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we need to solve for u1 first, right?
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
Vaidehi09
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our unknowns are u, v, A. start anywhere.
Boushra
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let's start with u
Boushra
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we should substitute eq 2 into 1?
Vaidehi09
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yea, u can do that or vice versa. whichever is convenient to u.
Boushra
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and the take the square root of it first?
Vaidehi09
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that way u can get rid of u.
Boushra
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do we go like this: 3 + v = 3 cos 45 + v cos A ??
Boushra
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?
Vaidehi09
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how did u get 3 + v ?
Boushra
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u^2 = 9 + v^2
Boushra
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I took the square root of that
Vaidehi09
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u need sqrt( 9+v^2 )
Vaidehi09
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no wait....sqrt( a^2 + b^2) is NOT = a + b
Boushra
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oh.
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?
Vaidehi09
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before that, let me confirm something. is the angle theta = 45 ?
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
Vaidehi09
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no i mean, do u have the answers?
Boushra
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oh yes, it's correct
Boushra
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theta = 45
Vaidehi09
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great...so i'm on the right path. ok proceed this way...
Boushra
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I just need to know the steps, I've having a hard time substituting..
Vaidehi09
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no prob, i'll help.
Boushra
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:)
Boushra
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can we start by knowing how to substitute and figure out theta?
Vaidehi09
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first, swap ur RHS and LHS of eq 3 and add it to eq 1.
Boushra
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what is RHS and LHS?
Vaidehi09
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right hand side and left hand side of an eq.
Boushra
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ok, I'd swap them and add them to the 1st equation
Vaidehi09
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so ur eq 3 becomes: 3sin45 = vsinA
Vaidehi09
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right. sow rite down what u got here.
Boushra
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ok good
Vaidehi09
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so u should have now,
u + 3sin45 = 3cos45 + vcosA + vsinA
Boushra
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yes! i got that too
Vaidehi09
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cancel out 3sin45 and 3cos45. so now u have, u = v( cosA + sinA)
Boushra
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is there a trig rule for cancelling them out that i should know of?
Vaidehi09
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what is sin45?
Boushra
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0.707106 which is the same is cos 45
Boushra
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so that's how we cancel them out.. ok!
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.
Vaidehi09
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ok, moving on...substitute that eq in eq2.
Boushra
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perfect, now we're left with u = v ( cos A + sin A)
Vaidehi09
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u should get : v^2(sinA+cosA)^2 = 9+ v^2
Boushra
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do we get: (v(cos A + sin A) ^2 = 9 + v^2 ?
Boushra
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yep, the same.
Vaidehi09
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that's right, now expand LHS and show me what u get.
Boushra
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i think there' a rule saying cos A + sin A = 1?
Vaidehi09
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no...that's when cosA and sinA both are squared terms
Boushra
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ok, then hold a sec
Boushra
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should we get v^2cos^2A +sin^2A = 9 + v^2 ??
Vaidehi09
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do u know the expansion of (a+b)^2 ?
Boushra
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yes... oh wait, I made a mistake.. just a sec
Boushra
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v^2 sin^2a + 2sinAcosA + cos^2 a = 9 + v^2
Vaidehi09
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thats right, except that, the trigo terms should be inside the bracket and v^2 outside it.
Boushra
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ok so : v^2 (sin^2a + 2sinAcosA + cos^2 a) = 9 + v^2
Boushra
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and then...
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
Vaidehi09
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sin2a = 2sinacosa...thats the sub i made.
Boushra
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wiat.. shouldn't we habe v^2 (1 + 2sinAcosA) = 9 + v^2 ??
Vaidehi09
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look at the second last post. i showed how I substituted.
Boushra
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ok hold on
Vaidehi09
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there's a formula : sin2A = 2sinAcosA. i used it.
Boushra
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aha... ok, got it.. and now what's next?
Boushra
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so the final formula here is v^2 sin2 a = 9
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)
Vaidehi09
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in the eq, the RHS is 9/3sin45
Boushra
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ok, the RHS is 9/3 sin 45 = 4.2426 ???
Vaidehi09
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don't go on to the final values just yet. it'll only complicate things further.
Vaidehi09
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continuing the above eq, we'll gte
vcosA = 3 sqrt(2)
Boushra
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ok
Vaidehi09
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gr8!
so now divide this eq with ur eq3
Vaidehi09
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u should get vcosA/vsinA = 3 rt(2)/ [3/rt(2)]
Boushra
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yes that's what i got
Vaidehi09
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which would lead us to cotA = 2 or tanA = 1/2
Vaidehi09
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so now u can find A = tan^(-1) [1/2]
Boushra
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wait cos a/sina = tan a?
Vaidehi09
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no, cosa /sina = cota = 1/tana
Boushra
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what's cota?
Vaidehi09
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cotA
Boushra
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yeah, i mean what's cot?
Vaidehi09
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u dont know ur trigo ratios?
Boushra
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this is the first time I see cot, I will definitely look up all the ratios after this
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.
Boushra
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so what do i need to look up in regards to trig in mechanics?
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.
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.
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 :)
Boushra
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yeah i know inverse ratios :)
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!
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
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