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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!

Physics
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We should use the perfect inelastic collision equation m1u1 + m2u2 = (m1+m2)v
can u post the diagram as well?
yep just a sec

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Other answers:

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

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