anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
We should use the perfect inelastic collision equation m1u1 + m2u2 = (m1+m2)v
anonymous
  • anonymous
can u post the diagram as well?
anonymous
  • anonymous
yep just a sec

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