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!

- anonymous

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- anonymous

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

- anonymous

can u post the diagram as well?

- anonymous

yep just a sec

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## More answers

- 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

|dw:1341315928417:dw|

- anonymous

the black ball is in motion after collision

- 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

oh... yeah right @Vaidehi09

- anonymous

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

- anonymous

m1 = m2 = m

- anonymous

|dw:1341316215450:dw|

- anonymous

should we write an equation based on components?

- 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

yes understood :) then

- 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

similar equation we can write it for vertical momentum also !

- anonymous

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

- anonymous

|dw:1341316752653:dw|

- anonymous

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

- anonymous

@meera_yadav ur first drawing....shouldn't we have vcos(theta) there...instead of sine?

- 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

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

- anonymous

supposing theta = A

- anonymous

yes, that's what I got @Vaidehi09

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

thats the eq for conservation of KE

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I think everythings known to u know !!

- anonymous

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

- anonymous

ok, hold on a sec

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

u mean v sin A ??

- anonymous

ah yes....that should be sine

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

do we substitue eq 3 into 1?

- anonymous

we need to solve for u1 first, right?

- 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

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

- anonymous

let's start with u

- anonymous

we should substitute eq 2 into 1?

- anonymous

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

- anonymous

and the take the square root of it first?

- anonymous

that way u can get rid of u.

- anonymous

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

- anonymous

?

- anonymous

how did u get 3 + v ?

- anonymous

u^2 = 9 + v^2

- anonymous

I took the square root of that

- anonymous

u need sqrt( 9+v^2 )

- anonymous

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

- anonymous

oh.

- anonymous

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

- anonymous

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

- 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

no i mean, do u have the answers?

- anonymous

oh yes, it's correct

- anonymous

theta = 45

- anonymous

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

- anonymous

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

- anonymous

no prob, i'll help.

- anonymous

:)

- anonymous

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

- anonymous

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

- anonymous

what is RHS and LHS?

- anonymous

right hand side and left hand side of an eq.

- anonymous

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

- anonymous

so ur eq 3 becomes: 3sin45 = vsinA

- anonymous

right. sow rite down what u got here.

- anonymous

ok good

- anonymous

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

- anonymous

yes! i got that too

- anonymous

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

- anonymous

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

- anonymous

what is sin45?

- anonymous

0.707106 which is the same is cos 45

- anonymous

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

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yep, the same.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok, then hold a sec

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

and then...

- 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

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

- anonymous

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

- anonymous

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

- anonymous

ok hold on

- anonymous

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

- anonymous

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

- anonymous

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

- 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

in the eq, the RHS is 9/3sin45

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

gr8!
so now divide this eq with ur eq3

- anonymous

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

- anonymous

yes that's what i got

- anonymous

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

- anonymous

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

- anonymous

wait cos a/sina = tan a?

- anonymous

no, cosa /sina = cota = 1/tana

- anonymous

what's cota?

- anonymous

cotA

- anonymous

yeah, i mean what's cot?

- anonymous

u dont know ur trigo ratios?

- anonymous

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

- 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

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

- 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

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

- 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

yeah i know inverse ratios :)

- 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

@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

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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