Collision question -- 2 balls colliding elastically

[SteveS]

Homework Statement

A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown. If the first ball moves away with angle 30° to the original path, determine

1. the speed of the first ball after the collision.
2. the speed and direction of the second ball after the collision.

** I realize there is another thread for this question, but I hadn't seen any activity on it for a month after asking a question so I'm posting again.

Homework Equations

m1v1 + m2v2 = m1v1' + m2v2'

1/2m1(v1)^2 + 1/2 m2(v2)^2 = 1/2m1(v1')^2 + 1/2m2(v2')^2

The Attempt at a Solution

[/B]
Ok. I have 3 unknowns in this equation. The v1', v2' and the angle of the second ball. For this I need 3 equations

For the X component I have:

V1 = v1' * Cos 30 + v2' * Cos θ

For the Y component I have:

v1' * Sin 30 = -v2' * Sin θ

Using the Conservation of energy I have:

V1^2 = v1'^2 + v2'^2

From here, I'm not sure where to proceed. I have tried some substitutions but I get into really long ugly math that turns circular and I never get anywhere. Any help would be appreciated. I know I need to use the trig relation of
Cos^2 θ + Sin^2 θ = 1, but I've not been able to get to any point where I can apply it.

Thanks in advance for your help.

 

[haruspex]

You can certainly use cos²+sin²=1 to eliminate theta. First, you need to get cos theta = (an expression not involving theta) and likewise for sin theta. Even if it does get a bit messy, please post your work as far as you get. It's the only way we can see where you are getting stuck.
You can avoid some of the mess by using Newton's Experimental Law, about ratios of velocities in the direction of impact, but in the context of the current problem it might not be much easier.

 

[SteveS]

Honestly I've got four pages of scribbles and I'm really turned around. I'm not sure what to do anymore. The last thing I did was

I rearranged the x component equation to v1' = va - v2' * Cos θ / Cos 30

I then substituted this into the energy equation which i got down to Cos 30 = Cos θ

is this right? Are the angles the same? I'm really not sure where to go from here.

I can rearrange the y component equation so that v2' = - v1' Sin 30 / Sin θ but putting this into an equation leads me to a no where. I'm really not sure what to do.

 

[PeroK]

Honestly I've got four pages of scribbles and I'm really turned around. I'm not sure what to do anymore. The last thing I did was

I rearranged the x component equation to v1' = va - v2' * Cos θ / Cos 30

I then substituted this into the energy equation which i got down to Cos 30 = Cos θ

is this right? Are the angles the same? I'm really not sure where to go from here.

I can rearrange the y component equation so that v2' = - v1' Sin 30 / Sin θ but putting this into an equation leads me to a no where. I'm really not sure what to do.

There's a neat shortcut to eliminating one of the angles by using the vector nature of momentum and the law of cosines. It amounts to the same as squaring things and using $cos^2 + sin^2 = 1$.

Using the law of cosines, you get:

$v_2^2 = u_1^2 + v_1^2 - 2uv_1cos(30)$

That plus your conservation of energy should get you on the right track.

(I used $u$ for the initial velocity, as it made the latex easier!)

 

[SteveS]

In my previous post I mentioned I was aware of the relation cos^2 θ + sin ^2 θ = 1... What I'm not sure is how you got to that equation you posted.

 

[PeroK]

In my previous post I mentioned I was aware of the relation cos^2 θ + sin ^2 θ = 1... What I'm not sure is how you got to that equation you posted.

Draw a vector diagram (triangle) involving the three velocity vectors involved and apply the law of cosines. Normally you would use the momentum vectors $mv$ but the masses cancel in this case - as you know.

 

[SteveS]

I have done so already which is what leads to the three equations that I laid out at the beginning.

 

[PeroK]

I have done so already which is what leads to the three equations that I laid out at the beginning.

Yes, but using the law of cosines is a significant shortcut to eliminating the unknown angle.

 

[SteveS]

I'm sorry i don't mean to be rude , but tell me something I've not already said... I know. I don't know how to use it. I can't ever get to a point where I can actually use the relation. In my circular calculations for the last three days I haven't been able to reduce anything to the point where I can get any sort of an answer. Really frustrated and I have half a mind to simply not do this question as I've already wasted 4 days trying to figure it out.

 

[PeroK]

I'm sorry i don't mean to be rude , but tell me something I've not already said... I know. I don't know how to use it. I can't ever get to a point where I can actually use the relation. In my circular calculations for the last three days I haven't been able to reduce anything to the point where I can get any sort of an answer. Really frustrated and I have half a mind to simply not do this question as I've already wasted 4 days trying to figure it out.

You've got:

$v_2^2 = u_1^2 + v_1^2 - 2uv_1cos(30)$

and

$u_1^2 = v_1^2 + v_2^2$

That gives you $v_1$ in terms of $u$. The rest should follow.

 

[SteveS]

The equation

u1^2 = v1^2 + v2^2 is the conservation of energy equation I already posted. The other equation you posted I do not know how to derive and the rest does not follow. Never mind I'm done with this crap. Pretty frustrated.

 

[chriskary]

The equation
u1^2 = v1^2 + v2^2 is the conservation of energy equation and Pithagoreio inverse theorem says v1 is vertical to v2...

 

[SteveS]

I'm not sure what that means or how that even helps. me. never mind no need for more replies. I can't waste any more time on this question.

 

[haruspex]

The equation u1^2 = v1^2 + v2^2 is the conservation of energy equation I already posted. The other equation you posted I do not know how to derive and the rest does not follow. Never mind I'm done with this crap. Pretty frustrated.

You don't understand how hard it is to help people who won't take the trouble to post all their working.
The equation you don't recognise is the cosine law, an essential tool in your toolbox. if you don't know it, learn it.

To finish the discussion on using cos^2 θ + sin ^2 θ = 1, which leads to the same equation:
You posted:
V1 = v1' * Cos 30 + v2' * Cos θ
v1' * Sin 30 = -v2' * Sin θ
From those we get ( v2')^2 = ( v2' * Cos θ)^2 + (v2' * Sin θ)^2 = (V1 - v1' * Cos 30)^2 + (v1' * Sin 30)^2 = (V1)^2 + (v1')^2 - 2 V1 v1'Cos 30

 

[chriskary]

SteveS read the ...image

 

[SteveS]

Ok. Sorry to those who have been trying to help me. I'm an accountant who graduated from university the first time about 12 years ago. I'm now a microbiology major and haven't had to use trig in a very very long time, but I have to take this one physics class in order to graduate. I took a walk around the block and came up with this solution that worked. I tried it before and must have made an error so that it didn't work. Here is what I did:

I took the x component equation I initially started with and rearranged it to isolate the θ on the right. This gave me:

v1-v1' * Cos 30 = v2' Cos θ

I then rearranged the y component equation to also isolate θ on the right. This gave me:

v1' * Sin 30 = v2' Sin θ

I then squared each of these equations and added them together. This then gave me use of the previously mentioned relation and simplified down to:

v1^2 + v1'^2 -2v1v1'*Cos 30 = v2'^2 ... I recognize now the equation previously posted from the cosine law.

I then rearranged the conservation of energy equation I previously had to isolate v2' on the right. I then subtracted this energy equation from the previously derived equation that eliminated θ.

This left me with an equation of:

v1' = v1 * Cos 30
v1' = 2.60 ( rounded answer)

feeding this back into the kinetic energy equation,

v2' = sqrt ( (9) - (2.60)^2)
v2' = 1.5 m/s

I can then use one of the component equations such as the x component equation and plug in the numbers to find θ:

Cos θ = v1 - v1' Cos 30 / v2' = (3) - (2.6)*Cos 30 / 1.5

θ = 60°

So the speed of mass 1 after the initial collision is 2.6 m/s and the speed of mass 2 after the initial collision is 1.5 m/s @ 60° below the horizontal x plane.

I think I've got it now.