Problem Involving Momentum & Collision

AI Thread Summary
The discussion revolves around a momentum and collision problem involving two identical balls. The initial momentum of the system is calculated to be 44.22 kg m/s, and the ratio of their final speeds is determined to be 1.37. The main challenge lies in calculating the final speed of the incoming ball after the collision, with participants discussing the correct approach to using horizontal momentum components. Clarifications emphasize the importance of considering the horizontal components of momentum and the relationship between the final velocities and their angles. The conversation concludes with an affirmation of the principles of conservation of momentum in vector form.
Pat2666
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Here's the problem :

A ball of mass m = 2.2 kg moving with a speed of v0 = 20.1 m/s strikes an identical ball which was initially at rest. After the collision, the incoming ball (ball 1) goes off at q1 = 24° relative to its original direction and the struck ball (ball 2) moves off at q2 = 34° as shown in the diagram.

I can't post the diagram, but ball 1 shoots off 24 degrees above the x-axis and ball 2 shoots 34 degress below the x-axis.

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a) What is the magnitude of the initial momentum of the two ball system?
pi = 44.22 kg m/s

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b) What is the ratio of the balls' final speeds?
|v1|/|v2| = 1.37

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c) What is the final speed of the incoming ball?
|v1| = m/s

Okay, I have another problem that I'm stuck on! I was able to figure out part A & B (answers above), but I can't seem to get part C.

I tried setting the horizontal momentum of the two objects after the collision equal to the initial horizontal momentum (44.22) and then solving for one variable.

44.22 = m1v1 + m2v2 or 44.22 = m (v1+v2)

Then I thought I would have to use the ratio of v1/v2 in order to solve for a variable. So I did the following:

v1/v2 = 1.37 ... v2 = v1/1.37

I then plugged that into the momentum equation above and solved for the horiztonal velocity and got 11.619m/s for v1.

Then I took that number and divided it by COS(24) to get the Vf since the ball shoots off at an angle. I got 12.72m/s, but it isn't right. Where have I gone wrong? :(
 
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OK. I think you may have made a calculation mistake while finding v_1. I am getting v_1=8.375m/s.

Check your calculations, again.
 
I'm still getting the same answer as I have before.

And I tried plugging your value into solve for V1final and it was still wrong. :-\
 
Pat2666 said:
44.22 = m1v1 + m2v2 or 44.22 = m (v1+v2)

Hi Pat!

(A very well laid-out answer, by the way … I wish other people were as clear! :rolleyes:)

Hint: You must take horizontal components of momentum! :smile:
 
Isn't that what I was doing though? I thought v1 and v2 that I was solving for in the equation you quoted were the two horizontal component?
 
tiny-tim said:
Hi Pat!

(A very well laid-out answer, by the way … I wish other people were as clear! :rolleyes:)

Hint: You must take horizontal components of momentum! :smile:

TT I think he means that v_1=horizontal component of momentum of the first ball.

Then he's saying:

v_1=v_{f1}\cos24^o

v_{f1}=\frac{v_1}{\cos24^o}

where v_f1 = final velocity magnitude.

Am I correct, Pat?
 
Yeah, I think so but I thought that's what I've been doing? I mentioned that I found v1 and then divided it by Cos24, but I must be getting the wrong value for v1.
 
v = v1cos24º + v2cos34º :smile:
 
tiny-tim said:
v = v1cos24º + v2cos34º :smile:


Now I'm a bit more confused sorry! lol

I don't know V do I? I thought you had to add the two horizontal components of momentum and set them equal to the initial momentum of the first ball? I wouldn't know V in your equation would I?
 
  • #10
v = 20.1​
 
  • #11
I finally got it! Thanks!

I have to ask though...do the horizontal components always equal the sum of the initial horizontal compents? I just want to understand what I did :)

And thanks again!
 
  • #12
Pat2666 said:
I have to ask though...do the horizontal components always equal the sum of the initial horizontal compents? I just want to understand what I did :)

Yup! :biggrin:

Good ol' Newton's second law, and conservation of momentum, are vector equations, so they're valid on all the components in anyone direction. :smile:
 
  • #13
Oh okay I understand now! Thank you so much :)
 
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