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Problem Involving Momentum & Collision

  • Thread starter Pat2666
  • Start date
  • #1
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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.

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

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

--------------------------------------------------------------------------------
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? :(
 

Answers and Replies

  • #2
G01
Homework Helper
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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.
 
  • #3
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I'm still getting the same answer as I have before.

And I tried plugging your value in to solve for V1final and it was still wrong. :-\
 
  • #4
tiny-tim
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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:
 
  • #5
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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?
 
  • #6
G01
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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 hes saying:

[tex]v_1=v_{f1}\cos24^o[/tex]

[tex]v_{f1}=\frac{v_1}{\cos24^o}[/tex]

where v_f1 = final velocity magnitude.

Am I correct, Pat?
 
  • #7
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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.
 
  • #8
tiny-tim
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v = v1cos24º + v2cos34º :smile:
 
  • #9
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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
tiny-tim
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v = 20.1​
 
  • #11
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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
tiny-tim
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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 any one direction. :smile:
 
  • #13
33
0
Oh okay I understand now! Thank you so much :)
 

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