Ball collision momentum problem

  • Thread starter petoknm
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  • #1
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Homework Statement



Hello. I was solving this problem about two balls in a plane colliding elastically. The first one had mass 100g radius 5cm and started with velocity of 3m/s to the right and hit the second ball with mass 50g radius 3cm(initially stationary). It hit the second ball such that the line directed by the velocity vector of the first ball going through the center of the first ball and the line parallel to it going through the center of the second ball (these two lines) are separated by a distance of 1cm.


Homework Equations



Conservation of energy AND conservation of momentum

The Attempt at a Solution



Well I started to identify the angles after the collision and I came up with the these angles:
first ball:
cos(alpha')=1/8; 0<=alpha'<=pi/2
and the second ball:
sin(alpha)=1/8; 0<=alpha<=pi/2
where alpha' is the angle above the x-axis to the velocity vector of the first ball and alpha is the angle below the x-axis to the velocity vector of the second ball.
Now all we need are the magnitudes of the velocity vectors after the collision. Because it is an elastic collision the energy is conserved and the momentum is conserved. So for energy we have
18=2|v1|^2+|v2|^2
And for the momentum we have
2*v0=2*v1+v2
2<3;0>=2a<cos(alpha');sin(alpha')>+b<cos(alpha);-sin(alpha)>
where a,b are the magnitudes of the vectors.

But this system has no solution. Where is the problem? Thank you!
 

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Answers and Replies

  • #2
Simon Bridge
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It is best practice to do the algebra first and put the numbers in as late as you can.
Why is there no solution?

i.e. do you have more unknowns than you have equations?

Aside: if the problem specifies that these are balls - are they rolling without slipping?
Do you need to account for conservation of angular momentum as well?
Or are these really just a couple of circular objects sliding on a table?
 
  • #3
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They are sliding. And I have one standard equation (energy conservation) and one two dimensional vector equation (momentum conservation) so basically I have three equations and just two unknowns and therefore I'm not able to find a solution.
 
  • #4
Simon Bridge
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Three questions and two unknowns means that one of the equations is surplus to requirements.
 
  • #5
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But I think that both energy and momentum are conserved in this situation... Or am I wrong?...
 
  • #6
bobie
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Well I started to identify the angles after the collision
You should start finding the angle of impact, which is the angle between the direction of the vector (the blue line in your picture) and the line joining the centres of the sphere.
After the collision ball B will move in that direction
 
  • #7
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That's exactly what I did...I imagined the situation at the moment of impact... There is a right triangle with hypotenuse 8 and opposite side 1...
 
  • #8
bobie
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and what is the angle of impact?
 
  • #9
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Sin(alpha)=1/8...alpha~7deg
 
  • #10
bobie
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Sin(alpha)=1/8...alpha~7deg
You need the cosine .992156 to find the x component of velocity.
then you know P = .1*3 and Ke =.45 and M/m =2, that's all you need.
Rewrite your equations clearly:

P =.3 = M*vM*cos α + mvm*cos 7.18
KE = .45 = M*vM2/2 + m*vm2/2

hint: m will go at 7.18°, v= 1.9843. (since v0=M+m, v=2cos7°)
 
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  • #11
Simon Bridge
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@bobie: this is what was done in post #1 - last equation.
@petoknm: the original problem statement does not claim that energy and momentum are conserved, no.
However, I have been advising you on the assumption that the collision is elastic.
Have you tried to use two of the three equations to get a solution, or did you just stop when your realized you had three equations and only two unknowns?
 
  • #12
bobie
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Because it is an elastic collision the energy is conserved and the momentum is conserved. ..
2<3;0>=2a<cos(alpha');sin(alpha')>+b<cos(alpha);-sin(alpha)>
@bobie: this is what was done in post #1 - last equation....

However, I have been advising you on the assumption that the collision is elastic.
Hi Simon, is that the equation you are referring to? I couldn't and cannot read it , and other lines.
I couldn't get what is his problem.

It seems he is assuming the collision is elastic.
 
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  • #13
Simon Bridge
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That's right - the equation is hard to read - <a;b> is a vector (a,b)t
Yes - OP is assuming the collision is elastic. Also assuming that the objects move on perpendicular trajectories after the collision.

I agree that it is very straight forward, I am trying to get OP to do the next step.
There are more equations than unknowns - so the next step is to solve the simultaneous equations.
The results will say more and allow OP to troubleshoot the answer further.
 
  • #14
bobie
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OP is assuming the collision is elastic. Also assuming that the objects move on perpendicular trajectories after the collision.
.

Didn't get that, if that's true, we must warn him that it is a gross mistake, probably that is the cause.
 
  • #15
Simon Bridge
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It'll come out in the wash.
 

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