Conservation of Momentum 2D - 2 marbles collide - find final velocities

In summary: I mentioned in my first post that I don't agree with your .004625 for the total energy.EDIT.004625 J = .0000125 kg (m^2/s^2) + .3 (m/s) + 9(vfB^2) + .015kg(vfB^2) Something else wrong here. You should have a middle term with a vfB in it from squaring the binomial: (a+b)^ = a^2 + 2ab + b^2Leave out the units so it is easier to read.Also, you can divide both sides by .005 to get numbers easier to work with. You have a right side that is about
  • #1
durandal995
6
0

Homework Statement


A 10.0 g marble slides to the left with a velocity of magnitude 0.500 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on elastic collision with a larger 30.0 g marble sliding to the right with a velocity of magnitude 0.150 m/s

Find the velocity of each marble (magnitude and direction) after the collision, the change in momentum of each marble, and the change in kinetic energy for each marble.


Homework Equations



B=bigger marble
S=smaller marble

(mB)(viB) + (mS)(viS) = (mB)(vfB) + (mS)(vfS)
.5(mB)(viB)^2 + .5(mS)(viS)^2 = .5(mB)(vfB)^2 + .5(mS)(vfS)^2

The Attempt at a Solution



This problem has been hard for me because I have not figured out how to substitute from one equation to the other. I have solved for the velocity of the Big marble (vfB) but when I substitute it into the kinetic energy equation I end up getting crazy numbers that are hard to simplify.

I solved for vfB and got:
(vfB) = .35(vfS)

When I substitute it though, I get:

.004625 J = .5(.01 kg)*(.05 m/s + 3vfB)^2 + .5(.03 kg)*(vfB)^2

So, I'm getting some pretty crazy numbers that are hard for me to work with. Once I find a final velocity this is going to be easy. Does anyone have a suggestion on how to proceed?
 
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  • #2
.004625 J = .5(.01 kg)*(.05 m/s + 3vfB)^2 + .5(.03 kg)*(vfB)^2
looks good except for the .004625 J. For the total KE I get
.5*.01*.5^2 + .5*.03*.15^2 = .0015875J

I didn't finish it, but from your step above you would just square the binomial in the brackets (results in 3 terms), combine everything into an ax^2 + bx + c = 0 and use the quadratic general solution.
 
  • #3
durandal995 said:

Homework Statement


A 10.0 g marble slides to the left with a velocity of magnitude 0.500 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on elastic collision with a larger 30.0 g marble sliding to the right with a velocity of magnitude 0.150 m/s

Find the velocity of each marble (magnitude and direction) after the collision, the change in momentum of each marble, and the change in kinetic energy for each marble.


Homework Equations



B=bigger marble
S=smaller marble

(mB)(viB) + (mS)(viS) = (mB)(vfB) + (mS)(vfS)
.5(mB)(viB)^2 + .5(mS)(viS)^2 = .5(mB)(vfB)^2 + .5(mS)(vfS)^2

The Attempt at a Solution



This problem has been hard for me because I have not figured out how to substitute from one equation to the other. I have solved for the velocity of the Big marble (vfB) but when I substitute it into the kinetic energy equation I end up getting crazy numbers that are hard to simplify.

I solved for vfB and got:
(vfB) = .35(vfS)

When I substitute it though, I get:

.004625 J = .5(.01 kg)*(.05 m/s + 3vfB)^2 + .5(.03 kg)*(vfB)^2

So, I'm getting some pretty crazy numbers that are hard for me to work with. Once I find a final velocity this is going to be easy. Does anyone have a suggestion on how to proceed?

Well, I just did a preliminary run through of your problem and you're right you will have to solve your conservation of momentum problem for one of your final velocities and substitute it into the conservation of energy equation so that you end up with a quadratic equation for the other velocity.

Now, your expression "(vfB) = .35(vfS) isn't correct. You should have another term which will be just a number.

Also, be aware --this is going to be a messy calculation. And, yes the numbers will look "crazy". The quadratic will give you two solutions one of which you will have to discard because it physically doesn't make sense.
 
  • #4
AEM said:
Now, your expression "(vfB) = .35(vfS) isn't correct. You should have another term which will be just a number.

Oh I see! I was wondering why my units of measure were getting messed up. I'll give it another run with this instead:

0.017 m/s + 0.333 m/s (vfS) = (vfB)

And thanks for the quadratic equation tip. I can't believe I forgot about that.
 
  • #5
I thought you just had a typo in the first equation because I agree with what you substituted in the second for vfB. At least I did before . . . this new value doesn't look right. I get -.01*.5 + .03*.15 = -.0005 for the total momentum.
So after the collision, -.0005 = .01Vs + .03Vb
and Vs = (-.0005 - .03Vb)/.01 = -.05 - 3Vb
That is what you substituted into your 2nd equation in the first post.
 
  • #6
Delphi51 said:
I thought you just had a typo in the first equation because I agree with what you substituted in the second for vfB. At least I did before . . . this new value doesn't look right. I get -.01*.5 + .03*.15 = -.0005 for the total momentum.
So after the collision, -.0005 = .01Vs + .03Vb
and Vs = (-.0005 - .03Vb)/.01 = -.05 - 3Vb
That is what you substituted into your 2nd equation in the first post.

Thanks for the save. I'm giving it another go now.
 
  • #7
Okay, I'm moving on from here and I must be doing something wrong. I'm getting a number that doesn't work:

.004625 J = .5(.01 kg)*(-.05 m/s - 3vfB)^2 + .5(.03 kg)*(vfB)^2

.004625 J = .0000125 kg (m^2/s^2) + .3 (m/s) + 9(vfB^2) + .015kg(vfB^2)

0 = 9.015(vfB^2) + .3 (m/s)(vfB) - .0046125 (J?)

so those are the A, B and C that I'm plugging into the quadratic formula. I'm getting:

0.0114413 m/s or 0.0447192 Neither seem correct. Where did I go wrong?
 
  • #8
.004625 J = .5(.01 kg)*(-.05 m/s - 3vfB)^2 + .5(.03 kg)*(vfB)^2
I mentioned in my first post that I don't agree with your .004625 for the total energy.

EDIT
.004625 J = .0000125 kg (m^2/s^2) + .3 (m/s) + 9(vfB^2) + .015kg(vfB^2)
Something else wrong here. You should have a middle term with a vfB in it from squaring the binomial: (a+b)^ = a^2 + 2ab + b^2
Leave out the units so it is easier to read.
Also, you can divide both sides by .005 to get numbers easier to write down.
 
  • #9
Delphi51 said:
Also, you can divide both sides by .005 to get numbers easier to write down.

Thanks for the tip. I'm getting 12, .3 and .1025 for A, B and C. Does that look right to you?
 
  • #10
I'm getting 12, -.3, -.4869.
 
  • #11
Delphi51 said:
I'm getting 12, -.3, -.4869.
Well, I'm about to give up. This question is worth 10% of my grade for this homework but I've been working on it for about 5 hours now. I'm getting vfB = -.189 m/s and vfS = .517 m/s.

vfB should be closer to -.1 m/s and vfS should be closer to +.5 m/s

I am finding those numbers at this site:
faculty.jsd.claremont.edu/jhigdon/phys33/phys33_hw_06_solutions_07.pdf

A marble problem with the same properties but different numbers is on page 6 of this PDF:

faculty.jsd.claremont.edu/jhigdon/phys33/phys33_hw_06_solutions_07.pdf

Delphi51, thanks so much for all your help. I appreciate you hanging in there with me. I hope that the above link can help anyone that has this same problem. Good luck.
 

What is conservation of momentum in 2D?

Conservation of momentum in 2D is a physical law that states that the total momentum of a system of objects will remain constant in all directions, as long as there are no external forces acting on the system. This means that if two objects collide in a 2D space, the sum of their momentums before the collision will be equal to the sum of their momentums after the collision.

How does conservation of momentum apply to a collision between two marbles?

In a collision between two marbles, the total momentum of the system before the collision will be equal to the total momentum of the system after the collision. This means that the sum of the momentums of the two marbles before they collide will be equal to the sum of their momentums after they collide.

What factors affect the final velocities of the marbles after the collision?

The final velocities of the marbles after the collision are affected by the mass and initial velocities of the marbles, as well as the angle and elasticity of the collision. The mass and initial velocities of the marbles will determine the initial momentum of the system, while the angle and elasticity of the collision will affect how the momentum is distributed between the marbles after the collision.

How do you calculate the final velocities of the marbles after the collision?

To calculate the final velocities of the marbles after the collision, you can use the conservation of momentum equation: m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the marbles, and v1, v2, v1', and v2' are the initial and final velocities of the marbles. This equation can be rearranged to solve for v1' and v2'.

Can the final velocities of the marbles ever be greater than their initial velocities?

No, according to the law of conservation of momentum, the total momentum of the system cannot change. This means that the final velocities of the marbles after the collision cannot be greater than their initial velocities. The only way for the final velocities to be greater is if there is an external force acting on the system during the collision.

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