Kinetic Energy, Momentum

  • Thread starter xaer04
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Homework Statement


From the text:
"Two identical billiard balls are initially at rest when they are struck symmetrically by a third identical ball moving with velocity (->v0) = v0i. Find the velocities of all three balls after they undergo an elastic collision."

..O
.OO

The top ball is the cue ball, the catalyst, and the bottom two are the targets. The top ball is moving toward them at velocity vo. Also, the targeted balls are touching eachother. My original diagram was sideways (because the equation stated the vo = voi, so it's really moving on the x axis), and ascii images aren't very fantastical, so please bear with me.


Homework Equations



NOTE: terms with a "->" in front refer to vectors. Kinetic energy doesn't get vectors because it's a scalar quantity. Also, v1o reads initial velocity of the first object, v2f reads final velocity of the second object... I don't take you for fools, but I want to cover all bases here.

Conservation of Kinetic Energy
Ki = Kf

(1/2)(m1)*(v1o)^2 + (1/2)(m2)*(v2o)^2 + (1/2)(m3)*(v3o)^2 = (1/2)(m1)*(v1f)^2 + (1/2)(m2)*(v2f)^2 + (1/2)(m3)*(v3f)^2

Conservation of Momentum
Pi = Pf

(m1)(->v1o) + (m2)(->v2o) + (m3)(->v3o) = (m1)(->v1f) + (m2)(->v2f) + (m3)(->v3f)


The Attempt at a Solution



There are no external forces spoken of, so I will neglect external forces and treat this as a perfectly elastic collision where all momentum and kinetic energy is conserved. I note that since the balls are all identical and during the moment of impact they are all in contact with one another, their centers of mass form an equilateral triangle, and since the impact was coming directly from the x axis i know that one ball will shoot off 30 degrees above the axis, and the other 30 degrees below. I know that because the ball that strikes the other two hits them symmetrically, it will bounce back because it is like hitting a single object of twice the mass (and if an object strikes an object with a larger mass than itself it will recoil).

I modified the relevant equations because i know all masses are equal, it's a perfectly elastic collision, and that balls 2 and 3 both begin with the same velocity and end with the same velocity. however, i didn't enter any values (even though i know that balls 2 and three begin from rest, and technically i could just ignore them since they're zero), because that ruins everything... trust me, i tried that first, hehe:) so, i'm left with these:

Modified Conserved K. Energy:

(v1o)^2 + 2(v2o)^2 = (v1f)^2 + 2(v2f)^2

Modified Conservation of Momentum:

(->v1o) + 2(->v2o) = (->v1f) + 2(->v2f)

I manipulated them so i could solve final velocities using only initial values.

1.) (v1o) - (v1f) = 2[(v2f)-(v2o)] //this will be used later

2.) (v1o)^2 - (v1f)^2 = 2[(v2f)^2 - (v2o)^2]
3.) [(v1o)+(v1f)]*[(v1o)-(v1f)] = 2 {[(v2f)+(v2o)]*[(v2f)-(v2o)]} //diff of 2 squares... i know, genius work here

considering step 1 applied to step 3:
4.) (v1o)+(v1f) = (v2o)+(v2f)

now i just have to decide which final velocity i'm solving for, set step 4 equal to the vf i'm not solving for, and substitute the terms into my modified conservation of momentum formula, and then solve the vf i AM solving for. so i get these:

[2(v1o)+(v2o)]/3 = (v2f) //for the final velocity of balls 2 and 3

[4(v2o)-(v1o)]/3 = (v1f) //for the final velocity of ball 1

ok, my questions:

does my work apply to the actual vectors, or just the components? also, is the v2f actually twice the velocity i need?
 

Answers and Replies

  • #2
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please? lol:smile:
 
  • #3
Doc Al
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I modified the relevant equations because i know all masses are equal, it's a perfectly elastic collision, and that balls 2 and 3 both begin with the same velocity and end with the same velocity.
They end with the same speed, not the same velocity--they go off in different directions.
Modified Conserved K. Energy:

(v1o)^2 + 2(v2o)^2 = (v1f)^2 + 2(v2f)^2
Good.

Modified Conservation of Momentum:

(->v1o) + 2(->v2o) = (->v1f) + 2(->v2f)
Not good, for reasons stated above.
 
  • #4
Doc Al,

Since they have equal mass shape and such... cant you just say that the cue ball will transfer all the momentum to both balls equally resulting in the two balls to have final speeds of 1/2 v_0? which means the cue ball final speed would be at rest..

Case 2: m_1 = m_2

thanks.
 
Last edited:
  • #5
Doc Al
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No. Why would you think that?

(Don't confuse this problem with that of a cue ball striking a single ball dead on.)
 

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