2D Collision of 3 Billiard Balls

In summary: Glad to be of assistance!In summary, in this conversation, the problem is to calculate the energy lost in a near elastic collision between a 500 g billiard ball going at 10 m/s [E] and 2 identical billiard balls. The initial and final momenta are equal and the masses are all the same, allowing for the cancellation of terms. By using conservation of momentum, the velocity of the third ball can be found and then used to calculate the energy lost in the collision. The assumption is made that there is no rolling involved in the collision.
  • #1
Europa
16
0

Homework Statement


A 500 g billiard ball is going at 10 m/s [E] when it impacts 2 other identical billiard balls. Afterwards, you observe 2 of the balls moving at 2 m/s [30 W of S] and 3 m/s[45 N of E] respectively. Calculate the Energy lost in the collision.

Homework Equations


KE = 1/2mv^2
p = mv
p(initial) = p(final)
Energy lost = final energy - initial energy

The Attempt at a Solution


Since there is some energy lost, this is only a near elastic collision.
Using,p(initial) = p(final), break the initial and final into components.
X: mv1 = mv1f +mv2f
Y: mv1 = mv1f + mv2f

Since the masses are all the same we can cancel them out.

X: 10 = 3cos45 - 2sin30
Y: 0 = 3sin45 - 2cos30

I don't even know what i am solving for, don't i already know every variable in the conservation of momentum equation? (Pretty sure I am a variable or something)
 
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  • #2
You're supposed to take the initial energy of the system and compare it with the final energy of the system using conservation of momentum as the link between.
 
  • #3
Europa said:

Homework Statement


A 500 g billiard ball is going at 10 m/s [E] when it impacts 2 other identical billiard balls. Afterwards, you observe 2 of the balls moving at 2 m/s [30 W of S] and 3 m/s[45 N of E] respectively. Calculate the Energy lost in the collision.

Homework Equations


KE = 1/2mv^2
p = mv
p(initial) = p(final)
Energy lost = final energy - initial energy

The Attempt at a Solution


Since there is some energy lost, this is only a near elastic collision.
Using,p(initial) = p(final), break the initial and final into components.
X: mv1 = mv1f +mv2f
Y: mv1 = mv1f + mv2f

Since the masses are all the same we can cancel them out.

X: 10 = 3cos45 - 2sin30
Y: 0 = 3sin45 - 2cos30

I don't even know what i am solving for, don't i already know every variable in the conservation of momentum equation? (Pretty sure I am a variable or something)

I think you might have forgotten the momentum of the first ball after the collision!
 
  • #4
BiGyElLoWhAt said:
You're supposed to take the initial energy of the system and compare it with the final energy of the system using conservation of momentum as the link between.
Can you explain how to link them?
Otherwise,
E1 = 1/2mv2, which is just 25 J, since only 1 ball is moving.
Do i not know E2 as well since i have their masses and their final velocities? where does momentum come in? I know it is not this simple.
 
  • #5
Europa said:
Can you explain how to link them?
Otherwise,
E1 = 1/2mv2, which is just 25 J, since only 1 ball is moving.
Do i not know E2 as well since i have their masses and their final velocities? where does momentum come in? I know it is not this simple.
As PeroK posted, there are three balls, and all have a velocity after the collision. How can you find the velocity of the third ball?
For the energy, the incoming ball is rolling, not sliding, I would have thought. However, it is not stated whether the observed velocities after collision are immediately afterwards or after rolling has been attained, so it's quite unclear whether you are supposed to allow for rolling.
 
  • #6
I would assume that there is no rolling, otherwise this problem would be impossible. Why don't you try using that p initial equals p final equation you posted was relevant. Whats p inital? What p final?
 
  • #7
Actually i semi retract that statement, the rolling is irrelevant.
 
  • #8
BiGyElLoWhAt said:
Actually i semi retract that statement, the rolling is irrelevant.
It's irrelevant if you take all post collision velocities and energies as being immediately after collision.
 
  • #9
Oh ok i see, i thought the 3rd ball just stopped. so finding the velocity of the third ball through momentum, and then plug into find energy loss?
 
  • #10
Yup
 
  • #11
Thanks i got it :D
 
  • #12
No problemo.
 

Related to 2D Collision of 3 Billiard Balls

1. What is the definition of a 2D collision?

A 2D collision is when two objects collide with each other in a two-dimensional space, meaning they can only move along the x and y axes.

2. How is the collision of three billiard balls simulated in a 2D space?

To simulate the collision of three billiard balls in a 2D space, we use a physics engine that takes into account the mass, velocity, and direction of each ball to calculate their movements and interactions when they collide.

3. What factors affect the outcome of a 2D collision between three billiard balls?

The outcome of a 2D collision between three billiard balls is affected by the masses, velocities, and angles of the balls, as well as external forces such as friction and air resistance.

4. Can the outcome of a 2D collision between three billiard balls be predicted?

With the use of advanced mathematical equations and precise measurements of all the factors involved, it is possible to predict the outcome of a 2D collision between three billiard balls. However, slight variations in the initial conditions can greatly affect the outcome, making it difficult to predict with absolute certainty.

5. How does the conservation of momentum and energy apply to a 2D collision of three billiard balls?

The conservation of momentum and energy applies to a 2D collision of three billiard balls because the total momentum and energy of the system before and after the collision remains the same. This means that the sum of the velocities and kinetic energies of the three balls will be equal before and after the collision.

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