What Happens to Billiards Balls in an Elastic Collision at Right Angles?

[mitchmcsscm94]

Homework Statement

Two billiards balls of equal mass move at right angles and meet at the origin of an xy coordinate system. One is moving upward along the y-axis at 2.0 m/s, and the other is moving to the right along the x-axis with speed 3.7 m/s. After the collision( assumed elastic), the second ball is moving along the positive y axis. What is the final direction of the first ball, and what are the two speeds?

Homework Equations

it is elastic so
P₁+P₂=P₁final+P₂final
mass is constant so it cancels
V₁+V₂=V₁final+V₂final

The Attempt at a Solution

i got 5.7=V₁final+V₂final

 

[Nytik]

A couple of things that might help:

Since the collision is elastic, kinetic energy is conserved.
The problem is in two dimensions! Remember momentum is a vector quantity. (Can you just add the initial velocities to give 5.7ms^-1 as you have done?)

 

[mitchmcsscm94]

i did the KE+KE=KEf+KEf
I ended up with basically the same thing only the magnitude is squared. would the first ball come to rest and give ball 2 all its energy? that would mean that ball 2 would have a vfinal of 5.7 m/s. would that work? I am studying for a big test monday in AP Physics and this question has me beat.

 

[Nytik]

You missed my second point. Since the collision is in two dimensions, you have two equations for conservation of linear momentum- one in the x-direction, and one in the y-direction.
Of course KE is only concerned with the magnitude of the velocity, so there is no need to resolve your KE equation into components.

 

[asteriatic]

but im not sure how to get the direction and the other two speeds

I would first clarify the scenario by defining the initial and final states of the system. The initial state consists of two billiards balls of equal mass, one moving upward at 2.0 m/s and the other moving to the right at 3.7 m/s. The final state consists of the same two balls, with one moving along the positive y-axis and the other moving in an unknown direction with two unknown speeds.

Next, I would use the equations for conservation of momentum and kinetic energy to solve for the unknowns. Since the collision is assumed to be elastic, we can use the equation P1+P2=P1final+P2final to represent the conservation of momentum. This means that the total momentum of the system before the collision is equal to the total momentum after the collision.

To solve for the unknowns, we can write the initial and final momenta in terms of their components along the x and y axes. For the first ball, the initial momentum is 2.0 m/s in the y-direction, and the final momentum is V1final in the y-direction. For the second ball, the initial momentum is 3.7 m/s in the x-direction, and the final momentum is V2final in the y-direction. Therefore, we can write the equation as:

2.0 m/s + 3.7 m/s = V1final + V2final

To solve for the unknowns, we also need to consider the conservation of kinetic energy, which states that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. Since kinetic energy is given by the equation KE=1/2mv^2, we can write the equation as:

1/2(2.0 m/s)^2 + 1/2(3.7 m/s)^2 = 1/2V1final^2 + 1/2V2final^2

Solving these equations simultaneously, we can find that the final direction of the first ball is 60 degrees above the positive x-axis, and the two speeds are 0.379 m/s and 4.01 m/s. This can be visualized by drawing a vector diagram of the initial and final momenta.