# Solving 2D Collisions: Find the Velocity of Eight Ball After Impact

• x86
In summary: V_f## in all the coordinate systems, you have to use ##\cos## in the Y coordinate system and ##\sin## in the X coordinate system.In summary, the cue ball travels at a speed of 5.3 m/s and after impact the eight ball moves at a speed of 4.4 m/s.

## Homework Statement

In a game of billiards the 0.165 kg cue ball is hit toward the 0.155 kg eight ball which is stationary. The cue ball travels at a speed of 6.2 m/s and after impact rolls away at an angle of 40.0 degrees counterclockwise from its initial direction with a velocity of 3.7 m/s. Find the velocity of the eight ball after the collision.

## Homework Equations

m1v1 + m2v2 = m1vf1 + m2vf2

## The Attempt at a Solution

I cut the coordinate system up into the x y and z axis. In the y axis, there is no movement, as the ball stays on the table. So, we have x and z to work with. I drew a Cartesian graph with the axis 'X' and 'Y' where 'Y' represents 'Z'. I will also interpret the cue ball being shot up the Y axis into the eight ball, instead of being shot horizontally.

So I cut the motion up like this, in the y-axis vectors:
0.165kg * 6.2 m/s + 0 = 0.155kgVf + 0.165kg*3.7m/s*sin40
Vfy = 4.4 m/s [up]

x vectors:
0 = 0.155kgVf + 0.165kg*3.7m/s * cos40
Vfyx = -3.0m/s

Now, I add the vectors using Pythagorean theorem and my result is 5.3 m/s [W 55 degrees N] or 5.3 m/s [35 degrees clockwise].

However, the book gives me the answer of 4.4 m/s [35.2 degrees clockwise].

This is the part that confused me, why is the book not taking the X forces into account? Its just using the answer I got in the Y coordinate system. This is confusing, because my teacher did the same thing in school and answered with "there is no movement in the Y axis".

The problem is that there are three dimensions: X,Y,Z and Y is set to 0, but we still have movement in the Z axis (which I represented as Y in this answer).

So I have no idea why its done like this, and am very confused. Did the book make a mistake, or am I wrong?

Pay attention... when you compute what you call ##V_f##, you have to consider that also the second ball goes away with an angle... therefore the correct equations should be

Along y:
$$m_c V_{0c}=m_c V_{fc}\cos(40)+m_8 V_f\cos\theta$$
Along x:
$$0=m_c V_{fc}\sin(40)-m_8 V_f\sin\theta$$

Now solve for $V_f$ and if you want also for ##\theta##. In this way indeed $V_f$ comes 4.4 =)

Last edited:
mmmh... sorry I noticed something just now... your way is also correct because, even if you would have better to call in different ways ##V_f## in the equations in the two directions, it is the same thing (you are leaving implicit the ##\sin\theta## and ##\cos\theta## I am writing explicitely).

The problem is in the definitions... along your Y you have to put the ##\cos## (measuring the angles leaving from the Y axis) while along X you put ##\sin##. In this way it works...

## What is a 2D collision in physics?

A 2D collision in physics refers to a situation where two objects collide with each other in a two-dimensional space, such as on a flat surface. This collision can be either elastic, where both objects bounce off each other, or inelastic, where the objects stick together after colliding.

## What is the conservation of momentum in 2D collisions?

The conservation of momentum in 2D collisions states that the total momentum of the system before and after the collision must remain the same. This means that the sum of the momentums of the two objects before the collision is equal to the sum of their momentums after the collision.

## How do you calculate the final velocities in a 2D collision?

To calculate the final velocities in a 2D collision, you can use the equations for conservation of momentum and conservation of kinetic energy. First, calculate the total momentum before the collision and set it equal to the total momentum after the collision. Then, use the equation for conservation of kinetic energy to solve for the final velocities.

## What is the difference between elastic and inelastic collisions in 2D?

The main difference between elastic and inelastic collisions in 2D is the amount of kinetic energy lost during the collision. In an elastic collision, the total kinetic energy of the system is conserved, while in an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound.

## How do you determine the type of collision in a 2D problem?

To determine the type of collision in a 2D problem, you can use the coefficient of restitution (e). This value represents the ratio of the relative velocity of the objects after the collision to the relative velocity before the collision. If e=1, the collision is elastic, and if e<1, the collision is inelastic.

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