Source: http://www.flickr.com/photos/ventsislav/2222807833

The physics behind billiards (or the physics behind pool), in large part, involves collisions between billiard balls. When two billiard balls collide the collision is nearly elastic. An elastic collision is one in which the kinetic energy of the system is conserved before and after impact. Therefore, for simplicity one can assume that for collisions involving billiard balls, the collision is perfectly elastic.

For collisions between balls, momentum is always conserved (just like in any other collision). For a simplified case assuming no friction (discussed below), we can combine this fact with the elastic-collision assumption to find the trajectory of two colliding billiard balls after impact. The figure below shows a collision between two billiard balls. For the general case, the collision is not head on, which is what the figure shows.

It is assumed that balls A and B have the same mass and that ball B is initially at rest (zero velocity). The initial velocity of ball A is

The line L

After impact at CP, ball B moves in the direction of the line joining the center of the two balls, as shown. This is because the force (impulse) delivered by ball A to ball B acts normal to the surface of ball B, assuming there is no friction between the balls (a good assumption since billiard balls are smooth). Thus, ball B moves in the direction of this impulse.

Notice that, after impact, ball A moves in a direction perpendicular to the direction of ball B. This interesting result can be proven as follows.

For the two colliding balls, the general vector equation for conservation of linear momentum is:

Since the masses

For an elastic collision kinetic energy is conserved, and the equation is:

Since the masses

By the Pythagorean theorem, this last equation tells us that the vectors

Thus, after impact ball A moves in a direction perpendicular to the direction of ball B. This is a very slick result.

There are two additional special cases to consider, involving ball collision.

For the case where the target ball B must be hit at an angle θ very close to zero (such as to sink it in the side pocket), ball A needs to be moving at a high speed

For the case where the impact is head on (θ = 90°) the above solution does not apply. In this case

For a more detailed and complete analysis, in which the trajectory of ball A is calculated (after impact), under the influence of friction between the ball and billiard table, see the problem, Cue ball trajectory with table friction.

The physics of billiards is similar to the Physics Of Hitting A Baseball, in that there is also a sweet spot on a billiard ball where you can strike with the cue stick so that no friction force develops between the ball and the billiard table. Knowing the location of this sweet spot can give you an idea of where to hit the ball so that it develops backspin or forward spin, which can be useful when making a shot.

Consider the figure below showing the position of the cue at height

We wish to find the height

In this analysis, we can represent the ball + cue system with a free-body diagram as shown below.

Where:

By Newton's Second Law, the general force equation in the x-direction is:

Where:

This equation becomes

Since

By Newton's Second Law, the general force equation in the y-direction is:

where

Since the billiard ball only moves in the x-direction

Therefore

We must now write the general moment equation for rotation of a rigid body about its center of mass

Where:

Since no frictional force develops between the ball and table, there is no relative slipping at point

In the above equation the negative sign is there to match the sign convention used in this problem.

The moment equation becomes

Combine equations (1) and (2) and we get

For a solid sphere

Therefore

This is the height to hit the ball so that no friction develops at point

In the cases where the cue strikes above or below this height

In the cases where slipping occurs we have the following inequality:

This means that there is relative motion between the ball and billiard table at point

In the case of pure rolling, the tangential velocity of the ball at point

Relative slipping between ball and billiard table is an interesting point of analysis. It's informative to understand how the ball moves depending on where it is hit relative to

Consider the figure below.

When the ball is hit sufficiently hard with a leftward force in region A

When the ball is hit sufficiently hard with a leftward force in region A

When the ball is hit sufficiently hard with a leftward force in region A

Thus, the nature of the slipping will change depending on which of the regions, (A

Note that for the three cases above, the frictional force that develops due to relative slipping is known as

On the other hand, when there is no relative slipping between two surfaces the frictional force between them is known as

This is a continuation of the previous section, with a more in-depth look at relative slipping. This section is optional, so you may skip it if you like.

The figure below shows a free-body diagram of a billiard ball experiencing a general case of relative slipping.

Where:

Let

Let

With no loss of generality we can assume

If the cue strikes the ball in region A

If the cue strikes the ball anywhere below height

Set

This is equal to +1 or -1. This factor accounts for the direction of relative slipping, which is important to know since we are dealing with kinetic friction. Such factors are mathematically very convenient when accounting for the direction of kinetic friction. (Note that |x| means the absolute value of x).

From before

This is the normal force acting on the ball at point

The general force equation in the x-direction is:

The kinetic friction acting at

where

Now,

The general moment equation for rotation of a rigid body about its center of mass

This becomes

From equations (3) and (4) we get:

The linear velocity of the ball is:

where

The angular velocity of the ball is:

We wish to find the time it takes for the ball to stop slipping, and begin pure rolling. Thus, using equations (5) and (6) we can formulate the following equality, which holds true when there is pure rolling:

From this we can solve for time

For a solid sphere

Therefore

Let's now find the distance traveled by the ball before pure rolling begins, using the time

The distance

A final consideration is finding the velocity of the ball before pure rolling begins. To do this we calculate the velocity

Note that the above three equations are only valid as long as there is relative slipping at point

Sample Calculation For Relative Slipping

This is a sample calculation using the results of the previous section.

Therefore the time it takes the ball to stop slipping is

The distance traveled by the ball during slipping is

The velocity of the ball just before pure rolling begins is

As you can see, the physics of billiards can get pretty involved when you start considering all the things that can happen in a typical game of pool. You can bet that professional players are very proficient in the practical usage of the physics presented here, as well as other aspects of the game not discussed here.

Return to

Return to

Confused and have questions? Check out Chegg Study. They've got answers. You can get step-by-step solutions to your questions from an expert in the field. If you'd rather get 1:1 study help, try 30 minutes of free online tutoring with Chegg Tutors.