The purpose of this study is to discover the physics involved in the game of pool. Using the principles behind Newton's laws, we are able to investigate how forces and motions can be applied on the pool table. The fundamental principles that will be tested include conservation of momentum, how a center of mass moves within a closed system, the effect of friction on the balls, and the angles of inelastic collisions between them.
Newton discovered the basic laws of mechanical physics through observations in real life. These laws can be applied to everyday activities, such as pool. Through gaining an understanding of Newton's laws, the actions and reactions between objects and forces, can be quantified in this game. The tracking of balls in collisions can yield the conservation of motion, or the center of mass. Due to the dense nature of the balls, the collisions between them are elastic. The angles of deflection can also be monitored. Lastly, the friction coefficient can be found for a pool ball rolling on a felt-covered table. Due to the linear nature of these interactions, pool exhibits an ideal situation for examining these forces.
CONSERVATION OF LINEAR MOMENTUM
Momentum, or the quantity of motion, is a function of the mass and velocity of an object. For momentum to be conserved in any given situation, all the forces acting on an object must be accounted for. The linear momentum of a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity.
p = mv
Law of conservation of linear momentum: Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant. This law tells us that the total momentum of an isolated system at all times equals it's initial momentum.
p1i + p2i = p1f + p2f
m1v1i + m2v2i = m1v2f + m2v2f
Notice there is no statement concerning the nature of the forces acting on the particle of the system. The only requirement is that the force must be internal to the system. In the case of pool, there is frictional force involved. The frictional force stops the motion of balls.
The total energy in the system is thus conserved.
Ke1i + Ke2i = Ke1f + Ke2f + Ffd
where d is the displacement of the objects traveled, F is the frictional force caused by gravitational force ( Ff = mg ).
CENTER OF MASS
When observing objects with a known mass in a closed system, the center of those masses can then be determined. If these objects are moving at a constant velocity, their center of mass will also move at a constant velocity. This holds true even it these objects collide. To determine the center of mass of a closed system, the positions of each object on a coordinate system must be known. The equation used for finding the center of mass is as such:
where m is the mass of the respective objects, and x is their position. The equation will determine the position of the center of mass. In our calculations, we used this equation twice to determine the x and y positions of the center of mass. Once the positions of the objects and center of mass were determined, they were depicted graphically.
COEFFICIENT OF FRICTION
In our studies of Newton's laws, we often used idealized situations in which we neglected the effects of friction. However, in practice, friction almost always exists. If friction could be considered a negligible force in the game of pool, the balls would never stop moving. The friction coefficient is a function of the force of weight and the normal force as described by the equation:
ANGLE OF DEFLECTION
Due to the density of the balls, the collisions between them are elastic. That is, momentum is conserved, opposed to an inelastic collision, where this is not the case (e.g. clay ball thrown at door). Because all of the balls have the same mass, that angle made by the balls after the collision is approximately 90 deg. This occurs because the sum of the momentum of the balls after the collision is equal to the momentum of the moving ball before the collision. The equation that was used in this case is:
Through our observations, we have shown that Newton's laws can be applied to the game of pool. First, the collisions between the balls are elastic, and therefore momentum is conserved within the system. This was studied through the tracking of their motions to find momentums before and after the collisions. We used the general equation Ptotal = (m1v1) + (m2v2) + (mnvn), to find each initial and final momentum. Since the initial and final values were statistically similar, with some variance, we were able to conclude that momentum was conserved for each elastic collision.
Conservation of momentum prompted the study of angles of deflection. Through the tracking of ball momentum and the angles between the balls after collisions, momentum is conserved yet again. One sedentary ball was hit by another ball with an initial momentum. The resulting vectors of momentum were 90 degrees apart because they add up to the initial momentum.
We studied the center of mass by colliding three balls and found that it moves at a constant velocity. This was accomplished by tracking the position of the three balls with respect to time and using the equation; . This equation yielded the position of the center of mass with respect to time. When these points were graphed, the center of mass had a linear relationship, thus having a constant velocity.
Billiard Ball Collision
In a game of billiards, a player wishes to sink a target ball in the corner pocket, as shown below. If the angle q1 to the corner pocket is 45o, at what angle is the cue ball deflected? Assume that friction and rotational motion are unimportant and that the collision is elastic.
Because the target ball is initially at rest, conservation of energy gives:
But all balls have the same mass, thus m1 = m2, so that:
(1) v1i2 = v1f2 + v2f2
Applying conservation of momentum to the two-dimensional collision gives:
(2) v1i = v1f + v2f
Note that because m1 = m2, the masses also cancel. if we square both sides and use the definition of the dot product of two vectors, we get:
Because the angle between v1f and v2f is q2 + 45o, v1fv2f = v1fv2fcos(q2+45o), and so
(3) v1i2 = v1f2 + v2f2 + 2v1fv2fcos(q2+45o)
Subtracting (1) from (3) gives
0 = 2v1fv2fcos(q2+45o)
q2+45o = 90o or = 45o
This result shows that whenever two equal masses undergo a glancing elastic collision and one of them is initially at rest, they move at right angles to each other after the collision.