Conservation of Linear Momentum | Center Of Mass |

Coefficient of Friction | Angle of Deflection |

Conclusion | A Physics Problem of Billiard |

**PURPOSE**

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.

**INTRODUCTION**

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.

p |

m |

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.

Ke |

where d is the displacement of the objects traveled, F is the
frictional force caused by gravitational force ( F_{f}
= 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:

After an initial force was applied on the ball by the pool stick, there were three forces acting on it: (1) the force of weight (mass*gravity) pulling in the negative y direction; (2) the normal force applied by the table acting on it in the positive y direction; (3) the force of friction between the ball and the table, causing the ball to have a negative acceleration.

**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:

where we understand that the vector of P1i is equal to the vector sum of P

**CONCLUSIONS**

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 P_{total} =
(m_{1}v_{1}) + (m_{2}v_{2}) + (m_{n}v_{n}), 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.

**Example Of
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 q

Solution:

Because the target ball is initially at rest, conservation of energy gives:

But
all balls have the same mass, thus m_{1}
= m_{2},
so that:

(1)
v_{1i}^{2}
= v_{1f}^{2}
+ v_{2f}^{2}

Applying conservation of momentum to the two-dimensional collision gives:

(2)
v_{1i}
= v_{1f}
+ v_{2f}

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 v_{1f}
and v_{2f} is q_{2}
+ 45^{o}, v_{1f}v_{2f}
= v_{1f}v_{2f}cos(q_{2}+45^{o}),
and so

(3)
v_{1i}^{2}
= v_{1f}^{2}
+ v_{2f}^{2}
+ 2v_{1f}v_{2f}cos(q_{2}+45^{o})

Subtracting (1) from (3) gives

0
= 2v_{1f}v_{2f}cos(q_{2}+45^{o})

0=cos(q_{2}+45^{o})

q_{2}+45^{o}
= 90^{o} or = 45^{o}

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.