SAMPLE PROBLEMS

Billards, the game of weekend night, smoked filled caverns…How could anyone imagine this game to be such a perfect representation of kinematics in physics? This intricate game involves a plethora of theoretical concepts, such as linear momentum, rotational kinetic energy, conservation of mechanical energy, angular acceleration, torque, all based on Newton’s laws. Here we will uncover the physics behind billiards by using several mathematical theories. It is impossible to play billards without at least knowing some aspect of physics:

LINEAR MOMENTUM

When you watch a game of billiards, you notice that the balls move with difference speeds with each ricocheting contact. Since the game is played on a flat surface, momentum is represented in a 2-D fashion. Using physics, you can calculate the relative speed of the balls, keeping in mind that we are dealing with elastic collisions, using the conservation of momentum theory.

PROBLEM 1

Ok, this may sound somewhat unreasonable for a weekend game of physics, I mean pool but you wanna get one ball into the corner pocket and you know the initial speed of the cue ball (white ball) to be about 20 cm/s and the other ball is pretty much stationary. You also happen to know that the mass of the cue ball is about 0.17 kg and the mass of the other balls to be about 0.15 kg. So, how cool is it that you can calculate how fast the ball you’re directly aiming for and the cue ball after the collision between the two? The conservation of momentum law for elastic collisions:

Conservation of momentum: m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

Conservation of kinetic energy : 0.5m₁v_1i²+0.5m₂v_2i²=0.5m₁v_1f ²+0.5m₂v_2f²

m₁(v_1i - v_2i) = m₂(v_1f - v_2f)

(0.17kg)(-0.2m/s-0.0m/s) =(0.15kg)(v_1f -v_2f)

-0.23kg*m/s= v_2f - v_1f

v_1i – v_2i = -(v_2i - v_2f)

0.2m/s-(-0.0m/s)=v_2f – v_1f

v_1f=0.0m/s v_2f=0.2m/s

*frictional force of the felt from the billiard table will decrease the velocity of the cue ball before it collides with the second billiard ball. We are assuming that there is not friction (see friction section).

GLANCING COLLISONS

Billiard balls do not always hit each other dead center. This previous example deals with the conservation of momentum only when the two balls hit each other directly in the center. If we could all do that with every shot we make, we’d all be professional pool players. Usually that is not the case and so we move on to glancing collisions. These are more realistic collisions that involve a ball striking another ball and then either one or the other moving at an angle to the horizontal.

This sort of problem must be solved by separating the x and y components of momentum:

x component: m₁v_1i + 0 = m₁v_1f cosθ+ m₂v_2fcosФ

y component: 0 + 0= m₁v_1f sinθ- m₂v_2fsinФ

PROBLEM 2

Let’s say that one billiard ball strikes another stationary ball of the same mass at 4 m/s. The final velocity of the initally moving billiard is 3.5 m/s at an angle of

25 degrees to the horizontal. So, what is the final velocity and angle to the horizontal of the second ball?

Let’s plug some numbers into the x and y components…

x-component: m₁v_1i + 0 = m₁v_1f cosθ+ m₂v_2fcosФ

v_1i=v_1f cosθ+v_2fcosФ

4.0m/s=3.5m/scos25+ v_2fcosФ

v_2fcosФ=0.83

y-component: 0+0= m₁v_1f sinθ- m₂v_2fsinФ

m₂v_2fsinФ= m₁v_1f sinθ

v_2fsinФ=(3.5m/s)sin25

v_2fsinФ=-1.48

0.83^2+-1.48^2=6.27

√6.27=2.50m/s

tanθ=(-1.48/0.83)

θ=tan^-1(-1.48/0.83)=60^o

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TORQUE & ANGULAR ACCELERATION

Torque is represented by observing a billiard ball in an isolated system. The force acting upon the ball makes the ball rotate around some axis. This is represented by:

Torque=Fd

“d” is the distance from the center of axis in which the force is applied.

Let’s say that the axis of rotation is at the radius of the billiard ball. There are infinite numbers of particles that are at different distances from this axis of rotation.

All of these particles travel at the same speed, therefore, travel with the same angular acceleration. There is also a direct relationship between this angular acceleration and the torque of a rotational system due to Newton’s Law.

Torque=mr²ά ά=a_t/r

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PROBLEM 3

If a cue stick struck a 0.15 kg billiard, 2 cm from the axis of rotation, with a tangential acceleration of 0.5m/s^2. What was the torque that was applied?

Torque=Fd

What if force from the cue stick was applied directly to the center of the billiard? Would it seem plausible that the ball will roll faster this way?

Torque=Fd

=(ma)d

=(0.15kg*0.3m/s^2)0.05m

=2.25e-3 N

Because the radius of the billiard ball depicts where the axis of rotation is, if a billiard was struck where radius is zero, then there is no torque (see torque equation). So what causes the ball to roll?

FRICTION

When the ball is struck directly in the center, the ball will not roll but rather slide until frictional force takes over. This will reduce the initial velocity of the billiard as well as cause the ball to lose kinetic energy out of the system. So, in order to get the maximum velocity, an initial spin or torque must be applied to the cue ball very close to the axis of rotation.

MOMENT OF INERTIA

Moment of Inertia of a billiard ball is represented by the theoretical moment of inertia for a solid sphere I=(2/5)MR^2

PROBLEM 4

What is the moment of inertia for a 0.15kg billiard ball that has a radius of 0.1m?

I=(2/5)MR^2

I=(2/5)(0.15kg)(0.1^2)

I=6.0e-4 kg*m^2

CONSERVATION OF ENERGY and ROTATIONAL KINETIC ENERGY

This theory involves the idea that energy is never lost but rather converted to different forms. If you observed one billiard ball in rotational motion, not only translational energy is represented but rotational energy as well. When an object is rotated around an axis of rotation, Newton’s law is represented but in the form of torque. Torque is used as a rotational force that is conserved. The conservation of energy is represented in the rotational kinetic energy equation:

(KE_t+KE_r+PE_g) _i=(KE_t+KE_r+PE_g) _f

Since the billiards are on an assumed 2-D theoretical surface, potential energy can be eradicated

(0.5mv²+ 0.5Iώ²) _i =(0.5mv²+ 0.5Iώ²) _f

*One must remember that the collisions of the billiard balls are not perfectly inelastic because energy in the system is converted to sound and heat. Also, since this is not perfectly elastic system, inelastic conditions are involved as well but can be ignored since billiards is a highly elastic system that should overcome inelastic conditions.