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

**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: m_{1}v_{1i}
+ m_{2}v_{2i} = m_{1}v_{1f }+ m_{2}v_{2f}

Conservation of
kinetic energy : 0.5m_{1}v_{1i}^{2}+0.5m_{2}v_{2i}^{2}=0.5m_{1}v_{1f}^{
2}+0.5m_{2}v_{2f}^{2}

m_{1}(v_{1i}
- v_{2i}) = m_{2}(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: m_{1}v_{1i} + 0 = m_{1}v_{1f }cosθ+
m_{2}v_{2f}cosФ_{}

y component: 0 + 0= m_{1}v_{1f
}sinθ- m_{2}v_{2f}sinФ

__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: m_{1}v_{1i}
+ 0 = m_{1}v_{1f }cosθ+ m_{2}v_{2f}cosФ

v_{1i}=v_{1f
}cosθ+v_{2f}cosФ

4.0m/s=3.5m/scos25+
v_{2f}cosФ

v_{2f}cosФ=0.83

y-component:
0+0= m_{1}v_{1f }sinθ- m_{2}v_{2f}sinФ

m_{2}v_{2f}sinФ= m_{1}v_{1f }sinθ

v_{2f}sinФ=(3.5m/s)sin25

v_{2f}sinФ=-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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visit: http://www.explorescience.com/activities/Activity_page.cfm?ActivityID=23

**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

Torque=mr^{2}ά ά=a_{t}/r

.

__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,

(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

(0.5mv^{2}+
0.5Iώ^{2})_{ i} =(0.5mv^{2}+ 0.5Iώ^{2})_{ 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.