Introduction to Impact Dynamics: the ideal collision when playing pool

While I was playing pool the other day, I remembered a lesson that I learned during my time at the University of Seville as well as at Cranfield University. It is related to the quite complicated subject of Impact Dynamics… But don’t be afraid, today I’ll just cover a simple case as an introduction. In particular, I’m going to write about how the ideal collision between the white ball and the other ball that we are aiming to move.

First of all, we need to differentiate between different types of collisions, depending on the loss of translational kinetic energy or, in other words, the conversion of kinetic energy to rotations, vibrations or heat. For this case, let’s consider two bodies: one in motion (impactor) and another in stationary conditions. Hence, we can distinguish between the following categories:

  • Elastic collision: all the energy is transmitted from one body to another, i.e. the impactor stops and the stationary mass starts moving at the same speed as the initial one from the moving mass.
  • Completely inelastic collision: the moving mass stops after hitting the stationary body. The stationary mass remains as it was.
  • Inelastic collision: the impactor suffers a decrease in speed after the collision, whereas the stationary mass starts moving at a certain speed.
  • Superelastic collision: if additional energy is provided to the stationary body during the impact, then it will start moving at a higher speed than the one at which the impactor hit it.

Once I’ve said that, it is time to introduce another concept: the coefficient of restitution (e). This parameter can be defined as the ratio of speed after impact to speed before the collision. Depending on the type of impact, the value of this coefficient will vary:

  • e=1: elastic collision.
  • e=0: completely inelastic collision.
  • e<1:  inelastic collision.
  • e>1:  superelastic collision.

Now let’s analyse a simple case: one shpere (impactor) moving at a certain speed towards another stationary sphere in a case known as head-on collision. The conditions before the impact are illustrated in Figure 1, where u is the initial velocity of the impactor:


Figure 1

Bearing in mind that this is a head-on collision, the situation after the impact would be the following:


Figure 2

The two states (before and after impact) can be related assuming that the linear momentum of the system is conserved. The reason for that assumption is that there aren’t any external forces acting on the system. Therefore, the amount of movement before the impact should be equal to the amount of movement after the collision:


Taking into account the definition of the coefficient of restitution, another equation can be written:


From those two equations, expressions for the velocities of each body after the collision can be found:


Now we can go back to the pool example. Ideally, we would like to transfer all the energy from the white ball to the coloured sphere (elastic collision, e=1). That would mean that the white ball should stop (v1=0) whereas the other body should start moving at the same speed at which the white ball was moving earlier (v2=u). Using the two expressions which have been derived above, it is quite straight forward to check this. In this case, the two spheres have the same mass (i.e. m1=m2), so substituting the masses and the coefficient of restitution it can be seen how the final velocities match the elastic collision conditions.

In reality, there will be a small amount of energy which is not transferred (inelastic collision). However, if you hit the balls with precission (I should probably avoid talking about precission in pool…), that energy would be so low that it would be very difficult to appreciate any movement of the white ball after hitting the other sphere. Obviously, there are many factors that influence the collision, such as the angle and rotation of the balls, but this example was presented in order to illustrate the physics behind an ideal case which can be (almost) observed in a daily basis. I’ll try to provide other examples where these conditions can be analysed using hand calculation in the near future!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s