# D-2 TWO DIMENSIONAL COLLISIONS ON AN AIR TABLE

1. PURPOSE:

To study conservation of energy and momentum in two dimensional collisions.

2. APPARATUS:

Air track, air table, camera, film, magnetic pucks, ordinary vacuum cleaner, air source, rubber, plastic or Velcro puck collars, and connecting hoses.

3. BACKGROUND:

1. General Instructions: Air Tracks and Air Tables
2. Techniques for Stroboscopic Photography Data Collection

4. GENERAL METHOD:

One puck is stationary near the center of the table. The other puck is propelled toward it. The collision events are photographed by strobe photography.

It is important to know the exact time of collision. This is done by closing the camera shutter before the pucks pass out of the camera field of view. Thus the "last" position of both pucks can be unambiguously identified. Call the initially moving puck No. 1. Label the first image of puck No. 1, after it left the launcher, as position No. 1. Sequentially number the successive positions of this puck up to the last one, N. (About 15 positions is optimal.) Also label the last position of puck no. 2 the same value, N; then work backwards labeling positions N-1, N-2, etc. back to the point of collision.

Elastic collisions are best obtained with magnetic pucks, which exert on each other forces of repulsion acting directly along the line joining the pucks. The pucks are not allowed to make actual contact during this kind of "collision." The radial repulsion ensures that the pucks do not transfer any angular momentum or angular kinetic energy to each other.

If magnetic pucks are not available, ordinary ones can be used, but tangential frictional forces at the point of contact will cause an energy loss of a few percent, and will cause transfer of angular momentum and energy between pucks.

Inelastic collisions are studied by placing a rubber, plastic, or Velcro collar on the circumference of one or both pucks.

5. THEORY:

Momentum is conserved in all collisions. A perfectly elastic collision is one in which kinetic energy is conserved—no mechanical energy is converted to heat. A perfectly inelastic collision is one in which the maximum amount of heat is produced (consistent with momentum conservation); this occurs when the colliding bodies stick to each other and move off together.

A useful parameter for describing collisions is the coefficient of restitution, defined by

 [1]
```
v' - v'
1    2
γ = - ———————
v  - v
1    2
```

Primed velocities are the velocities after collision, unprimed ones are before collision. The subscripts label the two bodies. γ = 1 for a perfectly elastic collision; γ = 0 for a perfectly inelastic collision. Contact collisions produce deformations of the bodies, and always some loss of energy to heat, so γ is less than one. For example, γ = 0.98 for a low velocity collision between hardened steel spheres. Because the amount of deformation depends on the relative impact velocity, the coe- fficient of restitution may also depend on this velocity.

6. PROCEDURE:

1. Study elastic collisions between magnetic pucks of equal mass, arranging the impact so that the motion is not confined to one dimension. (i.e. The impact is "off center.")

2. Study a one dimensional partially elastic collision between equal mass pucks. An air track is actually better for this part. Try the same with unequal masses. (Two dimensional inelastic collisions invariably set one or both pucks into rotation, getting us involved with angular momentum and kinetic energy. This will be studied in a later experiment.)

3. Study a one dimensional perfectly inelastic collision between equal mass pucks. Velcro collars around both pucks ensure that they will stick together when they collide. Again an air track may be used, with strips of Velcro on the bumpers, or double-sided cellophane tape.

4. If there is time, repeat the elastic collision with unequal masses.

7. ANALYSIS:

The momentum analysis is carried out most easily by establishing a x axis on the photo record along the initial velocity of the initially moving puck. Then draw a perpendicular y axis. Then calculate components of momentum along these axes, to see whether each component is conserved. Since kinetic energy is a scalar quantity, its analysis is straightforward.

Calculate the coefficient of restitution in each case, and verify conservation of momentum. Calculate the energy lost to heat in each case.

8. QUESTIONS:

1. Show (derive) that if a mass is dropped straight down from a height y1 onto a massive floor and rebounds to a height y2, that the coefficient of restitution is

γ = (Y2/Y1)1/2

2. Show that for a head-on, one dimensional collision between one stationary and one moving body, that if the bodies stick together, the coefficient of restitution is 0. What fraction of the incident energy is converted to heat?

Text and drawings © 1997, 2004 by Donald E. Simanek.