Glancing Collision in two Dimensions

## Question:

Consider two smooth disks, A and B each weighing 47lb. Disk B is
at rest and disk A has velocity 8 fps such that it will strike
disk B in a glancing collision. At the instant of impact the
velocity of A makes a 30 degree angle with the line connecting
the disk centers. Find the speed of each disk immediately after
collision, neglecting friction. The coefficient of restitution
between the disks is 0.8.
## Answer:

When two smooth cylindrically symmetrical objects collide the
forces acting to change the velocity of each can only act along
the line connecting their centers, in the line of sight so to
speak. Their smoothness prevents any other force component from
existing. So only the velocity components in the line of sight
are subject to change in such a collision. In the collision of
the question, the velocity in the line of site in the instant
before the collision,vils0 is
(vA0-vB0)*cos(30). The initial velocity of the A disk, vA0
is 8 fps. The initial velocity of the B disk, vB0 is 0. This
makes vils0=8*.866=6.928f/s. The
coefficient of restitution, Cr, is the ratio of the velocity of
separation to the velocity of approach. The velocity of
separation is the component in the line of sight, of the
difference between the velocities of the two disks after
collision, vB1-vA1. The velocity of approach is the component in
the line of sight, of the difference in velocities of the two
disks before collision, vils0=6.928f/s.
Let's call the velocity of separation -vils1. The minus sign
appears because approach and separation are in opposite
directions. So we have
Cr=0.8=-vils1/6.928 or
vils1=-5.5424. The change in velocity in the line of sight
then is 12.47f/s.
Since the masses are equal this change in velocity in the line
of site will be shared equally between the disks, each getting
6.24f/s. To get the speed of each disk after collision we need to
take the square root of the sum of the squares of the speed in
the line of sight and the speed across the line of sight. For
disk A the velocity across the line of sight is
8*sin(30)=4 fps. For disk B the speed across the line of
sight is zero. These speeds are unchanged by the collision. For
disk A the speed in the line of sight after the collision is
6.928-6.24=0.688f/s. For disk B the speed
in the line of sight after the collision is 6.24f/s. The total
speed of disk b is just that in the line of sight, 6.24f/s. The
total speed of A is the square root of
(.688^{2}+4^{2}) = 4.06 f/s

This information is brought to you by M. Casco Associates, a
company dedicated to helping humankind reach the stars through
understanding how the universe works. My name is James D. Jones.
If I can be of more help, please let me know.

JDJ