Elastic Collision Between Different
Masses

## Question:

Consider two disks of equal size but different mass sliding on a
smooth surface. Disk A has mass 250 grams and initial velocity of
2 m/s. Disk B has mass 175 grams and is initially at rest. The
disks collide directly, with the velocity of Disk A along the
line connecting their centers. The collision is totally elastic,
with a restitution coefficient of 1.0. Determine the velocity of
each disk immediately after the collision and show that the total
kinetic energy is the same before and after the collision.
## Answer:

The coefficient of restitution, Cr, is the ratio of the velocity
of separation to the velocity of approach in a collision. In the
case of an elastic collision it is 1.0. For two moving objects, A
and B, Cr = (vA-vB)/(vBi-vAi) where the i
indicates initial conditions. Here is a good opportunity to get
the signs messed up. The velocity of approach carries the
opposite sign from the velocity of separation so the order of the
terms in numerator and denominator must be reversed. In this case
(vBi-vAi) = (vA-vB) since Cr=1.0. We also
know that momentum is conserved so mA*vAi +
mB*vBi = mA*vA + mB*vB. This gives us two equations in the
two unknown post collision velocities, vA and vB.
Two solve this set of equations let's use the Cr
relationship to get vA in terms of vB. We know that (0 - 2 m/s)=(vA-vB) or vA = vB-2. Plugging this
into the momentum equation we get mA*vAi +
mB*vBi = mA*(vB-2) + mB*vB = (mA + mB)*vB -2*mA. So vB =
(mA*vAi+mB*vBi+2*mA)/(mA+mB). Putting in the known quantities we
get vB = (.25*2+0+2*.25)/(.250+.175) = 1/.425
=

2.35294117647058823529411764705882 m/s.

Pardon the absurd number of decimal places but we are
trying to make a point here.

Since vA=vB-2,
vA=0.35294117647058823529411764705882/s

The kinetic energy of the system before the collision was all
in disk A. KE=1/2*.25*2^{2} Joules = 0.5
Joules. After the collision the kinetic energy is shared
between the disks. The disk A kinetic energy,
KEA is

1/2*.25*0.35294117647058823529411764705882^{2} Joules
=

0.0155709342560553633217993079584775 Joules.

KEB is
1/2*.175*2.35294117647058823529411764705882^{2} Joules
=

0.484429065743944636678200692041522 Joules.

The total KE after the collision
is

0.499999999999999999999999999999522 Joules,

about as close to the original value as we might
expect.

This information is brought to you by M. Casco Associates, a
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understanding how the universe works. My name is James D. Jones.
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JDJ