An
Analysis of Newton's Cradle

Isaac Newton
Energy and Momentum In
Nuclear Reactors

Gravitational
Slingshot

__Collision
Analysis__

Assumptions:

- Spheres are perfectly smooth
- Collisions are head on.
- Energy is purely translational and not rotational or vibrational
- Collisions are perfectly elastic

Applying the Conservation of Linear Momentum to this situation we have:

m_{1}v_{1i}
+ m_{2}v_{2i} = m_{1}v_{1} + m_{2}v_{2}

where *v*_{i} is the initial
velocity of each object and v is the final velocity of each object.

Conservation of Kinetic Energy gives us the equation

m_{1}v_{1i}^{2}
/ 2 + m_{2}v_{2i}^{2} / 2 =
m_{1}v_{1}^{2
}/ 2 + m_{2}v_{2}^{2
}/
2

These may be recast::

m_{1} (v_{1i}
- v_{1}) = m_{2} (v_{2} - v_{2i})

for the momentum equation and

m_{1} (v_{1i}^{2}
- v_{1}^{2}) = m_{2} (v_{2}^{2}
- v_{2i}^{2})

for the energy.

Dividing the second equation by the first one yields

v_{1i} + v_{1}
= v_{2} + v_{2}_{i}

or

v_{1i} - v_{2i}
= v_{2} - v_{1}

In other words in a one dimensional elastic collision, the relative velocity of approach before the collision equals the relative velocity of separation after collision.

__Final
Velocities__

To get the final velocities in terms of
the initial velocities and the masses, solve the last equation for
v_{2} and substitute into the momentum equation: We then find:

v_{1} = v_{1i}
(m_{1} - m_{2}) / (m_{1} + m_{2}) + v_{2i}
(2 m_{2}) / (m_{1} + m_{2})

Likewise

v_{2} = v_{1i}
(2 m_{1}) / (m_{1} + m_{2}) + v_{2i} (m_{2}
- m_{1}) / (m_{2} + m_{1})

So we have the basis for a model of a one
dimensional elastic collision.

For initial conditions v_{1i}and
v_{2i},
if a collision happens, the final velocities depend on the masses as above.

Now in Newton's Cradle the two masses, m

v_{1} = v_{1i}
(m - m) / (m + m) + v_{2i} (2 m) / (m + m)

Likewise

v_{2} = v_{1i}
(2 m) / (m + m) + v_{2i} (m - m) / (m + m)

This reduces to the following expressions
for final velocities v_{1} and v_{2}.

v_{1} = v_{2i}
and v_{2} = v_{1i}

In the case of equal mass collisions, the two objects just exchange velocities. Of course if the initial velocity of one of the balls were zero, then the one colliding with it would just stop and the hit ball would take off at the original speed of the incoming ball.

Now
consider the action in the cradle.. It looks like there must be a series
of two ball collisions of the sort described above so that the last ball
in the line leaves the group with approximately the same velocity as the
ball which struck the group originally. something about the If the
contact between the balls were rigid, the incoming ball would impact a
single massive object comprised of several balls. It would then rebound
from the collision in accordance with the rules developed above for collision
between differing masses. We conclude that even though the balls in the
line appear to be in contact, there must be initially a very soft reaction
between adjacent balls.

For another application of this physics
, click here.