An Analysis of Newton's Cradle
Isaac Newton  Energy and Momentum In Nuclear Reactors
Gravitational Slingshot

In a Newton's Cradle , hard steel spheres hang at rest side by side. If one or more of the spheres is displaced and released , near elastic collisions ensue. See animation above.

Collision  Analysis

Consider two of the smooth spheres, one of mass m1 and the other of mass m2. They are approaching each other along the line joining their centers, a head on collision.

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

m1v1i + m2v2i = m1v1 + m2v2

where vi is the initial velocity of each object and v is the final velocity of each object.

Conservation of Kinetic Energy gives us the equation

m1v1i2 / 2 + m2v2i2 / 2    =    m1v12 / 2   +   m2v22 / 2

These may be recast::

m1 (v1i - v1) = m2 (v2 - v2i)

for the momentum equation and

m1 (v1i2 - v12) = m2 (v22 - v2i2)

for the energy.

 Dividing the second equation by the first one  yields

v1i + v1 = v2 + v2i


v1i - v2i = v2 - v1

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 v2 and substitute into the momentum equation: We then find:

v1 = v1i (m1 - m2) / (m1 + m2) + v2i (2 m2) / (m1 + m2)


v2 = v1i (2 m1) / (m1 + m2) + v2i (m2 - m1) / (m2 + m1)

So we have the basis for a model of a one dimensional elastic collision.
For initial conditions v1iand v2i, if a collision happens, the final velocities depend on the masses as above.

So What About Newton's Cradle Then ?
Now in Newton's Cradle the two masses, m1 and m2 are equal so we can replace both with  m:

v1 = v1i (m - m) / (m + m) + v2i (2 m) / (m + m)


v2 = v1i (2 m) / (m + m) + v2i (m - m) / (m + m)

This reduces to the following expressions for final velocities v1 and v2.

v1 = v2i and v2 = v1i

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.

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