Analysis of Newton's Cradle
Isaac Newton Energy and Momentum In Nuclear Reactors
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.
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.
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.
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.