Newton’s Cradle Demonstration
• Five or more balls of
equal size and mass
suspended in a line.
• Pull one back and
release  one exits the
far side
• Pull two back and
release  two exit the
far side
Why does this work?
Answer from Textbook:
“This is the only way to
Conserve both
Momentum and Kinetic
Energy.”
“This is the only way to Conserve both
Momentum and Kinetic Energy.”
• Answer is misleading
• What would happen if the first ball had twice
the mass of the other balls?
– Would two balls exit the far side with the same
initial speed? (This also conserves both
momentum and kinetic energy)
Analysis of Motion: One Ball
• Series of Elastic Collisions
– Conservation of momentum
– Conservation of Kinetic Energy
• Equal Masses
Conservation of Momentum:
mv1  m(0)  mv1 f  mv2 f
Conservation of Kinetic Energy:
v1  v1 f  0  v2 f
Final Speeds:
v1 f  0
v2 f  v
With multiple collisions momentum and energy
are transferred between each ball.
– Initially ball on the right has the momentum and
ball on left is at rest
– After collision ball on right is at rest and ball on
left has momentum.
Momentum is transferred through each ball
until last ball has no collision and moves away.
Analysis of Motion: Two Balls
• Series of Elastic Collisions
– Conservation of momentum
– Conservation of Kinetic Energy
• Equal Masses
Analysis of Motion: Two Balls
• Two balls move at same initial speed.
• No force between balls.
• Multiple elastic collisions
2 to 3, 3 to 4, 4 to 5  5 moves out with speed v
1 to 2, 2 to 3, 3 to 4  4 moves out with speed v
Analysis of Motion: Unequal Mass
• First Ball has mass M=2m
• Other Balls have mass m
• Series of Elastic Collisions
– Conservation of momentum
– Conservation of Kinetic Energy
Analysis of Motion: Unequal Mass
Consider the First Collision:
Conservation of Momentum
2mv1  m(0)  2mv1 f  mv2 f
Conservation of Kinetic Energy
v1  v1 f  0  v2 f
Final Speeds:
v1 f  13 v1
v2 f  43 v1
Analysis of Motion: Unequal Mass
• Energy/Momentum transferred to second ball
will transfer from it to all successive balls. Final
ball exits with speed 4v1/3.
• First Ball is not at rest, so it will collide again
with Second Ball at rest.
Analysis of Motion: Unequal Mass
Conservation of Momentum (2nd collision):
2m  13 v1   m(0)  2mv1 f  mv2 f
Conservation of Kinetic Energy
1
3 v1  v1 f  0  v2 f
Final Speeds:
v1 f  19 v1
v2 f  94 v1
Energy/Momentum transferred through each ball
and penultimate leaves at v=4/9v1
Analysis of Motion: Unequal Mass
First ball still not at rest, so it collides again, and
again.
Results: To conserve momentum and energy, each
ball will exit at successively smaller speeds:
v5  43 v1
v4  94 v1 v3 
4
27
v1
v2  814 v1
v1 f  811 v1
Conclusion: Newton’s Cradle
• When the balls are of equal mass, momentum and
energy are conserved through successive collisions.
In these collisions all of the momentum/energy of
the moving ball is transferred to a stationary ball.
• The same number of balls exit as were released not
“because this is the only way to conserve
momentum and energy” but because of the
successive collisions.
• Balls of unequal mass will not transfer all of the
momentum/energy so each ball travels with a finite
final speed.
A Few Other Interesting Scenarios
•
•
•
•
What if the final ball has twice the mass?
What if an interior has twice the mass?
What if the first ball has half of the mass?
What if the last ball has half of the mass?
Newton’s Cradle can produce many opportunities for
students to analyze the conservation of momentum
and energy in elastic collisions.
A Second Interesting
Demonstration
Drop two balls from a given
height with the small ball on top
of the larger ball.
The small ball will bounce back
to a much higher point than it
was dropped.
Analysis of Motion
1. Balls fall with the same acceleration and velocity
before hitting the ground.
2. Pre-collision velocity calculated from
conservation of energy or kinematic equations.
v   2 gh0
3. Consider two elastic collisions:
First: Lower Ball with Ground
Second: Upper and Lower Balls
Analysis of Motion
First Collision: Since Ground is much more
massive than the ball, the speed of the ball will
remain the same, but change direction.
v1  v
v1  v  2 gh0
Analysis of Motion
Second Collision: Balls are now moving at he
same speed, but in opposite directions.
Conservation of Momentum:
Mv  m  v   Mv1 f  mv2 f
Conservation of Kinetic Energy:
v  v1 f  v  v2 f
Final Speeds:
 M  3m 
v1 f  
v
 M m 
 3M  m 
v2 f  
v
 M m 
Analysis of Motion
Final Height of Second Ball is found from conservation
of energy:
2
 3M  m  v
 3M  m 
 h2 f = 
 

 h0
 M  m  2g  M  m 
2
mgh2 f = 12 mv22 f
Example: Basketball (M=0.623 kg)
Tennis Ball (m=0.043 kg)
h2 f  7.5h0
2
Questions?