Stacked Ball Drop  Lessons in Conservation of Energy and Momentum
posted on 20 Jul 2015 by guy
last changed 9 Jun 2016
Your vote (click to rate) ages: 10 to 99 yrs budget: $0.00 to $12.00 prep time: 0 to 5 min class time: 5 to 40 min 
Drop a stack of balls on the floor. By carefully choosing the relative masses of the balls, it is possible to send the top ball flying many times higher than the distance it fell. Analysis of the collision offers good lessons in relative motion and the conservation of momentum and energy. The same physics appears in the "gravity assist" method for accelerating space probes, in supernova explosions that blow off the outer shell of a star, and in the "AstroBlaster" toy. Deep technical understanding requires some facility with algebra, but even young children can gain a qualitative understanding of the phenomenon. Everyone can have fun. Lecture slides and spreadsheet are attached.

optional equipment: Oring, AstroBlaster
subjects: Astronomy, Mathematics, Physics
Balance a golf ball on top of bouncy ball on top of a basketball. Drop the whole stack. What do you think happens? Check out the flight path of the golf ball in the video above. The golf ball bounces 8 times higher than where it was released.
why does the golf ball rebound so fast?
Fig. 1: When a ping pong ball collides headon against a brick wall, the wall hardly moves and the ball reverses direction. As long as the collision does not dissipate much energy, the ping pong ball rebounds with almost the same speed as it had before impact.
Fig. 2: Basketball and golf ball collision from the point of view of the basketball. The golf ball approaches at speed 2v before the collision, and recedes at (almost) 2v after the collision.
When a golf ball collides with a basketball, the golf ball recoils at high speed in part because of the big difference in mass between the two balls. When a light object meets a heavy object, the heavy object hardly changes velocity. The light object recoils much quicker.
When you throw a golf ball straight at a brick wall, the brick wall does not move appreciably. The brick wall is so heavy and has so much inertia, it is hardly affected at all when it's hit by a tiny golf ball. The golf ball, on the other hand, completely reverses direction. Since not much energy is lost in the collision, the golf ball rebounds with almost the same speed as it went in.
The same is true when a golf ball has a headon collision with a stationary basketball. The basketball weighs 13 times more than the golf ball1, so the inertia of the golf ball is small compared to the inertia of the basketball. Consequently, the basketball doesn't recoil very fast as a result of the collision, and the golf ball recoils quite a lot.
The situation is just a little different in the stacked ball drop. (Assume here that we use a twoball stack, with a golf ball directly above a basketball.) In this case, the basketball is not stationary when it meets the golf ball. It's moving upward with about the same speed that the golf ball is moving down. To understand this statement, imagine that there is a small gap between the golf ball and the basketball when they are dropped. Both balls will accelarate the same (at 9.8 m/s^{2} on the surface of the earth) during the drop and will have the same downward velocity right before the basketball hits the ground. As soon as the basketball hits the ground, it will reverse direction (since the ground has a lot more inertia than the basketball) and head back up with almost the same speed it had right before the collision with the ground. An instant later, the basketball and golf ball will meet in a headon collision, with the same incoming speeds.
To understand the final speed of the golf ball in this situation requires just a little imagination. We know that if the basketball is stationary, the golf ball just reverses direction at the same speed. So imagine what the ball drop looks like from the point of view of an ant riding along on the basketball (figure 2). If the golf ball is headed down at speed $v$, and the basketball is headed up at (almost) speed $v$, then the two balls are approaching each other at approximately speed $2v$. From the ant's point of view, the golf ball is headed straight for the basketball at speed $2v$.
The ant is watching the collision between a golf ball moving at $2v$ and a stationary basketball (from its point of view). We already know what happens in that case. Before the collision, the ant sees the golf ball headed straight for the basketball at speed $2v$. After the collision, assuming little energy is lost in the collision, the ant will see the golf ball headed away from the basketball at almost the same speed $2v$.
To a person standing on the ground, the view looks different, but it's easy to figure out (figure 3). The basketball does not change velicity very much as it collides with the golf ball. After the collision, it's still headed upward at roughly speed $v$. In the meantime, the golf ball is headed away from the basketball (upward) at speed $2v$. Relative to the person standing on the ground, the golf ball must be moving upward at speed $3v$. The logic follows from the Galilean transformation: the velocity of the golf ball relative to the ground is equal to the velocity of the golf ball relative to the basketball, plus the velocity of the basketball relative to the ground. In equation form,
$$v_{golf2ground} = v_{golf2basket} + v_{basket2ground} = 2v + v = 3v.$$
The golf ball rebounds with a speed three times higher than its speed before the collision. So of course it bounces higher.
mathematics for geeks
Alternatively, we can understand the dynamics of the collision using the conservation principles for momentum and energy. Using this approach, we can calculate the recoil velocities for objects of any mass, not just very heavy objects colliding with very light objects.
Momentum is the product of mass times velocity ($p=mv$). The total momentum of all the balls added together is conserved (constant) during the collision if there is no net external force on the balls.2 In the case of the dropped ball stack, there IS an external force acting on the balls throughout the experiment: the force of gravity. However, during the very brief time that the balls are in collision, gravity does not alter the motion of the balls appreciably, and we can ignore the effects of gravity during that time. To a good approximation, momentum is conserved during the collision.
Conservation of momentum states that the total momentum before the collision is the same as the total momentum after the collision. For a headon collision between two objects, the algebraic representation is:
\begin{equation}m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\end{equation}
where the subscript $i$ refers to the initial state before the collision and the subscript $f$ refers to the final state after the collision.
Kinetic energy is the product of half the mass times the velocity squared ($E={1\over2}mv^2$). Total kinetic energy is the same before and after the collision as long as energy is not dissipated in other forms. In the case of the stacked ball drop, energy is certainly dissipated; sound waves carry some energy away, and friction among the various parts transform some energy into heat. However, the amount of energy lost is not very a large fraction of the total energy, and it is is reasonable to treat the kinetic energy as approximately conserved.
In a headon collision between two objects, conservation of kinetic energy gives us:
\begin{equation}{1\over2}m_1v_{1i}^2+{1\over2}m_2v_{2i}^2={1\over2}m_1v_{1f}^2+{1\over2}m_2v_{2f}^2\end{equation}
Side note: Using equations (1) and (2), it is possible to derive a third equation
\begin{equation}\tag{2'}v_{1i}v_{2i}=v_{2f}v_{1f}\end{equation}
which can replace equation (2) in further calculations. Equation (2') is linear in velocities rather than quadratic, and is therefore easier to work with.
Using equations (1) and (2') together, we can solve for the final velocities in terms of the masses and the initial velocities. (Selfprofessed geeks are invited to derive the following equations for themselves.)
\begin{equation}v_{1f}={(m_1m_2)\over(m_1+m_2)}v_{1i}+{2m_2\over(m_1+m_2)}v_{2i}\end{equation}
\begin{equation}v_{2f}={2m_1\over(m_1+m_2)}v_{1i}+{(m_2m_1)\over(m_1+m_2)}v_{2i}\end{equation}
When one object is much more massive than the other ($m_1>>m_2$), equations (3) and (4) simplify to
\begin{equation}v_{1f}\approx v_{1i}\end{equation}
\begin{equation}v_{2f}\approx 2v_{1i} v_{2i}\end{equation}
as we expected. (For the stacked ball drop, remember that $v_{1i}=v_{2i}$, which gives us $v_{2f}=3v_{1i}$, a factor of 3 increase in speed.)
how high does the golf ball go?
The recoil speed of the golf ball should be about 3 times as fast as its incoming speed (assuming little energy is lost during the collision). Since kinetic energy is proportional to the speed squared, the recoil kinetic energy is nine times as big as the incoming energy. As the golf ball travels upward, kinetic energy is transformed into gravitational potential energy, and the ball slows down. Since gravitational potential energy is proportional to height, the maximum height of the golf ball should be about nine times the height from which it was originally dropped.
more than two balls
It's impossible to get a full factor of three increase in speed for a twoball stack. One ball would have to be infinitely heavier than the other ball, and the collision would need to transfer energy without any losses whatsoever. For a golf ball on top of a basketball, the speed increase is a factor of 2.7 assuming no energy losses. In real life, deformations in the balls during the collision cause some energy to be lost in heat, and the factor is noticably lower, depending mostly on how well the basketball is inflated.
With three balls it's possible to do better. If the middle ball is very light compared to the bottom ball, and the top ball is very light compared to the middle ball, it's theoretically possible to get as much as a factor 7 increase in speed under idealized conditions (a very heavy bottom ball, a very light top ball, and no energy losses). For a golf ball on top and a basketball on bottom, the maximum speed increase is smaller; the golf ball is too heavy. In this case, a maximum factor 3.9 increase in speed is achieved when the middle ball weighs about 6 oz. Once again, the speed increase in real life is lower due to energy losses.
With a four ball stack in which each ball is much lighter than the one below, the theoretical maximum is a factor 15 increase in speed. For a five ball stack, the maximum is a factor 31. (Do you see a pattern?)
The attached spreadsheet (link at top) supplies velocity calculations according to equations (3) and (4) for stacks of 2, 3, 4, and 5 balls. Change the masses of the balls to calculate the best result for your own ball stack.
AstroBlaster  mass (grams) 
ball 1 (bottom)  65 
ball 2  28.8 
ball 3  10.2 
ball 4 (top)  3.7 
Fig. 4: Schematic of the "gravity assist". A space probe (blue dot) approaches at speed v and does a U turn around the back side of a planet (red circle) traveling at speed u. After the 'collision', the probe emerges at speed 2u+v.
AstroBlaster
It's hard to drop a stack of four balls in a perfectly straight line, but Fascinations Inc. has developed the AstroBlaster toy to do just that by threading an axle through the balls to hold them in line. The masses of the four balls are given in table I. With these masses, the top ball has a theoretical maximum speed of five times its speed before collision. In practice, because of energy losses the actual increase in speed is about a factor of 3.4. The paper "Velocity Amplification in Stacked Balls" analyzes the technical details of AstroBlaster dynamics and provides some experimental measurements for the enthusiast.
Check the equipment links at the top of the page if you want to get your own AstroBlaster.
the gravity assist
NASA uses the same collision dynamics in order to help accelerate space probes. The probe is launched so that it whips around the back side of a passing planet, picking up some of the planet's momentum in the process. The maneuver is called the "gravity assist" (also known as the "gravity slingshot"), and it works exactly like the collision of a heavy ball with a very light ball except that the space probe and the planet do not actually have to collide in order to transfer momentum from one to the other. Gravity acts at a distance in order to transmit the force.
Figure 4 shows the idealized arrangement, with a probe moving at speed $v$ directly towards a planet moving at speed $u$. The probe does a Uturn around the back side of the planet and emerges at speed $v+2u$, having picked up an additional speed equal to twice the speed of the planet. (In the case of the stacked ball drop, the speeds of the two objects were the same, i.e. $v=u$, which caused the small ball to emerge with speed $3v$.) As long as the probe does not enter the planet's atmosphere, there is virtually no energy loss in the encounter, and the velocity increase is close to the theoretical maximum. However, other constraints in the mission usually mean that the probe does not execute a complete U turn, and does not pick up a full $2u$ of speed.
One of the first spacecraft to make extensive use of the gravity assist was the Cassini probe, launched in 1997 to study Saturn and its moons. On its way to Saturn it made used gravity assists accelerations around Venus (twice), Earth (once) and Jupiter (once). See Figure 5.
tips and tricks
 Balancing balls on top of each other for the drop is difficult with two balls, and close to impossible with three or more balls. With two balls, hold one in each hand; they don't need to be pressing together when they are released. For more than two balls, use a ring of hot glue as Physics Girl demonstrates in the video above, or a small rubber Oring, in between each pair of balls.
 Launching a golf ball in a room with a low ceiling is asking for trouble. If you have to do the demonstration indoors, use a ping pong ball instead of a golf ball.
 For best results, slightly over inflate the balls to reduce energy losses.
questions to ponder
Why doesn't the maximum energy transfer between balls also produce the maximum recoil speed?
In a twoball collision, the maximum energy transfer occurs when the large ball comes to a complete stop during the collision, thereby giving up all of its kinetic energy to the small ball. We cannot generally achieve this condition and conserve momentum at the same time. Only one particular ratio of masses causes the large ball to stop: the small ball must be 1/3 the mass of the large ball. In that case, the small ball recoils at twice its incoming speed, not three times. (Try this case out in the attached spreadsheet to check.) If we make the small ball lighter still, it will recoil even faster, but its inertia is not able to completely stop the large ball. In order to achieve the highest recoil velocity in a ball stack, each ball should be very light compared to the ball below it. In order to get the maximum energy transfer in a ball stack, the second ball should be 1/3 the mass of the first (bottom) ball, the third ball should be 1/2 of the second ball, and the fourth ball should be 3/5 of the third ball. These ratios will insure that every ball (except the top one) will come to a stop. (Try them out in the attached spreadsheet.)
How do we know how much energy is lost in each collision?
The energy loss is very hard to predict, as it depends on the detailed structure of the materials involved. However, it might be possible to measure it with reasonable accuracy. To find out how much energy is lost in a collision between a basketball and a golf ball, try setting the basketball on the floor and dropping the golf ball straight onto it. It may take a few tries to get a good bounce straight off the basketball. Measure how far the golf ball falls and also how far back up it rebounds. If the golf ball rebounds to, for instance, 87% of its original height, then 13% of the energy was lost in the collision. Try measuring how this changes as you inflate the basketball a little more or less.
How does the calculation change if energy losses are significant? (question for geeks)
The energy loss in a collision is sometimes characterized by the "coefficient of restitution" $\epsilon_{12}$, which describes the change in speed due to energy losses and is defined according to
\begin{equation}\epsilon_{12}(v_{1i}v_{2i})=v_{2f}v_{1f}\end{equation}
Since the kinetic energy is proportional to speed squared, the fraction of energy lost in the collision is $(1\epsilon_{12}^2)$. In this case, equations (3) and (4) are modified to become
\begin{equation}\tag{3'}v_{1f}={(m_1\epsilon_{12}m_2)\over(m_1+m_2)}v_{1i}+{m_2(1+\epsilon_{12})\over(m_1+m_2)}v_{2i}\end{equation}
\begin{equation}\tag{4'}v_{2f}={m_1(1+\epsilon_{12})\over(m_1+m_2)}v_{1i}+{(m_2\epsilon_{12}m_1)\over(m_1+m_2)}v_{2i}\end{equation}
teaching notes
The description of balls and brick walls under "why does the golf ball rebound so fast?" works fine for students 10 years and older. Before you explain too much, challenge the class to think about what Figure 3 BEFORE looks like from the point of view of the basketball (namely Figure 2 BEFORE), and then what Figure 2 AFTER looks like from the point of view of the ground (namely, Figure 3 AFTER). Use the attached lecture slides if you think they will help.
The derivation of velocities using energy and momentum conservation is generally accessible to students 14 years and older. Algebraically adept students should be encouraged to derive equations (3) and (4) on their own. You may or may not want to tell them about the relative velocity equation (1') first. Really adept algebraists may want to derive equations (3') and (4') with energy losses. Ask students to try to consider how they should characterize energy loss. Do they expect it to be a constant fraction of the total energy or a constant amount?
Computer geeks might be encouraged to build their own spreadsheet along the lines of the attached spreadsheet, but incorporating energy losses à la equations (3') and (4'). Lead them to make measurements to estimate the coefficients of restitution following the suggestions under "questions to ponder" above.
further resources
The paper "Velocity Amplification in Stacked Balls" by a group at the University of Calgary derives the dynamics of the AstroBlaster 4ball stack, incuding energy losses.
The JAVA program by FuKwun Hwang at the National Taiwan Normal University simulates the 4ball stack drop. Adjust the various masses to see how the dynamics change. You may have to add a security exception to get the program to run.
NASA provides an introductory tutorial on the gravity assist at http://saturnarchive.jpl.nasa.gov/mission/missiongravityassistprimer/
 1. An NBA official basketball weighs 22 oz, while a USGA official golf ball is 1.62 oz.
 2. Newton's second law teaches us that the change in momentum of an object, or a collection of objects, is determined by the net external force on the object(s). If the external force is zero, the momentum does not change.
1 Comment
That's very interesting,
Submitted by Vistried1984 on 1 Dec 2017 03:12