ENERGY AND MOMENTUM: Stacked Ball Drop, Impulse, and the Galilean Transformation

Q: Three balls of mass m₁, m₂, and m₃ fall together towards the earth. They accelerate until impact, and once the three-ball system collides elastically with the earth’s surface, the balls within the system approach one another with an instantaneous velocity ( v ). The momentum ( p = mv ) = ( m₁ + m₂ + m₃ )( v ) of the system must be conserved, and since the uppermost ball ( m₃ ) is much less massive than m₂ and m₂ is much less massive than m₁, the recoil velocity of m₃ after impact will be great. What is the final velocity of m₃ relative to the earth’s surface?

A: Everything is relative! Observations made within the framework of the physical universe are dependent upon the frames of reference from which they are made. An observer located on the side of a busy freeway may measure a bus’s velocity to be 40m/s; however, a passenger traveling with the bus observes occurrences within the bus as if the bus is sitting still. Which frame of reference is most valid? According to the laws of physics, both points of reference are equally valid. This fact gives rise to an enormous number of real-world physics applications. The Galilean transformation allows us to conveniently change our point-of-reference to simplify problems involving objects whose acceleration is equal to zero ( constant velocity ). These conditions hold true until the relative motion between objects approaches the speed of light ( 3.0 x 10⁸ m/s ).

Consider a system within which three balls of differing mass are stacked atop one another, and they accelerate towards earth at a rate of 9.8 m/s². The acceleration ( a ) of the system is constant; however, upon impact, the accelerating system is brought to an abrupt halt. In an instant, the system ceases accelerating, and it possesses an instantaneous velocity ( v_f ) prior to decelerating until v = 0. The instantaneous velocity of the system upon impact is dependent on the time ( t ) that transpires as the system falls towards the earth. Let’s assume the system was initially at rest ( v_i = 0 ):

Since the magnitude of the instantaneous velocity is brought to zero over a brief period of time upon impact, the force that causes the system to decelerate is described as follows:

F = ma

a = Δv/Δt

F = ( m )( Δv/Δt )

Furthermore, a systemic change in momentum occurs upon impact: 

Δp = mΔv

Thus, the impulse ( J ) needed to decelerate the system is the product of the force exerted by the earth on the system over a period of time:

FΔt = Δp = mΔv = ( m )( v_f – v_i )

The same impulse needed to bring the system to rest would require less force if, for example, a cushion increases the amount of time ( Δt ) the system spends decelerating to rest.

We must now take the direction of travel of the system into account. Velocity is a vector quantity! Additionally, the instantaneous velocity ( v_f ) of the system upon impact must now be regarded as the initial velocity of the system prior to undergoing a deceleration. If we assign a negative value to the downward direction, the initial velocity of the system is ( – v_i ). If the system had been brought to rest upon impact without bouncing back upward, the change in momentum would have the following value:

FΔt = Δp = mΔv = ( m )( v_f – v_i ) = ( m )[ 0 – ( – v_i ) ] = mΔv

If the system hadn’t been involved in a collision, the change in momentum would be zero:

FΔt = Δp = mΔv = ( m )( v_f – v_i ) = ( m )[ – v_f – ( – v_i ) ] = ( m )( – v_f + v_i ) = ( m )( 0 ) = 0

If an impulse from behind the system caused the system’s momentum to double as it moved in the assigned negative direction, the change in momentum would be more negative:

FΔt = Δp = mΔv = ( m )( v_f – v_i ) = ( m )[ – 2v_f – ( – v_i ) ] = ( m )( – 2v_f + v_i ) = ( m )( – v ) = – mΔv 

In the elastic collision described in the question, the system changes direction and moves in the positive direction ( upward ). As a consequence, the ball at the bottom of the stack will “ inherit “ the initial velocity of the system:

FΔt = Δp = mΔv = ( m )( v_f – v_i ) = ( m )[ v_f – ( – v_i ) ] = ( m )( v_f + v_i ) = ( m )( 2v ) = 2mΔv 

In mathematics, a doubling oftentimes represents a 100% change in some quantity. 50% x 2 = 100%. In this circumstance, we must not mistakenly think that the velocity of the system has doubled! We have a system whose quantity of motion has completely reversed direction due to an elastic collision, and 2mΔv represents a 100% change.

From the vantage point of the earth’s surface, the ball at the bottom of the stack moves upward with a velocity ( v ), but the balls above the bottom-most ball still have a downward velocity vector as the collisions between the balls within the system begin to occur:

Here’s where the Galilean transformation comes into play. If m₂ elastically bounces off of m₁, its change in momentum is Δp = 2mΔv; however, Δp calculations could get tedious. For this reason, it will be much more convenient to evaluate the motion of m₂ from the vantage point of an observer riding upward on m₁. For convenience, we ignore the previous sign designations assigned to the downward and upward directions. If m₁ were sitting still, an observer on m₁ would see m₂ approaching at a speed of v. Since m₁ moves towards m₂ with a speed of v, an observer at rest with m₁ will see m₂ approaching at a speed of 2v:

After the collision between m₁ and m₂ is complete, from the vantage point of an observer on m₁, the recoil speed of m₂ is 2v in the opposite direction:

Relative to m₁, m₂ moves upward with a velocity of 2v. Remember, however, that m₁ moves upward at a speed of v relative to the ground. For this reason, in relation to the earth’s surface, m₂ approaches m₃ at a speed of v + 2v = 3v! 

From the vantage point of an observer riding upward on m₂, m₃ approaches at a speed of 3v + v = 4v:

Likewise, an observer on m₂ will observe m₃ moving upward at a speed of 4v after the collision between m₂ and m₃ has transpired:

The uppermost ball of mass m₃ moves upward at 4v with respect to m₂, and m₂ moves upward at a speed of 3v with respect to the ground. As a consequence, m₃ moves upward at a speed of 4v + 3v = 7v with respect to the earth’s surface!

The kinetic energy ( KE ) of a body in motion is ½ mv². Since KE is proportional to v², m₃ has ( 7v )² ≅ forty-nine times the kinetic energy it had as it fell with the system towards the surface of the earth. As m₃ rises upward, it loses KE and gains gravitational potential energy ( PE = mgh ):

KE = PE = mgh

The gravitational PE gained by m₃ is directly proportional to the height ( h ) that it will rise upward after the collision with m₂; therefore, m₃ will rise to a height that is forty-nine times its position atop the system prior to its collision with the earth’s surface. In real-world collisions between macroscopic objects, energy losses do occur, however, and the maximum height attained by m₃ will be less than the theoretical value. With a four-ball stack, a theoretical factor 15 increase in speed will occur. For a five-ball stack, a factor 31 increase in speed will occur. The numbers 1, 3, 7, 15, and 31 reveal an interesting 2x + 1 pattern worthy of further study.

Finally, consider a reversal of direction that occurs as a satellite revolves around a planet X traveling with a velocity = u:

An observer on planet X will see the satellite approach with a speed of v + u. After revolving around the planet, an observer on X will see the satellite leave with a velocity of v + u. Since planet X travels at a speed of u relative to an outside observer, the satellite’s final motion is 2u + v. This phenomenon is referred to as the “ gravity assist “, and it is used extensively in the study of planets within our solar system.