ELASTIC COLLISIONS: Kinetic Energy, Momentum, Two Equations, and Two Unknown Variables

If a + b = c, and if b = e + f, then it’s also true that a + e + f = c. Any time a variable is common to two or more similar equations, solving one of the two equations will enable us to substitute the common variable into the remaining equation. In regard to elastic equations, it is of interest to us that ( 1 ) kinetic energy ( KE = ½ mv² ) is conserved, ( 2 ) momentum ( p = mv ) is conserved, and ( 3 ) energy and momentum are related quantities that can be expressed in terms of one-another. In the aforementioned equations, m = mass in kilograms ( kg ), and v = velocity expressed in meters per second ( m/s ).

Q: How can it be shown that KE and p are related phenomena?

A: If p = mv, then p² = m²v². Dividing both sides by m gives us p²/m = mv². Finally, dividing both sides by 2 gives us the familiar equation for kinetic energy in terms of momentum: p²/2m = ½ mv².

Now that a clear relationship between kinetic energy and momentum has been mathematically established, we will expand our use of KE and p to relate the before-and-after scenarios of an elastic collision:

KE_i = KE_f [ initial ( i ) and final ( f ) kinetic energies ], and p_i = p_f [ initial ( i ) and final ( f ) momentum ].

A slightly different method of solving two equations with two unknowns may be viewed here. The subsequent method used here involves a process that, if practiced, will enable us to solve elastic collision problems without relying upon rote memorization or upon solutions that are more messy than necessary.

Let’s assume that the green ball headed to the right in the above attachment ( + direction ) has a mass ( m1 ) of 2kg. Let’s also assume that the purple ball moving to the left ( – direction ) has a mass ( m2 ) of 1kg. Additionally, each object moves with a constant velocity of v1_i = 3m/s, and v2_i = -5m/s.

VECTOR QUANTITIES, UNLIKE SCALAR QUANTITIES, ARE DESCRIBED USING MAGNITUDE AND DIRECTION. An example of a scalar quantity is mass ( kg ). For our discussion, mass always has a positive value. However, momentum ( p = mv ) is a quantification of the motion possessed by an object, and as we will see, the concept of momentum would be meaningless in a system in which collisions occur if the colliding masses are not appropriately assigned + or – values that indicate the direction in which they travel. Nonetheless, we will begin our analysis of the before-and-after system conditions using variables that initially have no + or – sign designations.

p_i = p_f [ initial ( i ) and final ( f ) momentum ].

Since p = mv, and we have two masses ( m1 and m2 ) in our system, we arrive at the following expanded expression. 

m1v1_i + m2v2_i = m1v1_f + m2v2_f. 

The first step in simplifying this expression involves aligning like masses on the left and right side of our equation.

m1v1_i – m1v1_f = m2v2_f – m2v2_i

We may now factor m1 and m2 on the left and right sides of our equation.

m1( v1_i – v1_f ) = m2( v2_f – v2_i )

Our equation is now more convenient, but the final velocities of m1 and m2 are still unknown. We will now use the Conservation of Kinetic Energy Theory to derive an equation with a structure similar to the one above.

KE_i = KE_f [ initial ( i ) and final ( f ) kinetic energies ]

½ m1v1_i² + ½ m2v2_i² = ½ m1v1_f² + ½ m2v2_f² 

One again, we will align like masses on the left and right side of our equation.

½ m1v1_i² – ½ m1v1_f² = ½ m2v2_f² – ½ m2v2_i² 

We may now multiply all terms by 2 to get rid of the ½ fraction, and we may also factor out m1 and m2 on the left and right sides of our equation.

m1( v1_i² – v1_f² ) = m2( v2_f² – v2_i² )

The KE_i = KE_f expression has now been simplified, but the squared terms of velocity are still problematic. Recall that we want our expression to be as similar to m1( v1_i – v1_f ) = m2( v2_f – v2_i ) as possible. We will now use the fact that a² – b² = ( a + b )( a – b ).

m1( v1_i – v1_f )( v1_i + v1_f ) = m2( v2_f – v2_i )( v2_f + v2_i )

THIS IS WHERE THINGS BEGIN TO GET INTERESTING!

Recall that our momentum equation simplified to m1( v1_i – v1_f ) = m2( v2_f – v2_i ). Notice how the left hand side of this equation is IDENTICAL to the first two factors of the term on the left-hand side of our simplified KE equation. We may now replace m1( v1_i – v1_f ) with m2( v2_f – v2_i ) in our KE equation.

m2( v2_f – v2_i )( v1_i + v1_f ) = m2( v2_f – v2_i )( v2_f + v2_i )

The satisfying result is that  we may now divide both sides by m2( v2_f – v2_i ).

v1_i + v1_f = v2_i + v2_f 

Since the masses of our system are known ( m1 = 2kg, and m2 = 1kg ), and our initial velocities were given ( v1_i = 3m/s, and v2_i = -5m/s ), it will be most useful to solve the above equation for either v1_f or v2_f. Once this has been done, we will have an expression within which numeric values can be placed.

v1_f = v2_i + v2_f – v1_i , and v1_f = -5m/s + v2_f – 3m/s. Further simplification gives us v1_f  = v2_f – 8m/s.

WE ARE NOW ABLE TO SUBSTITUTE v1_f with v2_f – 8m/s INTO THE RIGHT HAND SIDE OF OUR EQUATION FOR OUR SYSTEM’S MOMENTUM: p_i = p_f [ initial ( i ) and final ( f ) momentum ] = m1v1_i + m2v2_i = m1v1_f + m2v2_f. We may also substitute ( m1 = 2kg, and m2 = 1kg ) and ( v1_i = 3m/s, and v2_i = -5m/s ) into ALL of the values on the left-hand-side of m1v1_i + m2v2_i = m1v1_f + m2v2_f . 

( 2kg )( 3m/s ) + ( 1kg )( -5m/s ) = ( 2kg )( v2_f – 8m/s_i ) + ( 1kg )( v2_f ).

6kg*m/s + ( – 5kg*m/s ) = ( 2kg )( v2_f ) – 16kg*m/s + ( 1kg )( v2_f )

1kg*m/s = ( 3kg )( v2_f ) – 16kg*m/s

17kg*m/s = ( 3kg )( v2_f )

v2f = 6m/s ( from 5.6 )

Since the KE_i = KE_f expression was reduced to v1_f  = v2_f – 8m/s, we may solve for the final velocity of the first projectile.

v1f = 6m/s – 8m/s = -2m/s

Our kinetic energy equation was reduced to v1_i + v1_f = v2_i + v2_f . Substitution of all values into this equation gives us 3m/s + ( -2m/s ) = ( -5m/s ) + 6m/s

SUMMARY: THE EQUATION ABOVE SUGGESTS THAT THE FINAL ANSWER IS CORRECT. This is one of the more tedious solutions encountered in Sophomore-level collegiate physics. It is most practical to memorize ( 1 ) v1_i + v1_f = v2_i + v2_f , and ( 2 ) m1v1_i + m2v2_i = m1v1_f + m2v2_f . Solve ( 1 ) for v1_f , substitute numeric values into v1_f = v2_i + v2_f – v1_i , then substitute v1_f into the right-hand side of m1v1_i + m2v2_i = m1v1_f + m2v2_f . Substitute initial values of velocity and mass into the left-hand side of the equation, then solve the problem. Fortunately, solving problems involving inelastic collisions is MUCH simpler from an algebraic standpoint, as are most other solutions encountered in algebra-based physics.