ENERGY AND MOMENTUM: Elastic Collision Determination ( Part 2 )

Q: An elastic collision occurs between two objects of mass ( m₁ ) and ( m₂ ). Prior to the collision, mass ( m₂ ) is stationary and approached by ( m₁ ) with a velocity ( v₁ ). If only the values of ( m₁ ), ( m₂ ), and ( v₁ ) are known, how can we determine what happens to each mass after the collision occurs?

A: In order for a collision to be elastic, the kinetic energy ( KE ) and momentum ( p ) within a system must be conserved. Both kinetic energy and momentum are mathematically expressed in terms of mass and velocity:

p_i = p_f

p_1i + p_2i = p_1f + p_2f

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

KE_i = KE_f

KE_1i + KE_2i = KE_1f + KE_2f

½ m₁v²_1i + ½ m₂v²_2i = ½ m₁v²_1f + ½ m₂v²_2f

Since ( m₂ ) is initially at rest in the system, each of these equations can be reduced accordingly:

v_2i = 0 m/s

m₁v_1i = m₁v_1f + m₂v_2f

m₁v_1i – m₁v_1f = m₂v_2f

p derivation: ( m₁ )( v_1i – v_1f ) = m₂v_2f

v_2i = 0 m/s

½ m₁v²_1i = ½ m₁v²_1f + ½ m₂v²_2f

m₁v²_1i = m₁v²_1f + m₂v²_2f

m₁v²_1i – m₁v²_1f = m₂v²_2f

( m₁ )( v²_1i – v²_1f ) = m₂v²_2f

KE derivation: ( m₁ )( v_1i + v_1f )( v_1i – v_1f ) = m₂v²_2f

We are left with two equations and two unknown values ( v_1f and v_2f ). By forming a ratio of ( KE/p ), we end up with a simplified equation that has no mass terms:

[ KE derivation / p derivation ] 

[ ( m₁ )( v_1i + v_1f )( v_1i – v_1f ) ] / [ ( m₁ )( v_1i – v_1f ) ] = [ ( m₂v²_2f ) / ( m₂v_2f ) ]

v_1i + v_1f = v_2f

v_1i = v_2f – v_1f

We may now substitute ( v_2f ) into the [ p derivation ]. This will eliminate ( v_2f ) and enable us to solve for ( v_1f ): 

p derivation: ( m₁ )( v_1i – v_1f ) = m₂v_2f 

v_2f = v_1i + v_1f

( m₁ )( v_1i – v_1f ) = ( m₂ )( v_1i + v_1f )

m₁v_1i – m₁v_1f = m₂ v_1i + m₂v_1f

m₁v_1i – m₂ v_1i = m₂v_1f + m₁v_1f

( v_1i )( m₁ – m₂ ) = ( v_1f )( m₁ + m₂ )

( v_1i )[ ( m₁ – m₂ )/( m₁ + m₂ ) ] = v_1f

We now solve for ( v_2f ) via substitution into the [ p derivation ] once again:

v_2f = v_1i + v_1f

v_2f = [ ( m₁ – m₂ )/( m₁ + m₂ ) ]( v_1i ) + v_1i

v_2f = ( [ ( m₁ – m₂ )/( m₁ + m₂ ) ] + 1 )( v_1i )

v_2f = [ ( m₁ – m₂ )/( m₁ + m₂ ) + ( m₁ + m₂ )/( m₁ + m₂ ) ]( v_1i )

v_2f = [ ( m₁ – m₂ ) + ( m₁ + m₂ ) ]/[ ( m₁ + m₂ )( v_1i ) ]

v_2f = [ ( m₁ + m₁ + m₂ – m₂ ) ]/[ ( m₁ + m₂ )( v_1i ) ]

v_2f = [ ( 2m₁ )/( m₁ + m₂ ) ]( v_1i )

If the masses of the target ( m₂ ) and projectile ( m₁ ) are known, we only need to know the initial velocity ( v_1i ) of the projectile in order to determine what happens when the masses collide. There are three possible outcomes of the type of elastic collision described herein. The stationary target, once impacted, will always move in the direction in which the projectile was traveling; however, the direction the projectile travels after the collision depends on its mass:

( 1 ) If ( m₁ ) = ( m₂ ), the final velocity of ( m₂ ) will equal the initial velocity ( v_1i ) of ( m₁ ). In this case, the projectile and target possess equal quantities of inertia. The only way for both momentum and kinetic energy to be conserved is if ( v_f ) = ( v_i ).

( 2 ) If ( m₁ ) < ( m₂ ), the target, which has a greater measure of inertia, will acquire a smaller velocity as the projectile bounces back in the opposite direction.

( 3 ) If ( m₁ ) > ( m₂ ), the projectile and target both move in the same direction after the collision. The target will have a greater velocity than the projectile, because it possesses less inertia than the projectile.

If both masses had initially moved towards one another, it would have been convenient to use the center of mass velocity ( v_com ) to determine how the collision will transpire.