ENERGY AND MOMENTUM: Final Velocity of Target and Projection After Elastic Collision ( Part 1 )

Q: Within a given system, a projectile moves with a constant velocity ( v₁ ) prior to colliding with a stationary target of equal mass ( m₁ = m₂ ). Since the system is isolated from outside forces, momentum ( p ) and kinetic energy ( KE ) both are conserved during the collision. ( a ) Why doesn’t the collision alter the system’s kinetic energy? ( b ) How can we prove that kinetic energy has been conserved?

A: ( a ) Newton’s Third Law of Motion states that forces occur in pairs. For every force, there is an equal and opposite force. The force pair acting upon the projectile and target during the collision may be modeled as follows:

F₁ = – F₂

Force is a product of mass ( m ) and acceleration ( a ):

F = ma

An accelerating or decelerating object experiences a change in velocity ( Δv ) per unit time ( Δt ) transpired. As a consequence, the force equation may be rewritten in terms of velocity and time:

F = ( m )( Δv/Δt )

Each body is in contact for an equal period of time during the collision. For this reason, an equal and opposite impulse ( J = Δp ) is enacted upon each object:

J = Ft

F₁Δt = – F₂Δt

m₁Δv₁ = ( m₂ )( -Δv₂ )

We must now determine how much the velocity of each body changed. In order for ( KE ) to be completely transferred from projectile to target, the projectile’s velocity must decrease to 0.0 m/s. Conversely, the target’s velocity must increase proportionally:

Δv = v_f – v_i

( m₁ )( v_1f – v_1i ) = ( m₂ )[ – ( v_2f – v_2i ) ]

( m₁ )( v_1f – v_1i ) = ( m₂ )( v_2i – v_2f )

( m₁ )( 0.00 m/s – v_1i ) = ( m₂ )( 0.00 m/s – v_2f )

The masses cancel, and the velocity gained by the target is equal in magnitude to the velocity lost by the projectile during the collision:

v_1i = v_2f

Notice that each velocity value is positive. This indicates that the target will travel in the same direction as the projectile was traveling prior to the collision. Furthermore, the collision did not alter the ( KE ) of the system, because the force pairs that occur between non-accelerating bodies within isolated systems are balanced. In this example, a brief deceleration and subsequent acceleration do occur during the collision, but no permanent deformation occurs as a consequence of the interaction.

A stationary target with a large mass relative to a smaller projectile will inherit a relatively small velocity, and vice versa; however, using the method above to assess collisions involving two objects with different mass values would be tedious. In such scenarios, two equations must be derived using the premise that energy and momentum are conserved during elastic collisions. Subsequently, one of the equations must be solved in order for a variable to be eliminated from the other equation via substitution. 

( b ) The initial and final kinetic energy of the system is evaluated as follows: 

KE = ½ mv² 

KE₁ = ½ m₁v²_1i

KE₂ = ½ m₂v²_2f 

v_1i = v_2f

KE₁ = KE₂