ENERGY AND MOMENTUM: Elastic Collisions and the Center of Mass Velocity ( Part 3 )

Q: Consider a system in which a mass ( m₁ = 6kg ) moves in the +x-direction at ( 2 m/s ) as a mass ( m₂ = 2kg ) moves in the -x-direction at ( – 4 m/s ):

How may we use the center of mass ( c.o.m. ) of this system to determine what happens when the masses collide?

A: This problem is made more difficult by the fact that neither mass is originally at rest. For this reason, we must use an alternate analysis that doesn’t involve deriving two equations in order to solve for two unknowns [ ( v_1f ) and ( v_2f ) ]. It was previously shown that the center of mass of a system is determined via the use of the following expression:

x_com = ∑ mx/M

This expression works fine with masses that are at rest with respect to a chosen vantage point; however, if the masses within a system move with a constant velocity, the change in position over time ( v = Δx/Δt ) of each mass must be taken into consideration:

v_com = ∑ [ ( m )( Δx/Δt ) ]/M

v_com = ∑ ( mv )/M

v_com = ∑ p/M

Thus, we must determine what momentum ( p ) is possessed by each mass and the total mass ( M ) of the system to determine the value of v_com. Always be sure to assign the appropriate direction to the velocity of each mass prior to doing the ( p ) calculation:

p₁ = 12kg*m/s

p₂ = – 8kg*m/s

v_com = [ 12kg*m/s + ( – 8kg*m/s ) ]/M

v_com = ( 12kg*m/s – 8kg*m/s )/M

v_com = ( 4kg*m/s )/M

v_com = ( 4kg*m/s )/8kg

v_com = 0.5 m/s

The positive value of ( v_com ) lets us know that the ( c.o.m. ) travels in the +x-direction. Keep in mind that the velocity values assigned to each mass are measured from a vantage point outside of the system. If we take an imaginary ride along with the center of mass, ( v_com = 0 m/s ). For this reason, we must adjust the velocity values associated with ( m₁ ) and ( m₂ ). From the ( c.o.m. ) vantage point, the value of ( v₁ ) must decrease by ( 0.5 m/s ). If this adjustment isn’t made, ( m₁ ) would be able to “ catch up “ to the center of mass, which has a velocity that was decreased by ( 0.5 m/s ). Likewise, the value of ( v₂ ) must increase by ( 0.5 m/s ); otherwise, its motion relative to the ( c.o.m. ) would be inaccurate:

Since the center of mass velocity represents the net velocity of the system, the net momentum of the system ( from the c.o.m. vantage point ) is zero as well. Furthermore, in order for energy and momentum to be conserved in an elastic collision, the velocities associated with ( m₁ ) and ( m₂ ) will each retain their magnitude after being re-oriented in the opposite direction:

Let’s now quantitatively prove that p_net = 0 kg*m/s for the system:

p_1i = ( 6kg )( 1.5 m/s ) 

p_1i = 9kg*m/s

p_2i = ( 2kg )( – 4.5 m/s )

p_2i = – 9kg*m/s

p_net = 9kg*m/s + ( – 9kg*m/s )

p_{net i} = p_{net f} = 0

We are now ready to evaluate the final velocities of ( m₁ ) and ( m₂ ), once again, from outside of the system. Keep in mind that the ( c.o.m. ) velocity doesn’t change as the elastic collision transpires; therefore, the center of mass continues to move to the right at ( 0.5 m/s ) when observed outside of the system. In order for the value of ( v_1f ) to be correct relative to ( v_com ), its magnitude must decrease. Likewise, the value of ( v_2f ) must increase:

v_1f = – 1.5 m/s + 0.5 m/s

v_1f = – 1 m/s

v_2f = 4.5 m/s + 0.5 m/s

v_2f = 5 m/s

We are now ready to determine the ( p_{net f} ) as viewed outside of the system:

p_{net f} = p_1f + p_2f

p_{net f} = ( 6kg )( – 1 m/s ) + ( 2kg )( 5 m/s )

p_{net f} = – 6kg*m/s + 10kg*m/s

p_{net f} = 4kg*m/s

The initial momentum of the system was as follows:

p_{net i} = ( 6kg )( 2 m/s ) + ( 2kg )( – 4 m/s )

p_{net i} = 12kg*m/s + ( – 8kg*m/s )

p_{net i} = 4kg*m/s

Thus, the kinetic energy and momentum of the system were conserved.