ENERGY AND MOMENTUM: Elastic Collisions and the Center of Mass Velocity ( Part 2 )

In Part 1 of this entry, it was determined that the center of mass between objects moving with constant velocities is also constant ( v_cm ):

We will now use mass ( kg ) and instantaneous-distance coordinates along ( x ) to determine the system’s center of mass. Let’s begin using the x-coordinates of ( m₁ = 1 kg  ) and ( m₂ = 2 kg ) to determine the location of the center of mass ( c.o.m. ). The x/y-origen will serve as the reference from which x-coordinate measurements are made:

c.o.m = ∑m_tx_t/M

c.o.m. = ( m₁x₁ + m₂x₂ )/M

At ( t = 0 ), we have the following c.o.m. value:

x_com = [ ( 1kg )( 0 m ) + ( 2kg )( 30m ) ]/3kg

x_com = [ ( 2kg )( 30m ) ]/3kg

x_com = [ ( 60 kg*m ) ]/3 kg

x_com = 20m

Here are the remaining c.o.m. values at t₁, t₂, and t₃:

t₁ :

x_com = [ ( 1kg )( 3m ) + ( 2kg )( 23m ) ]/3kg

x_com = [ ( 3 kg*m ) + ( 46 kg*m ) ]/3kg

x_com = [ ( 49 kg*m ) ]/3kg

x_com = 16m

Furthermore:

Δx = ( 16m – 20m ) = – 4m

The change in position occurred over the course of ( Δt = 1s ), so v_com = ( Δx/Δt ):

v_com = -4 m/s

t₂ :

x_com = [ ( 1kg )( 6m ) + ( 2kg )( 16m ) ]/3kg

x_com = [ ( 6 kg*m ) + ( 32 kg*m ) ]/3kg

x_com = [ ( 38 kg*m ) ]/3kg

x_com = 13m

Δx = ( 13m – 16m ) = – 3m

v_com = ≅ -3 m/s

t₃ : 

x_com = [ ( 1kg )( 9m ) + ( 2kg )( 9m ) ]/3kg

x_com = [ ( 9 kg*m ) + ( 18 kg*m ) ]/3kg

x_com = [ ( 27 kg*m ) ]/3kg

x_com = 9

Δx = ( 9m – 13m ) = – 4m

v_com = ≅ -4 m/s

The second ( v_com = -3 m/s ) value is a little low due to rounding errors. For this reason, the average value will be tabulated:

v_avg = [ ( -4 m/s ) + ( -3 m/s ) + ( -4 m/s ) ]/3

v_avg ≅ -4 m/s

As we will see, the center of mass is a convenient vantage point from which we may “ observe “ collisions within the system’s frame.