Question

A 1200-kg station wagon is moving along a straight highway at 12.0 m/s. Another car, with mass 1800-kg and speed 20.0 m/s, has its center of mass 40.0 m ahead at the center of mass of the station wagon. Find the position of the center of mass of the system consisting two automobiles.

Answer 1

Solution:

$x_cm = \dfrac{sum(i = 1, N) m_i*x_i}{sum(i=1,N) m_i}$

We have N = 2 in this case.  Let's sit on car 2 and call our position x = 0 despite the fact that car 2 is moving with respect to the "outside" world.  Then car 1 appears to be moving away from car 2 and moving in the negative x direction.  We can describe this by:

$x(t) = (v_1 - v_2)*t - x_0$ and since $v1 < v2, x(t) < 0$

Now $x_{cm}$ can be computed taking $x_2$ = 0 and $x_{cm}= \dfrac{m1*((v1-v2)*t - x0)}{(m1 + m2)}$

Set $t=0$ and $x_{cm}(0)= \dfrac{m1*(v1-v2)}{m_1+m_2}$ -

$x_{cm} = -3.2m$ relative to $m_2$