Calculating Center of Mass for a Group on a Boat

[cookie monsta]

this question has really stumped me.

23 people on a boat rest on water without friction. Each person has an average mass of 70 kg, and the boat itself weighs 10^4 kg. The entire party walks the entire 8 m distance of the boat from bow to stern. How far (in meters) does the boat move?

i know i have to use this equation somwhere:

x_cm = ( m1x1 + m2x2 ) / ( m1 + m2 )

so far i think
m1 = 23 * 70
x1 = 0
m2 = ?
x2 = ?

can anyone help me?

thnx

 

[Zeno's Paradox]

You have data to calculate the center of mass before the people move. Since, there are no external forces acting on the c.m., will he change?

So,

$$x_{cm}_{i} = x_{cm}_{f}$$​

 

[vijay123]

is it like this?

in an isolated system, momentum is conserved. so, m1v1=m2v2, were m1 is the mass of boat, v1 is it's velocity, m2 is the mass of all the people, v2 is the velocity of them,

v2=(m1v1)/m2
d2/t=m1/m2 x d1/t
d2=m1/m2 x d1

since every1 covers d1, assume that the center of mass is over 1point that covers the 8meters...

does this make sense?

 

[cookie monsta]

no i don't think I am makes sense because we're not dealing with velocity here

 

[cookie monsta]

so i got this so far but i don't know which is which to plug in

$$m_{1}x_{1}_{i} + m_{2}x_{2}_{i} = m_{1}x_{1}_{f} + m{2}x_{2}_{f}$$​

am i on the right track?