# Center of mass question

1. Nov 11, 2006

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

Last edited: Nov 12, 2006
2. Nov 12, 2006

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,

$$x_{cm}_{i} = x_{cm}_{f}$$​

3. Nov 12, 2006

### 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?

4. Nov 12, 2006

no i dont think im makes sense cuz we're not dealing with velocity here

5. Nov 12, 2006

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