Calculating Center of Mass for a Group on a Boat

AI Thread Summary
To calculate the center of mass for 23 people on a boat weighing 10,000 kg, the equation x_cm = (m1x1 + m2x2) / (m1 + m2) is essential. The mass of the people (m1) totals 1610 kg, while the boat's mass (m2) is 10,000 kg. As the people walk 8 meters, the center of mass remains unchanged due to the conservation of momentum in an isolated system. The discussion highlights confusion about applying velocity and distance in the context of center of mass calculations. Ultimately, the key takeaway is that the center of mass does not shift despite the movement of individuals on the boat.
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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
 
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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}​
 
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?
 
no i don't think I am makes sense because we're not dealing with velocity here
 
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?
 
Last edited:

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