Cat walking on a log floating on water

[GreenTea09]

Homework Statement

A dog and a cat, are standing at opposite ends of a uniform log that is floating in a lake. The log is 11.0m long and has mass 460kg . The dog has mass 9.0kg and the cat has mass 7.0kg . Initially the log and the two pets are at rest relative to the shore.The cat then begins to walk to the dog's end of the log to get it.

Relative to the shore, what distance has the log moved by the time the cat reaches the dog? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log

Homework Equations

m1v1+m2v2=m3v3

The Attempt at a Solution

i tried to use conservation of momentum to solve this problem.
Mass of cat*initial velocity of cat + 0(initial momentum of the dog and the log) = Mass of (log+cat+dog)*velocity(3)
i get 2 unknowns, initial velocity of cat and velocity(3)
is there other clue that i can gather from the question since the question did not provide me with any of the pets nor the log velocity..

 

[Doc Al]

Hint: Where's the center of mass of the system before and after the cat moves?

 

[GreenTea09]

here's my attempt:

initial centre of mass:
{(460*5.5)+(9.0*0)+(7.0*11)/460+9.0+7.0=5.477m

final centre of mass:
(460*5.5)+(9.0*0)+(7.0*x)/460+9.0+7.0 = (2530+7x)/476

if i try to solve it using conservation of momentum, there is still an unknown- time taken for the cat to complete its journey..

 

[Doc Al]

here's my attempt:

initial centre of mass:
{(460*5.5)+(9.0*0)+(7.0*11)/460+9.0+7.0=5.477m

Good. That's the initial location of the center of mass as measured along the log.

final centre of mass:
(460*5.5)+(9.0*0)+(7.0*x)/460+9.0+7.0 = (2530+7x)/476

Do the same thing that you did for the initial center of mass--the only difference is that the cat has moved to the other side. So what is x?

Hint: The center of mass will shift position with respect to the log, but does it shift with respect to the shore?