# Cat walking on a log floating on water

1. Nov 3, 2013

### GreenTea09

1. The problem statement, all variables and given/known data
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

2. Relevant equations
m1v1+m2v2=m3v3

3. 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..

2. Nov 3, 2013

### Staff: Mentor

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

3. Nov 3, 2013

### 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..

4. Nov 4, 2013

### Staff: Mentor

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

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?