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?

 

[Nickie]

I would approach this problem by first considering the forces acting on the log and the pets. Since the log is floating on water, we can assume that the only forces acting on it are the weight of the log and the buoyant force from the water. The cat and dog also have their weights acting on the log.

Next, I would consider the motion of the log and the pets. Since the log and the pets are initially at rest, we can assume that the net force on them is zero. As the cat walks towards the dog, the log and the pets will start to move towards the dog's end of the log. The distance that the log has moved can be calculated by using the equation d = v*t, where d is the distance, v is the velocity, and t is the time. We can also use the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time.

To solve for the unknowns, we can use the conservation of momentum equation as you have mentioned. However, we also need to consider the direction of the motion. Since the cat is moving towards the dog, the momentum of the cat and the log will be in the same direction, while the momentum of the dog will be in the opposite direction. We can set up the equation as follows:

(mass of cat * velocity of cat) + (mass of log * velocity of log) + (mass of dog * velocity of dog) = 0

We know the masses of the cat, log, and dog from the given information. We can then solve for the unknown velocity of the log, which will give us the distance it has moved by the time the cat reaches the dog.

Additionally, we can also consider the conservation of energy in this system. Since the log and the pets are initially at rest and there is no external work being done on the system, we can assume that the total energy of the system remains constant. We can use this information to check our calculations and make sure that the final answer is physically reasonable.

In summary, as a scientist, I would approach this problem by considering the forces and motion of the log and the pets, and using conservation of momentum and energy to solve for the unknowns and determine the distance that the log has moved by the time the cat reaches the dog.