How Far is the Dog from Shore After Walking the Length of the Boat?

[roam]

Homework Statement

A dog weighing 5 kg is at the back of a boat of length 4m and weight 20 kg as shown below. Initially the dog is 8 m from shore. If the dog walks to the front of the boat, how far from shore will he be?

[PLAIN]http://img72.imageshack.us/img72/6067/70134062.gif

Correct Answer: 4.8 m

The Attempt at a Solution

I absolutely have no idea on how to apprach this problem. The velocity of dog and box are not given so it's impossible to use equations of momentum. I appreciate it if anyone could please give me some directions.

 

[Academic]

hmm, I am not entirely sure either. My first thought is to use center of mass. You can assume that the center of mass before and after the dog walks will be at the same location. Then, find the dogs position relative to that.

 

[Redbelly98]

Yes, center of mass is the key here. The tie-in with momentum is that there is no net external force on the dog+boat (considered together as a single object).

 

[roam]

to find the centre of mass

$$x_{CM}= \frac{12 \times 5 + 8 \times 20}{20 +5}= 8.8$$

* (I wrote "5×12" because the dog is at the back of the boat, thus 12 m from the shore. And I don't know if it was correct to write "8×20", because the mass of the boat is distributed through 4 meters).

If my calculation for the centre of mass is correct then, total momentum of the system of particles is given by

$$M \vec{v_{CM}} = 25 \times \frac{d\vec{r_{CM}}}{dt}$$

I guess I need to solve for "r_CM", right? Which is the vector position of the centre of mass of an extended object. But how can I solve this when I don't have the time?

 

[Redbelly98]

You calculated r_cm at the beginning of that last post. However, the dog is 8 m from shore, not 12 m, since it says "initially the dog is 8 m from shore."

Also, the boat's center of mass is located at the center of the boat.