1. Apr 13, 2010

### jeremygogan

1. The problem statement, all variables and given/known data
A dog of mass 10 kg is standing on a raft so that he is 20 m from shore. He walks 8 m along the raft towards shore and then halts. The raft has a mass of 40 kg, and we can assume that there is no friction between the raft and the water. How far is the dog from shore when he stops? [Answer: 13.6m

2. Relevant equations

3. The attempt at a solution
i can work out the solution by utilizing the momentum of the dog raft and then implement in into the vector addition no problem however according to my teacher no matter which way i explain how i've done it i am wrong. I know that the momentum of the dog / raft scenario works out to mass of the dog times the displacement divided by the mass of the raft is the same as the displacement of the raft backwards. you then take the two displacements, find their sum, minus that from the original distance. this results in 13.6m. not a problem, however there is apparently a flaw in my logic somewhere. apparently i am supposed to take the 20m and place it into the dog raft displacement equation before i solve for the momentum peice. PLEASE ANYONE WHO CAN HELP, IT WOULD BE GREATLY APPRECIATED!!!

2. Apr 13, 2010

### AtticusFinch

Show exactly how you think you solved the problem. I'm not quite sure what you are saying here so I can't tell you if you did something wrong.

3. Apr 13, 2010

### jeremygogan

let D=displacement vector

dog D ground = dog D raft + raft D dog

use the momentum of the dog and the raft to solve for one unknown variable

P=P'
let m1=dog, m2=raft
m1v1+m2v2=m1v1'+m2v2'

momentum before is equal to zero

0 = m1v1'+m2v2'
-m2v2'=m1v1 v=d/t
-m2d2'/t=m1d1'/t times both sides by time
-m2d2'=m1d1'
-d2=m1d1'/m2
-d2=10*8/50
d2= -1.6m

now put this back into the original equation

dog D ground = dog D raft + raft D dog
dog D ground = 8m + (-1.6m)
dog D ground = 6.4m

now minus this from the original 20 and it gives you the 13.6m

4. Apr 13, 2010

### AtticusFinch

I think your displacement equation is what is causing the confusion. For me you are switching reference frames way too many times. Here's how I would get a displacement equation.

The speed of the raft in the dog's reference frame, Vr-d, is equal to the speed of the dog in the shore's reference frame, Vd-s, minus the speed of the raft in the shore's reference frame, Vr-s. In equation form:

Vr-d = Vd-s - Vr-s

Multiplying through by time we get

Dr-d = Dd-s - Dr-s

Each side only uses one reference frame. Your equation uses two different frames on one side. How did you derive it?

Continuing the problem my way...

First consider an observer on the shore. By conservation of momentum the observer must see
10vdog=-40vraft

Consequently,

vraft = -1/4vdog

Now consider the dog's reference frame. To him the raft is moving by at a speed of:

vdog + 1/4vdog = 5/4vdog

He stops when he sees 8 m of raft go by, given by:

5/4vdogt = 8

t = 6.4/vdog

Obviously the time interval is the same for the observer. He sees the dog move forward by

vdog*6.4/vdog = 6.4 m = 20 - x

x = 13.6 m