Question

A dog weighing 5 kg is standing on a flat boat so that it is 10 m from the shore. The dog walks 4m on the boat towards the shore and then halts. The boat weighs 20 kg and one can assume that there's no friction between and the water. How far is the dog from the shore now?

Answer 1

Let x1 be the distance of center of mass of boat from the shore, and Xc be the distance of center mass of the system from the shore 

Xc.m = (5*10+20*1)/(20+5) 

= 2 + (4/5) x1 

 

Now let boat move x distance away from the shore when dog move 4m towards the shore  

So new distance of boat from shore = x1+x 

New distance of dog from shore = 10-4+x = 6+x 

 

Now   

x.cm = (5*(6+x) + 20*(x1+x))/(20+5) 

= 30+25x +20x1/25 

= 6/5 + x + 4/5 1 

Equate both xc.m 

6/5 + x + 4/5 1  

= 2 + (4/5) x1 

X= 2-6/5 

= 4/5 

So distance = 6+4/5 

= 34/5

Answer 2

Answer:

22/3 is the correct answer.

hope it will definitely help you!!