- #1

A 70 Kg person stands at the back of a 200 Kg boat of length 4 m that floats on stationary water. he begins to walk toward the front of the boat. When he gets to the front how far back will the boat have moved? (neglect the resistence of water)

The initial momentum is 0 both for the person and the boat since they are still. As the person begins to walk his momentum changes by an amount proportional to the force exerted on the boat to move forward:

§p=F*§t where §=change

p= momentum

t= time

§p=(70)(9.8)*§t

this §p has to be equal and opposite in sign to the §p of the boat for the law of conservation of momentum.

(70)(9.8)*§t=(200)a*§t where a is the acceleration of the boat.

a=3.43m/s2

now we can calculate the relationship between the space the person moves through and the space the boat moves

s=Vt+(1/2)at^2 that is, since the initial velocity is0 s=(1/2)at^2

s(person)=(1/2)(9.8)§t^2

s(boat)= (1/2)/(3.43)§t^2

therefore s(person)/s(boat)= 2.86

2.86=4m/(total s of the boat)

total distance the boat moves=1.4 m

the answer given in the book is 1.04 and not 1.4

Please, tell me what I did wrong!