Conservation of momentum of a boat

[Chiara]

can u please tell me what i did wrong?
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!

 

[sridhar_n]

Hi,

You have not considered the Centre of Mass of the system.
Consider the mass of the man to be m, mass of the boat to be M,
velocity of the man to be u and velocity of the centre of mass of the
man-boat system to be v.

The velocity of the boat gets adjusted to the velocity of the man walking
on the boat.

Therefore,

(M+m)*v = m*u.

Thus,

u = v*(1+(M/m)). Calculate the time taken for the man to walk the
length of the boat (4m) = t.

In this time the boat travels a distance of (v*t) m.

Note: use the ratio of u/v in order to solve the problem (since the value of the
boat velocity is not given)

Hope u got ur solution...


Sridhar

 

[HallsofIvy]

One major mistake you made was in calculating the force as the force of gravity. The man is walking horizontally, not vertically!
Since the force applied would depend upon the resistance of the water, which is not given, I don't see any good way of calculating the force applied.

sridhar_n's suggestion- Calculating the center of mass both before and after the motion- is what I would recommend. Since there is no external force, the center of mass should remain fixed.

 

[jamdr]

I know I'm probably a few months late and I doubt you need the answer anymore, but I'll post this in case someone else comes across a similar problem.

You were going in the wrong direction with your solution. This is not a conservation of momentum problem; it is a center a mass problem. HallsOfIvy was correct to say that since there are no external forces acting on the system, the center of mass remains the same. Therefore, you need to determine how the man walking across the boat changes the center of mass, and how the boat will move to compensate.

To calculate the center of mass of a system:

(m1*x1 +...+ mn*xn) / M

where n is the number of particles within the system and M is the total mass of the system.

The center of mass in your problem is (4m*70kg) / (200kg+70kg) = 1.04m from 0, the starting position. The boat therefore moves 1.04m.