Conservation of Momentum.Boat Question

[mfianist]

A boat of mass m is carrying a load of bricks with a total number of N and mass M, and is floating freely on water. SUPERMAN, in a show of strength, throws the entire load of bricks backward at once with a relative speed of v0. A GENIUS, in a show of intelligence, throws the bricks backward with a relative speed of v0 one at a time. Argue and explain how and why GENIUS achieves a much higher speed than SUPERMAN. Take N >> 1 and M >> m. Neglect the friction between the boat and water, and the mass of GENIUS and SUPERMAN.

I think i know how to get the velocity of the boat in the case where SUPERMAN throws the bricks all at once.

But the problem comes when the bricks are thrown one by one. In this case, the velocity of the brick,relative to the eart, is decreasing everytime the brick is thrown. Should i use the velocity of the brick relative to the earth, or relative to the boat to do this question?
Also the mass constantly decreases... for every throw, velocity and mass changes...
i have no clue on how to show that GENIUS achieves higher speed than SUPERMAN.

I need some help.... somebody.. please help me~~!!

[HallsofIvy]

Try starting with just two bricks and see if you can understand what is happening. Since the momentum of the boat and bricks together is 0 (relative to the surface of the water), the total momentum of the system will remain 0. If superman throws both bricks with total mass M with velocity v relative to the boat (which is the same as relative to the water_, he gives them momentum Mv and so the momentum of the boat must be -Mv: Taking u to be the velocity of the boat, mu= -Mv and u= -(M/m)v.

Throwing only one brick, of mass M/2, at velocity v relative to the boat (and water), "genius" give the brick momentum (M/2)v and so for the boat we have
u= -(M/(2(m+M/2))v= -(M/(2m+M))v (the "m+ M/2" is because the second brick is still in the boat).
Throwing the second brick, of mass M/2, at velocity v relative to the boat means that he is throwing it at velocity v- (M/(2m+M))v = (2m/(2m-M))v relative to the water: The boats increase in momentum is now mu= -(2Mm/(2m-M)) v so
u= -(M/(2m+M))v- (2M/(2m-M))v= -4mM/(4m2-M2).
Now show that that is always larger (in absolute value) than -(M/m)v.

[ehild]

But the problem comes when the bricks are thrown one by one. In this case, the velocity of the brick,relative to the earth, is decreasing every time the brick is thrown. Should i use the velocity of the brick relative to the earth, or relative to the boat to do this question?
Also the mass constantly decreases... for every throw, velocity and mass changes...
i have no clue on how to show that GENIUS achieves higher speed than SUPERMAN.

You feel the problem quite well :smile: . I think it is easier to solve the problem relative to the earth.
As N>>1 you can approximate the problem by applying differential calculus for the velocity of the boat in terms of the bricks still present.

Assume that Genius has thrown some bricks already so the boat has v velocity and there remained n bricks on the boat, when he throws the next.
He throws the brick with -vo velocity with respect to the boat, that is v-vo with respect to the earth. The mass of one brick is
m_B=M/N


The change of the velocity of the boat by throwing one brick is 1*dv/dn. We get the following equation for the momentum before and after the one next brick is thrown:

(m+n*m_B)v=[m+(n-1)*m_B]*(v+dv/dn)+(v-v_0)*m_B

Arranging this equation, some terms cancel and you get:



m_B*v_0=[m+(n-1)*m_B]*\frac{dv}{dn} ,


so the derivative of v(n) wit respect to n is

\frac{dv}{dn}=\frac{m_B*v_0}{m+(n-1)*m_B}

You have to integrate from n=N to n=1. The result is:

v=v_0\ln(\frac{m+(N-1)m_B}{m})\sim{v_0\ln(\frac{m+M}{m})}

ehild

[mfianist]

You feel the problem quite well :smile: . I think it is easier to solve the problem relative to the earth.
As N>>1 you can approximate the problem by applying differential calculus for the velocity of the boat in terms of the bricks still present.

Assume that Genius has thrown some bricks already so the boat has v velocity and there remained n bricks on the boat, when he throws the next.
He throws the brick with -vo velocity with respect to the boat, that is v-vo with respect to the earth. The mass of one brick is
m_B=M/N


The change of the velocity of the boat by throwing one brick is 1*dv/dn. We get the following equation for the momentum before and after the one next brick is thrown:

(m+n*m_B)v=[m+(n-1)*m_B]*(v+dv/dn)+(v-v_0)*m_B

Arranging this equation, some terms cancel and you get:



m_B*v_0=[m+(n-1)*m_B]*\frac{dv}{dn} ,


so the derivative of v(n) wit respect to n is

\frac{dv}{dn}=\frac{m_B*v_0}{m+(n-1)*m_B}

You have to integrate from n=N to n=1. The result is:

v=v_0\ln(\frac{m+(N-1)m_B}{m})\sim{v_0\ln(\frac{m+M}{m})}

ehild

i kinda get what you are saying...
i've never thought of looking at this problem with dv/dn...
but i don't get the integration part...
so what you are saying is the area under the curve when n=1 and n=N is the velocity achieved by GENIUS?...

[ehild]

i kinda get what you are saying...
i've never thought of looking at this problem with dv/dn...
but i don't get the integration part...
so what you are saying is the area under the curve when n=1 and n=N is the velocity achieved by GENIUS?...

Something like that. Haven't you learnt about definite integral or differential equations yet? In this case you should do what HallsofIvy suggested. I just wanted to show an alternative way.

ehild

[mfianist]

Thank you ehild!!

i got it!! i learned about integrals.. i just forgot that moment...

i spent 2 days and nights for this problem.. and couldn't found a way to solve the problem...

u r my life savior!!