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## Homework Statement

Two iceboats hold a race on a frictionless horizontal lake. The two iceboats have masses m and 2m. Each iceboats has an identical sail, so the wind exerts the same constant force

**F**on each boat. The two ice boats start from rest and cross the finish line a distance s away. The total work done to accelerate each of the boats from rest are the same (because the net force and displacement were the same for both). Hence both iceboats cross the finish line with the same kinetic energy.

a)Which iceboat crosses the finish line with greater momentum?

b)Can you show that the iceboat with mass 2m has √2 times as much momentum at the finish line as the iceboat of mass m?

## Homework Equations

**J = p**

p=mv

K=(1/2)mv

W=Fds

_{2}-p_{1}=ƩFΔtp=mv

K=(1/2)mv

^{2}W=Fds

## The Attempt at a Solution

I know that the boat with mass 2m will have greater momentum crossing the finish line by realizing the boat with the larger mass will take a longer amount of time for it to travel from rest to a distance s. Thus the impulse from the larger boat will be bigger. Since the iceboat starts

from rest, this equals the iceboat's momentum p at the finish line:

P=FΔt.

Im having trouble with part b of the problem. This is my thought process:

Both boats will cross the finish line with the same kinetic energy

**∴ (1/2)mv**

^{2}=(1/2)(2m)[(1\2)v^{2}**Half, of the heavier boat's, square speed**must be equal to the

**square**of the lighter one for the kinetic relation to be true. If this is the case then,

can't I make new relation of speeds? what I mean is:

**(1/2)v**

after some algebra → (v

_{heavier}^{2}=v_{lighter}^{2}after some algebra → (v

_{heavier})/√2 = v_{lighter}so

**p**

p

_{heavier}=2mv_{heavier}p

_{lighter}=mv_{lighter}so to find how much larger the momentum of the heavier boat is to

**divide**the two using the substitution of the lighter velocity:

**(2m)(v**

_{heavier})/[(mv_{heavier})/√2] = 2√2why is my quantity two times larger than it should be?