Can Dropping Objects Increase Momentum?

[icedoverfire]

Question:

A boy is sliding down a long icy hill on a sled. In order to decrease his mass and increase his velocity, he drops his heavy winter coat and heavy boots from the sled while he is moving. Will his strategy work?

A. No, b/c he loses the potential energy of the objects he leaves behind.
B. No, because although his kinetic energy increases, his momentum decreases.
C. Yes, because although his kinetic energy decreases, his momentum increases.
D. Yes, because although his momentum decreases, his kinetic energy decreases.

Thought Process:

I answered B, when the correct answer is A. I'm not seeing why.

I know that C and D are just wrong - not sure why. Know it intuitively but cannot explain it.

Thanks,

- ice

 

[Quantumpencil]

Speed down an ideal incline is independent of mass.

Take his mass at anytime

mgh=1/2mv^2

sqrt(2gh)=v

No mass dependence. Just think of him "losing" the momentum of the objects he throws away, or equivalently the force of gravity accelerating him less (because the potential field exerts a force proportional to the mass).

 

[razored]

Isn't his momentum increasing, mv ? He has a higher velocity on a lower point on the hill. It is A because the PE=mgh; decreasing the mass would decrease the PE. PE can be converted to KE. The more PE, the more work can be done and hence faster speed. I think my explanation is right.

 

[marindo]

how i see it isn't really momentum...

momentum is seen as mass x velocity = mv
impulse = F_avg * time = m (vf - vi)

essentially i don't see it really as any loss of momentum but loss in energy

because this kid is sliding down from an icy hill, this is obviously a potential energy question

potential energy = mgh

relating this with our work energy theorem, work = f x d = 1/2mv^2

the kid is attempting to increase his velocity by throwing off his jacket, however by throwing off his jacket he's actually losing mass, and therefore energy which therefore velocity decreases as well

if u take a look... mgh = 1/2mv^2 (it should be delta h [change in height] and delta v [change in speed] but w.e)

anyways as we can see, mass is on both sides of the equation and constant, therefore by decreasing mass on 1 side, velocity will subsequently also decrease because they are directly proportional to each other, if they were *inversely* proportional then by decreasing mass we would increase velocity

... hmm meh i think that's right...havent touched this stuff in about a month and a half after my finals sorry > < might be wrong but that's my thought proces anyways ><

 

[razored]

By saying it is icy, I don't think they imply that it is frictionless. Although friction also has an m in umgcos(A)d, you cannot assume all energy goes into friction or KE. Hence, mass remains. In fact, some can be converted to thermal for all we know.

Here is what I would assume: PE + Wo = KE.
Where Wo is work other, meaning anything like friction or thermal.

But you are essentially correct Marindo. I think.

 

[marindo]

By saying it is icy, I don't think they imply that it is frictionless. Although friction also has an m in umgcos(A)d, you cannot assume all energy goes into friction or KE. Hence, mass remains. In fact, some can be converted to thermal for all we know.

Here is what I would assume: PE + Wo = KE.
Where Wo is work other, meaning anything like friction or thermal.

But you are essentially correct Marindo. I think.

yeah nice to have a little support to the train of thought, and totally forgot about the energy lost to heat from friction x x