How Long to Slide Down an Icy Slope When Losing Mass?

[karnten07]

Homework Statement

a 5.4kg toboggan carrying 35kg of sand slides from rest down an icy 90m slope inclined at 30 degrees below the horizontal. The sand leaks from the back of the toboggan at a rate of 2.3kgsec^-1. How long does it take the toboggan to reach the bottom of the slope?

Homework Equations

previous equations involved rockets and the equation:

v(final) = v(initial) + u ln(mi/mf)
where u is the exhaust speed

not sure if this relates to this question though?

The Attempt at a Solution

I drew the situation but can't see how to find the time, any suggestions please?

 

[learningphysics]

Seems like a trick question. Is the sand just leaking out? or is it thrust out with a velocity relative to the toboggan? if it is just leaking then it has exactly the same velocity the moment it is leaked, as when it was still in the tobaggan...

so by conservation of momentum... the toboggan's velocity doesn't change when the sand is released... it is just like if you're traveling in a toboggan horizontally... and you start dropping objects to the ground... you won't speed up...

also although gravity and normal force change because of the decreased mass... the acceleration doesn't...

If it is just leaking out, then it won't affect the speed of the toboggan... it's just a purely kinematics problem... the 2.3kgs^-1 is unnecessary information.

Hope I'm not misunderstanding the question...

 

[karnten07]

Seems like a trick question. Is the sand just leaking out? or is it thrust out with a velocity relative to the toboggan? if it is just leaking then it has exactly the same velocity the moment it is leaked, as when it was still in the tobaggan...

so by conservation of momentum... the toboggan's velocity doesn't change when the sand is released... it is just like if you're traveling in a toboggan horizontally... and you start dropping objects to the ground... you won't speed up...

also although gravity and normal force change because of the decreased mass... the acceleration doesn't...

If it is just leaking out, then it won't affect the speed of the toboggan... it's just a purely kinematics problem... the 2.3kgs^-1 is unnecessary information.

Hope I'm not misunderstanding the question...

I think youre right because the lesson coordinator said something about it not changing, but i wasnt paying full attention. SO if acceleration remains constant could i just use the equation of motion s = 0.5at^2 ?

 

[learningphysics]

I think youre right because the lesson coordinator said something about it not changing, but i wasnt paying full attention. SO if acceleration remains constant could i just use the equation of motion s = 0.5at^2 ?

yes. be careful about the a... what is the acceleration here?

 

[karnten07]

yes. be careful about the a... what is the acceleration here?

ah yes you're right, is it 0.5 g because when you resolve the force of mg downwards, you get a force of 0.5mg down the slope. So by f = ma the acceleration = 0.5 g. Yay, thankyou so much, i had it all here in my notes but it just took a little prodding to understand it lol

 

[learningphysics]

yes. be careful about the a... what is the acceleration here?

ah yes you're right, is it 0.5 g because when you resolve the force of mg downwards, you get a force of 0.5mg down the slope. So by f = ma the acceleration = 0.5 g. Yay, thankyou so much, i had it all here in my notes but it just took a little prodding to understand it lol

yup. that's it.

 

[Dick]

yup. that's it.

This one worries me. Since mass is variable shouldn't we be using F=dp/dt? Physically the picture is that the escaping mass, even though it has no relative velocity, is not being accelerated, whereas the rest of the toboggan is. I'd say more, but I'm having trouble solving the resulting ODE.

 

[learningphysics]

This one worries me. Since mass is variable shouldn't we be using F=dp/dt? Physically the picture is that the escaping mass, even though it has no relative velocity, is not being accelerated, whereas the rest of the toboggan is. I'd say more, but I'm having trouble solving the resulting ODE.

When the mass is leaked... the momentum of the toboggan without the mass of the leaked sand doesn't change... I see it the same way as if you were traveling horizontally and dropping things out of a cart...

 

[learningphysics]

Ok, when you are dropping things out of a cart, the thing you are dropping is exerting a dragging force on you because you are accelerating. When you drop it, that force disappears. This is, in effect, a net thrust forward. This is fun, though it's giving me a headache! I still can't solve the ODE though.

but that's only in the case of friction right? this is frictionless...

 

[Dick]

You are right, of course. That's complete nonsense. How can something accelerate faster than g by passively falling apart? Nothing like one last great good idea before going to bed... Sorry.