Physics - conservation of momentum

[uJesternWind]

I'm in my last year of highschool, got a physics question for extra credit that I have no idea how to do. I'm mainly stuck because no numbers are given, obviously it's a conservation of momentum question. I've looked at it for a while but I'm still stumped, any pointers/help is appreciated


Consider the following scenario. You are sitting in a cart containing several bags of sand at the top of a hill. The path you want to take goes down the hill and then up the next, which is higher than yours. There is no friction between your cart and the path. To get started, you throw one bag of sand out the back of the cart. In order to make it as far as possible up the other hill, when should you throw the other bags out? All bags have the same mass and you always throw them at the same speed relative to the cart out of the back.


Edit: So basically because all bags have the same mass and you throw them at the same speed, Delta p for each bag thrown will be the same. Basically to maximize the momentum I have to think about the way that momentum changes if I understand it correctly...we didn't really go over it much in class other than the fact that something called impulse exists, but I don't know how to apply it/if it's even the right thing to apply.

 

[DRSX]

How steep is the hill? Youll have to take gravity and force into the calculations as well

 

[uJesternWind]

How steep is the hill? Youll have to take gravity and force into the calculations as well

No numbers are given, the idea is that conservation of energy states the cart would go up to the same height as before, but throwing bags of sand would help climb the steeper hill. I have to figure out (probably using conservation of momentum) when throwing bags of sand would help the most, and then prove it.

 

I'm in my last year of highschool, got a physics question for extra credit that I have no idea how to do. I'm mainly stuck because no numbers are given, obviously it's a conservation of momentum question. I've looked at it for a while but I'm still stumped, any pointers/help is appreciated


Consider the following scenario. You are sitting in a cart containing several bags of sand at the top of a hill. The path you want to take goes down the hill and then up the next, which is higher than yours. There is no friction between your cart and the path. To get started, you throw one bag of sand out the back of the cart. In order to make it as far as possible up the other hill, when should you throw the other bags out? All bags have the same mass and you always throw them at the same speed relative to the cart out of the back.


Edit: So basically because all bags have the same mass and you throw them at the same speed, ** for each bag thrown will be the same. Basically to maximize the momentum I have to think about the way that momentum changes if I understand it correctly...we didn't really go over it much in class other than the fact that something called impulse exists, but I don't know how to apply it/if it's even the right thing to apply.

The act of throwing a bag from the trolly imparts an equal and opposite force on both the bag and the trolly. Therefore for each bag thrown there is some force for some (small) length of time. This is called impluse and is another way of stating you get a change in momentum for each bag thrown.

Because each bag is identical in mass the change in momentum will be identical for each bag thrown. This is important to note. The impluse imparted to the trolly by throwing a bag is NOT dependent on the speed of the trolly. (Your teachers have the question wrong if this is what they think.) Impulse is the integral (area under curve) of force wrt time. The force imparted on each object (the trolly and the bag) by the throwing of the bag is identical each time and occurs for the same duration each time, therefore the impulse must be identical each time.

Momentum is a conserved vector quantity and so the direction that you throw the bag will matter. If you throw the bag up for example the imparted momentum to your trolly will be down. This in effect will impart the momentum into the earth.

To maximise your forward momentum (relative to the curve) you'll want to throw the bags tangent to the curve that you are following. So the question depends on the geometry of the curve. Wherever the curve is flattest is where you want to throw them all. This assumes you can't fly off the curve (for example the trolly is locked into a track). If you can fly off the curve then you'll probably want to throw them at even intervals.

 

[I Am Hunter]

edit: its already been said, didn't reload page for ages ;-;

 

[uJesternWind]

The act of throwing a bag from the trolly imparts an equal and opposite force on both the bag and the trolly. Therefore for each bag thrown there is some force for some (small) length of time. This is called impluse and is another way of stating you get a change in momentum for each bag thrown.

Because each bag is identical in mass the change in momentum will be identical for each bag thrown. This is important to note. The impluse imparted to the trolly by throwing a bag is NOT dependent on the speed of the trolly. (Your teachers have the question wrong if this is what they think.) Impulse is the integral (area under curve) of force wrt time. The force imparted on each object (the trolly and the bag) by the throwing of the bag is identical each time and occurs for the same duration each time, therefore the impulse must be identical each time.

Momentum is a conserved vector quantity and so the direction that you throw the bag will matter. If you throw the bag up for example the imparted momentum to your trolly will be down. This in effect will impart the momentum into the earth.

To maximise your forward momentum (relative to the curve) you'll want to throw the bags tangent to the curve that you are following. So the question depends on the geometry of the curve. Wherever the curve is flattest is where you want to throw them all. This assumes you can't fly off the curve (for example the trolly is locked into a track). If you can fly off the curve then you'll probably want to throw them at even intervals.

In all questions we've done so far you're stuck to the track so that's what I'll assume.
Throwing the bags tangent to the curve makes sense, thanks for the thorough explanation. Why does it matter that the bags are thrown out where the curve is flat, vs lets say throwing them out backwards with an angle of 45 degrees downwards if the hill is perfectly inclined at 45 degrees? Both would be throwing the bags tangentially to the curve.

edit: I see what you mean now, these two cases would be the same.

Also if it's at the flattest point (assuming hills that are never flat), the two possibilities are right at the start, and right at the bottom in between the two hills. Would it make a difference where you throw the bags off? Intuition tells me that it shouldn't matter where you throw it off, but one of my friends is telling me that the gravitational potential mgh will be greater at the top of the hill if you decide to throw them off at the bottom of the hill (because of a greater mass), leading to a greater speed.

And another way to think about why throwing it off at the bottom would be better is the fact that kinetic energy is 1/2mv^2, so a change in velocity with a higher velocity (i.e. when the cart is at the bottom) would have a greater overall net effect than when you throw them off right at the top when so much energy is still stored in gravitational potential energy.

 

In all questions we've done so far you're stuck to the track so that's what I'll assume.
Throwing the bags tangent to the curve makes sense, thanks for the thorough explanation. Why does it matter that the bags are thrown out where the curve is flat, vs lets say throwing them out backwards with an angle of 45 degrees downwards if the hill is perfectly inclined at 45 degrees? Both would be throwing the bags tangentially to the curve.

The "flattest part of the curve" doesn't refer to the closest to the horizontal (0 degrees) it refers to the radius of curvature. You want to throw them at the point on the curve with the largest radius of curvature.

edit: I see what you mean now, these two cases would be the same.

Also if it's at the flattest point (assuming hills that are never flat), the two possibilities are right at the start, and right at the bottom in between the two hills.

Consider if the cart were instead a sack of rocks you had to push up a hill. When the sack is at the bottom of a sharp valley and you are pushing it sideways rather than up, your force is split between the horizontal and the vertical. A lot of it will be going into the ground and being resisted by the normal force of the hill.

Would it make a difference where you throw the bags off? Intuition tells me that it shouldn't matter where you throw it off, but one of my friends is telling me that the gravitational potential mgh will be greater at the top of the hill if you decide to throw them off at the bottom of the hill (because of a greater mass), leading to a greater speed.

And another way to think about why throwing it off at the bottom would be better is the fact that kinetic energy is 1/2mv^2, so a change in velocity with a higher velocity (i.e. when the cart is at the bottom) would have a greater overall net effect than when you throw them off right at the top when so much energy is still stored in gravitational potential energy.

Yes your friend is right, you'd get an extra kick from throwing them at the bottom assuming the valley isn't too narrow. Momentum is area under the curve of the force-time graph. The weight force changing (from large to small) at the turning point in the curve would give you the greatest momentum. (My bad for not considering that at the start.) Essentially you'd be subtracting a smaller area from a larger area when you took the integral of the total weight-force time graph.

 

[uJesternWind]

Thanks a lot guys.