Momentum problem - answer doesn't work

[bigboba]

momentum problem - answer doesn't work!

Here's the problem: you've got two carts sliding along frictionless tracks at an equal speed, etc... as they move along 500g weights are regularly dropped onto the carts, again with equal regularity, etc...

on each cart there is a worker ( same mass, etc). on the first cart, the worker throws off the rocks sideways (that is perpendicular to the movement) as they come onto the cart, while the second worker does nothing. the question is, which cart goes furthest in a set time?

from a credible source, I understand the answer to be that the lazy worker's cart to go further. This is because, when the rocks land on the cart, they take up some of the carts momentum, so when they're thrown off the cart, the worker is actually throwing away this momentum. Momentum is mv, so part of the stuff he's throwing away is velocity. thus his cart goes slower and the lazy worker's cart goes faster and further in the time.

this all seems to make perfect sense until it is put into numbers...

let's say the cart has mass 1000kg, initial speed 10m/s and that the total mass of the weights added over the time is 250kg.

so, initially momentum for both carts is 10m/s x 1000kg = 10,000kg⋅m/s. when the rocks fall on (for simplicity let's say they all fall on at the same time) these numbers obviously change. although, i can't decide how exactly, but either way the numbers still don't work!

option A: by landing on the cart, and because there's no friction, they actually add to the momentum as the mass increases. so now it's 1250kg x 10m/s = 125000kg⋅m/s. when the rocks are thrown off, that's 250kg worth of of mass that's disappearing so things just revert to how they were initially: 1000kg x 10m/s = 10,0000kg⋅m/s. it's speed hasn't changed at all, both carts were always going 10m/s so neither would have gone further, which doesn't agree with the book's answer.

option b: by landing on the cart, and as the momentum is constant, the extra mass must lead to a decrease in velocity. i.e. we've still got 10000kg⋅m/s but now 1250kg of mass, so the velocity must now equal 100000/1250 = 8m/s. however, again, when the hard working person throws the rocks off, the mass decreases back to 1000 so the velocity just stays at 8m/s again, which means it's still going the same speed as the lazy cart. which again doesn't tally with the answer.

i can't help but feel I'm just making a stupid mistake somewhere, but no matter how many times i look, i simply can't find it. any ideas? PLEASE HELP ME!

btw, this is my first post, so hello to everyone!

 

[Doc Al]

from a credible source, I understand the answer to be that the lazy worker's cart to go further. This is because, when the rocks land on the cart, they take up some of the carts momentum, so when they're thrown off the cart, the worker is actually throwing away this momentum. Momentum is mv, so part of the stuff he's throwing away is velocity. thus his cart goes slower and the lazy worker's cart goes faster and further in the time.

Why would throwing something sideways affect the speed of the cart? What's your 'credible source'?

 

[bigboba]

it's not the throwing sideways so much as just getting rid of mass. it's specifically sideways though, because throwing it forwards or backwards would interfere with the momentum.

it's from a physics book written by an mit professor. his answer's are rubbish though as go into way too much detail with formulas while often glossing over the simplistic principles. they are always correct though and one thing he does state, is that the lazy workers velocity is always higher than that of the worker who does the sweeping.

 

[bigboba]

to clarify, the answer is all about momentum and how it is "being thrown away" when the rocks are removed. so I know this is in the right area, but I'm just trying to get it down into simple numbers. or at least a solid, and simple, explanation as to why it isn't possible to do that.

 

[Doc Al]

it's not the throwing sideways so much as just getting rid of mass. it's specifically sideways though, because throwing it forwards or backwards would interfere with the momentum.

My point: If you are riding along in a cart holding a rock then toss the rock sideways, how does that affect the speed of you and the cart?

it's from a physics book written by an mit professor. his answer's are rubbish though as go into way too much detail with formulas while often glossing over the simplistic principles. they are always correct though and one thing he does state, is that the lazy workers velocity is always higher than that of the worker who does the sweeping.

What book? What's his argument for his answer?

 

[Doc Al]

to clarify, the answer is all about momentum and how it is "being thrown away" when the rocks are removed. so I know this is in the right area, but I'm just trying to get it down into simple numbers. or at least a solid, and simple, explanation as to why it isn't possible to do that.

What matters is whether the cart speeds up or not. See my previous post.

 

[bigboba]

sorry, i missed something out in the initial question: both carts were given one initial push but are now just relying on momentum. so it matters because the rock, be being in the cart and moving in the cart, holds part of the carts total momentum. so if it's thrown out of the cart, then that momentum is lost. and as momentum is mv, part of that is speed.

the book is an old physics book of my dad's. the explanation, goes something like this:

as no interaction to lazy man's cart the horizontal momentum must be conserved.

the worker's cart: as weights fall onto it, they acquire the velocity of the cart and therefore some momentum. when they're removed, some of the total momentum in the cart is also removed. thus this cart is no longer isolated from surroundings and conservation of momentum can no longer be applied.

he then goes into some pretty odd formulas with exponential functions and what not, before climaxing with a graph that clearly show the lazy man's cart's velocity to always be higher than the other cart.

 

[cepheid]

I don't understand. If you throw a rock out of the cart in a direction perpendicular to the track, it *retains* its original velocity in the direction parallel to the track. Let's say that originally, the cart is moving along the track with speed v. So, if the total system before the ejection of the rock had mass M, and the rock thrown out had mass dm, then the forward momentum of the system before the loss of the rock would be:

M*v

and the forward momentum of the *system* after the loss of the rock would be the forward momentum of the cart PLUS the forward momentum of the rock:

(M-dm)*v + dm*v = M*v - dm*v + dm*v = M*v

So, as you can see, the momentum of the original system is conserved in the parallel direction *without* any change in the speed of the cart. Now, if you define the "system" to be just the cart and whatever is in it at any instant, then clearly the momentum of that is not conserved. But its forward speed *is* conserved in this particular situation.

 

[Doc Al]

sorry, i missed something out in the initial question: both carts were given one initial push but are now just relying on momentum. so it matters because the rock, be being in the cart and moving in the cart, holds part of the carts total momentum. so if it's thrown out of the cart, then that momentum is lost. and as momentum is mv, part of that is speed.

That doesn't make much sense. The 'cart+rock' has some momentum. They are both moving at the same speed along the track. If you toss the rock sideways, the momentum of the cart doesn't change (it's constrained to ride along the track). Momentum is conserved parallel to the track.

 

[Skrambles]

he then goes into some pretty odd formulas with exponential functions and what not, before climaxing with a graph that clearly show the lazy man's cart's velocity to always be higher than the other cart.

Is the author considering aerodynamic drag in that analysis? If both carts have the same drag force at a given velocity, the cart with less mass will slow down more quickly because it has less momentum. Frictionless tracks don't always mean frictionless air in these problems.

 

[Drakkith]

A: If the rocks have no forward momentum when dropped, both carts will slow down, as they must take energy from the carts to be sped up.

B: Ignoring friction from the tracks, air, ETC, and assuming the rocks have equal forward velocity to the carts, neither cart would change speed.

Adding or removing weight doesn't matter as long as the same thing is happening to both carts.

 

[Curl]

if you use logic, then its easy to see why both carts will be at the same speed all the time.

throwing the mass sideways has no effect. When the rock is in the air (after its thrown) then it still has same speed (downtrack direction) as before it was thrown. Doesn't matter if its touching the cart or not

 

[willem2]

The incoming rocks are what makes the difference. They won't have the velocity of the cart, but the velocity of the ground.

Suppose the mass of the cart is M, the mass of a rock is m, and $v_n$ is the velocity of the cart after n rocks are dropped on it.

If the rocks stay in the cart, conservation of momentum tells you that:

$$M v_0 = (M+ n m) v_n$$ so

$$v_n = \frac {M} {M + n m} v_0$$

and if M=1000kg, m=0.5 kg, n=500, v_0 = 10 m/s,
you have 250 kg of rocks, and v_n will be 8 m/s

Suppose the rocks do not remain on the car, and n rocks have already been
dropped on it and been thrown off, and the cart moves now with speed $v_n$.
Conservation of momentum for the collison of the cart with the next rock:

$$(M+m) v_{n+1} = M v_n$$ so

$$v_{n+1} = \frac {M} {M+m} v_n$$

when you throw off this rock sideways, the speed doesn't change

the solution of this recurrence relation is

$$v_n = (\frac {M} {M+m})^n v_0$$

if m<<M this is approximately equal to:

$$v_n = e^{\frac {-n m} {M}}$$

with the same parameters as in the first case, you now get

$$v_n = e^{-1/4} v_0$$ = 7.79 m/s

The difference between the two methods will become much bigger with more rocks.
If 9000 kg of rocks are thrown on the cart one by one, the cart still moves at 1 m/s if the rocks are left on it, but only $10 e^{-9}$ = 0.0123 m/s if the rocks are thrown off.

 

[Doc Al]

willem2 nailed it! I kept thinking in terms of one rock, but the rocks keep coming, adding to the mass of the moving 'cart+rocks'. D'oh! (My coffee is obviously not strong enough.)

Nice job.

 

[cepheid]

Yeah, I second the kudos to willem2. This is an interesting result. At first, it seemed very non-intuitive to me that the cart that becomes more and more heavily laden with mass would not slow down as quickly as the cart whose mass was in some sort of steady-state equilibrium. But then I thought about it some more, and I realized how, after each rock drop, the fractional change in speed of the two carts would differ.

For both carts, after each iteration, the new velocity is given by

$$v_{n+1} = \frac{M_n}{M_n + m} v_n$$

$$= \frac{1}{1 + \frac{m}{M_n}} v_n$$​

So, for both carts, the velocity decreases by a factor of 1 + (m/M_n). The difference is that, for the lazy man's cart, M_n steadily increases with each iteration, and so the factor by which the velocity is reduced each time gets smaller and smaller. In contrast, the for the worker's cart, M_n is always equal to M, since it doesn't gain mass, and so the velocity is reduced by a constant factor of 1 + (m/M) each time (a geometric sequence). That's why it decays so much more quickly.

I think that this is a good way to at least understand why the result is true, without delving into the full analysis. What do you guys think?

 

[Doc Al]

Looks fine to me, cepheid.