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conservation of momentum.

Posted on 2003-11-07
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Hi,
I am looking through past Physics exam papers and I am having trouble understanding a question. Here it is.

"Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass and are using the same type of toboggan. When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his toboggan, so he pushes a box containing their ice-skating gear off the side of his toboggan.

Explain, giving clear reasons, whether this will be a successful way for Jack to catch up to Jill and help him win the race"

The solutions to the exam say that "this question could have been answered using a conservation of momentum approach or by application of Newton's Second Law ....."

Firstly could someone please tell me what the Newton's second law states. Isn't the rule for conservation of momentium m1*u1 + m2*u2 = -m1*v1+m2*v2
where m = mass
u=init velocity
v=final velocity

I am not sure how this idea could be used to explain the situation. Could someone please help. I realise that the mass won't increase or decrease acceleration, but I have to prove it using the methods described in the solutions (stated above)

Thanks in advance.
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Question by:adam8
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by:Archie_Gremlin
ID: 9700898
Newton's second law is the one that's summarised as "f=ma". In otherwords, an object accelerates if a force is applied to it. If the same force is applied to two different objects, the lighter one will accelerate more.

The rule for conservation of momentum is that the value of mass times velocity is conserved for all objects in a closed system. Your equation isn't quite right. It shouldn't have a minus sign on the right hand side. In other words:
 m1*u1 + m2*u2 = m1*v1 + m2*v2

The toboggan puzzle is simple in theory but difficult in practice. Let us make a few simplifying assumptions.
1) Assume that the ice is frictionless
2) Assume that there's no drag caused by the air

In this case, Jack doesn't gain any benefit by pushing the box off the side of his toboggan. To get rid of the box, he pushes it at right angles to the direction he's going in. Remember that the direction of travel matters when you measure velocity and acceleration. (Unlike speed.)

Because of convservation of momentum, if the box has mass m2 and is if thrown out at velocity v2 then the toboggan must move at velocity (-m2*v2 / m). Unfortunately, this movement is sideways so it doesn't help him to win the race.

Likewise you could work the same result with Newton's laws. Newton's third law tells us that the box pushes back on Jack when Jack throws it out. Newton's second law tells us that this will cause Jack, and the toboggan, to accelerate away from the box. You need to do some horrible integration to work out how much velocity Jack gains. The maths is _much_ easier with conservation of momentum. :~)

So we know that throwing the box out sideways won't help. However, if Jack throws the box out the back of the toboggan he will gain velocity in the direction that matters. Secondly, the faster the box goes backwards, the more velocity Jack will gain. If he really wants to win, he should use a cannon. :~)

In reality, the relationship between mass, drag and sliding friction means that it's hard to know what'll happen. If wind drag is high compared to the friction from the ice then throwing out the box will slow Jack down. I'll explain some of these issues in more detail if you want me to.

regards,
Gremlin
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by:wytcom
ID: 9701788
Jack should heave the box to the rear as hard as he can.  This will increase his speed down the hill by conservation of momentum.  Then he could retain male dominance as he leaves Jill behind!  :-)
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by:rfr1tz
ID: 9702099
A long time ago, olympic bobsledders were really heavy because the extra weight gave them an advantage. Now, they have weight limits, so the bobsleders are pictures of health.

My point is, maybe throwing weight off the sidc actually works against you.
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by:GwynforWeb
ID: 9703976
Putting the above into maths if the ice cream is pushed of the back (so all velocities are in the same direction). If

mass ice cream + sledge  =  Mi + Ms

both travelling at velocity V1 and the ice cream is ejected directly off the back with a force so its new velocity is V3 then if Vf is the final velocity of the sledge

  V1(Mi + Mj) = V3*Mi + Vf*Ms

ie the final velocity of the sledge Vf is

   Vf = V1 + (V1-V3)Ms

which since V3 < V1, will always means the new velocity Vf of the sledge is greater than the initial V1.
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by:adam8
ID: 9704866
Well in the diagram it shows the box being pushed off the back, but I just read the question again and it states that he pushes it off the side of the toboggan.

I suppose that pushing it off the side would only add to the horizontal component of force, and not assist in the direction he is going.

I think the question is trying to really pose, does the mass of a toboggan assist in its acceleration. I dont think they would want friction or air resistance to be considerated.

I don't understand how this can now be related to conservation of momentum (if we are talking about the effects of change of mass on the toboggans velocity)

I do understand what you said (Archie_Gremlin) about the force used when pusing the box off (to the side) only affects the velocity to the side, which doesn't help him get down the hill any quicker, but i don't think this is what the examiners were after.
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by:GwynforWeb
ID: 9706124
ps my algebra is off a bit, the momentum equations give

Vi*Mi +Vi*Ms = VIf*Mi + VSf*Ms

where Vi initial velocity of both icream and sled with masses Mi and Ms. The final velociy of the sled VSf is  given by (VIf final vel of icecream)

   VSf =  Vi + (Vi-VIf)*Mi/Ms

ie has increased by (Vi-VIf)*Mi/Ms



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by:Archie_Gremlin
ID: 9706536
adam8 said:
> I think the question is trying to really pose, does the mass of a toboggan assist
> in its acceleration. I dont think they would want friction or air resistance to be
> considerated.

I'm not sure what's bothering you here. Perhaps you're wondering how the toboggans accelerate down the hill due to gravity. In that case...

Both toboggans are accelerating down the hill because of gravity. In the absence of friction, the mass of the toboggans doesn't matter. They will gain velocity down the hill at the same rate.

You can't prove this with the law of Conservation of Momentum. You need to use Newton's Second Law; f=ma. This makes me think that it's unlikely to be what the examiners are interested in. However, it's all educational. :~)

Here's why the mass of the toboggans don't matter.

a) The force F exerted on the toboggans due to gravity is F = kmG. Where
        G is the acceration due to gravity
        k is constant between 0 and 1 which depends on the steepness of the hill
       m is the mass of the sledge
b) The force due to gravity causes the sledge to accelerate. Using Newton's law again F = ma.
c) F and m are the same in both equations. so

kmG = F = ma
kmG = ma
kG = a

In other words, the acceleration only depends on the steepness of the hill and the acceleration due to gravity; G. The mass of the sledge doesn't matter.

This means that it doesn't matter whether Jack keeps hold of the box or drops it over the side. When I say drop, I mean that the box and Jack have exactly the same velocity when they part company.

Conservation of momentum tells us that the Jack gets a sudden boost in velocity if he throws the box out the back. Once he's let go of the box though, he will be back to accelerating at kG just like Jill.

Does this help?



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by:adam8
ID: 9706552
what is bothering me is that the exact words from the examiners report say that "this question could have been answered using a conservation of momentum approach ...."
I don't understand how if the box is pushed off the side of the toboggan how a conservation of momentum approach could explain this. It states in the problem that the box of gear was pushed off the side of the toboggan.
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GwynforWeb earned 200 total points
ID: 9706930
Looking at this vectorially if the icecream is pushed of the side then the yes the speed will increase but the sledge will now not be going straight it must be steered back on track, it will have a higher speed than it did before the icecream was pushed of. ( unless it is done whist turnig a corner in which case no further steering is required)

Since the force is applied (fairly instantaneously)at  right angles to the direction of travel then there is no change in the component of the momentun vector in the direction of travel, before being expelled the momentun of the systen perpendicular to the direction of travel is 0, if  Ss and Si are  the speeds of the ice cream perprdicular to the track respectively after expulsion then balancing momnetum before and after gives

        0 = Ms*Ss+Mi*Si

ie      Ss = Si*Mi/Ms

if the speed of the sledge before and after expulsion in the track direction is S then the new speed is

        sqrt ( S² +(Ss)²)  >  S  

ie the sledge is going faster but needs to be steered back onto track direction




where Vi initial velocity of both icream and sled with masses Mi and Ms. The final velociy of the sled VSf is  given by (VIf final velocity of icecream)
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by:GwynforWeb
ID: 9706958
oops ignore the last sentence at the bottom, its a remnant of cut and past.
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by:GwynforWeb
ID: 9707481
' if the speed of the sledge before and after expulsion in the track direction is S '

should read

' if the speed of the sledge before the ice cream expulsion in the track direction is S '
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