Jumping on a Train: Tips and Advice

[john007]

Homework Statement

imagine a train that is moving straight and a man jumps as it moves to the right around a sharp bend.
There are 4 point labelled A,B,C D where:
A=position of person before he jumps
B= position a certain distance to the left of A
C=position twice distance of B to the left of A(ie BC=AB).
D= a position behind A and AD=AB.

When the train turns right, which is the closest point where man will land.

I know the answer but don't understand why.

Relevant Equations

non

Hi,

In theory its a simple question. What am I missing?

 

[phinds]

What am I missing?

What you're missing right now is telling us what you think the answer is and your best effort at explaining why it's the right answer. Please read the rules. We can't help until you have shown an effort.

 

[john007]

ok.

The train is initailly moving staright(north) as the man jumps and then suddenly it moves to the right round sharp bend. The momentum of man keeps him moving North in the air. So both C and D are possibilities depending on the speed of the train.
Thanks

 

[PeroK]

ok.

The train is initailly moving staright(north) as the man jumps and then suddenly it moves to the right round sharp bend. The momentum of man keeps him moving North in the air. So both C and D are possibilities depending on the speed of the train.
Thanks

You're also missing a diagram and clear statement of the question.

 

[phinds]

One of the things that I learned long ago is that it will sometimes clarify a math problem if you take things to the extreme (within the parameters of the problem).

What happens if the train takes an 89 degree turn to the right and is now heading almost due East ?

NOTE: This would require a ridiculous acceleration of the train but as I said, you take things to an extreme within the bounds of the problem, which this is.

Also, I agree w/ Perok. Always draw a diagram where possible and clarify the problem statement relative to the diagram.

 

[john007]

sorry - just noticed attach files. This will make it easier. It is now attached. The option is either A,B,C or D.

Thanks

 

[jbriggs444]

So now that you have clarified the question (which, sadly, remains bogus), can you support your argumentation in favor of C or D being the right answer?

Can you support your argumentation against A and B?

 

[john007]

using the lettering in the diagram, not the one given in the first description, I actually think he would land on c or D. The reason is that initially, as shown in diagram the train is due North just before man jumps so the man travels due North in the air. The speed determines if he lands closer to C or D which makes the question very complicated and does not contain sufficient data for calculations. In fact the i think the moree confusing it gets because now i am thinking D is possible. The only on i can eliminate is A because the man moves North as the train turns right and A will always be behind him(not necessarily directly behind him).

 

[jbriggs444]

using the lettering in the diagram, not the one given in the first description, I actually think he would land on c or D. The reason is that initially, as shown in diagram the train is due North just before man jumps so the man travels due North in the air. The speed determines if he lands closer to C or D

Slow down and actually explain your reasoning this time.

The man travels due North in the air. Great. But that does not tell how the train moves beneath him.

How does the speed make a difference? What happens for a slow moving train? What happens for a fast moving train?

While you are busy changing the speed of the train, what other things are you holding constant? The radius of the curve? The angular rotation rate of the train? Something else?

 

[haruspex]

using the lettering in the diagram, not the one given in the first description, I actually think he would land on c or D. The reason is that initially, as shown in diagram the train is due North just before man jumps so the man travels due North in the air. The speed determines if he lands closer to C or D which makes the question very complicated and does not contain sufficient data for calculations. In fact the i think the moree confusing it gets because now i am thinking D is possible. The only on i can eliminate is A because the man moves North as the train turns right and A will always be behind him(not necessarily directly behind him).

Assuming the train's speed does not change, which way would the man accelerate if he stayed at the same point in the carriage? What force causes that acceleration?

 

[Lnewqban]

using the lettering in the diagram, not the one given in the first description, I actually think he would land on c or D.
...

According to posted diagram, C=position of person before he jumps.
Do you mean he would land on C?

When the train beginns turning to the right, what would happen to a luggage with wheels that is not restained respect to the floor of the train?

 

[john007]

Assuming the train's speed does not change, which way would the man accelerate if he stayed at the same point in the carriage? What force causes that acceleration?

It states 'sharp bend' so am assuming this means that the track curves very quickly to make a right angle ie a small curve. Irrespective of speed, when the man jumps and assuming the train quickly turns to the right , the speed of the man in the North direction is greater than that of the train so he will travel further than the train in that direction so he will move away from C. Depending on how long he is in the air he could land near B or A. Although I eliminated D during my earlier thoughts, that is now a possibility.
As fro the trolly/suitacase, if the train suddenly moves to the right then the momentum would keep the trolley moving due north. But this still does not help.

 

[john007]

Assuming the train's speed does not change, which way would the man accelerate if he stayed at the same point in the carriage? What force causes that acceleration?

The centripetal force which acts inwards. But the man is in the air and as he and train are moving North just before he jums he carries on moving North in the air.

 

[kuruman]

The centripetal force which acts inwards. But the man is in the air and as he and train are moving North just before he jums he carries on moving North in the air.

That is correct. How does that help you decide where, with respect to points ABCD, he is likely to land? Imagine an oriented frame of reference fixed on the ground. The labeled points on the train and the passenger are all moving relative to this frame. Can you draw the paths (seen from above) of the 4 points in that frame? What about the path of the passenger while he is in the air? How do you interpret what happens when two paths intersect?

 

[jbriggs444]

It states 'sharp bend' so am assuming this means that the track curves very quickly to make a right angle ie a small curve.

If the track makes a right angle turn and the train car makes this turn within the time that the jumper takes off and lands, what sort of acceleration would that imply for the front and rear of the car. Is that realistic? Does that, perhaps, temper what might reasonably be meant by "sharp"?

If the train car is to negotiate a 90 degree turn within the time frame for a jumper to take off and land, does this put any constraint on the speed of the train car?

 

[kuruman]

If the train car is to negotiate a 90 degree turn within the time frame for a jumper to take off and land, does this put any constraint on the speed of the train car?

It did not occur to me that the turn could be that sharp, certainly not for an iron horse train car on a track. However I can imagine a platform flying at constant speed due North; when the man jumps, he activates a switch that simultaneously reduces the northward speed from V to zero while increasing the eastward speed from zero to V in the 1-2 seconds the man is in the air.

 

[haruspex]

The centripetal force which acts inwards.

Inwards to the curve, yes, but which way in terms of the frame of reference of the train?
Had he not jumped, what would have provided that force?
So which way will the man accelerate relative to the train when that force is removed?

 

[john007]

friction provides centripetal force so if no friction the man would slide radially. Sill can't see how to figure it out. To me the question is vague and and have assumed the paths is a right angle for simplicity rather than a circle although a slight curve is obviously needed.
thanks

 

[haruspex]

the man would slide radially

Yes! Stick with that.
But I am convinced that the question has been incorrectly stated. It surely intended that the man starts at B.

Edit: see later posts.

assumed the paths is a right angle for simplicity

But you would have no idea whether that gives the right answer.

 

[john007]

The last reply says that the question is wrong and so can't be answered.
Can ~I have comments from physics experts on if this is a viable question as it stands and if so what the correct option is or if its impossible to answer?
I know the answer to the physics question but don't understand the reason.
If the experts can't agree then how can i answer it.
Once I get some answers on the option for correct answer I will confirm mine.
It will be interesting to get different views on this and how certain everyone is with their answer. I don't want to influence anyone with my answer so await yours.
Thanks
pos this should really take no more than 30sec to answer as it is a mcq.

 

[john007]

Hi,
Can I have a least a comment on if this question and options are viable or if the question was incorrectly set?

Thanks

 

[jbriggs444]

Can I have a least a comment on if this question and options are viable or if the question was incorrectly set?

I do not like the question much. But if one applies some real world constraints on how high one can jump, how wide rail cars are and how fast they move one can prune it down to two answers. Given how physics teachers think, one can prune that further and divine an intended answer.

1. Rail cars do not turn by 45 degrees during a jump. So choice D is out. The deflection will be mostly centrifugal.

2. Rail card are wider than the height of a jump. And they do not accelerate at 1 g. So choice A is out. The jumper won't deflect that far relative to the car.

That leaves choice B (significant movement outward) and C (negligible movement outward). A physics teacher who wrote "sharp bend" is not going for an answer saying that the effect is negligible. (Though in my estimation with a reasonable rail car at a reasonable speed around a reasonable curve, the deflection will be closer to zero feet than to three).

 

[haruspex]

I do not like the question much. But if one applies some real world constraints on how high one can jump, how wide rail cars are and how fast they move one can prune it down to two answers. Given how physics teachers think, one can prune that further and divine an intended answer.

1. Rail cars do not turn by 45 degrees during a jump. So choice D is out. The deflection will be mostly centrifugal.

2. Rail card are wider than the height of a jump. And they do not accelerate at 1 g. So choice A is out. The jumper won't deflect that far relative to the car.

That leaves choice B (significant movement outward) and C (negligible movement outward). A physics teacher who wrote "sharp bend" is not going for an answer saying that the effect is negligible. (Though in my estimation with a reasonable rail car at a reasonable speed around a reasonable curve, the deflection will be closer to zero feet than to three).

I doubt the question is intended to be that subtle. The most likely to me is that the man is supposed to start at B and finish at A.

Edit: I withdraw that - see later post.

 

[john007]

I doubt the question is intended to be that subtle. The most likely to me is that the man is supposed to start at B and finish at A.

So are you implying that the question is set wrong?

 

[kuruman]

I think that there is a clear cut answer to the problem as posed, under the following assumption: The turn is executed by an acceleration $\vec a=a~\hat x-a~\hat y$ with the speed of the train remaining constant at $v_0.$ This acceleration is "on" during the time interval $0 \leq t \leq v_0/a$. Over this interval the speed of the train changes from $v_0$ northward to $v_0$ eastward.

With only this assumption we can write kinematic equations for the man and points A, B, C, and D relative to an inertial frame at rest with the ground and with the origin at the position of A at t = 0. Let $x_0$ be the separation between nearest neighbors. Then, while the man is in the air, we have position vectors $$ \vec r_{man}=v_0 t~\hat y~;~~\vec r_{A}=\frac{1}{2}at^2~\hat x+\left(v_0 t-\frac{1}{2}at^2\right)~\hat y$$ $$ \vec r_{B}=\left(x_0+\frac{1}{2}at^2\right)~\hat x+\left(v_0 t-\frac{1}{2}at^2\right)~\hat y~;~~ \vec r_{C}=\left(2x_0+\frac{1}{2}at^2\right)~\hat x+\left(v_0 t-\frac{1}{2}at^2\right)~\hat y $$ $$\vec r_{D}=\left(x_0+\frac{1}{2}at^2\right)~\hat x+\left(-x_0+v_0 t-\frac{1}{2}at^2\right)~\hat y. $$ The distances between the man and each of the four points are $$d_A=\left[ \left(\frac{1}{2}at^2\right)^2 + \left(-\frac{1}{2}at^2\right)^2 \right]^{1/2}~;~~d_B=\left[ \left(x_0+\frac{1}{2}at^2\right)^2 + \left(-\frac{1}{2}at^2\right)^2 \right]^{1/2}$$ $$d_C=\left[ \left(2x_0+\frac{1}{2}at^2\right)^2 + \left(-\frac{1}{2}at^2\right)^2 \right]^{1/2}~;~~d_D=\left[ \left(x_0+\frac{1}{2}at^2\right)^2 + \left(-x_0-\frac{1}{2}at^2\right)^2 \right]^{1/2}$$Note that these are the distances at any time $t$, including the landing time which need not be equal to the time $v_0/a$ required for the train to complete the turn as the question does not state that it is. When the negative signs in the squared terms are taken care of, it should be obvious which distance is the shortest and addresses the original question "which is the closest point where (the) man will land?"

 

[john007]

The answer given by the person who set the question is...

 

[kuruman]

The answer given by the person who set the question is...

$\dots$ not necessarily correct without justification.

 

[john007]

That is what I wanted to confirm, as some of the responses wanted me to give an answer and justify it and I was getting more confused each time I looked at the question and thought that there was a correct option.

But I don't think everyone on the forum thinks that this question and option don't make sense. Btw the answer given by the person who set it is A.

Initially I narrowed it down to A or B but thought A and B are equivalent so to speak so was wondering how to distinguish between them.

The mathematical method is too much for me but it would be interesting which position the formulae give.

 

[jbriggs444]

But I don't think everyone on the forum thinks that this question and option don't make sense. Btw the answer given by the person who set it is A.

Initially I narrowed it down to A or B but thought A and B are equivalent so to speak so was wondering how to distinguish between them.

The mathematical method is too much for me but it would be interesting which position the formulae give.

You cannot get into mathematics until the interpretation is nailed down.

With @haruspex interpretation, A is the obvious choice.

 

[haruspex]

You cannot get into mathematics until the interpretation is nailed down.

With @haruspex interpretation, A is the obvious choice.

I withdraw my interpretation.

 

[haruspex]

That is what I wanted to confirm, as some of the responses wanted me to give an answer and justify it and I was getting more confused each time I looked at the question and thought that there was a correct option.

But I don't think everyone on the forum thinks that this question and option don't make sense. Btw the answer given by the person who set it is A.

Initially I narrowed it down to A or B but thought A and B are equivalent so to speak so was wondering how to distinguish between them.

The mathematical method is too much for me but it would be interesting which position the formulae give.

I withdraw my previous remarks in this thread.
Just plot the path the man will take and the paths points A to D will take. Remember to keep the speed about the same for the man and the train, except that point A will travel a bit further and point C not quite so far, depending on the radius.

 

[Steve4Physics]

Hi. Thought I’d throw in my tuppence-worth!

I can't read the diagram. Here’s a version of the question which I think is equivalent to what is intended, plus my answer. Maybe it will help.
_____

Question:

A platform (P) is moving at 2m/s north.
P contains an xy grid (initially the y-axis is north).
A man (M),stands at (0,0) on the grid.

At t=0, M jumps up and remains in the air for 1s (lands at t=1s).

At t=0.5s the platform instantaneously rotates 90º clockwise and now moves at 2m/s east (note the y-axis is now east.)

What are M’s coordinates on the xy grid when he lands?
_______

Answer:

At t=0.5s, M is at max. height and is above the point (0,0) (because his velocity relative to P is zero from t=0 to t=0.5s).

Between t=0.5s and t=1s, M has moved a further 2m/s*0.5s = 1m north (now the -x direction) while the platform has moved 1m east.

In the platform’s frame of reference, between t=0.5s and t=1s, M has moved 1m in the -x direction and 1m in the -y direction. So M’s final coordinates are (-1, -1)m.

 

[kuruman]

Yes, as I wrote in post #19 previously.

Can you then comment what is wrong with the simple mathematical model in post #25? It conforms to the parameters of the question as stated in that
The man takes off from point A.
It is possible for the train to complete the 90^o turn.
Regardless of the man's time of flight, there is only one of the listed points where the man lands the closest. This is answer A as OP has revealed.

Furthermore, the model checks out in the limiting cases:
As the time of flight goes to zero, the distance between the landing point and A goes to zero.
When the acceleration is zero, we have the ballistic cart solution, namely that the man lands at the take off point regardless of the time of flight unless, of course, he hits the train wall before landing.

Needless to say, this is not a unique model; it is the simplest one that I could conjure up.

 

[haruspex]

Can you then comment what is wrong with the simple mathematical model in post #25? It conforms to the parameters of the question as stated in that
The man takes off from point A.
It is possible for the train to complete the 90^o turn.
Regardless of the man's time of flight, there is only one of the listed points where the man lands the closest. This is answer A as OP has revealed.

Furthermore, the model checks out in the limiting cases:
As the time of flight goes to zero, the distance between the landing point and A goes to zero.
When the acceleration is zero, we have the ballistic cart solution, namely that the man lands at the take off point regardless of the time of flight unless, of course, he hits the train wall before landing.

Needless to say, this is not a unique model; it is the simplest one that I could conjure up.

I have since withdrawn that post and updated post #19. See also post #31.

Wrt post #25, I don't see how the acceleration can be modeled as constant.

Please note that the description in post #1 is incorrect. See the diagram in post #6. The man starts at point C, which is to the right.
In your post above, you seem to have A as start and finish... I guess you meant takes off from C.

We don't know where the centre of the turn is, neither how far to the right nor how far forward or back from point C, nor where the mass centre of the carriage is (the one point that we can be sure should travel the same arc distance as the man). We are mot told it is 90 degrees, merely that it is sharp, so a radius not too much greater than the separation of the points in the diagram, I guess.

Suppose the man moves distance y in the original direction.
If we assume point B of the train maintains roughly constant speed then it can't go y in the original direction, much less point C. It's less clear, but I cannot find a way for D to get there either. Point A can get there because it arcs at a greater radius, so moves faster.

 

[jbriggs444]

The man takes off from point A.

How can we agree on a solution when we cannot even agree on the problem statement. Post #6 specifies a take-off at point C.