## An easy physics problem…

Based on the popularity of something I posted last week, I thought some of you might find this little problem amusing. It’s from a Physics A-level paper I took in 1981. The examination comprised two papers in those days (and a practical exam); one paper had long questions, similar to the questions we set in university examinations these days, and the other was short questions in a multiple-choice format. This is one of the latter type, from the mechanics section.

And here is a poll in which you may select your answer:

### 83 Responses to “An easy physics problem…”

1. It’s not clear to me that one of two possibilities clearly stands out.

2. telescoper Says:

This is obviously not as easy as I thought – the vast majority of answers so far are wrong!

3. I suspect the vast majority of them would be correct if we observed the blob of mud in the frame of reference of the linear motion of the wheel.

• I wouldn’t count on it!

I think that question (the one where the axis of the wheel is fixed, as opposed to the rolling one Peter asked) is on some of the standard diagnostic tests used by physics education researchers. They find that scores on such simple-seeming questions are dismal before instruction (not surprisingly), and that they’re all too often virtually unchanged after instruction.

4. Out of interest how quickly did you have to answer questions like this – i.e. how many questions were there on the paper and what was the time limit?

• telescoper Says:

Physics Paper 1 consisted of 50 questions like this, with a time allowed of 2 1/2 hours i.e. 3 minutes per question.

5. stringph Says:

I voted for the second most popular option. But to be honest it seems to depend how the mud was attached to the rim: can we assume it was exactly ‘on the side’ such that none of the paths indicated would be impeded?

• None of the paths are impeded! Path D and E look like they go “inside the wheel”, but the left-most boundary of the wheel is also moving right. So even if mud is on outside of “rim”, on the tread, say, it follows initial trajectory E.

6. telescoper Says:

This sure is a popular post!

7. For those interested, take a look at the wiki page on cycloid motion.

http://en.wikipedia.org/wiki/Cycloid

8. I’m not with the vast majority of answers, so I guess I have a chance…

9. Disappointed Says:

Oh dear. I always assumed the majority of your readers were professional astronomers and physicists. Give the answers to your poll, I do hope I’m wrong…

• telescoper Says:

Most readers of this post are probably physics teachers…

• Bryn Jones Says:

I’d really like to know which professional astronomers and physicists got the answer wrong!

10. At the moment of disconnect, the particle is moving in the direction of C with respect to the wheel. And, all the arrows start on the edge of the wheel, so they must be with respect to the wheel, right?

I hate problems like this.

11. It seems clear to me that the question is asking in which direction the mud will move with respect to the fixed point P. The wheel is stated to be rolling, from left to right. The only fixed part of the diagram is the road itself.

And since both the speed at which the mud is launched “backward” and the speed at which it is moving forward are dependent only on the rotation of the wheel it is possible to determine which of the remaining possibilities is the path taken.

• telescoper Says:

the mud is launched “backward”?

• Backward relative to the wheel, of course.

It only occurred to me after I had thought it through that all I was doing was thinking about cycloid motion. It’s all the little connections and shortcuts like that that make me wish that I was still actively involved in math/physics. I just want to slap myself and say “Duh!” after being confused about something that would have been relatively obvious to me a few years ago.

And if you do read this, why is it not possible to reply to replies?

• telescoper Says:

It is possible for me to set an additional level of reply, but I can’t be bothered to fiddle with the settings.

12. Theemuts Says:

If you let something go it’ll initially travel with the velocity it has at its release. The direction of the velocity is the direction in which M travels at the moment it’s released, which is C, because it’sd tangential to the object.

• telescoper Says:

That’s its velocity relative to the wheel….but isn’t the wheel also moving relative to the ground?

13. I was thinking of it in terms of a ball attached to a string spinning in a circle when I answered it. But thats not quite right because there is also a translational motion of the wheel,

14. I was thinking in terms of a ball attached to a string in circular motion which would be C, but thats not right. There is also translational motion.

15. John Womersley Says:

*Much* easier than that bloody ball problem!

16. My answer was B, because it’s the velocity of the mud when it gets to point P and I think B always points to the left – I don’t see any information in the question that suggests you have to rotate the arrows by the angle between M and P before giving an answer.

17. It tends to travel in the direction of the E arrow so it has to recoil back. Since B is too idealistic a situation the answer should be A?

18. Anton Garrett Says:

ee, bah gum

19. The wheel has a larger velocity in the x direction (so one might expect there would be no recoil), however the velocity of mud has a positive y component which explains the downward reaction force exerted by the wheel.

20. telescoper Says:

I think this polll demonstrates why democracy isn’t necessarily a good idea…

• Katulobotomia Says:

In response to your reply to Michael Dicken’s comment: “Acceleration? Why is it accelerating?”

I think he meant by acceleration = centripetal-acceleration. If an object has a constant velocity AND is following a circular path like the mud in here, it has a centripetal force acting towards the center of the circular path, which allows the mud to chance its direction from being just a straight path.

• telescoper Says:

Not when it has detached though…

• It’s completely irrelevant whether or not it’s accelerating anyway. It’s asking for the initial path of the particle which is given by its velocity – the acceleration just tells you what the velocity will be compared to that shortly after.
Even if there’s an acceleration (and there is presumably, due to gravity) it doesn’t matter when it comes to answering.

21. I think it would be towards the center of the circle since that’s the direction of its acceleration. If it’s relative to the circle it would be C, and if it’s relative to the ground it would be towards O. (I put E because that was the closest answer.)

22. net_force_equals_ma Says:

Seems to me we have the tangential component of velocity that would point in the direction aligned with vector C and has a magnitude equal to the velocity of the tire. Besides that we also have a vector to the right aligned with the velocity of the tire. That vector is slightly smaller than the tangential vector because the mud is lower than midpoint of the tire. The top of the tire moves to the right at a velocity twice that of the center of the tire and the bottom of the tire is at rest relative to the ground. The vector in direction C and the velocity vector to the right are similar in size and at an angle of more than 90 degrees apart.

I think that makes E the correct answer.

23. At the instant of the mud leaving the wheel there will be a horizontal linear velocity in the direction of travel and a rotational component of the motion tangential to the wheel rim. Resolving them would give the initial direction of the travel of the mud.

24. I was thinking the way to actually calculate the vector would be to use the derivatives for the x and y components of the parametric equation for the cycloid? It’s been a long time since my last physics or math class, and neither are my profession so apologies for being slow.

• telescoper Says:

I think that’s a very long-winded way to do it, but it should work. I prefer just to think about the magnitude and direction of two vectors!

25. […] post: An easy physics problem… « In the Dark This entry was posted in Uncategorized and tagged axis, diagnostic-tests, peter, […]

26. I don’t like this question because it’s not 100% clear what reference frame the answer should be given in. The question should either state that the reference frame is that of the stationary ground or that of the wheel without rotation. (Are you standing on the sidewalk watching the car go by, or are you in a second car watching the wheel from the back seat?)

• telescoper Says:

I disagree – it’s clearly stated that the wheel is moving from left to right….

• Anton Garrett Says:

The problem specifies that the wheel *moves* (specifically, rolls, as the bent arrow indicates). All motion is relative, so the failure of this sentence to mention what the wheel moves/rolls relative to – ie, the ground – clearly implies that the ground is the reference frame for the answer.

• That’s probably what I should have pointed out to the person that got confused by the nature of the arrow showing the rotation of the wheel which is clearly drawn in the reference frame of the wheel. This ended up with me getting rather confused too, as above, but I settled on the frame of the road in the end.

27. This is a really nice question and makes the student think about translational and rotational motion. Its nice that it can be solved in the head and understood by writing down or two lines of basic formulae.

I suspect that this type of question wouldn’t appear on today’s A-level papers.

• telescoper Says:

I think it’s another illustration of the virtue of making use of *all* the information given. There’s a big hint in the phrase “without slipping”, for example…

28. Anton Garrett Says:

Peter: When did dumbing down start? It’s always nice to feel that one’s own generation is the crown, but I recall seeing A-level maths questions from a generation or two before ours and thinking that they were somewhat tougher than we did.
Anton

• telescoper Says:

I have the same impression…I think a teacher at school showed us some papers from the 60s and we all though they were fiendish.

29. John Peacock Says:

Anton is right: I remember as an undergraduate looking at past physics papers from the 1950s and 1960s and concluding that the people who managed to pass them must have been complete heroes. I presume this trend started back with Newton: certainly I find most arguments in the Principia really hard to follow.

• telescoper Says:

Is that because they’re in Latin?

• Anton Garrett Says:

Chandrasekhar’s wonderful book “Newton’s Principia for the common reader” is most certainly NOT for the common reader, but as a commentary on the Principia for physical scientists today it is superlative.

NB That’s Chandrasekhar the physicist, not the spin bowler. I don’t think either would have been much good at the other’s speciality.

30. Anton Garrett Says:

Whittaker and Watson’s book “Modern Analysis” includes some A-level questions from the Edwardian era as exercises at the end of each chapter, and sobering *they* are too…

31. If the forward motion of the wheel is a factor in this, since both D and E are forward of tangential line, without more information it wouldn’t be possible to tell which of them is correct.

I’m sticking with C.

• telescoper Says:

There is enough information to rule out one of D or E. C is clearly incorrect, as the wheel as a whole is moving from left to right.

• But, it says “initial direction”, at the instant the mud becomes detached, it will have a direction tangent to the wheel.

• telescoper Says:

No. The wheel is also moving from left to right. Its velocity will have two components, one tangent to the wheel and one parallel to the ground pointing to the right.

You need to think about the relative size of these vectors to decide whether the correct answer is D or E. For this, you need the information that the wheel does not slip….

• Ok, so since the velocity vector of the wheel is about the same length as the velocity vector of the mud, adding vectors C + P would give E.

This is relative to a bystander at rest relative to the ground, correct?

• But point M is still moving backward, no?

32. I think to some extent its not just the case that the difficulty of A-levels has changed, but also that their style and content are different. Or at least that was my impression when (~10 years ago) I was taking A-levels and looking at past papers from the early 90s.

One thing that sticks in my mind from doing Physics A-level was that the syllabus assumed no knowledge of maths beyond GCSE, so no logarithms or exponentials and very little ability to differentiate or integrate. This made some aspects of the Physics course (radioactive decay for example) a little farcical.

The problem above does look like the sort of thing that came up in the mechanics modules I did for further maths though. On that point, in my experience, the further maths A-level I did had far more in common with undergraduate physics than A-level physics did.

• Anton Garrett Says:

I’m sure you are right that the questions are ‘different’ today, but that doesn’t mean they can’t be compared, and trying to do physics without maths is surely dumbing down.

33. At what rate is the wheel rolling: 1 rev/sec? 1 rev/year? If the wheel is rolling at a rate of 1 revolution per decade, will the same apply?

• telescoper Says:

You do not need to know, because whatever rate it is rolling at also fixes the tangential speed of the particle when it detaches from the rim. If the wheel is rolling very slowly, the particle will also be moving very slowly; the direction of the resulting motion will be the same regardless …

• But if the tangential force of the particle when it detaches is orders of magnitude less than gravity now acting through the particle, the particle will not follow the path of the tangential force and will not be moving left-to-right. At the point of detachment, the particle is moving right-to-left and gravity is acting downward. My answer, A. Please disregard this if it is way off base; it’s been ages since I’ve used this terminology.

If the wheel was not moving forward, it would have launched the piece of mud in direction C.

But as it also has a horizontal (+x) speed component, it will neutralize the -x component of direction C and launch it directly upwards

• telescoper Says:

No. The horizontal component does not cancel.

• You are right…

• Nick Cross Says:

If the mud was launched at a point when the tangential direction was directly upwards, then there would still be the horizontal motion of the car. If the mud was launched at a slightly earlier point, the horizontal motion backwards from the motion of the wheel would be slightly larger, but the horizontal motion of the car would still be the same. The speed of the wheel is exactly the same as the speed of the car, so the net motion is zero at the base (no slippage). Then it is a simple relative velocity problem. E, is the only sensible answer.

• telescoper Says:

It may be the only sensible answer, but only about 17% have got it right!

• Anton Garrett Says:

17%? That’s less than by chance…

35. This question was one of the questions on an admissions test I took when at Cambridge, and I remember being a little confused until I remembered vector addition! Peter, I thought your audience were more educated!

36. Martin Oldfield Says:

Isn’t it simpler to consider the motion of the cylinder as a rotation only, albeit about an axis which moves ? This avoids the complication of adding vectors.

Given that it rolls without slipping, the line in contact with the floor is at rest. In the diagram above then, P is the instantaneous centre of the rotation. The mud at M must be moving perpendicularly to PM, and hence E is the answer.

• Cute solution. Same answer as how I did it, but faster. (Once you see it.)

37. I am not sure E is the answer! Obviously E is the theoretical instantaneous velocity of the particle, but it cannot be the actual “path followed by it.” that would depend on the speed of the wheel and where the particle gets detached. I think it is very intuitional that you should expect in most cases the particle falls off immediately, while E predicts an upward component!

38. Hang on! Does the question mean it’s attached at that moment in time pictured. I read it as the wheel rotates so that the mud at M is detached when it gets to point P.

😦

39. I mean: *detached at the moment in time

40. […] by the popularity of the little physics problem I posted last week, I thought some of you might be interested in this little conundrum which dates […]

41. John Sansom Says:

The answer is E. Lets simplify this, think of a dirtbike without a fender above it’s back wheel. Where is the tire going to sling the mud? All over the person riding it.

42. Jody Woodland Says:

I believe Chase was the first person to explain it well. There are two (and only two!) velocity vectors to consider – that of the wheel as a whole moving left to right and that of the rotation of the wheel. Both vectors are the same length (they have to be if the resultant for point P, given the wheel is not slipping, is zero). The resultant of these two would look like E.

43. […] a collection of short questions of multiple-choice type from which I’ve already posted one example on this blog.  This one is Paper 2 and, as you’ll see, it consists of longer questions with a freer […]

44. dilshan dhananjaya Says:

c

45. Is it E?

46. B,
The correct option according to me would be this ,as the object is rolling the sand at the point of detachment will start following the parabolic path in the opposite direction of its motion

47. […] moons ago I posted an `easy’ physics problem from the Physics A-level paper I took in 1981. The examination comprised two papers in those days […]