0
$\begingroup$

This is an example from Morin's Classical Mechanics text, in the section on momentum:

You are riding on a sled that is given an initial push and slides across frictionless ice. Snow is falling vertically (in the frame of the ice) on the sled. Assume that the sled travels in tracks that constrain it to move in a straight line. Which of the following three strategies causes the sled to move the fastest? The slowest?

A: You sweep the snow off the sled so that it leaves the sled in the direction perpendicular to the sled’s tracks, as seen by you in the frame of the sled.

B: You sweep the snow off the sled so that it leaves the sled in the direction perpendicular to the sled’s tracks, as seen by someone in the frame of the ice.

C: You do nothing.

The text indicates that we are to implicitly make the following assumptions:

  • Horizontal force due to sweeping snow can be neglected, as the normal force from the track is plenty to keep the sled from sliding off.

  • Vertical motion of the snow is also irrelevant, as the vertical normal force from the tracks keeps the sled from falling through the ice.

I am having difficulty understanding the solution in the text, which argues that in strategy B the snow is moving slower than the sled, and in strategy A the snow is moving faster than the sled.

My misunderstanding probably arises from not really knowing how to interpret the phrases "perpendicular in frame of the sled" and "perpendicular in frame of the ice." Presumably, the former means "as measured by someone on the sled" and latter means "as measured by someone still on the ice," but I'm struggling to visualize how these directions are different. If someone could clarify this and provide a more clearly written solution than the one in Morin's text, I'd really appreciate it!

$\endgroup$
0
$\begingroup$

My misunderstanding probably arises from not really knowing how to interpret the phrases "perpendicular in frame of the sled" and "perpendicular in frame of the ice." Presumably, the former means "as measured by someone on the sled" and latter means "as measured by someone still on the ice," but I'm struggling to visualize how these directions are different.

Yes, that's how to interpret it. I think it's easier to talk about everything in the "global" frame, which is this case is going to be identical to the "frame of the ice." Snow leaving perpendicular "in the frame of the sled" will still be moving forward with the sled to an observer standing next to the track, whereas moving perpendicular "in the frame of the ice" will be not moving forward relative to that observer.

This appears to be a conservation of momentum problem where we only really care about the forward direction. Let's look at everything from "frame of the ice" where the sled has forward momentum and the snow has no forward momentum as it falls. Look at how the snow is coming off in each case and if it steals any forward momentum from the sled:

  1. In case A, since the snow stays next to the sled in the perpendicular axis, it will have the same forward velocity as the sled. Since it started with no forward velocity, this will slow the sled down.
  2. In case B, since the snow will go backwards relative to the sled, and end up carrying no forward velocity. Since it started with no forward velocity, this will not slow the sled.
  3. In case C, it's fairly evident that the snow traveling with the sled will slow the sled by the same reasoning as case A.

The easy way to think of this from the perspective of being on the sled is to break the action into two parts. The snow lands on the sled and steals a little momentum. Now, you have the option to sweep the snow off directly to the side (case A), slightly backwards (case B), or do nothing (case C). Moving it backwards gives you some thrust that counteracts the drag from picking up the snow in the first place.

If you want to approach this rigorously, remember the total derivative F=dP/dt=mdv/dt+vdm/dt must equal zero (momentum being conserved), but here the dm/dt is positive so dv/dt must be negative so at any moment mdv/dt=vdm/dt. I won't go into the rest of that because it seems like you don't need it and it might get a bit long.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In the ice frame: A) Swept snow velocity same as sled. Same as it was kept on the sled. Sled velocity does not change. B) Swept snow does not carry forward momentum. To conserve momentum, sled velocity increases. $\endgroup$ – npojo Jan 7 '18 at 21:10
  • $\begingroup$ Yes, if you mean the sled slowed at first when the snow fell on it in the first place. If you say that the snow falls and the sled slows, THEN you push it to the side A) the sled will stay at it's lower velocity or B) it speeds back up to where it was. $\endgroup$ – admthrawnuru Jan 7 '18 at 23:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.