Railroad cars collide
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
Need help? Yes, please!
The following quantities appear in the problem:
Impuls \(p\) /
The following formulas must be used to solve the exercise:
\(\sum p_{\scriptscriptstyle\rm tot} \stackrel{!}{=} \sum p_{\scriptscriptstyle\rm tot}' \quad \)
No explanation / solution video for this exercise has yet been created.
But there is a video to a similar exercise:
In case your browser prevents YouTube embedding: https://youtu.be/APbBIWHbsoE
But there is a video to a similar exercise:
Exercise:
A .t railroad car mathcalA travelling at a speed of .kilometerperhour strikes an identical car mathcalB at rest. If the cars lock together as a result of the collision what is their common speed just afterwards?
Solution:
newqtym.ekg newqtyve % The linear momentum before and after the collision remains the same i.e. the linear momentum of one vehicle before the collision is the same as the linear momentum of both together after the collision. ImpulsSchritte PGleichungp_ p_' + p_' PGleichungm_v_ m_v_' + m_v_' PGleichungmv_ mv' + mv' AlgebraSchritte MGleichungmv' mv_ MGleichungv' fracv_ PHYSMATH Thus the their common speed is solqtywfracv_ven/ al v' wf fracve wTTTT wTTT. We can see that for two identical railroad cars this speed does not dep on their mass.
A .t railroad car mathcalA travelling at a speed of .kilometerperhour strikes an identical car mathcalB at rest. If the cars lock together as a result of the collision what is their common speed just afterwards?
Solution:
newqtym.ekg newqtyve % The linear momentum before and after the collision remains the same i.e. the linear momentum of one vehicle before the collision is the same as the linear momentum of both together after the collision. ImpulsSchritte PGleichungp_ p_' + p_' PGleichungm_v_ m_v_' + m_v_' PGleichungmv_ mv' + mv' AlgebraSchritte MGleichungmv' mv_ MGleichungv' fracv_ PHYSMATH Thus the their common speed is solqtywfracv_ven/ al v' wf fracve wTTTT wTTT. We can see that for two identical railroad cars this speed does not dep on their mass.
Meta Information
Exercise:
A .t railroad car mathcalA travelling at a speed of .kilometerperhour strikes an identical car mathcalB at rest. If the cars lock together as a result of the collision what is their common speed just afterwards?
Solution:
newqtym.ekg newqtyve % The linear momentum before and after the collision remains the same i.e. the linear momentum of one vehicle before the collision is the same as the linear momentum of both together after the collision. ImpulsSchritte PGleichungp_ p_' + p_' PGleichungm_v_ m_v_' + m_v_' PGleichungmv_ mv' + mv' AlgebraSchritte MGleichungmv' mv_ MGleichungv' fracv_ PHYSMATH Thus the their common speed is solqtywfracv_ven/ al v' wf fracve wTTTT wTTT. We can see that for two identical railroad cars this speed does not dep on their mass.
A .t railroad car mathcalA travelling at a speed of .kilometerperhour strikes an identical car mathcalB at rest. If the cars lock together as a result of the collision what is their common speed just afterwards?
Solution:
newqtym.ekg newqtyve % The linear momentum before and after the collision remains the same i.e. the linear momentum of one vehicle before the collision is the same as the linear momentum of both together after the collision. ImpulsSchritte PGleichungp_ p_' + p_' PGleichungm_v_ m_v_' + m_v_' PGleichungmv_ mv' + mv' AlgebraSchritte MGleichungmv' mv_ MGleichungv' fracv_ PHYSMATH Thus the their common speed is solqtywfracv_ven/ al v' wf fracve wTTTT wTTT. We can see that for two identical railroad cars this speed does not dep on their mass.
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Unelastischer Stoss by uz
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Unelastischer Stoss 1dim by TeXercises
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Unelastischer Stoss by pw