Repelling Particles
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
\(\LaTeX\)
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Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Two positive particles with masses m_ and m_ and charges q_ and q_ are placed at an initial distance r_. abcliste abc Derive a formal expression for the two particles' final velocities v_ and v_. abc Calculate the velocities for a proton and an alpha particle with an initial distance rO. abcliste
Solution:
abcliste abc The total energy of the system of the two charged particles corresponds to the initial potential energy which is the electric potential energy of particle in the potential of particle or vice versa: sscEpot k_Cfracq_ q_r_ The potential energy is converted to kinetic energy. In the final state the following relation holds true: sscEkin+sscEkin sscEpot fracleftm_ v_^+m_ v_^right k_Cfracq_ q_r_labelconsE From the conservation of momentum we know that m_ v_ m_ v_. We can therefore express v_ as a function of v_: v_ fracm_m_ v_ labelconsP Using this in refconsE we find fracleftm_+m_fracm_^m_^right v_^ k_Cfracq_ q_r_ fracleftfracm_+m_m_right m_ v_^ k_Cfracq_ q_r_ Solving this for v_ leads to v_ vaF With refconsP we find the corresponding expression for v_: v_ fracm_m_vaF vbF abc For a proton m_ma q_qa and an alpha particle m_mb q_qb we can calculate the numerical values of v_ and v_: v_ sqrttimesnckctimesfracqatimesqbrfracmbmatimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmbXmaXtimesmaX+mbXtimesncu resultvaP v_ sqrttimesnckctimesfracqatimesqbrfracmambtimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmaXmbXtimesmaX+mbXtimesncu resultvbP abcliste
Two positive particles with masses m_ and m_ and charges q_ and q_ are placed at an initial distance r_. abcliste abc Derive a formal expression for the two particles' final velocities v_ and v_. abc Calculate the velocities for a proton and an alpha particle with an initial distance rO. abcliste
Solution:
abcliste abc The total energy of the system of the two charged particles corresponds to the initial potential energy which is the electric potential energy of particle in the potential of particle or vice versa: sscEpot k_Cfracq_ q_r_ The potential energy is converted to kinetic energy. In the final state the following relation holds true: sscEkin+sscEkin sscEpot fracleftm_ v_^+m_ v_^right k_Cfracq_ q_r_labelconsE From the conservation of momentum we know that m_ v_ m_ v_. We can therefore express v_ as a function of v_: v_ fracm_m_ v_ labelconsP Using this in refconsE we find fracleftm_+m_fracm_^m_^right v_^ k_Cfracq_ q_r_ fracleftfracm_+m_m_right m_ v_^ k_Cfracq_ q_r_ Solving this for v_ leads to v_ vaF With refconsP we find the corresponding expression for v_: v_ fracm_m_vaF vbF abc For a proton m_ma q_qa and an alpha particle m_mb q_qb we can calculate the numerical values of v_ and v_: v_ sqrttimesnckctimesfracqatimesqbrfracmbmatimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmbXmaXtimesmaX+mbXtimesncu resultvaP v_ sqrttimesnckctimesfracqatimesqbrfracmambtimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmaXmbXtimesmaX+mbXtimesncu resultvbP abcliste
Meta Information
Exercise:
Two positive particles with masses m_ and m_ and charges q_ and q_ are placed at an initial distance r_. abcliste abc Derive a formal expression for the two particles' final velocities v_ and v_. abc Calculate the velocities for a proton and an alpha particle with an initial distance rO. abcliste
Solution:
abcliste abc The total energy of the system of the two charged particles corresponds to the initial potential energy which is the electric potential energy of particle in the potential of particle or vice versa: sscEpot k_Cfracq_ q_r_ The potential energy is converted to kinetic energy. In the final state the following relation holds true: sscEkin+sscEkin sscEpot fracleftm_ v_^+m_ v_^right k_Cfracq_ q_r_labelconsE From the conservation of momentum we know that m_ v_ m_ v_. We can therefore express v_ as a function of v_: v_ fracm_m_ v_ labelconsP Using this in refconsE we find fracleftm_+m_fracm_^m_^right v_^ k_Cfracq_ q_r_ fracleftfracm_+m_m_right m_ v_^ k_Cfracq_ q_r_ Solving this for v_ leads to v_ vaF With refconsP we find the corresponding expression for v_: v_ fracm_m_vaF vbF abc For a proton m_ma q_qa and an alpha particle m_mb q_qb we can calculate the numerical values of v_ and v_: v_ sqrttimesnckctimesfracqatimesqbrfracmbmatimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmbXmaXtimesmaX+mbXtimesncu resultvaP v_ sqrttimesnckctimesfracqatimesqbrfracmambtimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmaXmbXtimesmaX+mbXtimesncu resultvbP abcliste
Two positive particles with masses m_ and m_ and charges q_ and q_ are placed at an initial distance r_. abcliste abc Derive a formal expression for the two particles' final velocities v_ and v_. abc Calculate the velocities for a proton and an alpha particle with an initial distance rO. abcliste
Solution:
abcliste abc The total energy of the system of the two charged particles corresponds to the initial potential energy which is the electric potential energy of particle in the potential of particle or vice versa: sscEpot k_Cfracq_ q_r_ The potential energy is converted to kinetic energy. In the final state the following relation holds true: sscEkin+sscEkin sscEpot fracleftm_ v_^+m_ v_^right k_Cfracq_ q_r_labelconsE From the conservation of momentum we know that m_ v_ m_ v_. We can therefore express v_ as a function of v_: v_ fracm_m_ v_ labelconsP Using this in refconsE we find fracleftm_+m_fracm_^m_^right v_^ k_Cfracq_ q_r_ fracleftfracm_+m_m_right m_ v_^ k_Cfracq_ q_r_ Solving this for v_ leads to v_ vaF With refconsP we find the corresponding expression for v_: v_ fracm_m_vaF vbF abc For a proton m_ma q_qa and an alpha particle m_mb q_qb we can calculate the numerical values of v_ and v_: v_ sqrttimesnckctimesfracqatimesqbrfracmbmatimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmbXmaXtimesmaX+mbXtimesncu resultvaP v_ sqrttimesnckctimesfracqatimesqbrfracmambtimesma+mb v_ sqrttimesnckctimesfracqaXtimesqbXtimesnce^rfracmaXmbXtimesmaX+mbXtimesncu resultvbP abcliste
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Acceleration Voltage by by