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The Physics AS-Level Thread
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Q12 they meet in the middle as they are joined they both move the same amount.
Q7 zero resultant force implies two (or more) forces which add together to zero.
Q7 zero resultant force implies two (or more) forces which add together to zero.
Original post by old_teach
Q12 they meet in the middle as they are joined they both move the same amount.
Q7 zero resultant force implies two (or more) forces which add together to zero.
Q7 zero resultant force implies two (or more) forces which add together to zero.
In Q7. the constant force the question says the truck is propelled by is considered as the resultant force, right? As resultant force=forward force-force of friction.
Just asking for my understanding of forces and the formula f=ma.
"In Q7. the constant force the question says the truck is propelled by is considered as the resultant force, right?"
Yes, that's right - they say air resistance is negligible, so the propelling force is the only force they're interested in.
Here they are imlpying that there are no opposing friction forces. When people write these questions, they are just trying to tease your application of F=ma, it's not a realistic application. Try writing your own A level questions - it's surprisingly hard! It's very easy to put in extra details to cover all eventualities, and that causes students to get confused, and think that those extra parts are important .... And there will always be someone who criticises your question because it's not totally true, and of course, they're correct, but ... so I considered applying to write A level questions, but you had to submit some examples, and I gave up! So I tend to feel sorry for question writers - but I doubt you'd agree!
Yes, that's right - they say air resistance is negligible, so the propelling force is the only force they're interested in.
Here they are imlpying that there are no opposing friction forces. When people write these questions, they are just trying to tease your application of F=ma, it's not a realistic application. Try writing your own A level questions - it's surprisingly hard! It's very easy to put in extra details to cover all eventualities, and that causes students to get confused, and think that those extra parts are important .... And there will always be someone who criticises your question because it's not totally true, and of course, they're correct, but ... so I considered applying to write A level questions, but you had to submit some examples, and I gave up! So I tend to feel sorry for question writers - but I doubt you'd agree!
Original post by old_teach
"In Q7. the constant force the question says the truck is propelled by is considered as the resultant force, right?"
Yes, that's right - they say air resistance is negligible, so the propelling force is the only force they're interested in.
Here they are imlpying that there are no opposing friction forces. When people write these questions, they are just trying to tease your application of F=ma, it's not a realistic application. Try writing your own A level questions - it's surprisingly hard! It's very easy to put in extra details to cover all eventualities, and that causes students to get confused, and think that those extra parts are important .... And there will always be someone who criticises your question because it's not totally true, and of course, they're correct, but ... so I considered applying to write A level questions, but you had to submit some examples, and I gave up! So I tend to feel sorry for question writers - but I doubt you'd agree!
Yes, that's right - they say air resistance is negligible, so the propelling force is the only force they're interested in.
Here they are imlpying that there are no opposing friction forces. When people write these questions, they are just trying to tease your application of F=ma, it's not a realistic application. Try writing your own A level questions - it's surprisingly hard! It's very easy to put in extra details to cover all eventualities, and that causes students to get confused, and think that those extra parts are important .... And there will always be someone who criticises your question because it's not totally true, and of course, they're correct, but ... so I considered applying to write A level questions, but you had to submit some examples, and I gave up! So I tend to feel sorry for question writers - but I doubt you'd agree!
Alright. Thank you for your help.
I see, tough luck there. Just trying my best to clear everything up as much as possible so I'm not tricked easily by them.
By the way, should the force in f=ma always be the resultant force?
(edited 5 years ago)
Original post by old_teach
"In Q7. the constant force the question says the truck is propelled by is considered as the resultant force, right?"
Yes, that's right - they say air resistance is negligible, so the propelling force is the only force they're interested in.
Here they are imlpying that there are no opposing friction forces. When people write these questions, they are just trying to tease your application of F=ma, it's not a realistic application. Try writing your own A level questions - it's surprisingly hard! It's very easy to put in extra details to cover all eventualities, and that causes students to get confused, and think that those extra parts are important .... And there will always be someone who criticises your question because it's not totally true, and of course, they're correct, but ... so I considered applying to write A level questions, but you had to submit some examples, and I gave up! So I tend to feel sorry for question writers - but I doubt you'd agree!
Yes, that's right - they say air resistance is negligible, so the propelling force is the only force they're interested in.
Here they are imlpying that there are no opposing friction forces. When people write these questions, they are just trying to tease your application of F=ma, it's not a realistic application. Try writing your own A level questions - it's surprisingly hard! It's very easy to put in extra details to cover all eventualities, and that causes students to get confused, and think that those extra parts are important .... And there will always be someone who criticises your question because it's not totally true, and of course, they're correct, but ... so I considered applying to write A level questions, but you had to submit some examples, and I gave up! So I tend to feel sorry for question writers - but I doubt you'd agree!
Also, by indicating there is no air resistance, in your view did the examiner indicate there is no force of friction, so the constant forward force acting on the truck only is considered as the resultant force by candidates? Just being a bit cautious for my understanding.
(edited 5 years ago)
"constant forward force acting on the truck only is considered as the resultant force"
yes - the writer was trying to keep it simple. It would be hard to analyse if friction were increasing with speed (or speed squared!) while the mass is decreasing. Remember in exams you have very little time for multiple choice questions, so you must keep it simple, the examiner is unlikely to put too many complexities in questions, just testing basic ideas.
yes - the writer was trying to keep it simple. It would be hard to analyse if friction were increasing with speed (or speed squared!) while the mass is decreasing. Remember in exams you have very little time for multiple choice questions, so you must keep it simple, the examiner is unlikely to put too many complexities in questions, just testing basic ideas.
Original post by old_teach
"constant forward force acting on the truck only is considered as the resultant force"
yes - the writer was trying to keep it simple. It would be hard to analyse if friction were increasing with speed (or speed squared!) while the mass is decreasing. Remember in exams you have very little time for multiple choice questions, so you must keep it simple, the examiner is unlikely to put too many complexities in questions, just testing basic ideas.
yes - the writer was trying to keep it simple. It would be hard to analyse if friction were increasing with speed (or speed squared!) while the mass is decreasing. Remember in exams you have very little time for multiple choice questions, so you must keep it simple, the examiner is unlikely to put too many complexities in questions, just testing basic ideas.
Okay. Thank you.
Original post by old_teach
I'm going to be brave and say I think Eimmanual is a bit wrong.
"since (m1gh-m2gh) is negative"
The question says m1 is partly counterbalanced by m2, so m1 > m2 (m1 is a lift), so m1gh-m2gh is positive.
It's a mc question and I really think it's time to move on!
"since (m1gh-m2gh) is negative"
The question says m1 is partly counterbalanced by m2, so m1 > m2 (m1 is a lift), so m1gh-m2gh is positive.
It's a mc question and I really think it's time to move on!
Indeed, I was wrong to say that the quantity (m1gh ‒ m2gh) is negative based on the very first sentence of the question. Thanks for pointing it out.
I copy the solution of another elevator problem where the counterweight (or counterbalance in this question) is heavier than the passenger car that I had without paying much attention to the wording of the actual question raised by Clark20.
Original post by Clark20
Also, by indicating there is no air resistance, in your view did the examiner indicate there is no force of friction, so the constant forward force acting on the truck only is considered as the resultant force by candidates? Just being a bit cautious for my understanding.
By indicating that there is no air resistance, it does not guarantee that there is no frictional force. It really depends on what types of frictional forces are you referring to.
Again there will be someone going to argue that I am complicating the problem as this is an A-level Physics question, so keep things simple.
As the question indicate that there is track, so I would assume that there are wheels on the rocket-propelled truck. Thus, friction is required to turn the wheels for the truck to accelerate. I am bringing up this point is because I find that some students don’t know how to differentiate frictional forces that are dissipative (bad friction – term use in O level) and non-dissipative (good friction – term use in O level) and would lump everything together.
In most of the A-level Physics questions (I have not seen every question), when the questions indicate that frictional forces can be ignored or considered to be negligible, the frictional forces are referred to those that dissipate energy. However, within the question, good friction can appear and not stated explicitly. Be careful.
I dont understand this question.
I get that the term "interference pattern" means appearance of both dark and bright fringes. And on RS not both fringes, but only one type of fringes appear so there is no pattern. But, how to determine there wont be a pattern on RS?
(edited 5 years ago)
Original post by Clark20
I dont understand this question.
I get that the term "interference pattern" means appearance of both dark and bright fringes. And on RS not both fringes, but only one type of fringes appear so there is no pattern. But, how to determine there wont be a pattern on RS?
I dont understand this question.
I get that the term "interference pattern" means appearance of both dark and bright fringes. And on RS not both fringes, but only one type of fringes appear so there is no pattern. But, how to determine there wont be a pattern on RS?
You can try to imagine the production of two circular water waves at point P and Q (this is shown in the above-left picture)
The above right picture shows that RS is along the x-axis. This shows that the waves from P and Q have the same path difference (or phase difference), so there is no interference pattern (constructive and destructive interference) along RS.
Be careful that “define” interference pattern as bright and dark fringes only. There are a lot of interference patterns do not have bright and dark fringes.
Original post by Eimmanuel
You can try to imagine the production of two circular water waves at point P and Q (this is shown in the above-left picture)
The above right picture shows that RS is along the x-axis. This shows that the waves from P and Q have the same path difference (or phase difference), so there is no interference pattern (constructive and destructive interference) along RS.
Be careful that “define” interference pattern as bright and dark fringes only. There are a lot of interference patterns do not have bright and dark fringes.
You can try to imagine the production of two circular water waves at point P and Q (this is shown in the above-left picture)
The above right picture shows that RS is along the x-axis. This shows that the waves from P and Q have the same path difference (or phase difference), so there is no interference pattern (constructive and destructive interference) along RS.
Be careful that “define” interference pattern as bright and dark fringes only. There are a lot of interference patterns do not have bright and dark fringes.
What really makes the path/phase difference constant along RS? It cant alone be because of the fact both the sources are kept exactly in line and emit waves which start moving from the same point (not above or below each other) as the same waves emitted from the sources go on to form an interference pattern along XY.
Also, how did we figure out only constructive interference occurs on RS?
(edited 5 years ago)
Original post by Clark20
What really makes the path/phase difference constant along RS? ….
The distance travelled by the waves from P and Q are the same, so the path difference is zero.
Original post by Clark20
… It cant alone be because of the fact both the sources are kept exactly in line and emit waves which start moving from the same point (not above or below each other) as the same waves emitted from the sources go on to form an interference pattern along XY….
Don’t really understand what you are trying to describe.
Original post by Clark20
…. Also, how did we figure out only constructive interference occurs on RS?
It is deduced based on the conditions of constructive interference and destructive interference.
Constructive interference:
Path difference = m λ (m = 0, ±1, ±2, …)
Destructive interference:
Path difference = (m + ½ )λ (m = 0, ±1, ±2, …)
Since there is zero path difference for the two waves from P and Q along RS, it satisfies the condition of constructive interference, it would be constructive interference.
Original post by Eimmanuel
The distance travelled by the waves from P and Q are the same, so the path difference is zero.
Don’t really understand what you are trying to describe.
It is deduced based on the conditions of constructive interference and destructive interference.
Constructive interference:
Path difference = m λ (m = 0, ±1, ±2, …)
Destructive interference:
Path difference = (m + ½ )λ (m = 0, ±1, ±2, …)
Since there is zero path difference for the two waves from P and Q along RS, it satisfies the condition of constructive interference, it would be constructive interference.
Don’t really understand what you are trying to describe.
It is deduced based on the conditions of constructive interference and destructive interference.
Constructive interference:
Path difference = m λ (m = 0, ±1, ±2, …)
Destructive interference:
Path difference = (m + ½ )λ (m = 0, ±1, ±2, …)
Since there is zero path difference for the two waves from P and Q along RS, it satisfies the condition of constructive interference, it would be constructive interference.
Thank you.
Why cant B be the answer?
Take moments about the centre of the base (where R hits the ground).
In B there is a resultant moment. This is a bad thing!
In B there is a resultant moment. This is a bad thing!
Need help with this. What exactly does "how many images of the slit does he see?" mean?
Original post by old_teach
Take moments about the centre of the base (where R hits the ground).
In B there is a resultant moment. This is a bad thing!
In B there is a resultant moment. This is a bad thing!
Load W is anticlockwise. Load L was clockwise in a way it was perfectly balancing load W, but the torque it provided was later increased by moving it further away so clockwise moment>anticlockwise.
In option B, they added a horizontal force. Would that force not act in an anticlockwise direction too to the pivot which increases the anticlockwise moments and hence brings equilibrium again?
(edited 5 years ago)
In B the horizontal force has a zero distance from the pivot, therefore cannot balance the moments.
Q27.
You need to use d sin theta = n lambda
put theta as 90 degrees and see what n is.
The actual biggest n will be the whole number just less than the n you've calculated (n has to be a whole number, of course)
That's the number of orders you'll see on one side.
The total number of lines you'll see will be n on one side, the central order one (n=0), and n more on the other side.
I hope that makes sense!
You need to use d sin theta = n lambda
put theta as 90 degrees and see what n is.
The actual biggest n will be the whole number just less than the n you've calculated (n has to be a whole number, of course)
That's the number of orders you'll see on one side.
The total number of lines you'll see will be n on one side, the central order one (n=0), and n more on the other side.
I hope that makes sense!
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