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Physics: Post your doubts here!

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Clark20

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Hamnah Zahoor said:
T - m1g = m1(a)

As in the whole system the elevator is going upwards against gravity thus acceleration is negative
P = m(-a)v
P = (m1 + m2)(-(m2-m1)g/m1+m2)v
P = (m1-m2)gv ---------- option D
Click to expand...


Sorry, but can you please explain this bit again. Why are we considering the direction of acceleration of just the elevator when we are calculating the power of not just the elevator, but the combined power of both the masses/power of the system.
 
Hamnah Zahoor

Hamnah Zahoor

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Clark20 said:
Sorry, but can you please explain this bit again. Why are we considering the direction of acceleration of just the elevator when we are calculating the power of not just the elevator, but the combined power of both the masses/power of the system.
Click to expand...
Oh!!! .......yes you are right I made a mistake there.
 
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Clark20

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Need help understanding a concept in this. It's said the truck moves with constant force, doesnt that mean it moves with constant velocity as well and hence has 0 acceleration?
 

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Hamnah Zahoor

Hamnah Zahoor

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Clark20 said:
Need help understanding a concept in this. It's said the truck moves with constant force, doesnt that mean it moves with constant velocity as well and hence has 0 acceleration?
Click to expand...
In order to know whether the object is moving with constant velocity or not the resultant force should be equal to zero not only the forward force.
In this case a constant forward force is acting with the friction (between the tyres and the road) opposing it and the net force zero is not mentioned thus velocity is not constant.
 
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Clark20

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Hamnah Zahoor said:
In order to know whether the object is moving with constant velocity or not the resultant force should be equal to zero not only the forward force.
In this case a constant forward force is acting with the friction (between the tyres and the road) opposing it and the net force zero is not mentioned thus velocity is not constant.
Click to expand...
But there's no force of friction. Can only the constant forward force itself be the resultant force in this case?
Also, the constant force is used in the formula f=ma in the question and keeping in mind the force f=ma always gives the resultant, doesn't that mean the constant forward force of the truck itself is the resultant force? Of course the resultant would be a bit lesser if there was mention of friction. But in this case, is the forward constant force itself really the resultant too, forward force=resultant?

In cases where a body is moving with a constant driving force and has no friction acting against it, the constant driving force itself should be considered the resultant.

Assuming, there's 0 force of friction against the trucks motion (as we're told to ignore air resistance by which I assume the examiner means to tell us there is no friction acting against the truck), the constant force it's moving with should be equal to the truck's resultant?

Just asking to clarify my concepts of forces and the equation f=ma. Always the force we get from it is the resultant force, right? That's why I'm being a bit cautious of interpreting the resultant, as resultant only is used in f=ma and not forward force (unless friction is 0). In this question of course the resultant would equal the forward force in case theres really zero friction.
 
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Hamnah Zahoor

Hamnah Zahoor

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Clark20 said:
Assuming, there's 0 force of friction against the trucks motion (as we're told to ignore air resistance by which I assume the examiner means to tell us there is no friction acting against the truck), the constant force it's moving with should be equal to the truck's resultant?

Just asking to clarify my concepts of forces and the equation f=ma. Always the force we get from it is the resultant force, right? That's why I'm being a bit cautious of interpreting the resultant, as resultant only is used in f=ma and not forward force (unless friction is 0). In this question of course the resultant would equal the forward force in case theres really zero friction.
Click to expand...
Yes in this case the forward force will be taken as resultant force. ( when assuming zero resistance)
But in the case if acceleration is to be zero it will be only be if the resultant force is zero which is not in this case thus option D is not correct.
And the points you stated above are correct.
 
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KashishV

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hey guys so I had my physics practicals a few days ago. They asked us to measure a distance x and next to the dash they wrote the units as cm.
They then asked us to formulate a table and continue with different x values. However, since we stress so much on base units and what not, I converted the x values from cm to m in my table. Therefore the graph was plotted in terms of metres and not cm. Thus my gradient was 0.1836... where others got 15,20,22 etc but with cms. If u multiply my gradient by 100 for conversion, even I will get 18.36

Will Cambridge penalise me for converting the units? because I will then lose a lot of marks for this. They shouldn't be, right?
 
Hamnah Zahoor

Hamnah Zahoor

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KashishV said:
hey guys so I had my physics practicals a few days ago. They asked us to measure a distance x and next to the dash they wrote the units as cm.
They then asked us to formulate a table and continue with different x values. However, since we stress so much on base units and what not, I converted the x values from cm to m in my table. Therefore the graph was plotted in terms of metres and not cm. Thus my gradient was 0.1836... where others got 15,20,22 etc but with cms. If u multiply my gradient by 100 for conversion, even I will get 18.36

Will Cambridge penalise me for converting the units? because I will then lose a lot of marks for this. They shouldn't be, right?
Click to expand...
No marks will be deducted as long as you clearly mentioned that you were using meters in your calculation.
My friend did the same thing in her Cambridge exam and she still got an A in her practical paper.
 
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KashishV

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Hamnah Zahoor said:
No marks will be deducted as long as you clearly mentioned that you were using meters in your calculation.
My friend did the same thing in her Cambridge exam and she still got an A in her practical paper.
Click to expand...

wow, thank you so much! I mentioned units in the table headings, graph axes, gradient, y-intercept and the next question abt constants a and b. So I think that should work, right? :))
 
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