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Clark20 Try reading the explanation on that website I recommended I hope that method helps.
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Physics: Post your doubts here!
- Thread starter XPFMember
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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.
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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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.
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?
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?
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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.
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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Yes it will work.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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Btw what was the year that appeared in your exam...any idea?
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Oh..... you appeared in your Cambridge exam I thought it was your mock test.year?
This was the feb/march 2019 paper
No worries.
Yes, as long as I don't lose these marks lol
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Inshallah ..You won't lose any marks.Yes, as long as I don't lose these marks lol
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Is it B?
If so its because the moment has to be kept constant to keep the bar in equilbrium so if x is increased then F should decrease therefore F and x would be inversely proportional
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Surely you will need to take into account the changing mass of the rocket. In this case Newton's Law becomes F = d/dt(mv), the reducing mass of the rocket will imply an increasing value of v since momentum must be constant as force is constant.
Yes. Thank you.
I understand why option A and D are wrong. However, I dont get how option C is justified.
The er says under conditions where mass keeps reducing, the bodys speed can change and it can accelerate. My question is, when a body is moving with constant velocity (zero resultant force) and even if its mass starts reducing, resultant force still is 0 and hence acceleration also is zero even if mass is lost. Speed shouldnt change.
The er says under conditions where mass keeps reducing, the bodys speed can change and it can accelerate. My question is, when a body is moving with constant velocity (zero resultant force) and even if its mass starts reducing, resultant force still is 0 and hence acceleration also is zero even if mass is lost. Speed shouldnt change.
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I did. Still having this confusion.