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how was M1 42

Aamer Faiyaz

Aamer Faiyaz

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mhalvi said:
no you are absolutely wrong,you must have made a mistake I got 2 different values.And certainly weight component has to be taken into account!! then what is the use of including hill(inclined plain) in this question??
no brother you are mistaken
Click to expand...

Sorry to burst your bubble but the weight component shouldn't be taken. :) If you check may/june 2006 question 6 (iii), its pretty much the same math and only the resitive work done is taken.
 
mhalvi

mhalvi

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Aamer Faiyaz said:
Sorry to burst your bubble but the weight component shouldn't be taken. :) If you check may/june 2006 question 6 (iii), its pretty much the same math and only the resitive work done is taken.
Click to expand...
hehehe I have checked it and now let me burst your bubble :D this question has nothing to do with OCT/NOV 2012 Qs.6i. the question you are talking about is no way linked to it.We were to find out velocities at the top of the hill and at the bottom for which we had to find out the driving force for which it is necessary to take weight component because it was going up the hill.Whereas the question you have linked t it has nothing like that we are already given velocities at bottom and hill.It just asks "what is the work done by the driving force?" a 2 mark question and as easy as abc W.D by driving force=G.P.E+W.D against resistance-loss in K.E that is it. While if you had seen Qs. 6 ii of our paper it was also asked to find out the W.D by driving force which was also unlike the question you are referring to because there was no loss of K.E there was gain of K.E and that was W.D by driving force=G.P.E+W.D against resistance+ Gain in K.E.
Hope this will clarify your doubt.I dont wanna have a fight on a question,I very well know the correct way and I am not sticking to it because I have done that way but because it a basic MECHANICS concept and a clear fact.SO lets just wait for the MS :)
 
Hassaan Khalid

Hassaan Khalid

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Assad Ali said:
man simple as that. use the formula f=ma to calculate force. subtract it with 1000 N and the substitute the values in p=fv to calculate velocities at bottom and top. use them to calculate gain in kinetic energy. Hows that?
Click to expand...
u r probably right :D
 
Aamer Faiyaz

Aamer Faiyaz

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mhalvi said:
hehehe I have checked it and now let me burst your bubble :D this question has nothing to do with OCT/NOV 2012 Qs.6i. the question you are talking about is no way linked to it.We were to find out velocities at the top of the hill and at the bottom for which we had to find out the driving force for which it is necessary to take weight component because it was going up the hill.Whereas the question you have linked t it has nothing like that we are already given velocities at bottom and hill.It just asks "what is the work done by the driving force?" a 2 mark question and as easy as abc W.D by driving force=G.P.E+W.D against resistance-loss in K.E that is it. While if you had seen Qs. 6 ii of our paper it was also asked to find out the W.D by driving force which was also unlike the question you are referring to because there was no loss of K.E there was gain of K.E and that was W.D by driving force=G.P.E+W.D against resistance+ Gain in K.E.
Hope this will clarify your doubt.I dont wanna have a fight on a question,I very well know the correct way and I am not sticking to it because I have done that way but because it a basic MECHANICS concept and a clear fact.SO lets just wait for the MS :)
Click to expand...

ohh i thought you were saying that the equation should be W.D by DF=Change in M.E. + W.D. by R + W.D. by weight
 
D

doorman1995

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Aamer Faiyaz said:
Sorry to burst your bubble but the weight component shouldn't be taken. :) If you check may/june 2006 question 6 (iii), its pretty much the same math and only the resitive work done is taken.
Click to expand...
well but then the formula for power is differebt ib tht case its P=Fcos0 come on hows it possible to ignore he weight component
 
Yaasir Burthun

Yaasir Burthun

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Thanks a lot friend for the answers. Got around 43-46 points in this paper.
Made mistakes for the EK gain as I took EK+EP altogether. Can get 3 out of 5 from this part
The second part I got 1219375J so you round it to 1220000J.
And for the question number 7 the last part was a bit tricky so I can't do it.

But overall the paper was extremely easy.
MashAllah
 
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