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PHYSICS PAPER 3 DOUBT

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I wish I could help.... Doing Cambridge O Levels... But did search the paper you have problem in... Thought I could give it a try.... Can you specify the paper code??? Cant find Paper 31?
 
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First I suggest you find the answer for 1)d i()
The speed gained by the trolley of weight o.4N in 1.2s is = 1.2 x 0.5
Speed gained= 0.6 m/s
Then for 1)d(ii) You have to use the above value to find the distance based on the equation Distance = Speed x Time
so Distance = 0.6 m/s x 1.2 s
0.72 m
:)
The equation i used for 1 (d) (ii) is 100% correct, but the marking scheme gives an answer of 0.36 m :S For 0.72 m you only get 1 mark
I may have misunderstood the question
 
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Yes I agree that its 100% correct but its just used for CONSTANT speed... When speed is not constant you have to use EQUATIONs of motion....

Its

2as= v final square - v initial square

s is distance

2(.50)s= .6 square - 0

Find s which is .36 m....
 
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Or you can use that other one..

s= v initial into time + 1/2 a t^2

s= 0 into 1.2 + 1/2 (.50) 1.2 ^2

s= .5 into .5 into 1.44

s= .36 m
 
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To be honest I learned that a week or two ago... I wonder its in there in our Syllabus or not.... But teacher did explained it... And the junior class got a question about it in Physics (PRE MOCKS EXAMS) So probably its there in our syllabus.... Its in Cambridge IGCSE syllabus but I have not found a question regarding it in O level Physics paper.... But still there is a question about it in the pre mocks EXAMS...
 
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thanks guys for your help, but the motion equations aren't mentioned in the syllabus so are we supposed to know them or use them in our calculations?
 
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I guess just use them in your calculations... So you just need to remember when speed is not constant you have to use these equations...

And you are welcome... :)
 
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