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Mathematics: Post your doubts here!

applepie1996

applepie1996

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Silent Hunter said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_4.pdf

help required in this paper

its Mechanics 1 ... question 7 ?
Click to expand...
okay sooo .......... :)
part 1 ) you have to show that after a time 't' the distance between the particles is 2.6t
so you use the formula s=ut+1/2at^2 :)
then d= (1.3t + 1/2at^2) - (-1.3t + 1/2at^2)
solve this and you get d=2.6t m
we take speeds as negative as they are moving in opposite directions

part 2 ) now i attached a picture for part 2 so you could get a better understanding
the general formula you use to find acceleration in these questions is a=g sin alpha where g=10ms^-2
and sin alpha = O/H which you can see from the diagram O=1.6m and H=2.6t and t given in the question is 2.5 so H=2.6 x 2.5=6.5 m
now put in the values in the equation a=g sin alpha
you get a=10 x 1.6/6.5 =2.46 ms^-2

part 3) this part of the question asks us to find how much of distance did P move when Q was zero
so for Q >>>> when v=0m/s >>>>> 0=-1.3 + 2.46t then you solve this and get time as 0.528 s
distance moved by P at time of 0.528 s so d=1.3 x 0.528 + 1/2 2.46 x 0.528^2 = 1.03 m
i hope you got it :D
 
PANDA-

PANDA-

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5173b8b83b42a.png


I managed to get both M and A... but not C. Could anyone explain how to obtain the answer for C?
 
Silent Hunter

Silent Hunter

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applepie1996 said:
okay sooo .......... :)
part 1 ) you have to show that after a time 't' the distance between the particles is 2.6t
so you use the formula s=ut+1/2at^2 :)
then d= (1.3t + 1/2at^2) - (-1.3t + 1/2at^2)
solve this and you get d=2.6t m
we take speeds as negative as they are moving in opposite directions

part 2 ) now i attached a picture for part 2 so you could get a better understanding
the general formula you use to find acceleration in these questions is a=g sin alpha where g=10ms^-2
and sin alpha = O/H which you can see from the diagram O=1.6m and H=2.6t and t given in the question is 2.5 so H=2.6 x 2.5=6.5 m
now put in the values in the equation a=g sin alpha
you get a=10 x 1.6/6.5 =2.46 ms^-2

part 3) this part of the question asks us to find how much of distance did P move when Q was zero
so for Q >>>> when v=0m/s >>>>> 0=-1.3 + 2.46t then you solve this and get time as 0.528 s
distance moved by P at time of 0.528 s so d=1.3 x 0.528 + 1/2 2.46 x 0.528^2 = 1.03 m
i hope you got it :D
Click to expand...

thank you very much

and one thing : how do we know that the accelerations in the blue parts are both same? for both particles?
 
Rutzaba

Rutzaba

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unseen95 said:
please answer it for me its killing me
Click to expand...
i did this halfway only and attempting this wud break ma heart yet again... i think sumone else shud try it
il give u a hint... to find out perpendicular distance from one point to one plane we use the formula given in the question

where a b and c are where the point starts that is the coordinates of p... and x y and z wud be the normal coordinates
find two such equations one with m and one with n and equate them with each other
 
A

A star

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in some question where equation comes down to 1.3^n=39 or zomething to solve it if we use logs will it given credit?
 
unseen95

unseen95

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Rutzaba said:
i did this halfway only and attempting this wud break ma heart yet again... i think sumone else shud try it
il give u a hint... to find out perpendicular distance from one point to one plane we use the formula given in the question

where a b and c are where the point starts that is the coordinates of p... and x y and z wud be the normal coordinates
find two such equations one with m and one with n and equate them with each other
Click to expand...
thanks it helped :)
 
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