Mechanics M1: Post your doubt here

[elbeyon]

OK now let's just forget what happened to Paper 1 and carry on with our preparation for Mechanics 1 i.e Paper 4. I created this thread separately especially for the preparation for Oct/Nov 2012. So people with any doubt on Paper 4 can post it here and I or someone else will answer it as soon as it comes into eyesight. This years Math's Paper 1 was really a tough one with various unique question. So people let's just ace Paper 4 and make it terrific and fulfill the marks that we lost in P1 through P4. So post the Paper 4 questions in which you are doubtful.

 

[bamteck]

OK now let's just forget what happened to Paper 1 and carry on with our preparation for Mechanics 1 i.e Paper 4. I created this thread separately especially for the preparation for Oct/Nov 2012. So people with any doubt on Paper 4 can post it here and I or someone else will answer it as soon as it comes into eyesight. This years Math's Paper 1 was really a tough one with various unique question. So people let's just ace Paper 4 and make it terrific and fulfill the marks that we lost in P1 through P4. So post the Paper 4 questions in which you are doubtful.

Are you good in P4 ?

 

[elbeyon]

Are you good in P4 ?

I guess I am but no one is perfect. I'm still a student . Please post your doubts if you have any .

 

[daviruss]

can u help for this question http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_qp_41.pdf (question 3)

 

[elbeyon]

can u help for this question http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_41.pdf (question 3)

Its quite easy. Consider angle AP1X as α. Since angle P1XP2 = 90 so angle CP2X would be 90 - α . Now resolve forces on X vertically and find out an equation with cosα, sinα and W. And again resolve forces on X horizontally and find out an equation with cosα and sinα. Then compare those equations and simplify them then you will end-up with the accurate values. Hope it helped if not then I would be happy to explain further.

 

[daviruss]

Its quite easy. Consider angle AP1X as α. Since angle P1XP2 = 90 so angle CP2X would be 90 - α . Now resolve forces on X vertically and find out an equation with cosα, sinα and W. And again resolve forces on X horizontally and find out an equation with cosα and sinα. Then compare those equations and simplify them then you will end-up with the accurate values. Hope it helped if not then I would be happy to explain further.

haha i knew it is solved a question like it but my problem how did end up with resolving would u plzz draw it os smthng i would be grateful and if u have any more questions am glad to help')

 

[iKhaled]

haha i knew it is solved a question like it but my problem how did end up with resolving would u plzz draw it os smthng i would be grateful and if u have any more questions am glad to help')

you resolve the weight (7.3) parallel to the line p2x put in the opposite direction and prependicular to the line? did u get it ?

 

[daviruss]

you resolve the weight (7.3) parallel to the line p2x put in the opposite direction and prependicular to the line? did u get it ?

but that seems harder u can use as "ELBEYON" sai but thx anyways! and didnt rlly get it can u just reexplain or smthng

 

[elbeyon]

haha i knew it is solved a question like it but my problem how did end up with resolving would u plzz draw it os smthng i would be grateful and if u have any more questions am glad to help')

I guess this figure makes you clear. Note that lines AP1, BD and CP2 are parallel . Now apply the concept that I mentioned above. I guess it now helps.

 

[iKhaled]

but that seems harder u can use as "ELBEYON" sai but thx anyways! and didnt rlly get it can u just reexplain or smthng

why harder man? did u get the part where u resolve the components of the weight of 7 newton at least ? the lines drawn in red r the components of the 7 newton weight did u get this ?

 

[Gunner7]

Well Actually I'm better in P1 than P4,Help me here please

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w04_qp_4.pdf
(question 6) Same Idea came again in 2010 or 2011 (not sure) and I dont understand it at all.

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_41.pdf

Someone please help me for no. 4 !
I would much appreciate it if a sketch is provided !
Thanks

 

[GorgeousEyes]

A stolen car, travelling at a constant speed of 40 m/s, passes a police car parked in a lay by.
The police car sets off three seconds later, accelerating uniformly at 8 m/s^2. How long does the police car take to intercept the stolen vehicle and how far from the lay-by does this happen?
Please help me in this

 

[Gunner7]

A stolen car, travelling at a constant speed of 40 m/s, passes a police car parked in a lay by.
The police car sets off three seconds later, accelerating uniformly at 8 m/s^2. How long does the police car take to intercept the stolen vehicle and how far from the lay-by does this happen?
Please help me in this

distance moved by Stolen car in first 3 sec = 40 x 3 = 120 m, Stolen car u=40 v=40 a=o ,S= 40t + 0.5 x 0 x t^2 , S= 40t

Police car , u=o a = 8 ,S = 0 x t + 0.5 x 8 x t^2 ,S=4t^2

now we equal distance moved by two cars but add 120 m to distance moved by stolen car ( In first 3 sec)

4t^2 = 40t+120 , 4t^2 - 40t -120 = 0 , (quadratic) t=12.4 ( negative answer rejected)

so police car intercept stolen vehicle 12.4 sec later, to get distance from lay-by = use 4t^2 (distance of police car ) 4 x 12.4^2 = 615 m

the question, is are those answers right ? I don't know I just solved them now 1 If they are wrong,will I hope I gave u an Idea how to solve it

 

[GorgeousEyes]

distance moved by Stolen car in first 3 sec = 40 x 3 = 120 m, Stolen car u=40 v=40 a=o ,S= 40t + 0.5 x 0 x t^2 , S= 40t

Police car , u=o a = 8 ,S = 0 x t + 0.5 x 8 x t^2 ,S=4t^2

now we equal distance moved by two cars but add 120 m to distance moved by stolen car ( In first 3 sec)

4t^2 = 40t+120 , 4t^2 - 40t -120 = 0 , (quadratic) t=12.4 ( negative answer rejected)

so police car intercept stolen vehicle 12.4 sec later, to get distance from lay-by = use 4t^2 (distance of police car ) 4 x 12.4^2 = 615 m

the question, is are those answers right ? I don't know I just solved them now 1 If they are wrong,will I hope I gave u an Idea how to solve it

Your answer is right Alhamdulelah , except the second one is 617 . Thank youu for your help I appreciate it .

 

[daviruss]

i wish you could help my fir the ring question how to resolve them i just need to knw how to resolve a such question

 

[iKhaled]

i wish you could help my fir the ring question how to resolve them i just need to knw how to resolve a such question

can u give an example ?

 

[elbeyon]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_41.pdf

Someone please help me for no. 4 !
I would much appreciate it if a sketch is provided !
Thanks

First of all find out the values of α and β from right angled triangle ASP and then replace the values of α and β in angles APQ and SPR. Then resolve the forces and you will get your answers. This figure will help you:


Sorry for the figure. I could not draw better than that . Hope this helps.

 

[bamteck]

First of all find out the values of α and β from right angled triangle ASP and then replace the values of α and β in angles APQ and SPR. Then resolve the forces and you will get your answers. This figure will help you:
View attachment 16411

Sorry for the figure. I could not draw better than that . Hope this helps.

Thanks

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_42.pdf

please help me for no. 6 (ii) & (iii)
Thanks