Mechanics M1: Post your doubt here

[daviruss]

hahaha... but i still leave many mistakes while doing them under exam conditions

yup i hate exam tension

 

[GorgeousEyes]

Please I want Ju06 no.4 .
And Nov04 no.6 .

 

[AhShun]

Jun06 p4 no 4
(i)(a) Height of A=Area under graph for 1st 0.7s
=(0.5)(0.7)(7)
=2.45m

(b) Depth of liquid=Area under graph for t between 0.7s and 1.2s
=(0.5)(1.2-0.7)(5+7)
=3m

(ii) Acceleration= (5-7)/(1.2-0.7)
= -4ms^-2
Therefore,deceleration= 4ms^-2

(iii) Weight - Resistance = Mass x Acceleration
mg - R = ma
10m - 0.7 = -4m
14m = 0.7
m = 0.05Kg[/QUOTE]

 

[AhShun]

Nov04 p4 no 6

 

[Mr Me]

Did you all completed all of the past papers ?

 

[deepum]

hey guys what are the most difficult questions set by CIE?

 

[iKhaled]

hey guys what are the most difficult questions set by CIE?

all the questions in the paper

 

[deepum]

i mean those which gives u best practice

 

[AhShun]

Did you all completed all of the past papers ?

No

 

[shikha95]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_42.pdf

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

6(ii) v^2=u^2 +2*a*s u=0 so v^2=2as
6(iii)v=u+at again u=0 so v=at

 

[GorgeousEyes]

Nov04 p4 no 6

Thank youu a loot !

 

[GorgeousEyes]

June01 n0.7 and specimin no.7 please .

 

[AhShun]

June01 n0.7 and specimin no.7 please .

Jun01???

 

[GorgeousEyes]

Jun01???

Yeaah , you don't have it ?

 

[AhShun]

No sorry

 

[GorgeousEyes]

No sorry

No problem , It's okay thank u

 

[GorgeousEyes]

Jun06 no.7 ii and iii
Jun06 no.5 iii
Jun07 no.1 ii

 

[Mr Me]

June01 n0.7 and specimin no.7 please .

Is it possible for you to upload the June 2001 paper ?

 

[AhShun]

Jun06 no.7 ii and iii
Jun06 no.5 iii
Jun07 no.1 ii

Jun06 p4 no5(iii)

 

[iKhaled]

Jun06 no.7 ii and iii
Jun06 no.5 iii
Jun07 no.1 ii

june 06 no.7 ii and iii

7 (ii) since the particles r traveling on a smooth plane then 3latol t3rafe ano al downward acceleration is a = g sinθ
w howa medeke al height 1.6 so calculate the distance from the given time and it will be d= 6.5
now use sinθ = 1.6/6.5
a= 10(1.6/6.5)
a= 2.46

7 (iii) first u calculate the time taken when q reached the maximum height w ante al mafrod tkone 3rfa lama y2olo maximum height da m3ana ano its velocity is 0
so vQ = u+ at
0 = 1.3 -2.46t
t = 0.5285 s

this means it took 0.5285 s for particle q to reach its maximum height now lets see whats the distance traveled by p when q reached its maximum height

s = 1.3(0.5285) + 1/2(2.46)(0.5285)^2
s= 1.03 m