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Mechanics M1: Post your doubt here

shikha95

shikha95

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AhShun

AhShun

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Jun07 p4 no1(ii)
mg sin(alpha) = ma
sin(alpha) = a/g
= o.5/10
=0.05
alpha = 2.87
 
bamteck

bamteck

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shikha95 said:
OMG! I swear I replied to this an hour ago! Where did my post go?
6(ii) v^2=u^2 +2as and since u=0, v^2=2as. Therefore s=(v^2/2a)
6(iii) For t1 __v=u+at again u=0 so v=at and t=(v/a)
For t2__v=u+at, here v=0 and a=-g which is -10 so 0=u-10t
T=t1+t2
Click to expand...

Thank youuuuuuuuuu Shikha :)
 
hussamh10

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iKhaled said:
View attachment 16559
7 (i)
you have a right angle triangle with height 2m and base of 1.5 m and 2 unknowns angle. make one angle called theta and another angle called alpha as i showed u in the triangle i showed u

now lets to some trigs,

cos theta = 2/2.5
sin theta = 1.5/2.5

cos alpha = 1.5/2.5
sin alpha = 2/2.5

i resolved Ta and Tc and notice that the angle that Ta makes with the vertical is equal to alpha and the angle that Tc makes with the vertical is equal to theta. now its time to build up some equations!

Tc cos theta = Ta cos alpha
Tc (2/2.5) = Ta (1.5/2.5)
0.8 Tc = 0.6 Ta

Tc sin theta + Ta sin alpha = 8
Tc ( 1.5/2.5) + Ta (2/2.5) = 8
0.6 Tc + 0.8 Ta = 8

Ta = 0.8Tc/0.6

0.6Tc + 0.8(0.8Tc/0.6) = 8...solve this equation and u will get Tc = 4.8N

Ta = 0.8(4.8)/0.6
Ta = 6.4 N

(ii) i guess that's easy to solve after finding the tensions and everythin..anyway if u still need help in part 2 tell me and i am gonna help ya!
Click to expand...
thanx alot i get it :)
 
bamteck

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Yeah ! But how do you get 0.875 ?
I mean how is the frictional force calculated ?
 
leadingguy

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june 2012 variant 41 question 7 part 2 I am sorry I am nt having the link but th link has been posted above back some messeges.... u can see from there hussamh10 iKhaled
 
GorgeousEyes

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iKhaled said:
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
Click to expand...
Thanks Khalid , Appreciate it .
 
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