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Mechanics M1: Post your doubt here

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this is what i want to ask that why is tension in B acting downwards???? as they bth are attached to a single rod and the tension is a cting upwards in A so will in B ?????
dude try to use ur imagination in this..u have 2 boxes and between them is a string pulling each box. so how come the tension in box b will act upwards? Oo it should act downwards and the tension in A should act upwards..u need to use ur imagination in this one ! not gettin it yet? :S
 
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Hey Guys... If i get 65 in my P1 Paper.. and abt 30 in M1. Will I still be able to get an A grade??:confused:
i guess u will get the A but this depends on the threshold btw i have a question too if i get 45 in p1 and 45 in m1 will iget a B
 
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Hey!! A big thanks to everyone here for their help motivation, and good luck.
My best wishes for all the candidates giving exams!!
Good luck!!
 
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i guess u will get the A but this depends on the threshold btw i have a question too if i get 45 in p1 and 45 in m1 will iget a B
I guess so.. then again, it depends on the grade threshold.. usually 'a' grade is 58-63 in P1 and 37-43 in M1 AND 'b' grade is 50-55 in P1 and 30-37 in M1.
I'm sorry I couldn't help but nobody knows what the grade threshold will be...
But still GOOD LUCK! :D
 
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Can someone help me with june 2012 p42 Q 7?

so -F = ma
-0.12 = 0.15a
a= -0.8 m/s^2
v=u+at
v=3+(-0.8x2)= 1.4 m/s (velocity it arrives at Y)
at Y k.E = 0.5x0.15x1.4x1.4 = 0.147 J
it loses 0.072 J so k.E at Z = 0.147 - 0.072 = 0.075 J
0.5x0.15x(v^2) = 0.075
v = 1 m/s
since it is in the opp direction velocity becomes -1 m/s
at Z v= o
using v=u+at
0 = -1 + 0.8t
t = 1.25 s
it already took 2s to reach y
therefore total time taken = 2+ 1.25 = 3.25 m/s
 
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can someone explain to me why the constant value in q7 part 2 for the first integration is 5? y the time is zero and not 0.5!! :S:S:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_4.pdf
He y man its an equation for acceleration so when the time for the equation starts the particle has velocity 5
speed at A=5
so
v=10t-0.3t^2+C (we get the equation by integerating)
When t=0
V=5
soC=5

It is not 0.5 because the equation is applicable only after A as u read the question in every such question the equation is applicable only after a point as the question specifies if iam not clear plzz ask again:)
 
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