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All Mechanics DOUBTS HERE!!!

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Fenomenoe9:
GUys Please this question is messing with me i dont get even what it means. PLease Help.

It's 2007 January Q7 part (f).

Appreciate the help peace..

Fenomenoe9:
here m and thanks

Fenomenoe9:

--- Quote from: Fenomenoe9 on May 16, 2010, 07:14:21 pm ---here m and thanks oh yahh the answer is 0.51s

--- End quote ---

cooldude:

--- Quote from: Fenomenoe9 on May 16, 2010, 07:14:21 pm ---here m and thanks

--- End quote ---

okay for this basically u find out the time the for P to come to rest and then double that value---.
as the string is now slack for this period there is no tension in the string so the deceleration of P will be-->
F=ma, -30sin30=3a
a=-10sin30=-5
the time taken for Q to reach the ground is-->
0.8=0.5*0.98*t^2
t=1.28 sec
therefore final speed of P=
v=0+(1.28*0.98)
v=1.24 m/s
0=1.24-5t
t=0.248 sec
this is the time P takes to come to rest therefore the time for it to become taut again will be double that=0.248*2=0.496 sec, By the way there might be some inaccuracies in the final answer as i did not use exact values

cooldude:

--- Quote from: cooldude on May 16, 2010, 07:29:14 pm ---okay for this basically u find out the time the for P to come to rest and then double that value---.
as the string is now slack for this period there is no tension in the string so the deceleration of P will be-->
F=ma, -30sin30=3a
a=-10sin30=-5
the time taken for Q to reach the ground is-->
0.8=0.5*0.98*t^2
t=1.28 sec
therefore final speed of P=
v=0+(1.28*0.98)
v=1.24 m/s
0=1.24-5t
t=0.248 sec
this is the time P takes to come to rest therefore the time for it to become taut again will be double that=0.248*2=0.496 sec, By the way there might be some inaccuracies in the final answer as i did not use exact values

take t as 4root5/7 sec and ull get the exact answer


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