All Mechanics DOUBTS HERE!!!

[halosh92]

edexcel
jan 2005 Q5b

[cooldude]

edexcel
jan 2005 Q5b

cud u please post the link of the paper as i'm not familiar with edexcel so i might end up solving the wrong question.

[halosh92]

cud u please post the link of the paper as i'm not familiar with edexcel so i might end up solving the wrong question.

[cooldude]

a=3.2
for the 0.8 kg mass---->
8-T=0.8a
8-T=0.8*3.2
T=5.44 N

[melony]

Insert Quote
edexcel
jan 2005 Q5b

There you go!

[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]

here m and thanks oh yahh the answer is 0.51s

[cooldude]

here m and thanks

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]

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

[Fenomenoe9]

ok but why does the time taken for P to rest double?

[astarmathsandphysics]

cos it has to travel the same distance with the same acceleration - motion in reverse

[cooldude]

ok but why does the time taken for P to rest double?

cuz c wen P reaches rest, Q is still incapable of movement as the string is still slack, but then when P goes back down the plane and when it reaches the point where it initially was i.e. 0.51 sec previously, the string becomes taut again and Q rises again, we double the time because we consider both the period when P travels up the plane and back down. P moves down the plane because of its weight, there is no other force acting on P as the tension in the string is non-existent for the duration of the time the string is slack, i.e. the interval of 0.51 sec, hope u get it.

[cooldude]

cos it has to travel the same distance with the same acceleration - motion in reverse

sorry sir, didnt notice you'd already answered   

[Fenomenoe9]

haha i love u man thanks alot for helping and yeah i get it man, thanks again! bye