How was 9709 paper 4 (Mechanics 1)?

[Chris1]

Oh great thanks!

main marks are for main logic applied to the question !!! yes it is possible

 

[Sarosh Jameel]

I guess it will be 40 maybe 39 but probably 40..

yes .. i got a question !!!

 

[epic123456]

What answer did you guys get for finding V in question 4?? please tell me

 

[Sarosh Jameel]

What answer did you guys get for finding V in question 4?? please tell me

it was Q 2 !!! in which V and H was to be foound

 

[epic123456]

not that one..the question on particle P something like that ( the question about plane A,b..)

 

[Sarosh Jameel]

not that one..the question on particle P something like that ( the question about plane A,b..)

V was 0.966

 

[Sarosh Jameel]

Yeah

in Q2... what was V and H ??

 

[Ziki629]

2)V was 30 something
H was 60 Something
3)t was 60 and 40 or something and S was 4hundred something
4)V was something
5) A was 2/3 T was 1.07 and length of string was 1.9something
6) mew was 0.56something ii a was something
7) Were all something
Oh and 1) F was twentysomething and G was 56 ?

 

[Sarosh Jameel]

2)V was 30 something
H was 60 Something
3)t was 60 and 40 or something and S was 4hundred something
4)V was something
5) A was 2/3 T was 1.07 and length of string was 1.9something
6) mew was 0.56something ii a was something
7) Were all something
Oh and 1) F was twentysomething and G was 56 ?

in q2 v was 40 and h was 80 !!

 

[Haris.I]

Guys what was the answer to the very last question... it was something related to distance

 

[Sarosh Jameel]

Guys what was the answer to the very last question... it was something related to distance

i dont remember the answer .... but it was ... KE AT B = KE AT A + WDr

 

[Sarosh Jameel]

Guys what was the answer to the very last question... it was something related to distance

WDr = AB * resistance

 

[Chris1]

Displacment in question 3 was 440 right?

 

[Syed Mirak]

my answers
Q1> G=56
Q2> v=40m/s h=80m
Q6>co effiecient of friction = 0.571
Q?>length of string 1.42
t=1.066 a=0.6667
a=o at time t=40 and t=60 distance =440


ANYONE WITH SIMILAR ANSWERS?

 

[Sarosh Jameel]

my answers
Q1> G=56
Q2> v=40m/s h=80m
Q6>co effiecient of friction = 0.571
Q?>length of string 1.42
t=1.066 a=0.6667
a=o at time t=40 and t=60 distance =440


ANYONE WITH SIMILAR ANSWERS?

yeah !! all same but my length of string was about 1.66

 

[Sarosh Jameel]

my answers
Q1> G=56
Q2> v=40m/s h=80m
Q6>co effiecient of friction = 0.571
Q?>length of string 1.42
t=1.066 a=0.6667
a=o at time t=40 and t=60 distance =440


ANYONE WITH SIMILAR ANSWERS?

what was ur at C ??? acceleration of ring ?? . gain in KE ?? and AB distance ??

 

[Syed Mirak]

what was ur at C ??? acceleration of ring ?? . gain in KE ?? and AB distance ??

I don't remember my change in k.e in q7 but my final speed was around 22 and initial was 13 or something

 

[Nushad Nahue]

i first found the speed of B when it reached which came out to be 1.871. (square root of 3.5) then used v^2=u^2+2as to find the distance P travelled and then subtracted that distance from 2.5 to get the length of string.

The acceleration when it was attached to the

I just solved it and I'm getting 1.9524m for my length of string.

T=1.0666...
a=2/3 (for 0.7m), then 2.8 (after the string breaks)

v (when string breaks) = *root* (2 x 2/3 x 0.7) = 0.96609
then it increases to 2
So
2^2 = 0.96609^2 + 2(2.8)s
s = 0.5476

So the length of string will be 2.5-0.5476 = 1.9523

Um I got 1.94 , maybe because I rounded up some values so will I still get marks?

 

[Nushad Nahue]

in q2 v was 40 and h was 80 !!

yeaah. I got the same.

 

[sid2333]

what was distance AB in Q7?