-
Prepare by Xtremepapers!
🚀 NEW from the xtremepape.rs team: AI exam prep — 120,000+ worked solutions, and it marks your handwritten working from a photo. Sign in with your forum account & try it free → prepare.xtremepape.rs
How was 9709 paper 4 (Mechanics 1)?
- Thread starter TariqBhai
- Start date
- Messages
- 14
- Reaction score
- 10
- Points
- 13
Yeah same here.
- Messages
- 455
- Reaction score
- 7,676
- Points
- 503
What was the tension?Yeah the paper was very easy this time!
- Messages
- 455
- Reaction score
- 7,676
- Points
- 503
In q6. We had to find R and equate uR with the single horizontal force right? For first part and use the same u value for different R for frictional force in the second part right ?Yeah the paper was very easy this time!
Because I was getting some weird answers.
length of string for me came out to be 2.41m
Tell me the answers if you remember them.In q6. We had to find R and equate uR with the single horizontal force right? For first part and use the same u value for different R for frictional force in the second part right ?
Because I was getting some weird answers.
But what was the length of the ramp? I'm trying to figure it out right now
Length of the ramp was 2.50
- Messages
- 138
- Reaction score
- 31
- Points
- 38
In q6. We had to find R and equate uR with the single horizontal force right? For first part and use the same u value for different R for frictional force in the second part right ?
Because I was getting some weird answers.
it was not a exact ansswer right ??? and acceleration was also not exact answer ???
- Messages
- 138
- Reaction score
- 31
- Points
- 38
i got about 1.67 m
Hey how did you get this?
- Messages
- 138
- Reaction score
- 31
- Points
- 38
yes both R were different /// my coffecient of friction was about 0.571In q6. We had to find R and equate uR with the single horizontal force right? For first part and use the same u value for different R for frictional force in the second part right ?
Because I was getting some weird answers.
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.
- Messages
- 138
- Reaction score
- 31
- Points
- 38
what was the speed at C ????
9.16ms^-1???
- Messages
- 138
- Reaction score
- 31
- Points
- 38
Great bro !!!
- Messages
- 138
- Reaction score
- 31
- Points
- 38
did u calculate the distance for subsequent when string broke ??
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
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