How was 9709 paper 4 (Mechanics 1)?

[TariqBhai]

.

 

[Ashna Jeewoonarain]

I gave variant 42. I think it was moderately easy.

Yeah same here.

 

[sid2333]

Yeah the paper was very easy this time!

 

[Anum96]

Yeah the paper was very easy this time!

What was the tension?

 

[Anum96]

Yeah the paper was very easy this time!

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.

 

[Harsheys]

For question 5, what did you guys get for the length of the string?
Could someone also give me the length of the ramp. I know the height was 0.7m

 

[sid2333]

For question 5, what did you guys get for the length of the string?
Could someone also give me the length of the ramp. I know the height was 0.7m

length of string for me came out to be 2.41m

 

[sid2333]

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.

Tell me the answers if you remember them.

 

[Harsheys]

length of string for me came out to be 2.41m

But what was the length of the ramp? I'm trying to figure it out right now

 

[sid2333]

But what was the length of the ramp? I'm trying to figure it out right now

Length of the ramp was 2.50

 

[Sarosh Jameel]

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.

What was the tension?

it was not a exact ansswer right ??? and acceleration was also not exact answer ???

 

[Sarosh Jameel]

For question 5, what did you guys get for the length of the string?
Could someone also give me the length of the ramp. I know the height was 0.7m

i got about 1.67 m

 

[sid2333]

i got about 1.67 m

Hey how did you get this?

 

[Sarosh Jameel]

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.

yes both R were different /// my coffecient of friction was about 0.571

 

[sid2333]

i got about 1.67 m

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.

 

[Sarosh Jameel]

what was the speed at C ????

 

[sid2333]

what was the speed at C ????

9.16ms^-1???

 

[Sarosh Jameel]

9.16ms^-1???

Great bro !!!

 

[Sarosh Jameel]

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.

did u calculate the distance for subsequent when string broke ??

 

[Harsheys]

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