How was 9709 paper 4 (Mechanics 1)?

[sid2333]

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

Yes dude.

 

[Sarosh Jameel]

Yes dude.

what was the tension ?

 

[sid2333]

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

hey can u show me how u solved for tension and acceleration in the first place?

 

[sid2333]

what was the distance AB in the last question?

 

[Sarosh Jameel]

hey can u show me how u solved for tension and acceleration in the first place?

yes your tension and acceleration is right !! i was scared that my tension and acc is wrong ....

 

[sid2333]

yes your tension and acceleration is right !! i was scared that my tension and acc is wrong ....

no i dont think so, the other guy in this thread posted some different value of that. What was yours btw??

 

[Sarosh Jameel]

no i dont think so, the other guy in this thread posted some different value of that. What was yours btw??

same as yours !!!! dont worry our answers are 100% correct !!!

 

[Sarosh Jameel]

add me on fb ... [email protected]

 

[Sarosh Jameel]

no i dont think so, the other guy in this thread posted some different value of that. What was yours btw??

add me on fb !! [email protected]

 

[sid2333]

add me on fb ... [email protected]

Okayy..

 

[Harsheys]

same as yours !!!! dont worry our answers are 100% correct !!!

Haha I'm not wrong

Mass of A (on the ramp) was 0.5kg, mass of B was 0.1 kg
The ramp was 2.5m long and 0.7m high, so that means that sin(Theta)=7/25

Component of A down the ramp will be mgsin(theta), and hence you get 1.4N

Use newtons second law, F=ma

1.4 - T = 0.5a
T - 1 = 0.1a

Add to get rid of T,

0.4 = 0.6a, hence a = 2/3 and T = 1.06666....

 

[Sarosh Jameel]

Haha I'm not wrong

Mass of A (on the ramp) was 0.5kg, mass of B was 0.1 kg
The ramp was 2.5m long and 0.7m high, so that means that sin(Theta)=7/25

Component of A down the ramp will be mgsin(theta), and hence you get 1.4N

Use newtons second law, F=ma

1.4 - T = 0.5a
T - 1 = 0.1a

Add to get rid of T,

0.4 = 0.6a, hence a = 2/3 and T = 1.06666....

yes .. thatss right

 

[Recoil]

What will be the GT any idea?

 

[Sarosh Jameel]

What will be the GT any idea?

42 or 43 ... 35 m was from the floor in q2 ... right ??

 

[Sarosh Jameel]

hey !

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

 

[sid2333]

Haha I'm not wrong

Mass of A (on the ramp) was 0.5kg, mass of B was 0.1 kg
The ramp was 2.5m long and 0.7m high, so that means that sin(Theta)=7/25

Component of A down the ramp will be mgsin(theta), and hence you get 1.4N

Use newtons second law, F=ma

1.4 - T = 0.5a
T - 1 = 0.1a

Add to get rid of T,

0.4 = 0.6a, hence a = 2/3 and T = 1.06666....

OMG. This question is a disaster for me.
I don't know.. I'm losing that 8 marks for sure then. :'(

 

[Sarosh Jameel]

OMG. This question is a disaster for me.
I don't know.. I'm losing that 8 marks for sure then. :'(

what was ur tension and acceleration bro ??

 

[Chris1]

Is there error carried forward in question 2 of the freefall because in i) I wrote all the steps correct but i suppose i put the values incorrect on the calculator so ii) also will be wrong .i'm hoping that i'll lose 1 mark in this question only.Is it possible?

 

[Sarosh Jameel]

Is there error carried forward in question 2 because in i) I wrote all the steps correct but i suppose i put the values incorrect on the calculator so ii) also will be wrong .i'm hoping that i'll lose 1 mark in this question only.Is it possible?

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

 

[Chris1]

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

What will be the GT any idea?