how was M1 42

[daviruss]

how its 5? look f=ma rite so the downward weight for a was 6-t=0.6x2 so u get 4.8N

u frgt mg

 

[doorman1995]

how did u get the last part of the last question ? explain pls

i did it by the motion formulas we had distance OA 216m we had final velocity 0m/s we had acceleration -0.72m/s2 so apply v2=u2=2as

 

[daviruss]

no it was not 4.44 or either 15 i guess u qouted the accelerations as speed the ques had give initial and final accelerations not speeds

dude u use the accelerations in the equations df-mgsin(theta)-fr=ma for (4) and the same for(0.2) then p=d*v u get 4.44 and 15

 

[iKhaled]

i did it by the motion formulas we had distance OA 216m we had final velocity 0m/s we had acceleration -0.72m/s2 so apply v2=u2=2as

how come???? V^2 = U^2 +2as is only applied to CONSTANT acceleration question and this isn't a constant acceleration question :S:S

 

[iKhaled]

u frgt mg

explain how it is 5 pls? what do u mean by mg? Oo

 

[daviruss]

how come???? V^2 = U^2 +2as is only applied to CONSTANT acceleration question and this isn't a constant acceleration question :S:S

dude how much did u get ur speed i think i got it wrong after wt doorman said but dnt rlly knw and i thnk he is mistaken

 

[doorman1995]

how did u get that speed then if t = 0 so the velocity should be equal to zero according to the equation given ?

no not necessarily the ques it self said find velocity of p at 0 so it means there was a value surleyfor which it passed O it cant pass O with om/s

 

[doorman1995]

dude how much did u get ur speed i think i got it wrong after wt doorman said but dnt rlly knw and i thnk he is mistaken

well whts ur answer?

 

[daviruss]

explain how it is 5 pls? what do u mean by mg? Oo

dude ForA: mg-t=ma 10m-t=2m.... t=8m first equation
ForB: t-10(1-m)=ma t-10+10m=2m.... t=10-8m second equation ..... so now T=T u get m=0.625 then substitute m in first eq 8(0.625)= 5N

 

[iKhaled]

dude ForA: mg-t=ma 10m-t=2m.... t=8m first equation
ForB: t-10(1-m)=ma t-10+10m=2m.... t=10-8m second equation ..... so now T=T u get m=0.625 then substitute m in first eq 8(0.625)= 5N

but how did u get that mass either..cuz i got same values as doorman for this question ? so i wonder how u got these answers ? and dude how many marks i would lose for the kinetic question if i wrote a wrong value for the second velocty by using a= 0.4 instead of a= 0.2 ?

 

[iKhaled]

no not necessarily the ques it self said find velocity of p at 0 so it means there was a value surleyfor which it passed O it cant pass O with om/s

i wanna know how u used a constant acceleration formula for a question that has non constant acceleration formula?!

 

[doorman1995]

HOW?\

u r making me sad bro cuz that kinetic energy question is easy but i did a simple mistake that would take marks off me..i had one velocty 4.44 and the other one 13.1 instead of 15 for using the acceleration 0.4 instead of 0.2 arghhh

WELL GUYS FOR THIS QUES WE WERE GIVEN ACCELERATIONS AT TOP AND BOTTOM RITE FOR EG LETS TAKE ONE SPEED THE ONE AT BOTTOM WE HAD TO CALCULATE ITFIRST
LOOK WE APPLIED F=MA RITE BY WHICH F-1000(RESISTIVE FORCE)=1250x4 F=6000 THAT IS THE DRIVING FORCE NOW FORMULA POWER=FORCEXVELOCITY 30000W/6000=5M/S THAT IS THE VELOCITY AT BOTTTOM THEN FOR THE SAME AT THE TOP BUT THIS TIME USE A AS 0.2 THEN SOLVE IT

 

[doorman1995]

i wanna know how u used a constant acceleration formula for a question that has non constant acceleration formula?!

WELL I USED IT CS OF EXAM TRICK AS USUALY PARTS OF QUES ARE LINKED SO I LOOKED AT WE HAVE THE DISTANCE THE ACCELERATION THE FINAL VELOCITY SO Y NOT USE A FORMULA ALSO THAT SPEED WAS NON CONSTANT NOT THE ACCELERATION

 

[doorman1995]

HOW?\

WELL GUYS FOR THIS QUES WE WERE GIVEN ACCELERATIONS AT TOP AND BOTTOM RITE FOR EG LETS TAKE ONE SPEED THE ONE AT BOTTOM WE HAD TO CALCULATE ITFIRST
LOOK WE APPLIED F=MA RITE BY WHICH F-1000(RESISTIVE FORCE)=1250x4 F=6000 THAT IS THE DRIVING FORCE NOW FORMULA POWER=FORCEXVELOCITY 30000W/6000=5M/S THAT IS THE VELOCITY AT BOTTTOM THEN FOR THE SAME AT THE TOP BUT THIS TIME USE A AS 0.2 THEN SOLVE IT

IT was not 0.4 it was 4 at bottom man

 

[daviruss]

d

HOW?\

WELL GUYS FOR THIS QUES WE WERE GIVEN ACCELERATIONS AT TOP AND BOTTOM RITE FOR EG LETS TAKE ONE SPEED THE ONE AT BOTTOM WE HAD TO CALCULATE ITFIRST
LOOK WE APPLIED F=MA RITE BY WHICH F-1000(RESISTIVE FORCE)=1250x4 F=6000 THAT IS THE DRIVING FORCE NOW FORMULA POWER=FORCEXVELOCITY 30000W/6000=5M/S THAT IS THE VELOCITY AT BOTTTOM THEN FOR THE SAME AT THE TOP BUT THIS TIME USE A AS 0.2 THEN SOLVE IT

did u calculate mgsin(theta) the eq DF-mgsin(theta)-fr=ma thats it and u apply and why r u mad chill dude we r just talking here

 

[daviruss]

but how did u get that mass either..cuz i got same values as doorman for this question ? so i wonder how u got these answers ? and dude how many marks i would lose for the kinetic question if i wrote a wrong value for the second velocty by using a= 0.4 instead of a= 0.2 ?

u already got a=2 right then u apply normally dude like any pully equation and u get m nt tht hard gl and i dnt think u will lose alot ofc the examiner will consider tht

 

[Amy Bloom]

Hey guys i just want to know one thing. To get A* or A in Maths, what are the minimum grades required for P1, P3, M1 and S1?

 

[leadingguy]

dude no way 25.2 for he u th ,ax speedi s 6.4

I disagree a little as in the 3rd or frth part he has asked to calculate th acceleration remember... ND It WAS IN NEGATIVE MEANS THE CAR WAS SLOWING DOWEN AND STOPPED CERTAINLY. HOPE m right.

it is 5N for the tension

mine tension was 4.8 N too. my U was 17.something in the last part too.

 

[iKhaled]

man the last part is killing me cuz..cant stop thinking of it. i think there is somethin with the word subsequent cuz how come we use a formula for constant acceleration to find u and the motion was in non-constant acceleration. it is really killing me !

 

[hussamh10]

1. 1090J
2. a=2m/s2 m=0.6kg T=4.8N
3. 10m/s a=0.4m/s2
4. 39N direction 22.6' AC from x direction
5. 0=30' coefficient=0.693
6. gain in KE=344000J driving force workdone= 1220000J
7. distance OA=216m a=-0.72m/s2 u=17.6m/s

Yipeeeeeee