HOW WAS M1 MATHS PAPER???

[DumbIdiot]

can

can i ask how u got the value of p, i did not get 6.12

sorry, i forgot the question.

and by the way what value of alpha(angle) did you guys get?

 

[girmal5]

i got 19 point sth as the angle.
and in question 4 6.12

 

[Musab100]

thats all the answers i can remember

1. 2.5N

4. value of p = 6.12N

alpha = 19.4 degrees

distance travelled
part 1 - 8.33 m
part 2 - 117 m(TDT)
average speed - 2.33 ms^-1

value of k 6.25
steady vmax on incline plane - 2.08 ms^-1

6. workdone on B = 7.68j
greatest height = 3.12m ( 2.4 + .72)

7. lift problem - 9.4 m distance
t1- 9072N
t2 - 9000N
t3 - 8910N

force on mass due to floor
Max - 1008N
Min - 990N

how u got the last question 7 part iii answer ??? because mine was different !!!

 

[Aamal khan]

i got 19 point sth as the angle.
and in question 4 6.12

how did u guys get 6.12 in ques 4

 

[nabeelah]

i've 225 as total distance!!and 7(ii) is not the same as yours
but the others are more or less the same!

 

[DumbIdiot]

how u got the last question 7 part iii answer ??? because mine was different !!!

take the mass and lift as a single object ( 800 + 100 = 900). total weight = 9000N
during the first stage it was accelerating @ 0.08ms^-2 upwards. using f=ma, T -9000=.08(900) -> T =9072N
no acc during the second stage T=9000N
during the third stage it was acc downward @ 0.1ms^-2(acc upwards @ -0.1ms^-2). T-9000=-0.1(900) -> T=8910N

 

[DumbIdiot]

i've 225 as total distance!!and 7(ii) is not the same as yours
but the others are more or less the same!

question 7 had 4 parts right?
(i)distance (9.4)
(ii)??????
(iii) Tension during the three stages(9072,9000 and 8910)
(iv) force on mass due to floor(1008 and 990)

if so , i dont think that i included 7(ii) in my post ( i forgot the question )

 

[nabeelah]

ohh sry its 7(iv)

 

[nabeelah]

did u use graph paper?/

 

[nabeelah]

and for total dist i've done 8.33+100+117 which gives 225(not sure if its good)
7(iv) i've done T(min/max)+R(min/max)-1000=100x(0.08 or 0.1)

 

[DumbIdiot]

did u use graph paper?/

for what?

and for total dist i've done 8.33+100+117 which gives 225(not sure if its good)
7(iv) i've done T(min/max)+R(min/max)-1000=100x(0.08 or 0.1)

you must not include tension. tension acts on the lift, not on the mass.(using free body diagrams) there are only two forces acting on the mass namely its weight(W) and the reaction force(R). during first stage R-W=0.08m and during the third stage R-W=-0.1m ( or W-R=0.1m)

as for other question, TDT = 8.33 + 100 + 8.33 (117)
the second 8.33 can be found by integrating or you could just state that DT in first stage = distance travelled in third stage. this is true since it took the same time to accelerate to V from rest and to decelerate to rest from V.(acc = dec in magnitude)

 

[Saqlain Ashraf]

Ok so it looks like I lost the 8 marks on question 4 but I will probably get something for the working,but could somebody please explain how to do it?
I just hope the threshold will be around 40 or lower(more live way lower)

 

[girmal5]

Yeah, obviously the threshold Will be lower.

 

[Musab100]

how did u guys get 6.12 in ques 4

here u go !!!!

 

[Yeongson522]

are you talking about the variant 42?

nope, mine is variant 43. Anyway best of luck to you!

 

[vishruthiit]

Discussion after 24 hours only please.

Variant 42

chal sundu bhagg!!

 

[vishruthiit]

no dude

The force exerted by the floor? I got 1008 and 990 , it was probably wrong because I didn't make use of the tension

that is right!b proud most of em hve done wrong!

 

[Khan08]

for what?



you must not include tension. tension acts on the lift, not on the mass.(using free body diagrams) there are only two forces acting on the mass namely its weight(W) and the reaction force(R). during first stage R-W=0.08m and during the third stage R-W=-0.1m ( or W-R=0.1m)

as for other question, TDT = 8.33 + 100 + 8.33 (117)
the second 8.33 can be found by integrating or you could just state that DT in first stage = distance travelled in third stage. this is true since it took the same time to accelerate to V from rest and to decelerate to rest from V.(acc = dec in magnitude)

Can you tell me the equation of the last phase of travel i.e 45 to 50 seconds? Q5 part2

 

[nabeelah]

if i'm not mistaken its 9t-0.1(t^2) -200

 

[Khan08]

if i'm not mistaken its 9t-0.1(t^2) -200

If we integrate it, the distance doesn't come as 8.33. The integration constant value must be zero? How did you do it?