How was Mechanics Paper 42!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[Jigar]

it wasnt like that
find time using v=u+at
acceleration is 10*sin@ = 10*16/65 = 2.46
8=0+ 2.46*T
T=3.252
T/2= 1.626
for finding distance apply s=ut + 0.5at^2
s=0(t) +0.5*2.46*1.626^2
s=3.25/
please correct me if i am wrong

I agree......well what was ur work done against resistance

 

[talha196]

ok coming towards question six which part of the graph did you shade..... i did the last part from t=20 to 26

 

[talha196]

I agree......well what was ur work done against resistance

750 JOULES

 

[talha196]

in question six last part i did find the accelration and applied s=ut+1/2 at^2
but didnt knew what to do next what do you think how much marks will i get out of 6 ?

 

[Jigar]

750 JOULES

and how do u show the second part

 

[Jigar]

in question six last part i did find the accelration and applied s=ut+1/2 at^2
but didnt knew what to do next what do you think how much marks will i get out of 6 ?

same here

 

[saadgujjar]

What was speed in q4

 

[talha196]

and how do u show the second part

the gain in potential energy and work done against resistance was equal to 1150 so change in kinetic energy is zero and speed at B was equal to speed at A...
PLEASE CORRECT ME IF I AM WRONG

 

[Jigar]

what was the constant steady speed in q5

 

[talha196]

what was the constant steady speed in q5

33.33

 

[saadgujjar]

Plz tel me statmnt of q4

 

[Your-Blood]

the gain in potential energy and work done against resistance was equal to 1150 so change in kinetic energy is zero and speed at B was equal to speed at A...
PLEASE CORRECT ME IF I AM WRONG

It's fine I did the same.

in question six last part i did find the accelration and applied s=ut+1/2 at^2
but didn't knew what to do next what do you think how much marks will i get out of 6 ?

I think at least 1 mark would be given for calculating correct acc.

 

[Your-Blood]

Plz tel me statmnt of q4

Part a was to calculate speed at S/2

Part b was to calculate distance when T/2

 

[HunzaZ]

gt should be around 42

 

[Your-Blood]

33.33

Should be 33.3 not 33.33 ( 1 mark would be deducted)

gt should be around 42

Not at all not so high
According to me it would be like 39,40

 

[WayneRooney10]

Nah. The gt is gonna be around 36-38. Not more than that.

 

[FRENZYAMU]

in question 7 answer was 0.47 m
in q 4 last part speed was 2.65 and in first paprt distanc was 13 .tension in pb was 13..P was 12KW and steady speed was 13.3
w work done against resistance was 750 and the question we have to show kinetic energy
plz tell me other questions,i forgot them

my power came out to be 20 kW i think

 

[talha196]

It's fine I did the same.




I think at least 1 mark would be given for calculating correct acc.

i also found area under the graph till t=20
and wat about 2nd part of q6 which part did you shade?

 

[0Louis0]

Part a was to calculate speed at S/2

Part b was to calculate distance when T/2

No one part asked for S, and another for V and S/2
I'm sure of that but not sure if there was a iii)

(shading from 20t > 26t)
same here

same here as well but is it correct?

Should be 33.3 not 33.33 ( 1 mark would be deducted)

if his number rounds to 33.3 I don't think they're gonna deduct.. that's O.L stuff

 

[mania _ manal]

Did we hav to show that the energy ie ke=o or just show that the speed is same ??