how was M1 42

[hussamh10]

guys what was the spped at the top??

 

[hussamh10]

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

How much marks was this question ????

 

[Silver Wing]

i
think the last u was 25.2 m/s and not 17.6.

the overall ppr was easy accept for the kinetic energy which was really a tricky one.How was the paper worldwide? i believe the threshold would not be more than 38 for A.

Very true, just very true...... (about the speed part only)

 

[Z.S.M]

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

how did you get the value of coefficient of friction in Q5? mine was .08 something
and work done in Q6?

 

[Z.S.M]

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

i did the same. since it was just 2 marks, we didn't have to reflect the genius of some noble prize holder.

 

[Z.S.M]

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 !

relax bro. it was just 2 marks.

 

[Z.S.M]

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?

this senior o mine had an E in stats and A's in all the other components. He had an overall A*.

 

[AhShun]

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

For B: T - 10(1-m) = a(1-m) <--------you forgot this
T - 10 +10m = 2 - 2m
T = 12 - 12m

Replace in A
8m = 12 - 12m
20m = 12
m = 0.6 kg

 

[iKhaled]

For B: T - 10(1-m) = a(1-m) <--------you forgot this
T - 10 +10m = 2 - 2m
T = 12 - 12m

Replace in A
8m = 12 - 12m
20m = 12
m = 0.6 kg

how do u do the last part of the last question ?

 

[AAAAAA]

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

You missed (1-m)a for B.
you wrote ma instead.
Answer is 4.8N.

 

[AhShun]

You missed (1-m)a for B.
you wrote ma instead.
Answer is 4.8N.

That's what I just said -_-"

 

[AhShun]

how do u do the last part of the last question ?

the acceleration = 2ms^-2

A: mg - T= ma
10m - T = 2m
T = 8m

B: T - 10(1-m) = a(1-m)
T - 10 +10m = 2 - 2m
T = 12 - 12m

Substitute,
8m = 12 - 12m
20m = 12
m = 0.6 kg

T=0.6(8)
= 4.8 N

 

[AAAAAA]

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

Max speed was not 6.4 ms-1. It was actually max velocity which was 6.4 ms-1.
The answer was around 25. (velocity was -25 something)

 

[AAAAAA]

That's what I just said -_-"

sorry I did not notice

 

[khan0000]

hey can any 1 tell me if some one having problem in calculations while method of questions are correct well examiner give marks for this case?

 

[iKhaled]

the acceleration = 2ms^-2

A: mg - T= ma
10m - T = 2m
T = 8m

B: T - 10(1-m) = a(1-m)
T - 10 +10m = 2 - 2m
T = 12 - 12m

Substitute,
8m = 12 - 12m
20m = 12
m = 0.6 kg

T=0.6(8)
= 4.8 N

lol? thats not the last part of the last question. i am talking about question 7 the last part in it....

 

[iKhaled]

hey can any 1 tell me if some one having problem in calculations while method of questions are correct well examiner give marks for this case?

u have the M1 marks which is given for the correct method used and the A1 marks for ur accuracy which means the correct answer given in the exact sig/dec point givin so u basically get marks for anythin u write in the paper related to the question!

 

[khan0000]

?

 

[khan0000]

can anyone help me out for phy practical ppr 34?

 

[khan0000]

is there electricity Q in ppr 34?