[Mindcoil]

Did people get pretty nice numbers when working out the time in the longer questions, like t=20, t=40 etc.?

I cannot see the grade boundaries being any lower than 62 at a very minimum, which of course is a shame.

Any predictions as to the grade boundaries?

[nmudz_009]

(Original post by AntyCole)
just average 90+ in c3 and c4

Thank God!!!! u sure yea? cos I know for a fact I got much less than 90 in m1. Not that I'm confident about getting 90+ in c4, but its good to know it's still possible to get an A*

[Oromis263]

Ok, well I felt that this was an alright paper, had some nice questions, kept changing my assumptions on 5c) though! Quote me and I'll edit answers into here to make a unofficial mark scheme.

Here are answers I remember:
Q1. Boat travelling in water
a. 5.39m/s
b. 338 degrees

Q2. Momentum
Mass of B = 0.5kg

Q3. About the car (I think?)
ai. -2m/s^2
aii. 10s
aiii. 2800N
b. 2600N

Q4. About the mass held in equilibrium
a. theta = 60 degrees
b. W = 17.3N
c. mass = 1.77kg

Q5. Connected particles
a. 5.88m/s^2
bi. T = 122N
bii. R = 118
biii. mu = 0.735
c. Assumptions (that I've heard people give):
There is no air resistance.
The block doesn't collide with the peg, or the particle with the floor
The block had a uniform mass
g = 9.8m/s^2 (not so sure about this one, as it's given on the front page, but it's possible)

Q6. Towing a sled
a. Diagram of forces acting on sled
b. NR = 78.4 - 0.5T
c. 23.5

Q7. Vectors/coordinate working of a particle in motion
a. s = (-i+3j)+0.5(0.1i-0.2j)t^2
b. Time = 30s
c. Speed = 1.41m/s

Q8. Projectile Motion of a particle
a. Show sinx = 0.875
b. 19.6m
c. 43.4m
d. 3.45s
e. 10.8m/s

[x_Raman_96]

I'm not sure what some of these answers are for, but this is what I got: 338, 0.5, 1.77, 5.88, 0.73, 23.5, 3.45, 12.6. those are the answers i stored on my 1st calculator.

2nd calculator: 5.39, -2, 3000, 122.4, 30, 43.4.

That's all I stored.

[x_Raman_96]

(Original post by Oromis263)
Ok, well I felt that this was an alright paper, had some nice questions, kept changing my assumptions on 5c) though! Quote me and I'll edit answers into here to make a unofficial mark scheme.

Here are answers I remember:
Q1. Boat travelling in water
a. 5.39m/s
b. 338 degrees

Q2. Momentum
Mass of B = 0.5kg

Q3. About the car (I think?)
ai. -2m/s^2
aii. 10s
aiii. 2800N
b. 2600N

Q4. About the mass held in equilibrium
a. theta = 60 degrees
b. W = 17.3N
c. mass = 1.77kg

Q5. Connected particles
a. 5.88m/s^2
bi. T = 122N
bii. R = 118
biii. mu = 0.735
c. Assumptions (that I've heard people give):
There is no air resistance.
The block doesn't collide with the peg, or the particle with the floor
The block had a uniform mass
g = 9.8m/s^2 (not so sure about this one, as it's given on the front page, but it's possible)

Q6. Towing a sled
a. Diagram of forces acting on sled
b. NR = 78.4 - 0.5T
c. 20.8

Q7. Vectors/coordinate working of a particle in motion
a. s = (-i+3j)+0.5(0.1i-0.2j)t^2
b. Time = 30s
c. Speed = 1.41m/s

Q8. Projectile Motion of a particle
a. Show sinx = 0.875
b. 19.6m
c. 43.4m
d. 3.45s
e. 10.8m/s

I'm not so sure about your: 7c, 3b and 6b: I am sure there is supposed to be a cos or sin in q 6b, if that was give R in a expression using t.

[Oromis263]

(Original post by x_Raman_96)
I'm not so sure about your: 7c, 3b and 6b: I am sure there is supposed to be a cos or sin in q 6b, if that was give R in a expression using t.

6b doesn't have to involve the sin30, as that is simply 0.5. They're the same answer, so should both get the marks.

[mathsguy]

Hey,
On the projectiles question i proved that sintheta = 0.875 but then followed through into all of my equations with sin0.875. Will i lose all the marks to b,c,d and e or will i pick up follow through marks. Hope i at least get some.

[ryanboi]

(Original post by Oromis263)
Ok, well I felt that this was an alright paper, had some nice questions, kept changing my assumptions on 5c) though! Quote me and I'll edit answers into here to make a unofficial mark scheme.

Here are answers I remember:
Q1. Boat travelling in water
a. 5.39m/s
b. 338 degrees

Q2. Momentum
Mass of B = 0.5kg

Q3. About the car (I think?)
ai. -2m/s^2
aii. 10s
aiii. 2800N
b. 2600N

Q4. About the mass held in equilibrium
a. theta = 60 degrees
b. W = 17.3N
c. mass = 1.77kg

Q5. Connected particles
a. 5.88m/s^2
bi. T = 122N
bii. R = 118
biii. mu = 0.735
c. Assumptions (that I've heard people give):
There is no air resistance.
The block doesn't collide with the peg, or the particle with the floor
The block had a uniform mass
g = 9.8m/s^2 (not so sure about this one, as it's given on the front page, but it's possible)

Q6. Towing a sled
a. Diagram of forces acting on sled
b. NR = 78.4 - 0.5T
c. 23.5

Q7. Vectors/coordinate working of a particle in motion
a. s = (-i+3j)+0.5(0.1i-0.2j)t^2
b. Time = 30s
c. Speed = 1.41m/s

Q8. Projectile Motion of a particle
a. Show sinx = 0.875
b. 19.6m
c. 43.4m
d. 3.45s
e. 10.8m/s

can you show me your working for question 5a ? Thank you. Cuz I got 1.96 :s

[x_Raman_96]

(Original post by Oromis263)
6b doesn't have to involve the sin30, as that is simply 0.5. They're the same answer, so should both get the marks.

Ahh, I see sorry for my misconception. For 3b I got 3000 for 7c I got something like 5. something.

I'm pretty sure you get 40 seconds , by multiplying the j component by -1 to get south. then equate both i and j to get south east t = 40 and insert in to v = u + at and do Pythagoras to get 5. something

[shannonmcnamara]

(Original post by Oromis263)
Ok, well I felt that this was an alright paper, had some nice questions, kept changing my assumptions on 5c) though! Quote me and I'll edit answers into here to make a unofficial mark scheme.

Here are answers I remember:
Q1. Boat travelling in water
a. 5.39m/s
b. 338 degrees

Q2. Momentum
Mass of B = 0.5kg

Q3. About the car (I think?)
ai. -2m/s^2
aii. 10s
aiii. 2800N
b. 2600N

Q4. About the mass held in equilibrium
a. theta = 60 degrees
b. W = 17.3N
c. mass = 1.77kg

Q5. Connected particles
a. 5.88m/s^2
bi. T = 122N
bii. R = 118
biii. mu = 0.735
c. Assumptions (that I've heard people give):
There is no air resistance.
The block doesn't collide with the peg, or the particle with the floor
The block had a uniform mass
g = 9.8m/s^2 (not so sure about this one, as it's given on the front page, but it's possible)

Q6. Towing a sled
a. Diagram of forces acting on sled
b. NR = 78.4 - 0.5T
c. 23.5

Q7. Vectors/coordinate working of a particle in motion
a. s = (-i+3j)+0.5(0.1i-0.2j)t^2
b. Time = 30s
c. Speed = 1.41m/s

Q8. Projectile Motion of a particle
a. Show sinx = 0.875
b. 19.6m
c. 43.4m
d. 3.45s
e. 10.8m/s

How did you's get Q3 aiii) and b) ...
think I answered differently but can't really remember...

[1platinum]

(Original post by x_Raman_96)
Ahh, I see sorry for my misconception. For 3b I got 3000 for 7c I got something like 5. something.

I'm pretty sure you get 40 seconds , by multiplying the j component by -1 to get south. then equate both i and j to get south east t = 40 and insert in to v = u + at and do Pythagoras to get 5. something

That's what I got, but apparently there is another method and people have ended up with 1.41 instead.

[Oromis263]

(Original post by ryanboi)
can you show me your working for question 5a ? Thank you. Cuz I got 1.96 :s

Ok, 5a.

For the block, T = 12a (surface is smooth, no gravity parallel to direction of acceleration)
For the particle, 18g - T = 18a

Simultaneous equations from there to the answer of 5.88m/s^2

(Original post by shannonmcnamara)
How did you's get Q3 aiii) and b) ...
think I answered differently but can't really remember...

a = -2 m = 1400
Therefore, using F=ma, braking force = -2800N, therefore magnitude of braking force = 2800N

For the next part, you do F - 200 = ma, same results for ma, so F = 200 + -2800 = -2600N, therefore magnitude of braking force now = 2600N

[Oromis263]

(Original post by 1platinum)
That's what I got, but apparently there is another method and people have ended up with 1.41 instead.

Did you use the equation you wrote out before for the position vector, using s = ut + .5at^2, or did you rerun it with v = u + at? I double checked that one, maybe my brain just assumed I'd got the numbers right when it worked again D: nooo

[1platinum]

(Original post by Oromis263)
Did you use the equation you wrote out before for the position vector, using s = ut + .5at^2, or did you rerun it with v = u + at? I double checked that one, maybe my brain just assumed I'd got the numbers right when it worked again D: nooo

From part (a) I made i= -j to find the time. Then did v = u + at to find the velocity. And then sq. root.

[karchun]

(Original post by Oromis263)
Ok, 5a.

For the block, T = 12a (surface is smooth, no gravity parallel to direction of acceleration)
For the particle, T - 18g = 18a

Simultaneous equations from there to the answer of 5.88m/s^2



a = -2 m = 1400
Therefore, using F=ma, braking force = -2800N, therefore magnitude of braking force = 2800N

For the next part, you do F - 200 = ma, same results for ma, so F = 200 + -2800 = -2600N, therefore magnitude of braking force now = 2600N

I don't think you needed to use simultaneous equation cause you could figure it out with one equation but I did get that answer as well I think

This was posted from The Student Room's Android App on my GT-N7000

[Oromis263]

(Original post by 1platinum)
From part (a) I made i= -j to find the time. Then did v = u + at to find the velocity. And then sq. root.

I'm probably incorrect then. Your method finds when the particle is located south-east of the origin, rather than when it's velocity is in a south-east direction. However, it did seem rather ambiguously worded, I swear it asked for speed when it's motion was in that direction, rather than speed when it was located there. However, a lot of people have done either one or the other, so they may give marks for either method. Argh

[1platinum]

(Original post by Oromis263)
I'm probably incorrect then. Your method finds when the particle is located south-east of the origin, rather than when it's velocity is in a south-east direction. However, it did seem rather ambiguously worded, I swear it asked for speed when it's motion was in that direction, rather than speed when it was located there. However, a lot of people have done either one or the other, so they may give marks for either method. Argh

Yes either way will be credited most of the method marks I reckon. Just need someone to upload the paper so we can see! This paper today is distracting me from other revision!!

[Oromis263]

(Original post by 1platinum)
Yes either way will be credited most of the method marks I reckon. Just need someone to upload the paper so we can see! This paper today is distracting me from other revision!!

Same, I'm acting as if I don't still have 7 exams.. :P and M1 was probably my easiest out of all 8, don't want to be stressing about it now, what's done is done!

[1platinum]

(Original post by Oromis263)
Same, I'm acting as if I don't still have 7 exams.. :P and M1 was probably my easiest out of all 8, don't want to be stressing about it now, what's done is done!

If you include that question, I think I have only lost 8 marks max, which is ok. After all C3 and C4 determine the A*! Are you feeling prepared for them?

[Katee9528]

Hahaa well thats awkward that I got the mass answer wrong.. I got something like 10kg xD ahh well this shows I'll be doing well in my physics tomorrow (Y)... :|