Mechanics M1 May 24th 2012 AQA

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

(edited 11 years ago)

Reply 63

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.

Reply 64

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.

Reply 65

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.

(edited 11 years ago)

Reply 66

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. :frown:

Reply 67

ryanboi

13

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

Reply 68

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

Reply 69

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...

Reply 70

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. :s-smilie:

Reply 71

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 :smile:

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

(edited 11 years ago)

Reply 72

Original post by 1platinum

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

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 :frown:

(edited 11 years ago)

Reply 73

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 :frown:

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

Reply 74

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 :smile:

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

Reply 75

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 :frown:

Reply 76

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 :frown:

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!! :colondollar:

Reply 77

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!! :colondollar:

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!

Reply 78