AQA Mechanics 1B 24 May 2012 Unofficial Markscheme

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ALSO I THINK I MADE A MISTAKE on the last question

where you needed to use quadratic formula

i used 22.4sin61 but i think i messed up

how exactly do you operate the quad formula

Reply 121

Freddy12345

v=? u=22.4sin61 s=5 a=-4.9 t=?

5=ut +1/2 (at^2)

so using quad formula

Reply 122

lollizard123

for the last question I got this quadratic

4.9t2-19.6t+5<0

because sin0= 0.875 the vertical initial speed= 22.4 x 0.875 =19.6

(edited 11 years ago)

Reply 123

MSI_10

Original post by elanor95

did anyone get 1.96 for anything?!

The pulley q? Yup.

Equation of the particle: 18g-T=18a
Equation for block: T-12g=12a therefore T=12g+12a

Sub 2nd equation into one

18g-(12g+12a)=18a
18g-12g-12a=18a
6g=30a

a=6x9.8 / 30 = 1.96

Reply 124

Freddy12345

Original post by lollizard123

for the last question I got this quadratic

4.9t2-19.6t+5<0

because sin0= 0.875 the vertical initial speed= 22.4 x 0.875 =19.6

ARE YOU not suppose to.... 22.4sin61?

Reply 125

MSI_10

Original post by Freddy12345

ARE YOU not suppose to.... 22.4sin61?

Haha, thats the same as 19.6 :biggrin:

Reply 126

Butdidudie

Original post by MSI_10

The pulley q? Yup.

Equation of the particle: 18g-T=18a
Equation for block: T-12g=12a therefore T=12g+12a

Sub 2nd equation into one

18g-(12g+12a)=18a
18g-12g-12a=18a
6g=30a

a=6x9.8 / 30 = 1.96

Exactley what I put, but is it right?

Reply 127

Freddy12345

Original post by Butdidudie

Exactley what I put, but is it right?

Reply 128

MSI_10

Original post by Freddy12345

Elaborate?

Reply 129

jph12

Original post by MSI_10

The pulley q? Yup.

Equation of the particle: 18g-T=18a
Equation for block: T-12g=12a therefore T=12g+12a

Sub 2nd equation into one

18g-(12g+12a)=18a
18g-12g-12a=18a
6g=30a

a=6x9.8 / 30 = 1.96

Hmm, for the block, the acceleration is horizontal, so you don't include the block's weight in the forces. Weight doesn't have a horizontal component in this case.

Reply 130

karchun

Original post by Freddy12345

ARE YOU not suppose to.... 22.4sin61?

What are we talking about... the very last Q's about minimum speed: for that one I got 10.8/9 "22.4cos61..."

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

Reply 131

Oromis263

DISCLAIMER: I do not assume any of these to be correct, they're just what I got! :smile:

(I added a few other suggestions for 5c). If you have any amendments, just quote this post with a reasoning. :smile:

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 undergoing a braking force
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
The block acts as a particle
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. 23.5N

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 132

Freddy12345

Original post by MSI_10

Haha, thats the same as 19.6 :biggrin:

so now look.

what this the answer:

-22.4sin61 +or- squareroot of (22.4sin61^2)-4(-4.9x5^2)/2

= 383.8- 490 but that would be negative and cannot squareroot.

Reply 133

MSI_10

Original post by jph12

Hmm, for the block, the acceleration is horizontal, so you don't include the block's weight in the forces. Weight doesn't have a horizontal component in this case.

...

Annnnnnnnd there goes my 75/75 :ashamed2:

Reply 134

lollizard123

what value did anyone get for the coefficent of friction between the block and the surface?
i got 0.7

Reply 135

karchun

Can someone put up all the Q's (not answers) and the amount of marks for each part... I think I can figure out what I got - I definitely know at least 10 marks down the drain :mwuaha:

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

Reply 136

MSI_10

Btw I put no air resistance and the string was TAUT...

Was that alright? :s-smilie:

Reply 137

louisjevans

Original post by Oromis263

DISCLAIMER: I do not assume any of these to be correct, they're just what I got! :smile:

(I added a few other suggestions for 5c). If you have any amendments, just quote this post with a reasoning. :smile:

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 undergoing a braking force
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

6 c may be 23.5 but i have no proof :redface:

No WAIT!

R=8g - Tsin30

F=ma
Tcos30 - 0.3 [8g - Tsin30] = 8 * 0.05
Tcos30 - 23.52 + 0.3Tsin30 = 0.4
Tcos30 + 0.15T = 23.92
T (cos30 + 0.15) = 23.92
T = 23.92 / (cos30 + 0.15)
T = 23.5427...
T = 23.5 (3sf)
:tongue:

(edited 11 years ago)

Reply 138

MSI_10

Original post by Freddy12345

so now look.

what this the answer:

-22.4sin61 +or- squareroot of (22.4sin61^2)-4(-4.9x5^2)/2

= 383.8- 490 but that would be negative and cannot squareroot.

It was a quadratic equation

4.9t^2 -22.4sint-5=0
(maybe +5, cant remember)
T was a very small number and T was 3.69 (or summthin.)

Reply 139

Freddy12345

why did i not get the correct value of the coeffecient can someone please show me, also if i used the wrong tension value for my coefficient do i still get marks for working?