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AQA MM1B - Mechanics 1 -Tuesday 21st June 2016

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Original post by LRLR1998
I got t equalled twelve too but everyone was telling me i was wrong and the answer was 5. Can you remember all the numbers in the question?


I got that too! used 3 pages of working and took 20 mins to do the whole question though!
Original post by LRLR1998
I got t equalled twelve too but everyone was telling me i was wrong and the answer was 5. Can you remember all the numbers in the question?


You aren't wrong!
q8 for people that haven't done C4 yet!!Parallel velocities means that the gradient of the magnitude of the velocities are the same, so j/i=j/i for A and B's velocities. Do some rearranging, t^2 cancels, get t=12, use this in position vector to get 106
Original post by LiesandAlibis
Quite a few people did, so we're all in the same boat here. I have a feeling it's right though!


I got that too! Such a relief! was a very challenging question!
Can someone explain the Q7 seven marker, the Q6 tension part and the Q3 angle part to me? Thanks
Yep me got t=5
Was pure chance i took it as 5
Original post by QuantumSylar
yh i tried finding horizontal and got error sqrt(1.5^2 - 9.8^2)
got error using x= sin-1(9.8/1.5) too


yeah probly cus usually the hypotenuse is bigger than than the opposite
unofficial mark scheme

1a. 0.4ms^-1 (3)
b. 15KG (3)

2a. p-4 i q+12 j (1)
b. p=14 q=-17 (4)
c. distance 17.9m (3)

3a. accel= 1.5ms^-2 (3)
b. velocty= 1.8ms^-1 (2)
c. angle =9 (4)

4a. V=12ms^1 (2)
b. resultant= 90.8ms^-1 (2)
c. bearing =008 (3)

5a. accel= 3.92 (5)
b. velcoity =2.50 (3)
c. max height = 1.12m (4)

6a. equation (2)
b. Tension= 43.8m (6)

7a. time= 1.76 (5)
b. distance =13.6m (2)
c. velocity to hit back of box =13.2 (7)

8. time=35 seconds (10)
distance =227m
Reply 188
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32908
9
Original post by AQA-Disgrace
I got that too! used 3 pages of working and took 20 mins to do the whole question though!




Yay I got this so haps
Was T 12 or 5
If so how many marks would be lost using rong value of T but doing the rest right eg modulus
I think A1 mark M2 marks would be lost
Original post by Busted838
unofficial mark scheme

1a. 0.4ms^-1 (3)
b. 15KG (3)

2a. p-4 i q+12 j (1)
b. p=14 q=-17 (4)
c. distance 17.9m (3)

3a. accel= 1.5ms^-2 (3)
b. velocty= 1.8ms^-1 (2)
c. angle =9 (4)

4a. V=12ms^1 (2)
b. resultant= 90.8ms^-1 (2)
c. bearing =008 (3)

5a. accel= 3.92 (5)
b. velcoity =2.50 (3)
c. max height = 1.12m (4)

6a. equation (2)
b. Tension= 43.8m (6)

7a. time= 1.76 (5)
b. distance =13.6m (2)
c. velocity to hit back of box =13.2 (7)

8. time=35 seconds (10)
distance =227m


I think t=12s for the last one but the rest looks fine
Reply 191
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32908
9
Original post by wil_is_he
q8 for people that haven't done C4 yet!!Parallel velocities means that the gradient of the magnitude of the velocities are the same, so j/i=j/i for A and B's velocities. Do some rearranging, t^2 cancels, get t=12, use this in position vector to get 106


This is exactly what I did. Over the moon tbh as all the other aqa a2 exams I have done have gone shocking and I was dreading mechanics but I think its gone okay so hopefully that can boost my overallgrade
Original post by QuantumSylar
yh i tried finding horizontal and got error sqrt(1.5^2 - 9.8^2)
got error using x= sin-1(9.8/1.5) too


oh i dont know though as i said i didn't know what to do but someone else doing the exam did it and got an answer
Original post by LiesandAlibis
Quite a few people did, so we're all in the same boat here. I have a feeling it's right though!


How many marks would you say getting time = 12 using that method would be?
Can anyone remember this 10 mark question? I'll have a go at it
q8)
particle a:
v=u+at for i component
v=4+0.2t
particle b:
v=u+at for i component
v=6-0.2t
these velocities will be the same because travelling parallel to each other horizontally
4+0.2t=6-0.2t
0.4t=2
t=5 then sub it into the r=ut+1/2at^2 for each particle
i can't remember any other numbers but there was 32.5 and 46.25 something like that.
not sure if my answer is right but i got 80.6 in the end
Reply 196
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Arima
16
i would've said it was a good paper if it wasn't for that last question LOL... general consensus for the last answer was 106m tho
Original post by Busted838
unofficial mark scheme

1a. 0.4ms^-1 (3)
b. 15KG (3)

2a. p-4 i q+12 j (1)
b. p=14 q=-17 (4)
c. distance 17.9m (3)

3a. accel= 1.5ms^-2 (3)
b. velocty= 1.8ms^-1 (2)
c. angle =9 (4)

4a. V=12ms^1 (2)
b. resultant= 90.8ms^-1 (2)
c. bearing =008 (3)

5a. accel= 3.92 (5)
b. velcoity =2.50 (3)
c. max height = 1.12m (4)

6a. equation (2)
b. Tension= 43.8m (6)

7a. time= 1.76 (5)
b. distance =13.6m (2)
c. velocity to hit back of box =13.2 (7)

8. time=35 seconds (10)
distance =227m


2c0 was a displacement so wouldnt you also have to put an angle if you didnt leave it as I and J vectors
Original post by Busted838
unofficial mark scheme

1a. 0.4ms^-1 (3)
b. 15KG (3)

2a. p-4 i q+12 j (1)
b. p=14 q=-17 (4)
c. distance 17.9m (3)

3a. accel= 1.5ms^-2 (3)
b. velocty= 1.8ms^-1 (2)
c. angle =9 (4)

4a. V=12ms^1 (2)
b. resultant= 90.8ms^-1 (2)
c. bearing =008 (3)

5a. accel= 3.92 (5)
b. velcoity =2.50 (3)
c. max height = 1.12m (4)

6a. equation (2)
b. Tension= 43.8m (6)

7a. time= 1.76 (5)
b. distance =13.6m (2)
c. velocity to hit back of box =13.2 (7)

8. time=35 seconds (10)
distance =227m


I'm sure 2c was displacement, not distance and 5c was 1.19 (to 3sf)
Original post by postexamtalk
How many marks would you say getting time = 12 using that method would be?


Full as long as you got the distance to be 106 m.

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