AQA MM1B - Mechanics 1 -Tuesday 21st June 2016

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Reply 200

Peter Wilkinson

Did no one get around 07.6 for the bearing ? :|

Reply 201

Chickenslayer69

Original post by koolgurl14

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

Got the exact same but think the answer is t=12 and 106m :frown:

Reply 202

Busted838

Original post by GabbytheGreek_48

2c0 was a displacement so wouldnt you also have to put an angle if you didnt leave it as I and J vectors

i did think about that and im not 100% but usually when they ask for displacement they want the distance from the origin

Reply 203

Applicant75235

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

how r we meant to get this if we're not doing A2 :frown:

unfair really

Reply 204

Ano123

Original post by Busted838

unofficial mark scheme

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

Part C asked for displacement, not distance, and t=12 for Q8, so distance would be 106m.

Reply 205

Arima

Original post by Peter Wilkinson

Did no one get around 07.6 for the bearing ? :|

3 figure bearings are without decimal place was it'd round up to 008 :smile:

Reply 206

LiesandAlibis

Original post by Peter Wilkinson

I'm sure 2c was displacement, not distance and 5c was 1.19 (to 3sf)

I got 1.12m exactly for 5c :/

Reply 207

username2237201

Original post by georgiashayes

how long had the airplane been in the air for to travel 240 metres? think i read it wrong ffs

t=20

Reply 208

GabbytheGreek_48

Original post by Busted838

i did think about that and im not 100% but usually when they ask for displacement they want the distance from the origin

they usually say the magnitude of displacement if the want that or just distance

Reply 209

Ano123

Original post by koolgurl14

Velocities were not equal but parallel, so you cannot just equate them as I think you may have done. One was a multiple of the other, but not necessarily a multiple of 1.

Reply 210

Peter Wilkinson

Original post by Arima

3 figure bearings are without decimal place was it'd round up to 008 :smile:

Uh yeah I forgot to do that :/ 1 mark lost right?

Reply 211

GabbytheGreek_48

Original post by LiesandAlibis

I got 1.12m exactly for 5c :/

think i also got 1.12

Reply 212

OturuDansay

for question 7, was the horizontal velocity 12cos50 or something, I can't remember what the initial figures were when setting up the question. Someone help lol

Reply 213

besocansec

Original post by A Slice of Pi

Can anyone remember this 10 mark question? I'll have a go at it

I can do some if it... Hopeully others wilk fill in after.

2 helicopters A and B,
A is at origin, B is at position (maybe 50i 40j? Someone needs to help)
A has acceleration (-0.2i + 0.1j), B: (0.2i - 0.1j)
Cant remember inial velocity...
Q: What is the distance A to B when their velocities are parallel. Answer=106. If others could fill in gaps thatll help. You can either do c4 method or gradient method, t=12

Reply 214

Busted838

Original post by LiesandAlibis

I got 1.12m exactly for 5c :/

agreed

Reply 215

gcsekid

Ignore the pencil line
How are people getting 43 newtons for T

Reply 216

Applicant75235

Original post by Ano123

Velocities were not equal but parallel, so you cannot just equate them as I think you may have done. One was a multiple of the other, but not necessarily a multiple of 1.