MEI M1 exam discussion - Friday 1st June

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Reply 200

pip94

Original post by naman

i thought q4 was 070 degrees?

You get that angle but then take it away from 90 degrees to get the bearing from north

Reply 201

StopMotion

what was the journey, of the two boats racing, supposed to look like?

was it one line from the origin to (13,18) and another line from (3,5) to (13,18), making a kind of triangle without a base.

or whatever A and B were...

Reply 202

naman

Original post by pip94

You get that angle but then take it away from 90 degrees to get the bearing from north

reckon i'll get two out of 3 for that?

Reply 203

pip94

Original post by naman

reckon i'll get two out of 3 for that?

Probably as you only missed out the last step really.

Reply 204

supaabzz

I think I forgot about doing 800000xgx(1/18). I did 800000x(1/18) :/

Reply 205

chris_gloyne

Original post by nasira372

DO markscheme please, add whatever you remember and correct mistakes

1) 6marks true/false question

2)

3)

4) Bearing was 021

5)

6)

7) Train Question
i) 121000 - 22R =800,000*0.11 R=1500 (3)
ii) 121000-5R-T =40,000*0.11 T=109100N (4)
121000-2R-T=40,000*0.11 T=113600N (3)

8) Ship question
Distance travelled was (15 18)
Graph (3)
Displacement was 4metres

for distance travelled the original position was (3,2) wasnt it? so wasnt it (15,18) - (3,2) to give (12,16).. could be wrong but thats what i put

Reply 206

xiyangliu

Original post by pip94

You get that angle but then take it away from 90 degrees to get the bearing from north

it should be 021

Reply 207

chris_gloyne

Also for the SUVAT question, with AB=300m and AC=500m i got a=2 v=35 t=15..anyone else get this?

Reply 208

xiyangliu

Original post by chris_gloyne

for distance travelled the original position was (3,2) wasnt it? so wasnt it (15,18) - (3,2) to give (12,16).. could be wrong but thats what i put

i got
truck is 5900N
locomotive is 100300N

Reply 209

xiyangliu

for the tension in the coupling

Reply 210

xiyangliu

Original post by chris_gloyne

for distance travelled the original position was (3,2) wasnt it? so wasnt it (15,18) - (3,2) to give (12,16).. could be wrong but thats what i put

and (12, 16) should be right... what they are aasking is the distance between AB, but the distance of B from the origin so (12, 16)

Reply 211

xiyangliu

I ll do a mark scheme

Reply 212

chris_gloyne

Original post by xiyangliu

I ll do a mark scheme

alright cool :smile:

Reply 213

xiyangliu

1)
(A) FALSE
(B)True
(C)True
(D)False
2)
(i)
v=3t^2-12t+9
(ii)s=t^3-6t^2+9t-2
3)
(i)
(0,0)
(ii) (A): equilibrium (B): zero Displacement
4)
(i) using s=ut+o.5at^2
500=5x20 + 0.5a400
a = 2
(ii)v = 35
5)
(i) hope you got that right, tension, Reaction, friction and weight
(ii)25
(3)50cos30+R=10g solve for R
(IV) NOT SURE
6)
(i) 3.57 some thing like this therefore over
(ii) air resistance

SECTION A

Reply 214

suriya24

Do you have section B too?

Reply 215

xiyangliu

SECTION B

7)
(i) R =1500
(ii) (A) 5900N (B)100300N

(iii)121000-( 800000g x 1/80 + 33000) = - 0.0125 direction down the slope

(iv) beta = 0.241 (3sf)

8)
(i) Abs(AB) which use pythagoras (12,16)
(ii) differentiat r = (3,2) + (6,8)t should get (6,8) therefore constant velocity
(iii) cant remember ...
(iv) finishes the race at the same time
(v) differentiate get speed is (6,8t), sub t = 2 into it get (6, 16) pythag again get speed which is abt 17 bearing is 021
(vi) S1 - S2, horizontal displacement cancels out
Verticle you get 4t^2 - 8t... differentiate this get 8t - 8, make this to 0 get maximum point

get t = 1

sub bak in, get greatest distance is 4km