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M1 Edexcel MEchanics

username22

why would you only get two marks? the only thing missing is multiplying by 2.

Out of the six marks, there will probably be at least two accuracy marks for getting the right answer and one method mark for realising that it comes back down to its vertical displacement when t=1.5 so if you put o.43, you are likely to get a maximum of 3 marks, I would think its more like 2, as if they asked you a standart free fall under gravity question, it would probably be two marks, so they would probably give you 2 marks for doing that bit right.

Reply 81

fblade

14

Here are mine:

1. 35; 32.9
2. 3; 0.5
3. 2; 0.35
4. 11; -1.75
5. 1.68; 0.547
6. 3.5; can’t remember the rest
7. q = 6, p = 2; lambda = -2

I found it pretty easy took about 40-50mins... me of my asnwers dont mashc up to others :frown:

.

btw for question 7a you can check if yours is right because the very last question is to show B and C have same speeds :biggrin:

Reply 82

sng

1

fblade

Here are mine:

1. 35; 32.9
2. 3; 0.5
3. 2; 0.35
4. 11; -1.755. 1.68; 0.547
6. 3.5; can’t remember the rest
7. q = 6, p = 2; lambda = -2

I found it pretty easy took about 40-50mins... me of my asnwers dont mashc up to others :frown:

.

btw for question 7a you can check if yours is right because the very last question is to show B and C have same speeds :biggrin:

acceleration is -1.75, deccelartion was 1.75

Reply 83

vietdoan

10

fblade

Here are mine:

1. 35; 32.9
2. 3; 0.5
3. 2; 0.35
4. 11; -1.75
5. 1.68; 0.547
6. 3.5; can’t remember the rest
7. q = 6, p = 2; lambda = -2

I found it pretty easy took about 40-50mins... me of my asnwers dont mashc up to others :frown:

.

btw for question 7a you can check if yours is right because the very last question is to show B and C have same speeds :biggrin:

almost got the same as mine... except question3b you got wrong// question6 i got half wrong.. think i should get more than 90%
Q1: a. T= 35.086N
b. W= 32.97N

Q2: a. m= 0.5kg
b. I= 3Ns

Q3: a. R= 2N
b. S=0.6m

Q4: a. V=11m/s
b. a=-1.75m/s^

Q5: a. 1.68
b. 0.55

Q6: a. a=2.8m/s^
b. T= 3.5N
c. 5/18 shown
d. t= 0.43s ( i think i got wrong here)

Q7: a. 2i+6j
b. (3+2T)i + (-4+6T)j
c. &= -2
d. 2¬10 km/h ( is this the proper answer for the question?)

Reply 84

fblade

14

Damn I really wanted 85+ :frown:

Reply 85

lally

the 6e question was on Jan 07 which is how i knew exactly how to do it. In the practice paper, i worked out the time it took the right hand particle (Q?) to stop, and then doubled it as it needed to return to the position when T=0. However the easiest method is to say displacement = 0, so T=6/7.

I thought that paper was quite simple, I finished it in 30 minutes, redid the whole paper in pencil, and had only one arithemtic error which i changed, and the rest of my answers all corresponded, so thats OK. I know a lot of my friends struggled with parts.

Im confused about the first question. For finding the weight, did you have to incorporate the 9.8 for gravity acting or not?

when you resolve, mgcos20=T (its possible sin 20, i cant remember), and mgsin20=12. You dont need to equate for g because mg = W, so in my numbering, 12/sin20 = W.

Reply 86

vietdoan

10

lally

the 6e question was on Jan 07 which is how i knew exactly how to do it. In the practice paper, i worked out the time it took the right hand particle (Q?) to stop, and then doubled it as it needed to return to the position when T=0. However the easiest method is to say displacement = 0, so T=6/7.

I thought that paper was quite simple, I finished it in 30 minutes, redid the whole paper in pencil, and had only one arithemtic error which i changed, and the rest of my answers all corresponded, so thats OK. I know a lot of my friends struggled with parts.

when you resolve, mgcos20=T (its possible sin 20, i cant remember), and mgsin20=12. You dont need to equate for g because mg = W, so in my numbering, 12/sin20 = W.

how about my method of using I=Ft... is it wrong...!?!? i got till 3/7 by using t=I/F???

Reply 87

surajsamtani

can someone please remind me what the questions were;
as far as i know/remember:
1. weight of the particle
2. collision of two particles
3. moments
4. drawing the graphs n stuff.
5. a particle being pulled by a rope or something
6. pulley
7. vectors of boats

please correct, as appropriate.

Reply 88

talentless

5

for the last part of question 7 it said show so i dont think they are looking for the speed per se i just showed that the magnitude of the velocities i.e their speeds were equal. that should be enough.

Reply 89

EnglishPride

2

If i did badly on M1, but half decent on C1 and C2, how much will M1 drag my maths grade down? What percentage of the course does the exam cover?

Reply 90

Liar

12

Damn how did everyone find this so easy. I thought it was terrible. Granted our mechanics teacher is awful but still... I hope to get over 95% for C1 and C2 but this M1 is gonna drag my UMS right down.

Reply 91

Les_Miserables

StudentBusiness12

If i did badly on M1, but half decent on C1 and C2, how much will M1 drag my maths grade down? What percentage of the course does the exam cover?

It depends on what you're doing (full A-level, just AS...), but they all have the same weighting...

Reply 92

desmondmac

1

I dont get why peopel times the first time by 2 for 6e

I worked out that when P hit the ground Q was travelling at 4.2 ms-1 (Note- it has travelled 3.15m).

Then worked out the time for it to go to 0 velocity- u =4.2 a=-9.8 v= 0
v=u +at

T= 3/7 (marked that as T1)

then i worked out the distanced it travelled in those 3/7 seconds used v^2= U^2 + 2as- i got 0.9 for the distance

As the string becomes taut again when it reaches its starting point i added the total distances i.e 0.9 m + 3.15 m= 4.05

now i worked out the time it would take to make that distance downwards: u=0 s=4.05 a=9.8

s= ut+0.5at^2

i then got t= 0.90913729 (T2)

so in my mind the total time shoudl be t1 + t2

3/7 + 0.90913729= 1.33771

then i put my answer as 1.34 3.s.f

I wouldnt be surprised if its wrong- i was expecting a MUCH harder paper- question 7 was porbably the easiest especially as with the last show question u could pretty muhc know if u had got it write or not. However with question 6 the steps i took seemed logical and worth a whole 6 marks...anyone know where i haev gone wrong?

Reply 93

onetime

lost out on a good few easy marks screwed up vectors by a sign error! god! and by sounds of things every1 did quite well was a pretty text book exam i thought the grade boundaries are likely to be sky high :eek:

Reply 94

lawrywild

Time was definately 6/7 = 0.857!

You find final velocity of of the particle which hits floor = 4.2 m/s
so initial velocity for other particle is 4.2m/s it travels to final velocity 0 m/s, acceleration -9.8 m/s/s so time was 3/7 then it does a parabola, so it's 2t = 6/7... 100% sure

I think I may have gotten 100% in this paper. Got 35.1N for T and 33.0N for weight in question 1 and can't really remember the rest I got. I'm more thinking about physics right now :frown:

Reply 95

lljaneth

desmondmac

I dont get why peopel times the first time by 2 for 6e

I worked out that when P hit the ground Q was travelling at 4.2 ms-1 (Note- it has travelled 3.15m).

Then worked out the time for it to go to 0 velocity- u =4.2 a=-9.8 v= 0
v=u +at

T= 3/7 (marked that as T1)

then i worked out the distanced it travelled in those 3/7 seconds used v^2= U^2 + 2as- i got 0.9 for the distance

As the string becomes taut again when it reaches its starting point i added the total distances i.e 0.9 m + 3.15 m= 4.05

now i worked out the time it would take to make that distance downwards: u=0 s=4.05 a=9.8

s= ut+0.5at^2

i then got t= 0.90913729 (T2)

so in my mind the total time shoudl be t1 + t2

3/7 + 0.90913729= 1.33771

then i put my answer as 1.34 3.s.f

I wouldnt be surprised if its wrong- i was expecting a MUCH harder paper- question 7 was porbably the easiest especially as with the last show question u could pretty muhc know if u had got it write or not. However with question 6 the steps i took seemed logical and worth a whole 6 marks...anyone know where i haev gone wrong?

u didnt need to work out the distance because the distance Q would have travelled up and then come back down would have been equal-thus displacement is 0.

Reply 96

lljaneth

desmondmac

I dont get why peopel times the first time by 2 for 6e

I worked out that when P hit the ground Q was travelling at 4.2 ms-1 (Note- it has travelled 3.15m).

Then worked out the time for it to go to 0 velocity- u =4.2 a=-9.8 v= 0
v=u +at

T= 3/7 (marked that as T1)

then i worked out the distanced it travelled in those 3/7 seconds used v^2= U^2 + 2as- i got 0.9 for the distance

As the string becomes taut again when it reaches its starting point i added the total distances i.e 0.9 m + 3.15 m= 4.05

now i worked out the time it would take to make that distance downwards: u=0 s=4.05 a=9.8

s= ut+0.5at^2

i then got t= 0.90913729 (T2)

so in my mind the total time shoudl be t1 + t2

3/7 + 0.90913729= 1.33771

then i put my answer as 1.34 3.s.f

I wouldnt be surprised if its wrong- i was expecting a MUCH harder paper- question 7 was porbably the easiest especially as with the last show question u could pretty muhc know if u had got it write or not. However with question 6 the steps i took seemed logical and worth a whole 6 marks...anyone know where i haev gone wrong?

''As the string becomes taut again when it reaches its starting point i added the total distances i.e 0.9 m + 3.15 m= 4.05''

-it wouldnt reach its starting point since the string between the pulley and Q would be shorter now that P has moved down.