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M1 mechanics edexcel modular exam

how was your paper?


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I am expecting a D. This exam was harder than all the previous papers but I blame myself - most of the long questions that I couldn't answer were ones I have done a long time ago and never revised them because I though they would never come.

I just need to forget about M1. Its over. Also, there is S1, C2, C3, and C4 remaining. Do you have S1 this Fri? Are you ready for it?

I'm not taking S1 : I only Have C2 left! I just hope I don't screw up like i did in M1 :mad:. I agree with you, the exams was hard no doubt, and getting a U is plausible . My moment is wrong, I guess i messed up my calculation because my logic was right! :)_____Good luck man and study hard! don't screw up again :)
 
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I'm not taking S1 : I only Have C2 left! I just hope I don't screw up like i did in M1 :mad:. I agree with you, the exams was hard no doubt, and getting a U is plausible . My moment is wrong, I guess i messed up my calculation because my logic was right! :)_____Good luck man and study hard! don't screw up again :)
Yup i also have C2 after 11 dayz.
Good luck guyz !
 
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i hate da assumptionsss total liesssss!!!!i dont know how dis m1 can help in practical life... how can u hav a frictionless surface..a smooth pulley , no resitance,, and so on....why r dey teaching us liesss...????

IF You knew this, why did you take the course?
 
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LMAO............frictionless surface??? lmao......honestly i have spend a disastrous evening during the exam.........

"frictionless surface" sounds creepy. You can make up for it; watch a movie or do something crazy and clear your mind off M1 -Once in a while you need nightmares like M1 to keep you alive xD
 
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Answers For M1.
Q1.
a. 9 Ns b. 2 kg
Q2.
a. 7800 N b. 390 N
Q3.
15.5 something N
Q4.
a. u= 14 m/s b. I think 6s
Q5.
a. Graph b. T= 75 c. 50 s d. a= 0.792

Q6.
a. Mass= 25 kg AM= 6 m c. 7.5 m

Q7.
a. ? b.? c. t=0.5 , t=2/3

Q8.
a. a=5.88 b. 11.76m N c. 16.63m N on a bearing of 225 degree.
 
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Answers For M1.
Q1.
a. 9 Ns b. 2 kg
Q2.
a. 7800 N b. 390 N
Q3.
15.5 something N
Q4.
a. u= 14 m/s b. I think 6s
Q5.
a. Graph b. T= 75 c. 50 s d. a= 0.792

Q6.
a. Mass= 25 kg AM= 6 m c. 7.5 m

Q7.
a. ? b.? c. t=0.5 , t=2/3

Q8.
a. a=5.88 b. 11.76m N c. 16.63m N on a bearing of 225 degree.
Think my answers were the same. Not sure what I wrote for the Q2b though. Also I just wrote south west for 8c :( silly move on my part
 
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Answers For M1.
Q1.
a. 9 Ns b. 2 kg
Q2.
a. 7800 N b. 390 N
Q3.
15.5 something N
Q4.
a. u= 14 m/s b. I think 6s
Q5.
a. Graph b. T= 75 c. 50 s d. a= 0.792

Q6.
a. Mass= 25 kg AM= 6 m c. 7.5 m

Q7.
a. ? b.? c. t=0.5 , t=2/3

Q8.
a. a=5.88 b. 11.76m N c. 16.63m N on a bearing of 225 degree.
are you sure question 5d acceleration = 0792 ?
isnt it supposed to be (22-0)/50 which equals to 0.44 m/s ???
 
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Guys would there be ECF on the last question. I got the acceleration wrong (I forgot to put g with the mass) and so the tension and force on the pulley are wrong (the ethod is all correct but only answers are wrong) . Would there be ECF for this question??
 
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Guys would there be ECF on the last question. I got the acceleration wrong (I forgot to put g with the mass) and so the tension and force on the pulley are wrong (the ethod is all correct but only answers are wrong) . Would there be ECF for this question??
I would imagine so, or some sort of method marks anyhow.
 
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This is what i got
1) impulse is 9N, mass is 2kg
2) tension of lift is 7800N, NR is 390N
3)Tension = 15.5 N
4) u=14ms^-1, T=6sec
5)T=70sec, a=0.309
7) last, parallel j= t is 1, parallel to -i-3j t=3/2
6) moments no idea...
8) a=5.88 /T= 17.6m/R=16.6m
And thats it!
 
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are you sure question 5d acceleration = 0792 ?
isnt it supposed to be (22-0)/50 which equals to 0.44 m/s ???
Sister who told u that final speed of motorcyclist is 22m/s. It was the speed of car. The motorcyclist was in constant acceleration. We don't know its final speed. It does not mean that car is having 22m/s motorcycle would have same. This is speed-time graph not distance-time graph. You interpret it wrong. If the graphs intersect it does not mean they are equidistant from the first sign of traffic. This only means that they both have same speed for a particular moment.

t= 50s u= 0 m/s s= 990 m a=?

s= ut + 0.5at^2
990= (0)(50) + 0.5(a)(50)^2
990= 0 + 0.5 (a) (2500)
990= 1250a
1250a= 990
a = 990/1250 = 0.792 m/s^2
 
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Sister who told u that final speed of motorcyclist is 22m/s. It was the speed of car. The motorcyclist was in constant acceleration. We don't know its final speed. It does not mean that car is having 22m/s motorcycle would have same. This is speed-time graph not distance-time graph. You interpret it wrong. If the graphs intersect it does not mean they are equidistant from the first sign of traffic. This only means that they both have same speed for a particular moment.

t= 50s u= 0 m/s s= 990 m a=?

s= ut + 0.5at^2
990= (0)(50) + 0.5(a)(50)^2
990= 0 + 0.5 (a) (2500)
990= 1250a
1250a= 990
a = 990/1250 = 0.792 m/s^2
ohh that makes more sense
how many marks does it carry ???
 
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