Edexcel M3 15th May Post- Exam (UK Spec)Watch

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1 week ago
#21
I think they even out in terms of difficulty and the grade boundaries won't change too much tbh, i wish they'd go down though oh and btw, only 4 marks???
(Original post by supalape)
Q1 was 4 marks. What do we think about grade boundaries then? Honestly while it wasn’t too bad, I still think 2017 and 2018 were easier
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1 week ago
#22
I got 0.452 to 3.s.f :/. That was like 5 marks right?

(Original post by supalape)
it was the multiple of the mass of the particle attached to the object when it was suspended
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#23
(Original post by ilazakristudent)
I think they even out in terms of difficulty and the grade boundaries won't change too much tbh, i wish they'd go down though oh and btw, only 4 marks???
Rip my A* then yeah I remember thinking it was a lot of work for 4 marks
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1 week ago
#24
You take moments about A right? Work out the distance from the vertical for each force, multiply by mass, and since its in equilibrium you equate?
(Original post by supalape)
it was the multiple of the mass of the particle attached to the object when it was suspended
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1 week ago
#25
right so. no one got 3/2h for the first question? it was the easiest question in the paper, idk how i managed to get it wrong? ill probably get at least 2 for drawing the dumbass COM table though. i guess.
1. 3/2h
2. 9/11R
3. 4.2, 1.7
4. T is uh mroot3{(w^2a/2)-g/root3)} and then u showed it. what was up with giving the angle from the top of the circle. this question took me more than the whole paper combined?
5. 2 and 71.6, its 71.6 because you had to realise it was resting on the surface from the circular face but we worked out the COM from the vertex, so it was arctan 3/1 not 3/2
it was something like (1cos10-3sin10)cos 10Mg=6sin10KMg?) i took moments about the point where the vertical went through
7? root (3gr)/2
and yeah k is 1/3
Last edited by dalliant; 1 week ago
0
1 week ago
#26
Bruh, I forgot to bloody multiply (1-3tan10) by cos10 to get the perpendicular distance, given that k isnt 0.890 like most are claiming it to be. I think the correct working in my eyes is as follows,

((1-3tan10)cos10)M=6sin10(kM).

And yeah the first bloody question was piss easy IDK how in gods name I got it wrong in the Exam, I mean i got it afterwards
(Original post by dalliant)
right so. no one got 3/2h for the first question? it was the easiest question in the paper, idk how i managed to get it wrong? ill probably get at least 2 for drawing the dumbass COM table though. i guess.
1. 3/2h
2. 9/11R
3. 4.2, 1.7
4. T is uh mroot3{(w^2a/2)-g/root3)} and then u showed it. what was up with giving the angle from the top of the circle. this question took me more than the whole paper combined?
5. 2 and 71.6, its 71.6 because you had to realise it was resting on the surface from the circular face but we worked out the COM from the vertex, so it was arctan 3/1 not 3/2
it was something like (1cos10-3sin10)cos 10Mg=6sin10KMg?) i took moments about the point where the vertical went through
7? root (3gr)/2
and yeah k is 1/3
0
#27
(Original post by dalliant)
right so. no one got 3/2h for the first question? it was the easiest question in the paper, idk how i managed to get it wrong? ill probably get at least 2 for drawing the dumbass COM table though. i guess.
1. 3/2h
2. 9/11R
3. 4.2, 1.7
4. T is uh mroot3{(w^2a/2)-g/root3)} and then u showed it. what was up with giving the angle from the top of the circle. this question took me more than the whole paper combined?
5. 2 and 71.6, its 71.6 because you had to realise it was resting on the surface from the circular face but we worked out the COM from the vertex, so it was arctan 3/1 not 3/2
it was something like (1cos10-3sin10)cos 10Mg=6sin10KMg?) i took moments about the point where the vertical went through
7? root (3gr)/2
and yeah k is 1/3
I got everything there except for 5b and c, how did I screw up that question that bad fml. Q1 I remember seeing a 9/8 somewhere but 3/2 also looks familiar, I honestly can’t remember what I got for the final answer. Also I think that tension is wrong because you would end up with a root of root 3 for W but I’m sure you just did a typo or something
Last edited by supalape; 1 week ago
0
1 week ago
#28
Im hoping question 1 is only 4 marks
(Original post by supalape)
I got everything there except for 5b and c, how did I screw up that question that bad fml. Q1 I remember seeing a 9/8 somewhere but 3/2 also looks familiar, I honestly can’t remember what I got for the final answer. Also I think that tension is wrong because you would end up with a root of root 3 for W but I’m sure you just did a typo or something
(Original post by dalliant)
right so. no one got 3/2h for the first question? it was the easiest question in the paper, idk how i managed to get it wrong? ill probably get at least 2 for drawing the dumbass COM table though. i guess.
1. 3/2h
2. 9/11R
3. 4.2, 1.7
4. T is uh mroot3{(w^2a/2)-g/root3)} and then u showed it. what was up with giving the angle from the top of the circle. this question took me more than the whole paper combined?
5. 2 and 71.6, its 71.6 because you had to realise it was resting on the surface from the circular face but we worked out the COM from the vertex, so it was arctan 3/1 not 3/2
it was something like (1cos10-3sin10)cos 10Mg=6sin10KMg?) i took moments about the point where the vertical went through
7? root (3gr)/2
and yeah k is 1/3
0
1 week ago
#29
Does anyone remember how many marks the last part of the SHM question was worth? Also does anyone remember exactly what it was/how to do it?
0
1 week ago
#30
WAIT ARE YOU SERIOUS ABOUT THE 3/2H THING. wasnt 9/8 for the distance of the cone from the bottom of O? or something? and my tension is 100% right cause i managed to prove b/ so i probably just couldnt use the keyboard to write what i meant. does anyone want a diagram for the COM k question because im insane like that and remember what i drew. i need reassurance because i REALLY need 100 in this.
(Original post by supalape)
I got everything there except for 5b and c, how did I screw up that question that bad fml. Q1 I remember seeing a 9/8 somewhere but 3/2 also looks familiar, I honestly can’t remember what I got for the final answer. Also I think that tension is wrong because you would end up with a root of root 3 for W but I’m sure you just did a typo or something
0
#31
(Original post by TheBigBrick)
Does anyone remember how many marks the last part of the SHM question was worth? Also does anyone remember exactly what it was/how to do it?
First part was 5 and the second part must’ve been 7ish. I found the period (S) using my value of W and then the time taken for P to travel 0.4m from O using x = asinwt. Then I multiplied it by 4 because in one oscillation it’s within 0.4m of the origin 4 times. I then divided that answer by my value of S, as it said that the time in one oscillation that it’s within 0.4m was k x S
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1 week ago
#32
****. i just drew the diagram again and you're right. it's ****ing tan not sin. how many marks would i lose if i wrote sin instead of tan. and listen its (1cos10-3tan10)cos 10Mg=6sin10KMg
you have to times it by cos again to get the perpendicular
if i wrote that but instea of tan i wrote sin. that's 2 lost marks right. one error and one for final answer. **** jesus. was the first question REALLY not 3/2h. are yall sure. ive never got those types of questions wrong. what the **** man.
(Original post by ilazakristudent)
Bruh, I forgot to bloody multiply (1-3tan10) by cos10 to get the perpendicular distance, given that k isnt 0.890 like most are claiming it to be. I think the correct working in my eyes is as follows,

((1-3tan10)cos10)M=6sin10(kM).

And yeah the first bloody question was piss easy IDK how in gods name I got it wrong in the Exam, I mean i got it afterwards
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#33
(Original post by dalliant)
WAIT ARE YOU SERIOUS ABOUT THE 3/2H THING. wasnt 9/8 for the distance of the cone from the bottom of O? or something? and my tension is 100% right cause i managed to prove b/ so i probably just couldnt use the keyboard to write what i meant. does anyone want a diagram for the COM k question because im insane like that and remember what i drew. i need reassurance because i REALLY need 100 in this.
Please do draw a diagram! I remember I definitely got it right as I worked backwards after and substituted my answer multiple times, but I can’t remember at all. It could be that you’re right
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1 week ago
#34
pretty sure it was 5 marks, and you had to find the amplitude by using the speed which you work out from impulse mv-mu, since v at the end of the amplitude is 0, so I=mu, and then u use that u with v^2=w^2a^ to find out amplitude, find the period and then u use x=asinw(t/4) because it's 0.4m? away 4 times. and you see that the time it's 0.4m within the centre of oscillation is 1/3 of the period.
(Original post by TheBigBrick)
Does anyone remember how many marks the last part of the SHM question was worth? Also does anyone remember exactly what it was/how to do it?
0
1 week ago
#35
Height of the cone is 100% 9h/8.

(3h/4)(M+(2M/5))=(M)(h/2)+(2M/5)(h+H/3), where H is the height of the cone.
(Original post by dalliant)
WAIT ARE YOU SERIOUS ABOUT THE 3/2H THING. wasnt 9/8 for the distance of the cone from the bottom of O? or something? and my tension is 100% right cause i managed to prove b/ so i probably just couldnt use the keyboard to write what i meant. does anyone want a diagram for the COM k question because im insane like that and remember what i drew. i need reassurance because i REALLY need 100 in this.
0
1 week ago
#36
Yeah bro, these callous mistakes are killing me aswell
(Original post by dalliant)
****. i just drew the diagram again and you're right. it's ****ing tan not sin. how many marks would i lose if i wrote sin instead of tan. and listen its (1cos10-3tan10)cos 10Mg=6sin10KMg
you have to times it by cos again to get the perpendicular
if i wrote that but instea of tan i wrote sin. that's 2 lost marks right. one error and one for final answer. **** jesus. was the first question REALLY not 3/2h. are yall sure. ive never got those types of questions wrong. what the **** man.
0
#37
(Original post by ilazakristudent)
Height of the cone is 100% 9h/8.

(3h/4)(M+(2M/5))=(M)(h/2)+(2M/5)(h+H/3), where H is the height of the cone.
yeah you’re right, I remember now I divided my left side (3/20) by 2/15 to get 9/8. Thank fk
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1 week ago
#38
When you say the second part do you mean the bit asking to prove it was SHM? Or the last part, which was asking for the time?

AI didn’t realise that it was asking for kS, I thought the S meant seconds and it was asking for the actual time spent. But I think the time I got was wrong anyway
(Original post by supalape)
First part was 5 and the second part must’ve been 7ish. I found the period (S) using my value of W and then the time taken for P to travel 0.4m from O using x = asinwt. Then I multiplied it by 4 because in one oscillation it’s within 0.4m of the origin 4 times. I then divided that answer by my value of S, as it said that the time in one oscillation that it’s within 0.4m was k x S
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1 week ago
#39
I WROTE THIS ON MY EXAM AND STILL GOT THE WRONG ANSWER, FUMING !!! XD
(Original post by supalape)
yeah you’re right, I remember now I divided my left side (3/20) by 2/15 to get 9/8. Thank fk
0
#40
Ok so realistically lads we all bottled it somewhere in this exam, let’s just pray that it’s a common occurrence and the grade boundaries are alright
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