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Edexcel IAL Maths - Mechanics 1 - WME01 25 Jan 2016

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Reply 40
Original post by omar5478
Honestly, I would say that it is a really annoying paper, but maybe thats just because I am pretty bad at practical stuff.Good luck anyways.


I agree wholeheartedly and share your sentiments. I'm not expecting a good grade tomorrow.
Original post by Zacken
I agree wholeheartedly and share your sentiments. I'm not expecting a good grade tomorrow.

I guess we will have to hope that other candidates score low so that the boundaries are get for us lol.
(edited 8 years ago)
Original post by Zacken
I agree wholeheartedly and share your sentiments. I'm not expecting a good grade tomorrow.


me neither
Original post by omar5478
I guess we will have to hope that other candidates score low so that the boundaries are get for us lol.


i was saying the exact same thing to my friends
Guys i also made simple mistakes! i couldn't get the range of the moment and by chance my value for Q.2 came 73/10u due west! i dont know why this value is wrong, as i did the question with utter care! just bad luck it is :frown:
Reply 45
Original post by syed.shahriyar
Guys i also made simple mistakes! i couldn't get the range of the moment and by chance my value for Q.2 came 73/10u due west! i dont know why this value is wrong, as i did the question with utter care! just bad luck it is :frown:


A lot of people seem to be getting this answer, but I think I suspect where you went wrong. It seems to be a fairly common "trap" in the question - but I might also just be wrong myself.
Original post by syed.shahriyar
Guys i also made simple mistakes! i couldn't get the range of the moment and by chance my value for Q.2 came 73/10u due west! i dont know why this value is wrong, as i did the question with utter care! just bad luck it is :frown:

Can you please describe how you got this answer because I am really curious to how the direction could be due west.Moreover,what was your value for Q2)a ??
(edited 8 years ago)
Original post by omar5478
Can you please describe how you got this answer because I am really curious to how the direction could be due west.Moreover,what was your value for Q2)a ??


I=33/5 mu, m=2m, u=4u

=> I=mv-mu
=> 33/5mu=2mv-(2 x 4u)
=> 33/5mu=2mv-8mu
=> 33/5mu + 8mu = 2mv
=> 73/5u = 2v
=> 73/10u = v
v = 73/10u

This is what i did! i hope i did correctly or is it wrong then plz correct! :/ confused
(edited 8 years ago)
Reply 48
Original post by syed.shahriyar
I=33/5 mu, m=2m, u=4u

=> I=mv-mu
=> 33/5mu=2mv-(2 x 4u)
=> 33/5mu=2mv-8mu
=> 33/5mu + 8mu = 2mv
=> 73/5u = 2v
=> 73/10u = v
v = 73/10u

This is what i did! i hope i did correctly or is it wrong then plz correct! :/ confused


Thing is, the magnitude of the impulse is 33/5 mu, you need to work out the direction yourself.
Original post by syed.shahriyar
I=33/5 mu, m=2m, u=4u

=> I=mv-mu
=> 33/5mu=2mv-(2 x 4u)
=> 33/5mu=2mv-8mu
=> 33/5mu + 8mu = 2mv
=> 73/5u = 2v
=> 73/10u = v
v = 73/10u

This is what i did! i hope i did correctly or is it wrong then plz correct! :/ confused

According to your working, the final velocity is positive(thus same direction as original) and bigger than the original velocity.This can only happen if the object was pushed which clearly was not the case. Your mistake here was that you assumed that the impulse was positive,while the negative value of impulse's magnitude should have been used.If were you, I won't be worried because a lot of people seem to have made this mistake which will lower the grade boundaries.
Original post by Zacken
Thing is, the magnitude of the impulse is 33/5 mu, you need to work out the direction yourself.


Exactly.
Well happy about the overall exam! :smile:
hey guys for the speed time graph of the rock projection , is the graph a straight line starting from a speed of 11 to a -14 ??
Reply 53
Original post by Mohammad_besani
hey guys for the speed time graph of the rock projection , is the graph a straight line starting from a speed of 11 to a -14 ??


A straight line start from 11.2 to -14 and cross the time-axis at (14/g, 0) or (1.43, 0).
Original post by Zacken
Before telling you how many marks you are going to lose, I'm afraid one of us have read the question wrong. I was fairly sure the question had said the mass of the block is 88 kg and not 55 kg. I can only ask the others to confirm:

Spoiler


By the way, Omar - what did you get for Q2?


Hey, sorry I just saw it and the mass is not 5 kg because I remember how I did it and I got a different answer, and I am sure I did it right because I revised the numberes so I think @Zacken is the right one

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Original post by Zacken
Fewer questions than usual, so more marks per question, more structure than usual. There wasn't the easy speed-time things with distance between two places, etc... but the paper was fairly standard, no thinking required. There'll be high boundaries.

1. (a) F = 2,200 N
(b) M = 800.
(c) Same acceleration for system/trailer as truck.

2. (a) 7u/10 due east
(b) 6u/5 due east.

3. T=16g53+1\displaystyle T = \frac{16g}{5\sqrt{3} + 1}

4. (a) hmax=10h_max = 10
(b) ttotal=28g2.86\displaystyle t_{\text{total}} = \frac{28}{g} \approx 2.86

5. (a) (i) TA=9410xT_A = 94 - 10x
(ii) TB=66+10xT_B = 66 + 10x
(b) 1x1.81 \leq x \leq 1.8

6. (a) 68=2178.25\sqrt{68} = 2\sqrt{17} \approx 8.25
(b) (5+2t)i + (5t-3)j
(c) (7-3t)i + (5-15t)j
(d) t = 2/5 so time 2:24 PM
(e) 29/5i - j

7 (a) T - g/2 = 2a
(b) 3g - T = 5a
(c) (i) a = 5g/14
(ii) T = 17g/14
(d) d= 17/14 (by letting g=1 in c(ii)).


Hey I got the same answers as you!

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Reply 56
Original post by PlayerBB
Hey I got the same answers as you!

Posted from TSR Mobile


Everything exactly the same? If so, congratulations! :smile:
Original post by Zacken
A straight line start from 11.2 to -14 and cross the time-axis at (14/g, 0) or (1.43, 0).


I started the line from 7.sth because I thought they asked from the instant when it passes A

Posted from TSR Mobile
Original post by Zacken
Everything exactly the same? If so, congratulations! :smile:


Yeah supposedly haha and thank you mate!! ^.^

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Reply 59
Original post by PlayerBB
I started the line from 7.sth because I thought they asked from the instant when it passes A

Posted from TSR Mobile


The question said that at A, the vertical velocity upwards was 11.2

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