Poll: Tough,hard,medium or easy
Tough (4)
22.22%
Hard (5)
27.78%
Medium (6)
33.33%
Easy (3)
16.67%
Asgard
Badges: 0
Rep:
?
#1
Report Thread starter 6 years ago
#1
What do you think the M1 exam will be?

If you have any problems in M1, post here to let other TSR students help you.

Good Luck Guys,
0
reply
Mesopotamian
Badges: 0
Rep:
?
#2
Report 6 years ago
#2
M1 is fairly straightforward if you put enough practice in. I found S1 last year much more confusing. As long as you learn to draw the correct diagram to go with the question I think it is quite easy to get an A. Questions 7 and 8 on each paper are always the hardest ones so if you can master those then you've pretty much got everything under control
0
reply
The H
Badges: 1
Rep:
?
#3
Report 6 years ago
#3
I find mechanics ok, better than C3 at least
0
reply
frony0
Badges: 2
Rep:
?
#4
Report 6 years ago
#4
The biggest problem is stupid mistakes, there's nothing really "hard"


This was posted from The Student Room's iPhone/iPad App
0
reply
.raiden.
Badges: 9
Rep:
?
#5
Report 6 years ago
#5
(Original post by The H)
I find mechanics ok, better than C3 at least
I find M1 is harder than C3
0
reply
Sarabande
Badges: 15
Rep:
?
#6
Report 6 years ago
#6
I'm expecting a devil of a paper after June 2012.
0
reply
The H
Badges: 1
Rep:
?
#7
Report 6 years ago
#7
(Original post by raiden95)
I find M1 is harder than C3
Tbh, I find them pretty much the same, but if I was to say one was harder it would be C3
0
reply
.raiden.
Badges: 9
Rep:
?
#8
Report 6 years ago
#8
Can someone explain how to do q3a) please?
Attached files
0
reply
MedMed12
Badges: 21
Rep:
?
#9
Report 6 years ago
#9
(Original post by raiden95)
I find M1 is harder than C3
(Original post by Sarabande)
I'm expecting a devil of a paper after June 2012.

I agree D:
0
reply
Joshmeid
Badges: 5
Rep:
?
#10
Report 6 years ago
#10
(Original post by raiden95)
Can someone explain how to do q3a) please?
The way I think of it is parallel = same gradient which you should know.

Gradient = y/x, in this case coefficient of j over coefficient of i.

a + \lambda b = i - 2j + \lambda (-3i + j)

\rightarrow (1 - 3\lambda)i + (\lambda - 2)j

\rightarrow \frac{\lambda - 2}{1 - 3\lambda} = \frac{-3}{-1}

And then solve.
0
reply
Jullith
Badges: 0
Rep:
?
#11
Report 6 years ago
#11
I quite like M1, the only thing I'm really struggling with at the moment is vectors. :P

Could someone explain part c of this question to me please:
Name:  Screen Shot 2013-01-20 at 15.01.08.png
Views: 178
Size:  21.2 KB
Thanks!
0
reply
MedMed12
Badges: 21
Rep:
?
#12
Report 6 years ago
#12
(Original post by Jullith)
I quite like M1, the only thing I'm really struggling with at the moment is vectors. :P

Could someone explain part c of this question to me please:
Name:  Screen Shot 2013-01-20 at 15.01.08.png
Views: 178
Size:  21.2 KB
Thanks!
that means the j components are equal to 0
2
reply
Sarabande
Badges: 15
Rep:
?
#13
Report 6 years ago
#13
(Original post by MedMed12)
that means the I components are equal to 0
No if it's moving parallel to i, then j must equal 0 as it will only have a velocity in the i direction for it to be parallel to i.
1
reply
MedMed12
Badges: 21
Rep:
?
#14
Report 6 years ago
#14
(Original post by Sarabande)
No if it's moving parallel to i, then j must equal 0 as it will only have a velocity in the i direction for it to be parallel to i.
woops sorry thats what I meant I ought to have double checked before posting
ahh silly me
I guess the way to remember it is:

parallel to i, j is 0
parallel to j, i is 0

opposites I guess
0
reply
MedMed12
Badges: 21
Rep:
?
#15
Report 6 years ago
#15
anyone got the june paper? x
0
reply
Jullith
Badges: 0
Rep:
?
#16
Report 6 years ago
#16
(Original post by MedMed12)
woops sorry thats what I meant I ought to have double checked before posting
ahh silly me
I guess the way to remember it is:

parallel to i, j is 0
parallel to j, i is 0

opposites I guess
Thanks!

And I've attached the June 2012 paper to this post. But I couldn't find the blank version, so this one has answers with the questions. :P
Attached files
1
reply
Jullith
Badges: 0
Rep:
?
#17
Report 6 years ago
#17
And could someone help me with part c of this question please?

Name:  Screen Shot 2013-01-20 at 23.43.46.png
Views: 338
Size:  36.7 KB
0
reply
MedMed12
Badges: 21
Rep:
?
#18
Report 6 years ago
#18
(Original post by Jullith)
Thanks!

And I've attached the June 2012 paper to this post. But I couldn't find the blank version, so this one has answers with the questions. :P

thank you!
is there a MS?

ignore this ^ theres answers on the other one
0
reply
Joshmeid
Badges: 5
Rep:
?
#19
Report 6 years ago
#19
(Original post by Jullith)
And could someone help me with part c of this question please?

Name:  Screen Shot 2013-01-20 at 23.43.46.png
Views: 338
Size:  36.7 KB
Draw a picture of what it says first. Resistant force F acts in opposite direction of motion. Thrust acts towards to car and trailer from the towbar, Acceleration will be acting in the opposite direction of motion to the car as it is decelerating. Resistance to motion is unchanged from previous parts. There is no longer a driving force.

You can then treat the car and the trailer as separate particles, to find the acceleration and then to find the magnitude of F.
0
reply
MedMed12
Badges: 21
Rep:
?
#20
Report 6 years ago
#20
anyone want to explain in jan 2012 why on question 5c the time is taken away from one another :l I just just left the two possible times i found :P
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Surrey
    Postgraduate Open Afternoon Postgraduate
    Wed, 23 Oct '19
  • University of Bristol
    Undergraduate Open Afternoon Undergraduate
    Wed, 23 Oct '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Wed, 23 Oct '19

Would you turn to a teacher if you were being bullied?

Yes (47)
25.54%
No (137)
74.46%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise