Competition - post a photo you've taken of nature >>

M1 23/1/2013 Thread.

Poll

Tough,hard,medium or easy

Tough Hard Medium Easy

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, :smile:

Scroll to see replies

Reply 1

Mesopotamian

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

Reply 2

The H

I find mechanics ok, better than C3 at least

Reply 3

bo536

The biggest problem is stupid mistakes, there's nothing really "hard"

This was posted from The Student Room's iPhone/iPad App

Reply 4

.raiden.

Original post by The H

I find mechanics ok, better than C3 at least

I find M1 is harder than C3

Reply 5

Sarabande

I'm expecting a devil of a paper after June 2012.

Reply 6

The H

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

Reply 7

.raiden.

Can someone explain how to do q3a) please?

image.jpg387.5KB

Reply 8

MedMed12

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:

Reply 9

Joshmeid

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.

(edited 11 years ago)

Reply 10

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:
Screen Shot 2013-01-20 at 15.01.08.png

Thanks!

Reply 11

MedMed12

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:
Screen Shot 2013-01-20 at 15.01.08.png

Thanks!

that means the j components are equal to 0

(edited 11 years ago)

Reply 12

Sarabande

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.

Reply 13

MedMed12

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 :tongue:

I ought to have double checked before posting
ahh silly me :tongue:

I guess the way to remember it is:

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

opposites I guess

Reply 14

MedMed12

anyone got the june paper? x

Reply 15

Jullith

Original post by MedMed12

woops sorry thats what I meant :tongue:

I ought to have double checked before posting
ahh silly me :tongue:

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

12 May M1 RA.pdf911.3KB

Reply 16

Jullith

And could someone help me with part c of this question please? :smile:

Reply 17

MedMed12

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! :smile:

is there a MS?

ignore this ^ theres answers on the other one

Reply 18

Joshmeid

Original post by Jullith

And could someone help me with part c of this question please? :smile:

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.