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

Mechanics 2 (M2) Exam 22nd June 2012 - OCR (Not MEI)

Scroll to see replies

Hopefully I'll catch someone. This is a simple enough question but there's something confusing me:

A skier of mass 80 kg is pulled up a slope which makes an angle of 20◦ with the horizontal. The skier is subject to a constant frictional force of magnitude 70 N. The speed of the skier increases from 2 m/s at the point A to 5 m/s at the point B, and the distance AB is 25 m. By modelling the skier as a small object, calculate the work done by the pulling force as the skier moves from A to B.

I did this but I thought you would take into account the weight of the skier as that is a resistive force. Is it because he's being pulled up that his weight component isn't relevant? :s

Reply 41

SharpSchool

http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61433_pp_11_jan_gce_472901.pdf?

would anyone be able to help on 5 i) and 7 iii)? Thanks in advance

Firstly on 5, i take moments from O, so i can see that 0.09cos(45) x mg is the anti clockwise moment. No for the clockwise, i formed a right angle triangle to find the distance from O to T, and did the sqaure root of 0.6^2 + 0.6^2 and then multiplied by 2. However the mark scheme adds 0.6 from somewhere?

And in 7 iii) they say the final speed of B is positive or negative but surely in a perfectly elastic collision direction has to be reversed?

Reply 42

SharpSchool

Original post by goggy

Do you have the answer? Ill try it :smile:

Ok i got 9294J, is this right?

(edited 11 years ago)

Reply 43

hevlar.kelmet

Original post by goggy

The weight certainly needs to be taken into account. Can you give a link to the question?

Reply 44

goggy

9294J is right.

Original post by hevlar.kelmet

The weight certainly needs to be taken into account. Can you give a link to the question?

It's in the January 2007 paper, question 4: http://blogs.thegrangeschool.net/maths/files/2012/03/M2-Past-Paper-Booklet-2010.pdf

There are four components but maybe I'm mistaking weight for something else in the markscheme. :s

Also, if anyone is up for it, question 6 on that paper. I keep getting the COM of the triangle to be 3 when it's supposed to be 4 x-wise. I think I'm over-complicating how I'm working it out and so doing it wrong. How do I figure out the triangle part's CoM in this instance?

(edited 11 years ago)

Reply 45

crazymanny00

good revision video on M2, scroll down to OCR and you'll see it, i'd recommend if you're worried http://www.furthermaths.org.uk/onlinerevision.php

Reply 46

SharpSchool

For the slope one, use suvat equations to find a. Then calculate the force required to produce that acceleration. This is 33.6N. You now know that there are 33.6 more newtons acting up than down. The components acting down are the resistance of 70N and the component of the weight down. Add 33.6 to these 2, then use the equation work done = force x distance :smile:

. Hope this helps

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

Reply 47

crazymanny00

is work done the net force? this has now confused me... why would you add force up the slope to force down the slope...

Reply 48

SharpSchool

The object is accelerating up so we need to know how many newtons of force cause this acceleration. Once we know this, we know that the resistive forces + force to provide acceleration = total force. Now we know the force pulling up the slope, we use work done = total force pulling up X distance moved

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

Reply 49

crazymanny00

Original post by SharpSchool

i'm not sure but i would say it is because they tell you the SPEED is 0.25, which is scalar and so you can't assume the direction. I don't know whether perfectly elastic actually implies they go in opposite directions

Reply 50

SharpSchool

Original post by crazymanny00

Ahh i see, thank you :smile:

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

Reply 51

Anon 17

Original post by SharpSchool

Firstly on 5, i take moments from O, so i can see that 0.09cos(45) x mg is the anti clockwise moment. No for the clockwise, i formed a right angle triangle to find the distance from O to T, and did the sqaure root of 0.6^2 + 0.6^2 and then multiplied by 2. However the mark scheme adds 0.6 from somewhere?

I made the same mistake - Take moments from the point directly below O on the ground, so you have the same anticlockwise moment but the clockwise one you have a distance of 0.6 + sqrt(0.6^2 + 0.6^2). This is because the extra 0.6 is the distance from the point on the ground to O, as the hemisphere has a radius of 0.6m.

goggy:

The centre of mass of a triangle is 2/3 the distance from the median of the hypotunese to the opposite vertex (the right angle corner). Moving 1/3 from the vertex gives a distance of 4cm X wise.

Jan 12 mark scheme, pretty please?

Reply 52

goggy

Thanks!

Argh, this exam. I am awful at mechanics. Provided C4 went reasonably well, I don't need to do terribly well to get a B overall (need a B for uni).

Reply 53

crazymanny00

Original post by Anon 17

I understand what you've done here, but why do you HAVE to take moments from the ground?

Reply 54

Anon 17

I'm not 100% sure but thinking about it, I'm pretty certain it's because the mass is rotating about the point on the ground thus you take moments about that point.

Reply 55

crazymanny00

ah that does make sense, thanks!

Reply 56

hevlar.kelmet

Original post by goggy

9294J is right.

It's in the January 2007 paper, question 4: http://blogs.thegrangeschool.net/maths/files/2012/03/M2-Past-Paper-Booklet-2010.pdf

There are four components but maybe I'm mistaking weight for something else in the markscheme. :s

Also, if anyone is up for it, question 6 on that paper. I keep getting the COM of the triangle to be 3 when it's supposed to be 4.

First calculate the forces:

[br]a=\dfrac{v^{2}-u^{2}}{2s}=\dfrac{5^2-2^2}{2*25}=0.42[br][br]80gsin(20)+70+80*0.42=371.74379....[br][br]25*371.74379....=9293.5948...[br][br]9294 J

Good luck everyone！

Original post by TheRenaissanceMan

Does anyone know how to do this maths help.png

Sorry for the late reply, only just done this:

i) You need to find the magnitude of the reaction force, so first we need to find the angle that the reaction force makes to the vertical. The distance from P to O is 5m (the radius of the hemisphere) and the distance from P to the top of the hemisphere is 3m. Thus using trigonometry, we get an angle of arccos(3/5) = 53.13 degrees. Then resolve vertically:

mg = Rcos(53.13)
R = (0.2 x 9.8) / cos(53.13) = 3.27N.

ii) First, we need to calculate the radius of the horizontal circle that P moves in. Again using trigonometry, we can work out the distance from P to the centre of the hemisphere (horizontally) to be 5sin(53.13) = 4m. Now resolve horizontally:

m(w^2)r = Rsin(53.13)
w^2 = (3.27 x sin(53.13)) / (0.2 x 4)

Which gives w = 1.81rad/s.

iii) The only difference between this part and the last part is the light, inextensible string attached to P. The first thing to do is work out what angle to the horizontal (or vertical) that the string makes with P. We already have the distance of P to the centre of the circle as 4m, and the distance vertically from P and the lowest point of the hemisphere is 5 - 5cos(53.13) which is 2. So the distance is arctan(2/4) = 25.565... degrees. Now we can resolve vertically:

mg = Tsin(25.565) = Rcos(53.13)

T is simply the tension in the string, so 0.1g. Substituting the values in gives us R = 3.997N.

Finally, we can resolve horizontally:

Rsin(53.13) + Tcos(25.565) = m(w^2)r

Rearranging and substituting in the values gives us w = 2.26rad/s.

Hope this helps!

Reply 59

Ttawwab

Original post by SharpSchool

Do you have the answer? Ill try it :smile:

Ok i got 9294J, is this right?

For this question, I did work done to increase kinetic energy, added to the work done against gravity (mgh) added to the work done against resistive forces I.e. (80gsin20+70)*25. What have I done wrong here? :frown: