OCR MEI M2 (14/01/13) Discussion Thread

I'm sure that if they wanted you to justify it, they would have said so. "State, with a reason, whether or not..." or something like that.

Reply 21

Ben McKenna

Original post by Jamesm66

Nice one, didn't know that. reckon I would've got a mark or two for saying y-coordinate stays the same? I said its stable also. And that there was twice the weight on the right edge of the thing. Don't think many people would of got that?

Yeah, I'm sure that y-coordinate statement would have been one of the marks. There probably would have been an answer mark for saying it's stable, but it might be one of those method-dependent marks, where you have to show some vaguely relevant working to get it. I think they're shown as "B1" in mark schemes.

What do you mean by twice the weight on the right edge?

Reply 22

Tedfoo25

OP

6

Original post by Ben McKenna

Yeah, I'm sure that y-coordinate statement would have been one of the marks. There probably would have been an answer mark for saying it's stable, but it might be one of those method-dependent marks, where you have to show some vaguely relevant working to get it. I think they're shown as "B1" in mark schemes.

Or you could have done y=mx+c to show it was in the region.

Reply 23

Jamesm66

Original post by Ben McKenna

Yeah, I'm sure that y-coordinate statement would have been one of the marks. There probably would have been an answer mark for saying it's stable, but it might be one of those method-dependent marks, where you have to show some vaguely relevant working to get it. I think they're shown as "B1" in mark schemes.

What do you mean by twice the weight on the right edge?

Just when looking at the plan view, folded under the right edge was 4 squares and folded under the left edge was 2 squares? As uniform I thought twice the weight was on the right so thought it had something to do with that.

Reply 24

MrNeilPatel

Original post by Ben McKenna

Do you mean that the velocity of P is 2.7ms^-1?

Yeah sorry :P does anyone have a copy of the paper or an unofficial mark scheme ?

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

Reply 25

Ben McKenna

Original post by Jamesm66

Just when looking at the plan view, folded under the right edge was 4 squares and folded under the left edge was 2 squares? As uniform I thought twice the weight was on the right so thought it had something to do with that.

Ahh right. Well that's true; it does have an effect. But it was already taken into account by the person who wrote the paper when they gave you that the x-coordinate was 1.7, and by you when you said the y-coordinate was unchanged. So that's it, now you have the position of the centre.

Reply 26

Ben McKenna

Original post by Tedfoo25

Or you could have done y=mx+c to show it was in the region.

Oh, how did you do that? Some kind of inequality to describe the area between the two lines?

Reply 27

Ben McKenna

Original post by MrNeilPatel

Yeah sorry :P does anyone have a copy of the paper or an unofficial mark scheme?

I don't, but I can do you one worse! I remember the question.

An object that is at rest on a horizontal plane when it explodes into two objects, P and Q. P has a mass of 0.4kg and a velocity of 6m/s in the opposite direction of Q. Q has a mass of 0.5kg and separates from P with a speed v.

1. Calculate v.

P rebounds elastically off a vertical wall, 0.75m away. The contact between P and the surface of the plane is smooth. Q, however, has a constant resistance to its motion and decelerates at a constant rate. At a time of 2/3 seconds after the explosion, P collides with Q.

2. Show that when P and Q collide, Q has a speed of 2.7m/s.

3. Hence, calculate the resistance to the motion of Q, and the coefficient of friction between Q and the surface of the plane.

P collides with Q with a coefficient of restitution of 1/8. The collision is instantaneous.

4. Calculate velocity of Q immediately after the collision.

Let me know if I've missed anything...

Reply 28

jjlbateman

I found question 4 impossible, I completely forgot we had the hypotenuse so could work out the value of (theta) so how did you prove that tan(theta) = 0.6?

(edited 11 years ago)

Reply 29

MrNeilPatel

Original post by jjlbateman

I found question 4 impossible, I completely forgot we had the hypotenuse so could work out the value of (theta) so how did you prove that tan(theta) = 0.6?

I got tan theta equals 0.6 by resolving forces parallel and perpendicular to the slope and then used that the system was in equilibrium so F = mew X R

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

Reply 30

BlueCanary

2

I found the latter half of q1 completely impossible, and the last part of q4 I probably got wrong as a mu value of 3 probably isn't right! The rest I feel like I did reasonably well on, though.
I am really hoping for lower boundaries though as I know i've effectively chucked 13 marks away by screwing up the 2nd half of q1. I couldn't even take a educated guesstimate at it.

Reply 31

Tedfoo25

OP

6

Original post by Ben McKenna

Oh, how did you do that? Some kind of inequality to describe the area between the two lines?

Well after joining the edges of the two table 'legs' to find the area where it is stable yields two lines. These can be given an equation in terms of y=mx+c.
Then, set y=0 (bottom of table top) to show that x<1.7 i.e. the line reaches the edge of the table before x=1.7 and then do the same for the upper line.

In hindsight, I probably should have put the y-coordinate in to my equation to show x<1.7 and vice versa for the other line. This would have saved time.

(edited 11 years ago)

Reply 32

MrNeilPatel

Original post by Ben McKenna

*snip*

Thank you Ben, could I ask did you know how to get the 2.7/ms for the velocity of Q before they collide ?

Reply 33

BeatlesFan96

Original post by Tedfoo25

Well after joining the edges of the two table 'legs' to find the area where it is stable yields two lines. These can be given an equation in terms of y=mx+c. Then, set y=0 (bottom of table top) to show that x<1.7 i.e. the line reaches the edge of the table before x=1.7 and then do the same for the upper line.

Surely this isn't necessary for 4 marks? The diagram, stating that the y-coordinate is the same (or working it out again), then saying whether it topples or not, then maybe a weird methody type mark?

I made a bit of a mistake on this question, I didn't shade all the way down to the bottom of the vertical edge, only to the edge of the square, so I guess I'll lose a mark there. I was the only one taking it at my college so it's interesting to see lots of people found it difficult.

Reply 34

Tedfoo25

OP

6

Original post by BeatlesFan96

Surely this isn't necessary for 4 marks? The diagram, stating that the y-coordinate is the same (or working it out again), then saying whether it topples or not, then maybe a weird methody type mark?

I made a bit of a mistake on this question, I didn't shade all the way down to the bottom of the vertical edge, only to the edge of the square, so I guess I'll lose a mark there. I was the only one taking it at my college so it's interesting to see lots of people found it difficult.

I'm pretty sure it said that the CoM was within the 3x3 square.

Reply 35

BeatlesFan96

Original post by Tedfoo25

I'm pretty sure it said that the CoM was within the 3x3 square.

Oh did it? I don't remember. If it did that's super!

it was horrific!!!

What did you lot get for the very last question - the 9 marker with the coefficient of friction between the beam and the floor?

I made a mess of it by using cos and sin wrong when resolving, and only realised it in the last few minutes, but I managed to rush my way to an answer of mew=0.99 .

Reply 38

iCiaran

11

The majority of people at my school (me included) came out saying they will most likely have to resit this, it was bad...

i managed to answer all the questions just, but some of my answers were questionable at best.

Reply 39

BeatlesFan96

Original post by dimension088

What did you lot get for the very last question - the 9 marker with the coefficient of friction between the beam and the floor?

I made a mess of it by using cos and sin wrong when resolving, and only realised it in the last few minutes, but I managed to rush my way to an answer of mew=0.99 .

I only just managed to do this in time.

Firstly, from basic trig, cos(theta) =0.8, sin(theta)=0.6

So, taking moments about A, 0.75*30*0.8=1.5*(Reaction at the block)
Therefore, the reaction at the block is 12N

Resolving vertically gives 12*0.8+(Reaction at floor)=Weight=30
Therefore, reaction force from floor = 20.4N

Resolving horizontally gives Friction = 12*0.6 = 7.2N

Object is on point of sliding, so F=mew*R; substituting the above results gives mew as 6/17 I believe, or 0.3529411765....

OCR MEI M2 (14/01/13) Discussion Thread

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