OCR M1 - 31st May 2012 :)

I got the answer in the end: T = 15.2.

My resolved components were:

Parallel: 10 sin 20 + 0.364R = T cos 45
Perpindicular: 10 cos 20 + T sin 45 = R.


My confusion is over why the mark scheme writes:
Perpindicular: 10 cos 20 + T cos 45 = R

Although this doesn't affect your final result since sin 45 = cos 45, why do they chose to write T cos 45?

The paper is Jun 09.

I see where my error has been.. i did s-2 in the wrong place so it ****ed up my equation! you legend! have an upvote.

Anybody else got any questions they think are decent to try out? I'm feeling confident, just sometimes i screw up last bits of questions.. like that one

Haha, yeah it's pretty demoralising when you have just a little mistake but can't figure out where :P No worries, and cheers xD

Same here guys, any help just post here and I'll give you a hand, if nobody else has first

Here you go:

From part ii, we know the the projection speed of Q is 5.6.

iii) P: (taking up to be positive)
s=s
u=8.4
v=
a=-9.8
t=t

Q:
s=s-2
u=5.6
v=
a=-9.8
t=t

use s=ut + 0.5at^2 for both of these.

Ie p: s=8.4t-4.9t^2
q: s-2 = 5.6t - 4.9t^2
q: s= 5.6t - 4.9t^2 +2

equate the s as you know they are at the same height:
8.4t-4.9t^2 = 5.6t - 4.9t^2 +2

cancel the -4.9t^2:
8.4t=5.6t+2
get t=5/7...
then you have this:
P:
s=
u=8.4
v=
a=-9.8
t=5/7

Q:
s=
u=5.6
v=
a=-9.8
t=5/7

Then use v=u+at for P to get v=1.4 and then use it for Q to get -1.4. Hence, they both have the same speeds, but are travelling in opposite directions, with Q travelling downwards and P travelling upwards.

I hope that's the question you meant anyways xD

Hope it helps!

Since they are going opposite directions, one is going towards the ground so surely one of the accelerations is positive 9.8 in this example Q should have a=9.8

Do we need to know the sine/cosine rules in M1 ?

Sometimes the cosine

Since they are going opposite directions, one is going towards the ground so surely one of the accelerations is positive 9.8 in this example Q should have a=9.8

I'm not sure if this applies, but when i solve these 2 types of equations, you flip one of them around to change the sign.. maybe this is what happens here?

Ah I'll make sure I have a look over it before the exam. Thanks

also i have a quick question about finding the angle of the resultant force, when it says to solve it from the horizontal and vertical, do you guys form triangles? or tan(y/x)... basically does anybody have an easy method lol (is this a stupid question? )

No, and no sorry guys :P

They are not going opposite directions..... at the beginning at least. We set upwards as positive right? so then we have that both P and Q have positive speeds, and negative accelerations as they are accelerating towards the ground... They only change direction after a certain amount of time..

In essence, what I'm saying is Always set one direction as positive and STICK TO IT.... that way, it almost always works out

And Tyles, I'm not really sure what you're saying, but if it's what I think, how would you know which equation to flip around unless you know the answer? I'm a bit confused by this tbh

Hope that helps

Thank you so much I understand Now , I over analyzed it , pretty much like momentum make one direction positive other negative,-> they are both accelerating towards the ground<- I had the concept wrong always perceived as deceleration by gravity going up and acceleration by gravity going down, when in reality they are both accelerating towards the ground.

also i have a quick question about finding the angle of the resultant force, when it says to solve it from the horizontal and vertical, do you guys form triangles? or tan(y/x)... basically does anybody have an easy method lol (is this a stupid question? )

I form triangles, then work out the angle they want and use Tan depending on angle you want to find out, this way when you are double checking it you can glance and do it quickly instead of starting from the top.

yeh i think thats the way forward, thanks

also, http://pdf.ocr.org.uk/download/pp_08_jan/ocr_18635_pp_08_jan_l_gce.pdf?

question 6, the last part. is the answer 2.45 because f < uR ? because if otherwise it would be 3.486 right?

Whenever I see justify I do the force they want, in this case friction and force opposing it which is the 4.9sin30.

Since Friction is more than the opposing force the object remains stationary.

Yeah your right, are you going through all the past papers? I am going to finish off in an hour or 2 then do the hard questions in the textbook

For this question, why is the friciton up and the reaction to the left ? I thought the reaction opposed the weight ?



And for this one, why is the normal reaction R is not included in the markscheme ? If it is threaded onto a string, then it's weight acting down is countered by the normal reaction plus the tensions ?