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M1 OCR (Not MEI) Exam - 9/06/2015

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June 09 question 5i.

How come the particles coalesce? It doesn't state it in the question, how would I know they did?
June 2013's paper is a gift. I hope we get a paper like that tomorrow.

Does anyone know where to find the June 2014 paper?
Original post by Breo
Hi all, I always seem to struggle when there is a block going down a slope involving friction and things like that (Almost all Q7s.) Any tips? Also, has anyone done Jan 2010 Q7. ii.b? really struggling and would appreciate a worked solution or some help. Thanks a lot.


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Original post by Inkozzo
I found that friction must act in the direction of the 3N force because if it acting is against it (i.e. in the direction of T in a or in the direction of Tcos30 in b) then T=0N. T exists as a force so it can't be 0N, so friction must be acting in the direction of the 3N force.


Cheers, I get that bit.. but what about part (b)? Normally I would see whether the magnitude of tcos30 is greater than 3N to decide which way friction acts, but we don't know what 't' is...!
Could someone explain b) please?june 10 q6.png
Can somebody please show me where to read the June 2014 paper?
Original post by Minecraft27
Can somebody please show me where to read the June 2014 paper?


http://www.dotmaths.com/ocr/
Need help with last part of the last question from June 09 as well :/
Thanks :smile:
Reply 228
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Inkozzo
Original post by RosuRourensu
Cheers, I get that bit.. but what about part (b)? Normally I would see whether the magnitude of tcos30 is greater than 3N to decide which way friction acts, but we don't know what 't' is...!


If we pretend friction acts in the direction of Tcos30 then:
3 - Tcos30 - Friction= 0 (as it is in equilibrium)
Friction=uR and R=8-Tsin30
so Friction is 0.375(8-Tsin30)
Thus 3- Tcos30 - 0.375(8-Tsin30)=0
3 - Tcos30 -3 + 0.375Tsin30 = 0
-Tcos30 + 0.375Tsin30 = 0
0.679T = 0
T=0N
But obviously T exists so it can't be 0N
Reply 229
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c223
2
Original post by Super199
Need help with last part of the last question from June 09 as well :/
Thanks :smile:


Split the graph up into 3 sections - accelerating, constant velocity, decelerating. You already found the distance travelled in the first section (18m) then you can use s=ut to find the distance at constant velocity (171m). Then do 200-18-171 to find the distance which needs to be travelled in the final section, which comes out as 11m. You can then use suvat with s=ut + 1/2at2 to find t, which you add to the 22s you already have for the first two sections.
(edited 8 years ago)
Reply 230
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MsFahima
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20
Hey guys. Hope your revision is going well. Will be here later to answer questions.

Gonna take a nap at the moment. :tongue:

Posted from TSR Mobile
January 2012
Q7 ii)

Since P is in contact with the ground why is there not a reaction force acting upwards on P?
E.g I thought it would be T+ mg(reaction) - mg(weight) = ma

Any help would be greatly appreciated and good luck for tomorrow everyone! :smile:
Original post by alexann95
Yeah, its the resultant of any forces that the surface is creating on the particle (hence the reaction force and the friction). I can't think of a better way to word it!


Does it include the componet of the objects weight going down a slope then? im really confusing myself!
Original post by lauren_2
Does it include the componet of the objects weight going down a slope then? im really confusing myself!


No because the weight of the object isn't caused by the surface, think about it. If the surface wasn't there, the weight of the object would still be exactly the same since its mass multiplied by 9.8 :smile:
I suck at m1.....so bad, feel like im gonna get a C. Should I do an all nighter, considering I have six exams this week
Contact Force - The resultant of friction and normal reaction force? Right?
Original post by chloe-jessica
Split the graph up into 3 sections - accelerating, constant velocity, decelerating. You already found the distance travelled in the first section (18m) then you can use s=ut to find the distance at constant velocity (171m). Then do 200-18-171 to find the distance which needs to be travelled in the final section, which comes out as 11m. You can then use suvat with s=ut + 1/2at2 to find t, which you add to the 22s you already have for the first two sections.


ah thank you :smile:
Original post by chloe-jessica
Split the graph up into 3 sections - accelerating, constant velocity, decelerating. You already found the distance travelled in the first section (18m) then you can use s=ut to find the distance at constant velocity (171m). Then do 200-18-171 to find the distance which needs to be travelled in the final section, which comes out as 11m. You can then use suvat with s=ut + 1/2at2 to find t, which you add to the 22s you already have for the first two sections.


What about Jan 10 1iii?
Original post by Jon1234321
January 2012
Q7 ii)

Since P is in contact with the ground why is there not a reaction force acting upwards on P?
E.g I thought it would be T+ mg(reaction) - mg(weight) = ma

Any help would be greatly appreciated and good luck for tomorrow everyone! :smile:


mass R has started to descend so P won't be on the ground anymore :smile:
so no reaction force...remove it and that's correct!

Good luck to you too! :smile:
Reply 239
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c223
2
Original post by Super199
What about Jan 10 1iii?


You want the distance from O to 5m above the ground, which will be 5m less than the distance you just calculated, so 17.3-5=12.3m. Let this equal s, they gave you u in the question, you know a=g as it's free fall, and you want v. So using suvat equation v2=u2 + 2as, v2=4.22+2*9.81*12.3. V should come out as 16.1ms-1 if I've done that calculation correctly.

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