[louisjevans]

(Original post by tom472)
Are you sure the coefficient of friction was 0.3? I'm not sure but with those values I get:

I get:

R=8g - Tsin30

F=ma
Tcos30 - 0.3 [8g - Tsin30] = 8 * 0.05
Tcos30 - 23.52 + 0.3Tsin30 = 0.4
Tcos30 + 0.15T = 23.92
T (cos30 + 0.15) = 23.92
T = 23.92 / (cos30 + 0.15)
T = 23.5427...
T = 23.5 (3sf)

SAME !

[Miyata]

(Original post by jph12)
I think the grade boundaries will be slightly lower than the past average, because there were a few questions that people didn't understand. However, I can't imagine it being like 50 for an A.

For the braking force use F=ma, and take right as positive, with an acceleration of -2.
Constant braking force, R: -R=ma: -R=1400x-2: =2800N.
Cant braking force with air resistance, AR: -R-AR=ma: -R-200=1400x-2:
-R=-2800+200: R=2600.

Thank you. Definitely didn't get that. I really hope its going to be lower, I know for mechanics 50/75 is impossible, but hopefully it will be unexpectedily low, so that as many people as possibe would get A. I think I did okay, didn't make too many mistakes, hopefully enough for an A.

[ChrisE2]

I think the second assumption is that the block moves in the direction of the pulley, (just had a flick through the text book),
Im unsure though, this is an obvious assumption.

[tom472]

I think the breaking force values are negative

http://answers.yahoo.com/question/in...9211214AA2rczT
http://answers.yahoo.com/question/in...8220700AAvzKHW

[ChrisE2]

(Original post by tom472)
I think the breaking force values are negative

http://answers.yahoo.com/question/in...9211214AA2rczT
http://answers.yahoo.com/question/in...8220700AAvzKHW

I have just looked up whether the braking force will be +ve or -ve, and the text book answers are showing +ve resistive forces (look at chapter 5).

[jph12]

Whether resistive forces should be negative or positive is irrelevant when they ask for their magnitude. The magnitude of -3 is 3. The magnitude of -12 is 12. The magnitude of 3 is 3. The magnitude of 12 is 12. The magnitude of -2800 is 2800, and the magnitude of -2600 is 2600. It is simply the size of the number. That is why if you're asked to find the magnitude of a vector, you don't put it in minus' or anything.

[Danatlive]

Im sure I got an inequality for 8)d)

[karchun]

(Original post by Danatlive)
Im sure I got an inequality for 8)d)

It asked the time the particle is above 5 metres not at which times! sorry to burst your bubble

This was posted from The Student Room's Android App on my GT-N7000

[Amy-Margarita]

(Original post by tom472)
I think the breaking force values are negative

http://answers.yahoo.com/question/in...9211214AA2rczT
http://answers.yahoo.com/question/in...8220700AAvzKHW

The question asked for the magnitude of the breaking force, which would always be positive. By using the equations you will get a negative value for it, but the magnitude is the size of the force, and therefore is like the mod of it - which is always positive.

[Hippokrates]

For 8d I got 3.52s

[NickW-Moores]

Anyone get co efficient of 0.413 or something like that

[tom472]

Would they give the marks regardless if it's + or - ? Seems a bit stingy.

[johnadams121212]

(Original post by louisjevans)
Use v=u +at, sub in u and a

i component equals negative j component because its SE.

So t = 20

Therefore v= i - j

Speed = root (1^2 + 1^2)
= 1.41





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

I equated i to j. How many marks do you think I lost?

It's pissing me off, cuz somehow i got the write answers first but then crossed them out.

[karchun]

(Original post by NickW-Moores)
Anyone get co efficient of 0.413 or something like that

It was something around 0.7---

This was posted from The Student Room's Android App on my GT-N7000

[karchun]

(Original post by Hippokrates)
For 8d I got 3.52s

Got 3.45

This was posted from The Student Room's Android App on my GT-N7000

[EAP]

(Original post by louisjevans)
Use v=u +at, sub in u and a

i component equals negative j component because its SE.

So t = 20

Therefore v= i - j

Speed = root (1^2 + 1^2)
= 1.41





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

How many marks was this worth?

[karchun]

(Original post by EAP)
How many marks was this worth?

6

This was posted from The Student Room's Android App on my GT-N7000

[shunn]

Did no body get 6.3 for t acceleration of the block/particle?

[karchun]

(Original post by shunn)
Did no body get 6.3 for t acceleration of the block/particle?

I don't think so..

This was posted from The Student Room's Android App on my GT-N7000

[EAP]

(Original post by karchun)
6

This was posted from The Student Room's Android App on my GT-N7000

Bugger. I used s=ut+1/2 at^2 (from the position vector we worked out earlier) and put that i= -j and got t=40 and then got like 3i-5j and did sq root of 5^2 + 3^2 and got 5.83.
Reckon I could get any method marks?