For iv, there was no sideways movement, and the force up (1.2) was less than gravity (0.4g)

For iii, there was sideways movement, but it was less than friction, so no movement.

Reply 426

MsFahima

Original post by xsanda

Xsanda's solutions:

Falling question
1i) v=28 [2]
1ii) s=6.384 [2]
1iii) t=0.598 [3]

Collision question
2i) u=6 [4]
2ii) m=0.2 [4]

Cycling question
3i) 20km, 5hrs [3]
3ii) 7 hrs [3]
3iii) 50km [2]

110° tick
4i) 9.74N, 35.4° [6]
4ii) 10.3N [1]
4iii) 54.6° [2]

θ force
5i) 1.04 [3]
5ii) 1.06 [3]
5iii) a=0 [3]
5iv) a=0 [2]

Coalescing
6i) v=68 [4]
6ii) s=78 [4]
6iii) k=132 [1]
6iv) v=7 [5]

Prism
7i) a=0.7 (show that), T=1.68 [5]
7ii) v=1.26 [2]
7iii) -a = 0.9 [2]
7iva) T=1.2 [3]
7ivb) Fr = 1.12 [3]

For q5. The force was applied so those answers are wrong... that's what I think anyway.

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Reply 427

wat a wizard

7)iv)b 0.3gsin30 -fr = 0.3*-0.9

fr = 1.47 + 0.27 = 1.74?

Reply 428

xsanda

Original post by lolkik

iit must be 13.3 for the resultant force

13.3 was the wrong diagonal: the diagonal from the ends of both of the vectors, rather than the diagonal from the start to the end tip-to-tail. You needed to move one vector to the end of the other.

Was q5 pulling?

T=7 hours
K=132

Is deffo correct

Reply 431

username1967507

Original post by seizetoday

T=7 hours
K=132

Is deffo correct

how did you get k to be 132? so i can see where i went wrong

Reply 432

Chenice

Original post by wat a wizard

7)iv)b 0.3gsin30 -fr = 0.3*-0.9

fr = 1.47 + 0.27 = 1.74?

You've missed out the tension I think

Reply 433

Minecraft27

What the **** was that paper? Questions 1-3 and 6-7 were fine but 4 and 5 were abysmal! I will be lucky to have picked up 5 of those 20 marks. Urgh.

I think 56/72 will be an A. Question 3 was a bit hard, but I got 7 hours BY GUESSING.

I guess overall the difficulty balances to medium, as all questions were easy apart from 4 and 5.

Reply 434

seizetoday

Predicted grade boundaries?

Reply 435

AsphodelMist

Original post by ChoccyPhilly

A lot of m friends are saying that the 1.2 force was pushing down onto the block. I thought it was pulling

This was the strangest question in terms of wording. 'A 1.2N force is applied to P' there was never any indication for pushing or pulling so it required an assumption to be made. I assumed push, but loads assumed pull. Some are saying that it can't have been push because the angle would have to be below the horizontal. But that's a load of rubbish, since when was that a law? However, I know I got the last two parts wrong because I did a silly mistake, which I'm kicking myself for now. Just hoping that's all I got wrong, which I think it is, in which case this stupid question doesn't matter to me, I can afford to lose it.

Reply 436

NatRichmond

i second that motion...i got k as 96 i think. how did you get 132?

Reply 437

seizetoday

Original post by Sid1234

how did you get k to be 132? so i can see where i went wrong

They gave you are formula which was displacement of Q= k-2t^3 I think
Cant remember exactly, but you plug in t=3
So you get k-54 but you equal this to what you get in your previous answer which was the displacement OP (OP=OQ at t=3)

So K-54=78
K=132

Hope this helps.

Reply 438

olliecass1