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# M2 projectiles

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1. Where am I going wrong 5i.
http://www.ocr.org.uk/Images/62242-q...echanics-2.pdf
http://m.imgur.com/VxkJVCr
2. You've used the approximation , leave your answer in terms of . Otherwise, it's fine.
3. (Original post by Zacken)
You've used the approximation , leave your answer in terms of . Otherwise, it's fine.
Hmm. If thats the case where am I going wrong in the next part.

Its a bit messy lol.
http://m.imgur.com/V7LRq6s
4. (Original post by Super199)
Hmm. If thats the case where am I going wrong in the next part.

Its a bit messy lol.
http://m.imgur.com/V7LRq6s
The line beginning with 3x^2 - 40xtan 30 + ...

Check your signs, why has it suddenly become -40xtan30 etc...
5. (Original post by Zacken)
The line beginning with 3x^2 - 40xtan 30 + ...

Check your signs, why has it suddenly become -40xtan30 etc...
What;s wrong with it. I've just taken everything to the other side to get of the -3 in front of the x^2.
6. (Original post by Super199)
What;s wrong with it. I've just taken everything to the other side to get of the -3 in front of the x^2.
Then why is the number before the x^2 +3 and not -3...?
7. (Original post by Zacken)
Then why is the number before the x^2 +3 and not -3...?
I don't get what you mean sorry.

y=xtan30 - 3/40x^2

Sub 0.6 in and multiply by 40.

24 = 40xtan30 - 3x^2
Make x^2 positive
3x^2-40xtan30 +24 =0
8. (Original post by Super199)

24 = 40xtan30 - 3x^2
Make x^2 positive
3x^2-40xtan30 +24 =0
Urgh, I'm being an idiot. Your mistake was in the first part.

You're dividing by (for the x^2 term) so you should multiply by . You're multiplying by .

That is:
9. (Original post by Zacken)
Urgh, I'm being an idiot. Your mistake was in the first part.

You're dividing by (for the x^2 term) so you should multiply by . You're multiplying by .

That is:
No worries
Do you mind helping with the third part. What does it mean by 'and the particle is rising' what does that effect?
10. (Original post by Super199)
No worries
Do you mind helping with the third part. What does it mean by 'and the particle is rising' what does that effect?
It just means that there are two moments where y=0.6, one when the particle is going up, passes the y=0.6 mark and continues going up to the max height before falling back down, passing the y=0.6 mark again and then falling down to the ground. The question is specifying to work with the first moment when the particle crosses the y=0.6 mark on its way up.
11. (Original post by Zacken)
It just means that there are two moments where y=0.6, one when the particle is going up, passes the y=0.6 mark and continues going up to the max height before falling back down, passing the y=0.6 mark again and then falling down to the ground. The question is specifying to work with the first moment when the particle crosses the y=0.6 mark on its way up.
So it has nothing to do with ii?
12. (Original post by Super199)
So it has nothing to do with ii?
Yep. It's just normal SUVAT.

(Although there is an alternative method where you can differentiate the answer to (i), blah blah but it's more complicated)
13. (Original post by Zacken)
Yep. It's just normal SUVAT.

(Although there is an alternative method where you can differentiate the answer to (i), blah blah but it's more complicated)
yh fair enough. So what was the point of putting that particle rising bs. Like that ****ed with me and I thought I had to use x= 1.73 from the previous part or some ****.
14. (Original post by Super199)
yh fair enough. So what was the point of putting that particle rising bs. Like that ****ed with me and I thought I had to use x= 1.73 from the previous part or some ****.
Like I said, it specifies that it wants the direction when the particle is going up and not at the second moment when the particle is going down. Because there are two points on the parabolic trajectory where y=0.6.
15. (Original post by Zacken)
Like I said, it specifies that it wants the direction when the particle is going up and not at the second moment when the particle is going down. Because there are two points on the parabolic trajectory where y=0.6.
would you use u=7sin30 if the particle was going down? Or is that only for up
16. (Original post by Super199)
would you use u=7sin30 if the particle was going down? Or is that only for up
You'd use it for both.
17. (Original post by Zacken)
You'd use it for both.
Cheers.
With q1 I never understand how these energy qs work.
Why don't we consider the resistance to motion * 30 as an energy?
18. (Original post by Super199)
Cheers.
With q1 I never understand how these energy qs work.
Why don't we consider the resistance to motion * 30 as an energy?
Where did 30 come from?
19. (Original post by Zacken)
Where did 30 come from?
Sorry 20m. As that is the work done by the resistance?
20. (Original post by Super199)
Sorry 20m. As that is the work done by the resistance?
Yes, it's the work done against resistance, but the energy of the sledge and load has nothing to do with the resistance of the motion, the energy of the sledge and load is purely K.E - G.P.E. The work done against resistance is used in the second part.

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