Mechanics 2 Projecticles Question (OCR)

Hi, I can't seem to find what I'm doing wrong on this question. I've got the values for s, v, a and t (for vertical and horizontal), but can't seem to get an equation that gets rid of the theta.

Can anyone point out how do the first part to this question? It'll be much appreciated!

Thanks :smile:

ocr m2 june 2013 q7.jpg195.1KB

bump

Original post by ChrissM

Note that at point of impact with plane, horizontal velocity is 14cos20 and vertical velocity is 14sin20

Reply 3

ChrissM

Original post by brianeverit

Note that at point of impact with plane, horizontal velocity is 14cos20 and vertical velocity is 14sin20

I'd already done that, but I still can't make it work.

Reply 4

ChrissM

Original post by brianeverit

Note that at point of impact with plane, horizontal velocity is 14cos20 and vertical velocity is 14sin20

Not that you care, but I finally got it (after at least two hours of trying). I forgot to put the vertical velocity at P as negative. I don't know whether to be happy or whether I wwant to die. :P

Reply 5

Sataris

Original post by ChrissM

Not that you care, but I finally got it (after at least two hours of trying). I forgot to put the vertical velocity at P as negative. I don't know whether to be happy or whether I wwant to die. :P

You know, I just tried to identify this paper and stumbled across the examiner's report - they said that many students made that mistake, so don't feel too bad! :biggrin:

Pretty interesting question though, I like it. I thought my answer for v was far too low, but I guess not.

Reply 6

ChrissM

Original post by Sataris

You know, I just tried to identify this paper and stumbled across the examiner's report - they said that many students made that mistake, so don't feel too bad! :biggrin:

Pretty interesting question though, I like it. I thought my answer for v was far too low, but I guess not.

Thanks! The question itself wasn't that bad provided you don't make that mistake. It just sucks how the rest of the questions are dependent on you finding theta, which is a lot of marks!

Reply 7

brianeverit

Original post by ChrissM

Not that you care, but I finally got it (after at least two hours of trying). I forgot to put the vertical velocity at P as negative. I don't know whether to be happy or whether I wwant to die. :P

Glad to hear you got it out o.k.in the end.

Reply 8

Sataris

Original post by ChrissM

Thanks! The question itself wasn't that bad provided you don't make that mistake. It just sucks how the rest of the questions are dependent on you finding theta, which is a lot of marks!

Oops, I just realised I forgot to work out theta :colondollar:

But the rest is fine with only the information given in the question.

Reply 9

ChrissM

Original post by Sataris

Oops, I just realised I forgot to work out theta :colondollar:

But the rest is fine with only the information given in the question.

What was your method without theta?

Reply 10

Sataris

Original post by ChrissM

What was your method without theta?

ii) just sub in the value for the final vertical velocity into S = ut + .5at^2.

iii) horizontal component is -vcos20. You can work out the horizontal distance between A and P with trig and part ii. Then use d=st horizontally and rearrange into an expression for w in terms of t.

Vertically, you can use S = ut + .5at^2 again - S is -2.90 from ii, and u is wsin20. Sub in the expression for w, and solve for t. Then sub t into your expression for w, and you get w! No theta required! :biggrin: