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m2 impulse exam question (edexcel)

I am really confused with q5 on this paper, I've tried approaching the question with vectors and for part a worked out:

mom before: 0.25(30i) = 7.5i
mom after: 0.25v

but not sure how to get an answer from there, and am clueless with part b :colondollar:

would appreciate some advice on how to tackle this thanks :smile:


Question paper: https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/June%202012%20QP%20-%20M2%20Edexcel.pdf

mark scheme: https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/June%202012%20MS%20-%20M2%20Edexcel.pdf
Reply 1
Original post by Katiee224
I am really confused with q5 on this paper, I've tried approaching the question with vectors and for part a worked out:

mom before: 0.25(30i) = 7.5i
mom after: 0.25v

but not sure how to get an answer from there, and am clueless with part b :colondollar:

would appreciate some advice on how to tackle this thanks :smile:


Question paper: https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/June%202012%20QP%20-%20M2%20Edexcel.pdf

mark scheme: https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/June%202012%20MS%20-%20M2%20Edexcel.pdf


Read this and was wondering why you were saying mum before and mum after. :rofl:

Anyways, vectors is slightly awkward here, let's resolve instead (which is almost the same thing as vectors)

If we consider only horizontal motion, we get:

Impulse = change in momentum. The horizontal component of impulse is 12.5sinα12.5 sin \alpha pointing to the left.

The speed of B horizontally is (let's say v):

The 12.5sinα=0.25v0.25(30)12.5\sin \alpha = 0.25v - 0.25(-30) where we use - since the ball is moving 30 metres per second opposite to the impulse and we gave impulse a +12.5 sinalpha.

Anyways, since sinα=0.6\sin \alpha = 0.6 we have: 12.5(0.6)=0.25v+7.5    7.5=0.25v+7.512.5(0.6) = 0.25v + 7.5 \iff 7.5 = 0.25v + 7.5, this gives v=0v=0.

So there is no horizontal component of B, thank god, that makes life simple!

Now - let's consider vertical motion:

This gives us: 12.5cosα12.5\cos \alpha (impulse acts upwards)

And the change in momentum is 0.25u00.25u - 0 (u is our vertical speed) and there is no vertical motion previous to the impulse.

So 12.5cosα=0.25u    u=4×12.5×0.8=4012.5\cos \alpha = 0.25u \iff u = 4\times 12.5 \times 0.8 = 40. So the speed is 40 vertical and 0 horizontal, and hence 40 overall.

And from this, since the 40 overall is acting vertically upwards, then the direction of travel of B is "perpendicular to the original direction". i.e: it went from pure horizontal to pure vertical.
Reply 2
Original post by Zacken
Read this and was wondering why you were saying mum before and mum after. :rofl:

Anyways, vectors is slightly awkward here, let's resolve instead (which is almost the same thing as vectors)

If we consider only horizontal motion, we get:

Impulse = change in momentum. The horizontal component of impulse is 12.5sinα12.5 sin \alpha pointing to the left.

The speed of B horizontally is (let's say v):

The 12.5sinα=0.25v0.25(30)12.5\sin \alpha = 0.25v - 0.25(-30) where we use - since the ball is moving 30 metres per second opposite to the impulse and we gave impulse a +12.5 sinalpha.

Anyways, since sinα=0.6\sin \alpha = 0.6 we have: 12.5(0.6)=0.25v+7.5    7.5=0.25v+7.512.5(0.6) = 0.25v + 7.5 \iff 7.5 = 0.25v + 7.5, this gives v=0v=0.

So there is no horizontal component of B, thank god, that makes life simple!

Now - let's consider vertical motion:

This gives us: 12.5cosα12.5\cos \alpha (impulse acts upwards)

And the change in momentum is 0.25u00.25u - 0 (u is our vertical speed) and there is no vertical motion previous to the impulse.

So 12.5cosα=0.25u    u=4×12.5×0.8=4012.5\cos \alpha = 0.25u \iff u = 4\times 12.5 \times 0.8 = 40. So the speed is 40 vertical and 0 horizontal, and hence 40 overall.

And from this, since the 40 overall is acting vertically upwards, then the direction of travel of B is "perpendicular to the original direction". i.e: it went from pure horizontal to pure vertical.


thanksss so much, your explanation was really helpful, i understand it all now! can you be my personal maths tutor please :biggrin::laugh:

how come you are always so quick to reply to my problems hehe :blushing:
Reply 3
Original post by Katiee224
thanksss so much, your explanation was really helpful, i understand it all now! can you be my personal maths tutor please :biggrin::laugh:

how come you are always so quick to reply to my problems hehe :blushing:


Oh, phew! Glad it makes sense. :biggrin:

It's 'cause I'm secretly madly in love with you. :laugh:
Original post by Katiee224
thanksss so much, your explanation was really helpful, i understand it all now! can you be my personal maths tutor please :biggrin::laugh:

how come you are always so quick to reply to my problems hehe :blushing:


27 minute turn around is slow for him loool. He just likes helping people and his latex skills are unmatched

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