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# edexcel M2 Jan 2010 question help?

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1. edexcel M2 Jan 2010 question help?
Here is the paper: http://www.kingsleyschoolbideford.co..._20100129.aspx

In Q4 why is the impulse split into 2 parts?

In Q6 why does the mark scheme have friction acting at an angle?

And how do you do Q8?

Thanks
2. Re: edexcel M2 Jan 2010 question help?
For Q8 you apply the equation s = ut + 0.5at^2

To find the time, you need to apply this.
The horizontal speed is constant as is of value u.
The horizontal distance is x.
Speed = Distance x Time
So time = u/x

Now we have the time we consider the vertical components.
SUVAT:
s = y
u = uc
a = -g ( = -9.8 )
t = u/x

So s = ut + 0.5at^2

Sub in and simplify and you'll get the correct answer.
Part B is straight forward.
Part C, all you do is use trigonometry.
If you need help with part B or C quote the post and I'll reply when I'm on.

For question 4, it's not split into two parts. Initially u = 30, then it's hit at an angle with an impulse giving it velocity in the j direction too.
For Q4 it would be easiest to change the components into i, j form then find the magnitude of the impulse after.

So u = 30i
v = 40cos(60)i + 40sin(60)j = 20i + 20*root(3)j
I = mv - mu

For question 6, Friction is at an angle relative to the rod, not to the ground though.
Last edited by JonathanM; 30-05-2012 at 14:24.
3. Re: edexcel M2 Jan 2010 question help?
(Original post by JonathanM)

For question 6, Friction is at an angle relative to the rod, not to the ground though.

Thanks,
but why is friction at an angle?
4. Re: edexcel M2 Jan 2010 question help?
Sorry for the late reply but we aren't taking moments relevant to the ground. We are taking them relative to the rod. So the component of friction relative to the rod wouild be equal to Length*Fsin(angle).

I'll draw a diagram to make sure. Triangle resolving .

http://img703.imageshack.us/img703/7...945fe00002.jpg

Red line on this diagram is the perpendicular component of F relative to the rod.
5. Re: edexcel M2 Jan 2010 question help?
(Original post by JonathanM)
Sorry for the late reply but we aren't taking moments relevant to the ground. We are taking them relative to the rod. So the component of friction relative to the rod wouild be equal to Length*Fsin(angle).

I'll draw a diagram to make sure. Triangle resolving .

http://img703.imageshack.us/img703/7...945fe00002.jpg

Red line on this diagram is the perpendicular component of F relative to the rod.

would you explain Q10 c) in more detail
6. Re: edexcel M2 Jan 2010 question help?
(Original post by NightKings)
would you explain Q10 c) in more detail
There is no 10)c), did you mean 8)c)?
7. Re: edexcel M2 Jan 2010 question help?
(Original post by JonathanM)
There is no 10)c), did you mean 8)c)?
yes. you can see, how much I dislike this paper. I can't even remember how many questions were actually on the paper
8. Re: edexcel M2 Jan 2010 question help?
I think there might be an error in the mark scheme for question 4.

I got the impulse as -2.5i+5(root3)j.

if you do this, the angle between this and AB is tan=5root3/2.5. You shouldn't have to subtract this value from 180 like it has done in the mark scheme. Please someone correct me.

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Last updated: January 5, 2013
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