Edexcel M2/M3 June 6th/10th 2013

HOW THE HELL DO YOU DO M2 JUNE 2012 QUESTION 4B , i almost spent 2 hours on this question and i still dont get what they have done, can you guys help me please .. heres the link to the question
https://e8c8970d-a-dbffc5af-s-sites.googlegroups.com/a/virtualfoxdistribution.com/sci18-02-13/dir/e/m2/2012%20Jun%20QP.pdf?attachauth=ANoY7cqLHWKWe67wUpSxeJ4hKnXVCPdOPQ12l39rIMk3GUwCdbPWOGOvxfoGu3N77nABXdrwvq-quWiAId__eb3CIryVrYjUsP_ospBr66DdRQeSoxhpSD77IdxEGX-rC2YiprbI5vJrSLNofnHaTfLj_6BV0TdCi8j36T21BTu4t5HbbTvkm2fx_9OczE7zKX-1pmxGgKwJAh9fZ-blpi32tiZOYs7-Jd-bxiZuAYTnWiPCshGUSOz9fWs1wVdWncDuGXc4Z1QU&attredirects=0

and heres the markscheme

https://e8c8970d-a-dbffc5af-s-sites.googlegroups.com/a/virtualfoxdistribution.com/sci18-02-13/dir/e/m2/2012%20Jun%20MS.pdf?attachauth=ANoY7cqQKjnUtFsh2F7LowBZseqcM-_XTL594BSY-43WkIDh8Dv6DJYFgtKuD5g-ZtEhBBkEtVveYbgzYttsg0gYlFIPS9zLTpR8UdIVg79LoZB9sJZbl415I-TOmmMS-3glGqsY_w_wckRaNkfXCK9rWS-wRHSOXwVYb-TqMTp1x9yjhcZi3EMdbwwIZ2IC7YVrPCOPgjhfjh2-QGHCFzg31j-NvRKgxZ8SRBvUkm-65M3cdc87FnLyGdFr5ryIdGzJDMdD4Q2P&attredirects=0

thanks .

Reply 202

KamilS

1

im having the exact same problem on question 4 june 2012, its disgusting. help would be appreciated

Reply 203

Boy_wonder_95

15

Original post by RYRK

HOW THE HELL DO YOU DO M2 JUNE 2012 QUESTION 4B , i almost spent 2 hours on this question and i still dont get what they have done, can you guys help me please .. heres the link to the question
https://e8c8970d-a-dbffc5af-s-sites.googlegroups.com/a/virtualfoxdistribution.com/sci18-02-13/dir/e/m2/2012%20Jun%20QP.pdf?attachauth=ANoY7cqLHWKWe67wUpSxeJ4hKnXVCPdOPQ12l39rIMk3GUwCdbPWOGOvxfoGu3N77nABXdrwvq-quWiAId__eb3CIryVrYjUsP_ospBr66DdRQeSoxhpSD77IdxEGX-rC2YiprbI5vJrSLNofnHaTfLj_6BV0TdCi8j36T21BTu4t5HbbTvkm2fx_9OczE7zKX-1pmxGgKwJAh9fZ-blpi32tiZOYs7-Jd-bxiZuAYTnWiPCshGUSOz9fWs1wVdWncDuGXc4Z1QU&attredirects=0

and heres the markscheme

https://e8c8970d-a-dbffc5af-s-sites.googlegroups.com/a/virtualfoxdistribution.com/sci18-02-13/dir/e/m2/2012%20Jun%20MS.pdf?attachauth=ANoY7cqQKjnUtFsh2F7LowBZseqcM-_XTL594BSY-43WkIDh8Dv6DJYFgtKuD5g-ZtEhBBkEtVveYbgzYttsg0gYlFIPS9zLTpR8UdIVg79LoZB9sJZbl415I-TOmmMS-3glGqsY_w_wckRaNkfXCK9rWS-wRHSOXwVYb-TqMTp1x9yjhcZi3EMdbwwIZ2IC7YVrPCOPgjhfjh2-QGHCFzg31j-NvRKgxZ8SRBvUkm-65M3cdc87FnLyGdFr5ryIdGzJDMdD4Q2P&attredirects=0

thanks .

You know the centre of mass of L from the previous part and that its mass is m, and you know that the centre of mass of the particle is at P and its mass is km, find the centre of mass of them combined first.

Reply 204

RYRK

Original post by Boy_wonder_95

You know the centre of mass of L from the previous part and that its mass is m, and you know that the centre of mass of the particle is at P and its mass is km, find the centre of mass of them combined first.

and then what do you do?

Reply 205

Here is my working out. I have also explained it below.

[1][2][3]

Total Mass = Km + m.
Since;
> tan(a) = 5/6
> The vertical passes through the new COM
> S to the mid point of the circle is 4a

5/6 must equal x/4a, because tan(a) is opposite side over adjacent side. The adjacent side is 4a. The opposite side is X. The ratio must be the same as 5/6. So X = 4a x 5/6. X = 10a/3.

10a/3 is the horizontal distance from the centre of the circle to the new COM. That means it is 4a-10a/3 from P, because the radius of the circle is 4a. The COM is 2a/3 from P.

Take moments;
> (14a/3 x m) + (0 x km) = km + m (2a/3)
> (14a/3 x m) + (0 x km) = m(k + 1) (2a/3)

The m's cancel;
> 14a/3 = k + 1 (2a/3)
> K + 1 = 7
> K = 6

(edited 10 years ago)

Reply 206

RYRK

Original post by Boy_wonder_95

You know the centre of mass of L from the previous part and that its mass is m, and you know that the centre of mass of the particle is at P and its mass is km, find the centre of mass of them combined first.

and also do you take the origin to be p or s?

P is the origin.

Original post by Riley F.

Here is my working out. I have also explained it below.

[1][2][3]

Total Mass = Km + m.
Since;
> tan(a) = 5/6
> The vertical passes through the new COM
> S to the mid point of the circle is 4a

5/6 must equal x/4a, because tan(a) is opposite side over adjacent side. The adjacent side is 4a. The opposite side is x. The ratio must be the same as 5/6. So x = 4a x 5/6. x = 10a/3.

10a/3 is the horizontal distance from the centre of the circle to the new COM. That means it is 4a-10a/3 from P, because the radius of the circle is 4a. The COM is 2a/3 from P.

Take moments;
> (14a/3 x m) + (0 x km) = km + m (2a/3)
> (14a/3 x m) + (0 x km) = m(k + 1) (2a/3)

The m's cancel;
> 14a/3 = k + 1 (2a/3)
> K + 1 = 7
> K = 6

woww!! that was quick, thanks alot, i finally get it now. :smile:

Reply 209

JenniS

OP

M3 help please!

A light elastic string has natural length 4m and modulus of elasticity 58.8N. Particle P of mass 0.5kg attatched to one end of the string, other end attatched to a vertical point A, particle is released from rest at A and falls vertically.

a) find the distance travelled before P instantaneously comes to rest for the first time.
b)Find the speed of the particle when the string becomes slack for the first time

HELP! anyone...

Reply 210

LShirley95

Original post by JenniS

M3 help please!

A light elastic string has natural length 4m and modulus of elasticity 58.8N. Particle P of mass 0.5kg attatched to one end of the string, other end attatched to a vertical point A, particle is released from rest at A and falls vertically.

a) find the distance travelled before P instantaneously comes to rest for the first time.
b)Find the speed of the particle when the string becomes slack for the first time

HELP! anyone...

Spoiler

(edited 10 years ago)

Reply 211

Rohit.15

3

Original post by Boy_wonder_95

Only sitting M2 this summer! :biggrin:

, I'm taking it your doing both? How's revision going?

I don't have any exams. I hope you enjoy yours :biggrin:

.

Reply 212

Can anyone help me with M2 June 2012 paper question number 4(b) and 5?
Thanks

Reply 213

Already posted a solution for June 2012 4(b).

For Q5, you need to resolve the Impulse. You know what tan(a) is, so you can work out what sin(a) and cos(a) are. The whole angle is 180 degrees, because it is a straight line. You already have (90 + a). The remaining chunk must be (90 - a). The angle between the downward vertical and the impulse is therefore a.

12.5sin(a) for the vertical component and 12.5cos(a) for the horizontal part.

Once you have the impulse in vector form, you just use I=m(V-U) to work out V.

(edited 10 years ago)

Reply 214

koolkid1011

is it me or are the older papers(pre-2007) in m3 harder than the new papers?? obviously june 2012 was hard but the older ones tend to be more difficult..

Reply 215

Original post by Riley F.

Already posted a solution for June 2012 4(b).

For Q5, you need to resolve the Impulse. You know what tan(a) is, so you can work out what sin(a) and cos(a). 12.5sin(a) for the vertical component and 12.5cos(a) for the horizontal part.

Once you have the impulse in vector form, you just use I=m(V-U) to work out V.

oh! so we have to use vertical and horizontal of impulse and work out velocity after? then find the resultant velocity?
dont we have to used sina for horizontal? coz mark scheme has kept sina of impulse to find velocity after.

lol i just saw that the solution for 4b was like one or two post above :colondollar:

(edited 10 years ago)

Reply 216

Original post by AS01

oh! so we have to use vertical and horizontal of impulse and work out velocity after? then find the resultant velocity?
dont we have to used sina for horizontal? coz mark scheme has kept sina of impulse to find velocity after.

lol i just saw that the solution for 4b was like one or two post above :colondollar:

Well this is my working out, so you can take a closer look at how I went about it. And yes my bad, 12.5cos(a) is for the vertical, not horizontal.

(edited 10 years ago)

Reply 217

Original post by Riley F.

Here is my working out. I have also explained it below.

[1][2][3]

Total Mass = Km + m.
Since;
> tan(a) = 5/6
> The vertical passes through the new COM
> S to the mid point of the circle is 4a

5/6 must equal x/4a, because tan(a) is opposite side over adjacent side. The adjacent side is 4a. The opposite side is X. The ratio must be the same as 5/6. So X = 4a x 5/6. X = 10a/3.

10a/3 is the horizontal distance from the centre of the circle to the new COM. That means it is 4a-10a/3 from P, because the radius of the circle is 4a. The COM is 2a/3 from P.

Take moments;
> (14a/3 x m) + (0 x km) = km + m (2a/3)
> (14a/3 x m) + (0 x km) = m(k + 1) (2a/3)

The m's cancel;
> 14a/3 = k + 1 (2a/3)
> K + 1 = 7
> K = 6

you know while taking moments the mark scheme has kept cos of distance can u please explain why. the thing is I am not getting the picture of what is actually happening to the lamina! the point P will not be horizontal right? the diameter QP will be a bit slanted right?

Reply 218

Original post by Riley F.

Well this is my working out, so you can take a closer look at how I went about it. And yes my bad, 12.5cos(a) is for the vertical, not horizontal.

Oh okay that makes sense now! we found I in terms of i and j coz there is come parts of velocity in vertical and horizontal component so the magnitude alone was not the impulse.
Thanks a lot!!

Reply 219