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The Official Edexcel M2 Revision Thread! (24/06/05 AM) watch

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    The exam is approaching...post questions here.
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    Q6.(b) June 2001 - soz have got the answer but don't really understand the method. Take a look at that if you like
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    (Original post by Revenged)
    Q6.(b) June 2001 - soz have got the answer but don't really understand the method. Take a look at that if you like
    Supposedly for (a) of that question, you got two equations, one from the law of conservation of momentum and one from the coefficient of restitution. You just substitute 'speed of B = 3u/2' into the momentum equation to find the speed of A.
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    can someone have a look at this question?

    Q. A smooth uniform sphere P of mass m is at rest on smooth horizontal ground at a distance d from a vertical wall. Another smooth uniform sphere Q has the same radius and the same mass as P and is moving with speed u in a straight line along the ground towards the wall in a direction which is perpendicular to the wall. Sphere Q strikes sphere P directly. The coefficient of restitution between P and Q is e, where e>0.

    (a) Show that the speeds of P and Q after the collision are u(1+e)/2 and u(1-e)/2 respectively.

    this bit is fine. its the next part that i'm confused about.

    (b) Sphere P then goes on to strike the wall. The coefficient of restitution between P and the wall is also e. By considering the time interval between the first and second collisions of P and Q, or otherwise:
    show that P and Q collide for a second time at a distance 2de²/(1 + )

    i'd be grateful for an outline about what to do, i've been trying impulse=Ft but i don't know how that would help in finding the time interval between the two collisions? anyone got any pointers, would be muchly appreciated.
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    btw. which year's M2 is the most difficult?
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    anyone have the mark schemes for Jan 2005 paper???
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    (Original post by lilbug)
    can someone have a look at this question?

    Q. A smooth uniform sphere P of mass m is at rest on smooth horizontal ground at a distance d from a vertical wall. Another smooth uniform sphere Q has the same radius and the same mass as P and is moving with speed u in a straight line along the ground towards the wall in a direction which is perpendicular to the wall. Sphere Q strikes sphere P directly. The coefficient of restitution between P and Q is e, where e>0.

    (a) Show that the speeds of P and Q after the collision are u(1+e)/2 and u(1-e)/2 respectively.

    this bit is fine. its the next part that i'm confused about.

    (b) Sphere P then goes on to strike the wall. The coefficient of restitution between P and the wall is also e. By considering the time interval between the first and second collisions of P and Q, or otherwise:
    show that P and Q collide for a second time at a distance 2de²/(1 + )

    i'd be grateful for an outline about what to do, i've been trying impulse=Ft but i don't know how that would help in finding the time interval between the two collisions? anyone got any pointers, would be muchly appreciated.
    Ok first you need to use the 'old skl' formula s=d/t. so in the cases below, s = speed not distance.

    I'll start by determining the time it took for P to reach the wall.
    t = d/s = d/0.5u(1+e)

    Now i'll find how far Q has moved in this time
    d = st = 0.5u(1-e) * d/0.5u(1+e) = d(1-e)/(1+e)

    The speed of P after it rebounds from the wall is v = eu = e(0.5u(1+e))

    So now we have both particles moving towards each other with a distance of d - d(1-e)/(1+e) between them. This can be simplified to 2ed/(1+e)

    We now need to find the distance from the wall where they meet. so we need [(distance/Vp+Vq) * Vp]
    so [2ed/(1+e) ÷ 0.5u(1+e)+0.5eu(1+e)] * 0.5eu(1+e)

    This then simplifies to give 2de²/(1+e²)

    Sorry i havent put the working out but its too long. If you dont understand wat iv done feel free to ask
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    here's da jan 2005 mark scheme...
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  1. File Type: pdf 12 M2 January 2005 mark scheme.pdf (66.8 KB, 162 views)
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    thanx

    i hate ladder and rod questions
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    i need a little help with projectiles..can someone show me how you find the speed of a projectile immediately before it is about to hit the ground..here's an eg..

    A stone is projected from a point O at the top of a vertical cliff OA. the point O is 14.7 m above the sea.the speed of projection is 19.6 ms-1 and the angle of projection is 30° above the horizontal.the stone moves.freely under gravity and T secs after projection hits the sea at the point X.
    Now i found T which is 3 secs and distance AX is 50.9 m. i couldnt do part c which says find the speed of the stone immediately before it hits the sea.the answer is 25.9ms-1.
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    Horizontal velocity of stone immediately before it hits the sea:

    19.6cos30 (constant throughout flight)

    Vertical velocity of stone immediately before it hits sea:

    Use SUVAT
    v^2 = u^2 + 2as
    v^2 = (19.6sin30)^2 + (2)(-9.8)(-14.7)
    v^2 = 9.8^2 + 288.12
    v^2 = 96.04 + 288.12
    v = 19.6

    Now use Pythagoras to get the speed:

    sqrt((19.6cos30)^2 + 19.6^2) = 25.9ms^-1 (to 3sf)

    Hope that's clear enough.
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    ya tht explains it..thnx
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    What does ranges of e mean?? I neva got those questions. Also, I got a stonker question i need helping with. Its question 1 in the Revise for M2 book first revision exercise of chapter 5.

    No teachers could do it so hopefully some bright spark here can. PLEASE HELP!!!
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    (Original post by jr2007)
    What does ranges of e mean?? I neva got those questions. Also, I got a stonker question i need helping with. Its question 1 in the Revise for M2 book first revision exercise of chapter 5.

    No teachers could do it so hopefully some bright spark here can. PLEASE HELP!!!
    post the question please

    range of e is usually by considering the speed of one of the spheres..
    i.e at times u can tell its greater than/ less than zero or some other value
    by getting e in terms of that speed u can determine the range for it

    hav a look at this n it shud get clearer

    http://www.thestudentroom.co.uk/t124251.html
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    I've got it:

    Draw a diagram - it always helps.

    I'll do my best to explain it without one though!

    OK, you've got 3 forces to consider: the weight of the ladder, the reaction of the wall at A and the reaction of the plane at B.

    Let the mass of the ladder = M, and the length of the ladder = 2L.
    Let the reaction at A = S, and the reaction at B = T.

    Resolve vertically: M.g = T.costheta

    Resolve horizontally: S = T.sintheta

    Take moments about B: L.sinphi.Mg = 2L.cosphi.S
    Cancel the Ls: sinphi.Mg = 2cosphi.S

    Substitute your first and second expression into the third: sinphi(Tcostheta) = 2.cosphi(Tsintheta)
    Cancel the Ts: sinphi.costheta = 2.cosphi.sintheta
    Divide by cosphi: tanphi.costheta = 2.sintheta
    Divide by costheta: tanphi = 2.tantheta

    Which is the answer you're looking for!
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    After all this, I better not have done something stupid along the way...

    Question: The diagram shows a unifrom ladder AB resting in equilibrium with end A in contact with a smooth vertical wall and end B in contact with a smooth inclined plane which make an angle theta with the horizontal. Given that the ladder makes an angle phi with the vertical, show that tanphi = 2tantheta.
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    (Original post by jr2007)
    What does ranges of e mean?? I neva got those questions. Also, I got a stonker question i need helping with. Its question 1 in the Revise for M2 book first revision exercise of chapter 5.

    No teachers could do it so hopefully some bright spark here can. PLEASE HELP!!!
    I know it's weird and these find the range of e questions only appeared in the exam...

    something i never knew before was that if a particle collides with a wall with a speed of u and the coefficient of restitution is e then the speed of rebound would be eu.

    e.g. if the coefficient of restitution of the ball after the collision was 1/2 then the speed after the collision would be 1/2 u

    also something that i taught myself with these questions is that if two particles collide, then one particle hits a wall rebounds and collides with the 1st particle again the coefficient of restitution is still the same...

    that may help you with the question...
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    just a quick question about centre of masses and equilibriums on a plane.
    Heinemann M2 book Ex2C, Q7
    a uniform rectangular lamina ABCD where AB=10 and BC=5 is place on a rough inclined plane.
    a) determine where rectangle can remain in equilibrim when the plane is angled at an angle of 25
    b) calculate the maximum possible angle of inclination...


    basically, i can do the question easily but i always do it by doing part b) first (ie i find max poss angle of inclination, and the plane must therefore be less than this).
    i just felt that it seems very odd to put these questions in this order - maybe there's a much easier way of doing part a) i am missing out on???
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    For the m2 exam, do you get an answer booklet in the question paper like c2 and c3?
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    (Original post by Widowmaker)
    For the m2 exam, do you get an answer booklet in the question paper like c2 and c3?
    yes
 
 
 
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