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# m4 and p6 watch

1. 1.
A rugby player is running due North with speed 4m/s. He throws the ball horizontally and the ball has an initial velocity relative to the player of 6m/s in the direction bearing 180+¤ where tan¤=4/3. Find the magnitude and the direction of the initial velocity of the ball relative to a stationary spectator.

2.
A small bead B of mass m can slide on a smooth circular wire of radius a which is fixed in a vertical plane. B is attached to one end of a light elastic string, of natural length 3a/2 and modulus of elasticity mg. The other end of the string is attached to a fixed point A which is vertically above the centre C of the circula wire with AC=3a.
Show that there is a stable equilibrium position of the system when ¤, the angle which the radius through B make with the downward vertical, satisfies cos¤=-1/6

3.

the equation of motion of a particle moving on the x-axis is
d^2y/dx+2kdy/dx+n^2x=0
Given that k and n are positive constants, with k<n show that the time between two successive maxima of x is constant

p6
1.
A= 101
311
427

the transformation T:R^3->R^3 is represented by the matrix A.
Find the catesian equations of the line which is mapped by T onto the line
x=y/4=z/3

I dont know how to start or do this kind of question at all so please explain me in detail

Again too many questions
>-<
2. For the first one, write the velocity of the ball, relative to the rugby player in vector form. It is already moving at 4ms due north, so if the balls velocity, relative to the rugby player is xi + yj, then relative to the spectators it will be xi + (y+4)j.
3. I still cant get the right answer
which is 4.82m/s and bearing 275deg
it isnt that simple i guess
4. I havent done any of the q's but i can give you some general pointers which may be useful...

1) For relative velocity questions it is essential that you first write down the correct vector equation, ie that bVp = Vb - Vp. Then use this information to construct a vector triangle, and use geometry to find the unknown length/angle. Often you have to use the cosine or sine rule.

2) Split the energy up into PE and EPE, with each one expressed as a function of ¤. Find dV/d¤, and it should be 0 when cos¤ = -1/6

3) These ones can be very tricky. Start by finding the general solution (P4) as x = f(t). Maxima occur when dx/dt = 0, so you need to find an expression for the values of t at the Nth and (N+1)th maxima. Almost always involving pi. Then take away and you should be left with something independant of N and t.

4) If the original vector was p = (x y z) [should be a column], and the result was q, then Ap = q ie p = A^-1 q basically find the inverse

Sorry if it was brief post back if you need more help
5. Yea, i get that answer the way i suggested.
6. (Original post by It'sPhil...)
I havent done any of the q's but i can give you some general pointers which may be useful...

1) For relative velocity questions it is essential that you first write down the correct vector equation, ie that bVp = Vb - Vp. Then use this information to construct a vector triangle, and use geometry to find the unknown length/angle. Often you have to use the cosine or sine rule.

2) Split the energy up into PE and EPE, with each one expressed as a function of ¤. Find dV/d¤, and it should be 0 when cos¤ = -1/6

3) These ones can be very tricky. Start by finding the general solution (P4) as x = f(t). Maxima occur when dx/dt = 0, so you need to find an expression for the values of t at the Nth and (N+1)th maxima. Almost always involving pi. Then take away and you should be left with something independant of N and t.

4) If the original vector was p = (x y z) [should be a column], and the result was q, then Ap = q ie p = A^-1 q basically find the inverse

Sorry if it was brief post back if you need more help

Sorry
i still cant do any of the questions
Except last one i know how to do them but i cant get the answer right. for number 1 is that right picture?
Attached Images

7. (Original post by totaljj)
Sorry
i still cant do any of the questions
Except last one i know how to do them but i cant get the answer right. for number 1 is that right picture?
For q1), you have bvp = 6m/s??
its v(b) that's 6m/s.
The rest is ok.
I got the same as jamesf.

Edit: the angle is wrong!
¤ is the acute angle at the top between the 4m/s velocity and the 6m/s velocity.
8. I've attempted Q4, but I'm not too sure about it
Better wait, perhaps, until someone confirms it !

Edit: problem with the image attachment. Looking at it !

Edit2: I had to split the image file into two bits. Mathtype can only handle so much - not too much

Edit3: changed some of the arithemetic on image_2 after fixing an element in the inverse matrix A^-1.
Attached Images

9. Hmmm, i got a slightly different inverse matrix to that, lemme check my workings.
10. I got A^-1 = 1/7*
(5,2, -1
-17, 3, 2
2, -2, 1)

Which gives the answer [x,y,z] = [10/7, 1/7, -3/7]

Plug this back and you get [1,4,3]
11. (Original post by JamesF)
I got A^-1 = 1/7*
(5,2, -1
-17, 3, 2
2, -2, 1)

Which gives the answer [x,y,z] = [10/7, 1/7, -3/7]

Plug this back and you get [1,4,3]
Yeah, I did a subtraction, instead of an addition on that central element - it should be 3/7
Fixing it now.
12. For question 2
is this picture ok and is it possible to get the length of string

always thx
Attached Images

13. (Original post by totaljj)
For question 2
is this picture ok and is it possible to get the length of string

always thx
T=λx/l

At equilibrium,as an approximation, let tension in string be that from weight of bead, therefore
(you'll actually have to resolve forces here, at the bead)

T=mg

also,

l=3a/2
λ=mg

then,

x=Tl/λ
x = mg*(3a/2)/(mg)
x = 3a/2 (extension) (remember this is only an approx)
=====

So, length of stretched string is l'=l+x

l'= 3a/2 + 3a/2
l'=3a
===
So, AB = 3a (an approx)

I think the picture should therefore look more like the one I attached.
Attached Images

14. The length of the string is found using the cosine rule. If the length of the string is d, then

d² = 9a² + a² - 6a².cos(pi - x) [i'll call my angle as x]

=> d² = a²(10 + 6cosx)

Therefore the EPE is mg/3a(d - 3a/2)²

= mg/3a (d² + 9a²/4 - 3ad)

=mg/3a (10a² + 6a²cosx + 9a²/4 - 3a².(10 + 6cox)^1/2)

Also PE is -amgcosx

So V = mga[ cosx - (10 + 6cosx)^1/2 ] + constant (after simplifying!)

dV/dx = mga[ sinx - 3sinx(10 + 6cosx)^-1/2 ]

dV/dx = 0 => mga.sinx(1 - 3(10 + 6cosx)^-1/2) = 0

Sinx = 0 is unstable, so 1 = 3(10 + 6cosx)^-1/2

Squaring gives 1 = 9/(10 + 6cosx)

or 10 + 6cosx = 9 => cosx = -1/6

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Updated: June 11, 2004
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