User:Kalaamitava

''' We build too many walls and not enough bridges. '''

Sir Isaac Newton

Physics- Gravity
First I will start with a straight strip with negligible thickness and depth. It has length L and mass per unit length M. The other object has mass "m" and has negligible dimension. The other object lies at a distance R from the strip on a line perpendicular to the strip and going through the mid of the strip. Now I want to study the nature of gravitational force by the strip on the object "m". Let us consider an infinitesimal length dx on this strip at a distance x from mid point. The gravitational pull of this infinitesimal portion on the mass “m" will have two components. One is straight towards the strip and another is perpendicular to it. The second force gets cancelled out if we consider both sides of the mid point. Only the first one will add up. Now writing the equation here for the force which is directed towards the strip.

(GmMdx/(R²+ x²))*(R/√(R²+ x²)). Now integrating it over the half length (L/2) we get

∫(GmMdx/(R²+ x²))*(R/√(R²+ x²)) = (GmMR)∫dx/(R²+ x²)3/2

Considering x/R = tanθ, we get dx = Rdθ/cos2θ,

substituting and rearranging

(GmMR)∫dx/(R²+ x²)3/2 = (GmMR)*(1/R²)∫cosθdθ = (GmMR)*(1/R²)*sinθ =(GmM)*(1/R)*sinΦ,

considering θ changes from 0 to Φ over L/2, Now multiply and divide by L/2, we get Gm(Ṁ/2)/(RṘ), Where Ṁ = M*L And (L/2)/Ṙ= sinΦ, considering both sides of the mid point it becomes GmṀ/(RṘ). Now comparing with normal equation GmṀ/R², we can see GmṀ/R² > GmṀ/(RṘ), as Ṙ > R.

2nd case - Now the second mass "m" lies on the line of the strip but at some distance from the strip. In this case the integration is from R1 (nearest to "m") to R2 (farthest from "m") and very straightforward.

∫GmMdx/x2 = -∫(GmM*(1/x) = ∫(GmM/(1/R1 - 1/R2)= GmM(R2 - R1)/(R1R2)= GmṀ/(R1R2)

and GmṀ/(R1R2) > GmṀ/R2, where R = (R1 + R2)/2

3rd case - using the result from first case. now suppose the negligible thickness is dR (towards the direction of R). Now integrating GmṀ{dR/(RṘ)} over length L will give total force acting by the bigger object with area L*L on the smaller object.

∫GmṀ{dR/(RṘ)} = ∫GmṀ{dR/(R√(R2 +(L/2)2)} as Ṙ = √(R2 +(L/2)2)

dṘ=RdR/Ṙ, the integration becomes GmṀ∫{dṘ/(Ṙ2 -(L/2)2)}

or, (GmṀ/L)∫{(1/(Ṙ - L/2) - (1/(Ṙ + L/2)}dṘ

or, (GmṀ/L){ln(Ṙ - L/2) - ln(Ṙ + L/2)}

or, (GmṀ/L)ln{(Ṙ - L/2)/(Ṙ + L/2)}, integrating Ṙ from Ṙ1 to Ṙ2, we get

(GmṀ/L)ln[{(Ṙ2 - L/2)(Ṙ1 + L/2)}/{(Ṙ2 + L/2)(Ṙ1 - L/2)}]

or (GmḾ/L2)ln[{(Ṙ2 - L/2)(Ṙ1 + L/2)}/{(Ṙ2 + L/2)(Ṙ1 - L/2)}], where Ḿ = ṀL, and Ṙ2 = √(R22 +(L/2)2), Ṙ1 = √(R12 +(L/2)2) and R2 - R1 = L

or, (GmḾ/L2)ln[{(1 - L/(2Ṙ2)(1 + L/(2Ṙ1))}/{(1 + L/(2Ṙ2))(1 - L/(2Ṙ1))}] or arranging

(GmḾ/L2)(ln{(1 - L/(2Ṙ2)/(1 + L/(2Ṙ2)} - ln{(1 - L/(2Ṙ1)/(1 + L/(2Ṙ1)} now expanding both the logs and neglecting higher terms we get.

(GmḾ/L2)[ {-2(L/2Ṙ2)+ --} + {2(L/2Ṙ1)+ --} = (GmḾ/L2){L/Ṙ1 - L/Ṙ2 + --} > (GmḾ/L){1/Ṙ1 - 1/Ṙ2} = {GmḾ/(Ṙ1Ṙ2)}{(Ṙ2- Ṙ1)/L}

Now it can be shown that {GmḾ/(Ṙ1Ṙ2)}{(Ṙ2- Ṙ1)/L} > GmḾ/R2, where R = (R1 +L/2) and Ḿ is the total mass of the object with area LL.

4th case- Now consider a ring (whose cross section looks like a rectangle, not like a circle). Whose radius is "r" and density is as usual "M". Now the small mass "m" lies on the axis of the ring with distance "L" from the center of the ring. Thickness of the ring towards the direction of ring is dr (infinitesimally small) and the other thickness is dz (which I will not mention immediately, but assume it is there). Now the portion rdθdr will exert force GmMrdθdr/(L2 + r2), but towards the direction of the axis it will be

{GmMrdθdr/(L2 + r2)}(L/√(L2 + r2)), integrating it we get {GmM2Πrdr/(L2 + r2)}(L/√(L2 + r2)),again integrating it wrt r, from r = 0 to r = b we get GmM2ΠL(1/L - 1/H), where we assumed √(L2 + r2) = h (the distance between the portion rdθdr and the mass "m"), which will be L at r = 0 and H at r = b.


 * note: GmM2ΠL(1/L - 1/H) = GmM2Πr2[{(H-L)/H}/r2]= Gm(MΠr2)2[{(H-L)(H+L)/((H+L)H)}/r2]

=GmṀ{2/(L2 + r2 +HL)}<GmṀ/L2

Finally we got the attraction force between the disc with radius "b" and mass "m" which is GmM2ΠL(1/L - 1/H). Now consider the disc is part of a sphere of radius "R" and the thickness is dz (=dL) which I mentioned earlier. Now "L" (as well as "H") is variable and "L" has two parts, one is "a" which is outside the sphere and thus constant and another is "b" inside the sphere, so "b" is variable depending on the position of disc from the mass "m". Now if we integrate, we get

∫GmM2Π(1 - L/H)dL= ∫GmM2Π(1 - (a +b)/√((a+b)2 + r2)db

noting that r= √(R2 - (R-b)2), we get

∫GmM2Π(1 - (a + b)/√(2(a+R)b + a2)db, using earlier method we finally get the integration in the form

∫GmM2Π(db - (a/(a+R))dH - (1/(2(a+R)2)) (H2-a2)dh) where H changes from a to a + 2R, b changes from 0 to 2R. Now performing the integration and putting the initial and final values of "b" and "H" we get

GmM2Π{(2R3 - (4/3)R3)/(a+R)2} or Gm(M4ΠR3/3)/(a + R)2 or {GmḾ/(a + R)2}, which is exactly what we assume that all the mass is concentrated at the center. This equation is valid only for a ≥ 0. (why ? what should we do for a < 0?).

Okay now the "m" what we assumed as a point mass (negligible dimension)so far, also becomes a sphere of appreciable dimension, how do we calculate the resultant force now? Surprisingly we do not need to calculate anymore. By the property of sphere we already saw that for a sphere (of any radius) we can assume all the mass is concentrated at the center, so for a concentrated mass we can assume that the mass is uniformly distributed over a sphere (of any radius). The equation will remain same!


 * a popular proverb "When you try to do something from scratch, keep in mind that somebody already might have scratched that before."

A slight diversion - What is space and how space can be curved – Our perception of space comes from what we see daily. We can assume a plane paper kept on a flat table, a two dimensional space. Suppose there are three points on the paper equal distance from each other like vertices of equilateral triangle and we are trying to connect them with straight lines. Now our perception of straight line is the shortest distance between two points. On a flat plane a straight line (segment) is the shortest distance between two points. So if we connect the three points on the flat plane with the lines which represent the shortest distance between them then they will form perfect straight lines. But on a curved plane (like on football) the shortest distance between two points is no more an absolute straight line. Also the angle subtended by any two lines will no more be 60 deg. Neither the space (area) bounded by the three lines will be same as original area. Moreover to us the path of light is straight line and represents the shortest distance between two points. If by any means (like due to gravity) the path of light is bent then the space (here area) bounded by the lines (path of light) will also be bent (like the surface of the football). And the curvature of the path of light due to gravity also in some way represents the curvature of space due to gravity.


 * okay the diversion ends here. back to the main topic.

The 4th case is a very useful one. (The technique I used in 4th and 5th case is based on the technique I used several years back to calculate volume of sphere, as my friend insisted. I hardly read first few pages of integral calculus at that moment- my friend did not know this. He found the proof which was in the book was more than one page long, too long, and he was confident that I could shorten that. Although I was not as confident as him.)

Let us take the disc and now consider it a part of cylinder. Now integrate it as usual

∫GmM2Π(1 - L/H)dl = GmM2Π{(L2- L1) - (H2- H1)} where L and H have the same meaning earlier. Now rearranging it

GmM2Π{(L2- H2) + (H1- L1)} and letting L2 →∞ so also H2 →∞, we get GmM2Π(H1- L1), which is the upper bound! that means no matter how much we increase the size of the cylinder by increasing its length (only) it can not be more than (or equal to) GmM2Π(H1- L1). where H1 and L1 both are finite and L1 is the distance between the disc and the mass "m"

Now in the original case GmM2Π{(L2- L1) - (H2- H1)} and letting R (the radius of the cylinder) →∞ that means H →∞ we get GmM2Π(L2- L1)!!, where L2- L1 is the length of the cylinder (which has now become a disc because of shape).

This is outright spooky to me as this gravitational force "GmM2Π(L2- L1)" does not depend on the distance between the infinite disc (with finite thickness) and the mass "m".

Finite Population Correction (FPC)
Here we are trying to find out the relation between population standard deviation σ and standard deviation of sample mean s, which is also called standard error. For a population of N if we take sample size n, then combination of all such sample is NCn.Then we calculate (∑ai/n -x)2, where ai is individual observations of a particular sample, ∑ai/n is sample mean and x is population mean as well as mean of sample means, sum it up over all NCn samples to get total variance multiplied by number of samples. Now

(∑ai/n -x)2 =(∑ai-n.x)2/n2 = ((∑ai)2 -2(∑ai)(n.x) + n2.x2)/n2

=( (∑ai2) + 2.(∑ai.aj) - 2.(∑ai).n.x +n2.x2 )/n2

where  ai and aj both are individual observations of a particular sample of size n, i = 1 to n and j = 1 to n but i ≠ j and ai.aj = aj.ai, so ∑ai.aj contains all the combinations of two observations out of n observations.

when summing over all NCn sample we get

(N-1Cn-1∑ai2 + 2.N-2Cn-2(∑ai.aj) - 2.n.x.N-1Cn-1∑ai + NCn.n2x2)/n2

Here ai and aj both are individual observations of whole population of size N, i = 1 to N and j = 1 to N but i ≠ j and ai.aj = aj.ai

The above expression can be further modified to

(N-1Cn-1∑ai2 + 2.N-2Cn-2(∑ai.aj) - NCn.(n2/N2)(∑ai)2)/n2

devide by NCn we get the variance of the sample mean s2

∑ai2/(N.n) + (2.(n-1)/((N.n.(N-1))).(∑ai.aj) - (∑ai)2/N2

multiply by n.(N-1) we get

( (N-1)∑ai2 + 2(n-1)∑aiaj – {n(N-1)/N}(∑ai)2 )/N =( (n-1)∑ai2 + 2(n-1)∑aiaj +(N-n)∑ai2 – {n(N-1)/N}(∑ai)2 )/N =( (n-1)(∑ai)2 +(N-n)∑ai2 – {n(N-1)/N}(∑ai)2 )/N

=( (N-n)∑ai2 – {(N-n)/N}(∑ai)2 )/N

Divide by N-n we get ( ∑ai2 - (∑ai)2/N )/N

Now ∑ai2 - (∑ai)2/N is equal to ∑(ai - x)2 where summation is over population N(i = 1 to N). Where x is population mean as well as mean of sample mean, x = ∑ai/N. and ∑(ai - x)2/N is variance of population mean. So we see the variance of sample mean s2 multiplied by n.(N-1)/(N-n) we get variance of population mean σ2. so finally we got

s2= σ2.(N-n)/(n.(N-1)) or s = (σ/√n).√((N-n)/N-1))

√((N-n)/(N-1)) is called the finite population correction or fpc.

the standard deviation of the difference(and sum) between two independent populations
If we subtract one normally distributed data from another normally distributed data, we get a new data set. Suppose we create a population say C which is created by the difference of two independent populations say A and B of size, M and N, we then see that the mean of C is x = x1- x2, Where x1 is the mean of A and x2 is the mean of B. Now if ai is individual observations in A and bj is individual observations in B, where i= 1 to M and j = 1 to N, we calculate {(ai – bj) – x}2, sum it up over all observations M.N to get total variance*number of observations. Now

∑∑ { (ai – bj) – x}2 =∑ { (ai – b1)2  +(ai – b2)2  + ……+(ai – bN)2   - 2x∑{ (ai – b1) +……+(ai – bN) +N∑x2

Where ∑ from i = 1 to M, and can be modified further

N∑ai2   -2∑ai∑bj +  M∑bj2 -2xN∑ai +2xM∑bj + MNx2 , ( ai is summed from i = 1 to M, bj is summed from j = 1 to N )

= N∑ai2   - 2MNx1x2 + M∑bj2 – 2MN xx1 + 2MNxx2 + MNx2

divide by MN (number of observations in C) we get variance of set C as σ2. so

σ2 = ∑ai2 /M  - 2x1x2 + ∑bj2/N  - 2xx1 + 2xx2 + x2

= ∑ai2/M + ∑bj2/N + (x – (x1 – x2) )2 - x12 – x22

= ∑ai2/M - x12 + ∑bj2/N – x22 + (x – (x1 – x2) )2

= σ12 + σ22 + (x – (x1 – x2) )2, where σ1 and σ2  are standard deviation of A and B respectively

= σ12 + σ22   as x – (x1 – x2) = 0,

So we proved σ2= σ12 + σ22

Note 1: for population A The variance σ12 is ∑(ai – x1)2/M where ∑ from i = 1 to M which can be written as ∑ai2/M  - 2x1∑ai/M  + Mx12/M

= ∑ai2/M - 2x1x1  + x12

= ∑ai2/M - x12

Similarly for population B we can write The variance σ22 is

∑bj2/N - x22

Note 2: prove x = x1- x2

x = ∑∑ (ai – bj)/(M.N) = ∑∑ai/(M.N) – ∑∑bj/(M.N) = N.∑ai/(M.N) - M.∑bj/(M.N) = ∑ai/M - ∑bj/N = x1- x2

Exactly in similar way we can prove that if we add one normally distributed data A to another normally distributed data B, then for the new set x = x1+ x2, and σ2= σ12 + σ22.

Now this can be generalized; if there are multiple sets A,B,C,D and a new set is formed by A+B-C+D. Then x = xA+ xB - xC+ xD and σ2= σA2 + σB2 + σC2 + σD2

Estimation of average squared deviation of population mean from multiple samples
Suppose we have k number of samples of different sizes nk from same populations, with sample means xk,average squared deviation of sample means from population mean ssk, average squared deviation of sample observations from respective sample mean ssnk and total population mean x. we need to prove that

∑ssk + ∑ssnk = ss (average squared deviation of population mean) for clear understanding

∑ssnk = ∑(xj – xk)2    (inside individual samples)

∑nkssk = ∑nk(xk – x)2

∑ss = ∑(xi – x)2           (over entire population)

∑ssnk = ∑∑xi2 - ∑(2(∑xi)xk) + ∑nkxk2 where first ∑ is over nk and second ∑ is over k.

= ∑∑xi2  - ∑2nkxk2 +  ∑nkxk2

= ∑∑xi2 - ∑nkxk2

= ∑∑xi2 - ∑nkx2 - ( ∑nkxk2 - ∑nkx2 )

Now we take A = ∑∑xi2 - ∑nkx2

and B = ∑nkxk2 - ∑nkx2  we see that A can be written as

= ∑xi2 - Nx2 (where N=∑nk =total population and ∑ is over total population

= ss

Now B = ∑nkxk2 - ∑nkx2

= ∑ssnk

so, ∑ssk + ∑ssnk = ss

- If the sample variance of a population is given as √ ∆2/(n-1)  where  ∆2 = ∑(xi - xj)2, xj is sample mean. We can try to see if we take the average variance of all such possible samples of size n, what is the relation between the average variance of samples of size n and population variance.

∑(xi - xj)2 is for a single sample. if we do summation over all such samples (There are total NCn ) we get NCn((n-1)/N)∑(xi2 - (2/n)N-2Cn-2∑xixj  . To get the average variance we need to divide it by NCn and n (as we did not  not divide it by n before) we get.

{(n-1)/Nn}∑(xi2 -2{(n-1)/(Nn(N-1))} ∑xixj

Variance of total population is given by ∑xi2/N - (∑xi)2/N2

3d elastic collision & Cartesian coordinate transformation
Cartesian coordinate transformation

Suppose in 3d we represent a vector U in one Cartesian coordinate frame (x,y,z). Now we want to represent the same vector in a different Cartesian coordinate frame (x1,y1,z1). We see that U is represented by Ux, Uy and Uz (along x,y,z direction respectively).

Let us first create the new frame first from scratch.

Let A is the unit vector along the new x (+ve direction) axis, B is the unit vector along the new y (+ve direction) axis. Then by dot product we see

AxBx + AyBy + AzBz = 0; (here subscript x,y,z denotes old axis)

So for a given A we can choose B in many ways. If we choose Bx and By then Bz will be calculated accordingly. We then divide all the component with √( Bx2 + By2 + Bz2) to get the unit vector B. Once we have A and B we get the new unit vector C along the new z (+ve direction) axis, by cross product of A and B

Cx = (AyBz – AzBy) Cy = (AzBx – AxBz) Cz = (AxBy – AyBx)

Now we will represent U in new frame. We can use again dot product for that.

Ux1 = UxAx + UyAy + UzAz ---(1) Uy1 = UxBx + UyBy + UzBz ---(2) Uz1 = UxCx + UyCy + UzCz ---(3)

In matrix form we write it like $$ \begin{vmatrix} A_x & A_y & A_z\\ B_x & B_y & B_z\\ C_x & C_y & C_z \end{vmatrix} \begin{vmatrix} U_x\\ U_y\\ U_z \end{vmatrix} = \begin{vmatrix} U_{x1}\\ U_{y1}\\ U_{z1} \end{vmatrix} $$ Above Ux, Uy and Uz are known and we calculated Ux1, Uy1 and Uz1. Conversely if we know Ux1, Uy1 and Uz1 we can calculate Ux, Uy and Uz from above three eq. In this way we can represent a vector back from the new frame to old frame. We can solve it by elimination method, or we can use program or code to use matrix operation to solve it.

Elastic collision in 3d

If we have a line (straight), we see we can draw multiple two dimensional flat surface containing that line. If we have two intersecting lines or two parallel lines we see that we can draw only one flat surface, containing those two lines. But lines which are neither intersecting nor parallel cannot be in the same plane. So we need to consider them in three dimensional space (3d).

Suppose two spheres with mass, velocity, M1,U1 and M2,U2(in (x,y,z) frame) collided elastically with each other where the velocity vectors are neither parallel nor intersecting. If we form a plane by U1 and line of collision, U2 will not lie in the same plane. We also note that line of collision goes through the center of the spheres, so there is no rotation involved here.

In this collision total momentum and total energy, both will be conserved. As we know there will be change in the component of velocity vectors only along the line of collision. The component of velocity vectors which are perpendicular to line of collision will not change. Assume at the time of collision the first sphere(its center) has coordinates X1,Y1,Z1 and second sphere has coordinates X2,Y2,Z2 then line of collision is a vector(assume) defined by its component (X2-X1),(Y2-Y1),(Z2-Z1). We make it a unit vector by dividing the components by √{ (X2-X1)2+(Y2-Y1)2+(Z2-Z1)2 }. Assume this as unit vector A. Then

U1x1 = U1xAx + U1yAy + U1zAz (along the line of collision, so will change) U1y1 = U1xBx + U1yBy + U1zBz (will not change) U1z1 = U1xCx + U1yCy + U1zCz (will not change) And U2x1 = U2xAx + U2yAy + U2zAz (along the line of collision, so will change) U2y1 = U2xBx + U2yBy + U2zBz (will not change) U2z1 = U2xCx + U2yCy + U2zCz (will not change)

Note that we need to align A (actually any one of A,B,C) with line of collision, but no such need for B or C. So we choose B and C arbitrarily as described in the previous section.

Suppose after collision the final velocities are V11 and V21 in new frame.

Writing down the energy eq.

(1/2)M1U12 + (1/2)M2U22 = (1/2)M1V112 + (1/2)M2V212 Or, M1(U1x12 + U1y12 + U1z12) + M2(U2x12 + U2y12 + U2z12) = M1(V1x12 + U1y12 + U1z12) + M2(V2x12 + U2y12 + U2z12) Or, M1U1x12+ M2U2x12 = M1V1x12 + M2V2x12

And the momentum eq.

M1U1x1+ M2U2x1 = M1V1x1 + M2V2x1

Solving these two eq.s as one dimensional collision eq.s we get

V1x1 = 1/(M1+M2) { U1x1(M1-M2) + 2M2U2x1 } V2x1 = 1/(M1+M2) { U2x1(M2-M1) + 2M1U1x1 }

Now we need to transfer all the components of V11 and V21 to old frame (x,y,z). And we already discussed how to do that.

--

Let us do an exercise.

Let during collision the coordinate of first sphere is (2,2,3) and coordinate of the second sphere is (3,3,1). So we get (X2-X1) = (3-2) = 1,(Y2-Y1) = (3-2) = 1, (Z2-Z1) = (1-3) = -2.

And √{ (X2-X1)2+(Y2-Y1)2+(Z2-Z1)2 } = √{ (1)2+(1)2+(-2)2 } = √6. We divide all the components with √6 to get the unit vector A.

So, Ax = 1/√6, Ay = 1/√6, Az = -2/√6.

Let us find vector B from following eq.

AxBx + AyBy + AzBz = 0;

Let Bx = 1, By = 1, then Bz = -(Ax + Ay)/ Az or, Bz = 1. We make B a unit vector by dividing all the components with = √{ (1)2+(1)2+(1)2 } = √3. So, Bx = 1/√3, By = 1/√3, then Bz = 1/√3 Let us find C

Cx = (AyBz – AzBy) = { (1/√6)(1/√3) – (-2/√6)(1/√3) } = 1/√2 Cy = (AzBx – AxBz) = { (-2/√6)(1/√3) – (1/√6)(1/√3) } = -1/√2 Cz = (AxBy – AyBx) = { (1/√6)(1/√3) – (1/√6)(1/√3) } = 0

Let U1 = (2,2,4) and U2 = (3,-3,-2)

U1x1 = U1xAx + U1yAy + U1zAz = 2/√6 + 2/√6 - 8/√6 = - 4/√6 U1y1 = U1xBx + U1yBy + U1zBz = 2/√3  + 2/√3  + 4/√3 = 8/√3 U1z1 = U1xCx + U1yCy + U1zCz = 2/√2  - 2/√2  + 0  =  0

U2x1 = U2xAx + U2yAy + U2zAz = 3/√6  -3/√6 +  4/√6 =  4/√6 U2y1 = U2xBx + U2yBy + U2zBz = 3/√3  -3/√3  + -2/√3 = -2/√3 U2z1 = U2xCx + U2yCy + U2zCz = 3/√2  +3/√2  + 0  =  6/√2

For simplicity let us take M1 = M2. Then

V1x1 = 1/(M1+M2) { U1x1(M1-M2) + 2M2U2x1 } = U2x1 = 4/√6 V2x1 = 1/(M1+M2) { U2x1(M2-M1) + 2M1U1x1 } = U1x1 = -4/√6

Let us solve the eqs.

V1x1 = V1xAx + V1yAy + V1zAz, or 4/√6 = V1x/√6 + V1y/√6 - 2V1z/√6, or 4 = V1x + V1y - 2V1z U1y1 = V1xBx + V1yBy + V1zBz, or 8/√3 = V1x/√3 + V1y/√3 + V1z/√3, or 8 = V1x + V1y + V1z U1z1 = V1xCx + V1yCy + V1zCz, or 0 = V1x/√2 - V1y/√2 + 0, or 0 = V1x - V1y

Solving the above eqs. We get V1z = 4/3, V1x = 10/3, V1y = 10/3, or V1 = (10/3,10/3,4/3)

V2x1 = V2xAx + V2yAy + V2zAz, or -4/√6 = V2x/√6 + V2y/√6 - 2V2z/√6, or -4 = V2x + V2y - 2V2z U2y1 = V2xBx + V2yBy + V2zBz, or -2/√3 = V2x/√3 + V2y/√3 + V2z/√3, or -2 = V2x + V2y + V2z U2z1 = V2xCx + V2yCy + V2zCz, or 6/√2 = V2x/√2 – V2y/√2 + 0, or 6 = V2x – V2y

Solving the above eqs. We get V2z = 2/3, V2x = 5/3, V2y = -13/3, or V2 = (5/3,-13/3, 2/3)

So now we have U1 = (2,2,4) and U2 = (3,-3,-2) and V1 = (10/3,10/3,4/3), V2 = (5/3,-13/3, 2/3)

Let us see whether it obeyed conservation of energy or not.

½( M1U12 + M2U22)   ? ½( M1V12 + M2V22) Or, U12 + U22    ? V12 + V22 Or, 22+ 22 + 42 + 32 + -32 + -22   ? (10/3)2 + (10/3)2 + (4/3)2 + (5/3)2 + (-13/3)2 + (2/3)2 Or, 46 ? 414/9 Or, 46  =  414/9, So it obeyed conservation of energy.

Let us see whether it obeyed conservation of momentum in original coordinate system or not. As M1 = M2 let us cancel them first and write

U1x + U2x ? V1x + V2x or, 2 + 3 ? 10/3 + 5/3 or, 5 ? 5 U1y + U2y ? M1V1y + M2V2y or, 2 – 3 ? 10/3 – 13/3 or, -1 ? -1 U1z + U2z ? M1V1z + M2V2z or, 4 – 2 ? 4/3 + 2/3 or, 2 ? 2

So we see conservation of momentum was obeyed in all 3 axis of original coordinate system.