Recent content by LASmith

I have the same problem with the limits, how is it that you are able to approximate it as two rectangles? Also, if you use the limits 2 and -2 for the x, I am unable to work how how you would find the limits for the y.

From this can I re-write the RHS of the equation to be = (m2u32c2 + m2c4)1/2 Squaring both sides I obtain m12(u12+c2) + m22(u22+c2) + 2(m12m22(u12u22+u12c2+u22c2+c4))1/2 = m2(u32+c2) Is this correct? So does momentum hold the same in special relativity as it does in general...

Homework Statement two particles of masses m1 and m2 move at speeds u1, u2 respectively collide and fuse. If α is the angle between the two directions of motion before the collision, find an expression for the new mass m, in terms of m1 m2 u1 u2 and α Homework Equations Etot2 =...

Yes, I see it now, if x=0, then the |v|=1 still so we can use the equation w-x=1. This implies that w=1, and as x=y=z=0 We get the second unit vector (1,0,0,0) So there are two unit vectors for this answer. Thank you for all your help :)

therefore we get w=1/2 therefore x=-1/2 y=-1/2 and z=-1/2 Also using x=0 we get w=0, y=0 & z=0 so when it asks for all the unit vectors there must only be one, as the second one has a modulus of zero. So are they no more unit vectors apart from (0.5,-0.5,-0.5,-0.5)?

Yes, I obtained this too, the you get x(w+x)=0 Therefore you get either x=0 or w+x=0 which leads to my answer w=-x

Okay, sorry I got the algebra wrong, I do not obtain w2-wx-x2=0 instead, I get w=-x. Which would be a solution, however, then fact vela pointed out w=1+x contradicts this, so what value am I supposed to use?

I haven't yet considered the unit vector part, I was just going to divide the answer by the modulus. I obtained this results as |u|cos\Theta is 2*1/2=1 for all the vectors A,B&C therefore right hand side of the equation for all the vectors is |v|, so we get v.u=|v| and I took |v| =...

Homework Statement Find all unit vectors in R4 making an angle of \pi/3 with the three vectors A=(1,1,-1,-1) B=(1,-1,1,-1) C=(1,-1,-1,1) Homework Equations u.v=|u||v|cos\ThetaThe Attempt at a Solution using V=(w,x,y,z) as the vector we are trying to find, I solved the above equation for all...

Therefore p will be 1 for both parts of the function, regardless of where the two parts of the function lie, between this interval?

I thought p represented the value for which we integrate between, as all the examples so far have been where the limit is between p and -p, however, because these functions are between -1 and 0 & 0 and 1, I would presume it was the non-zero number, although I do not understand where the 1/p at...

Can this only be done the long winded way of calculating a0, an and bn for each equation, therefore doing 6 integrals? Or will some cancel for being odd/even functions, which I cannot see at first glance? Also would the value of 'p' in my 'relevant equations' be -1 for x+1 and 1 for x-1?

Homework Statement f(x) = x+1 for -1,x<0 x-1 for 0<x<1 0 for x=0 expand it in an appropriate cosine or sine series Homework Equations f(x) = a0/2 + \sum [ancos (n\pix/p) + bn sin (n\pix/p) a0 = 1/p \intf(x).dx an = 1/p \int f(x)cos...

Expanding the brackets and then integrating I obtained the correct answer, so thanks for that. However, looking back I had no idea how I did not get a factor of 1/5, I did not expand out the brackets, and just used the chain rule to obtain (1-x2)3/-18x Then putting in the limits 0 and 1...

\int y3\sqrt{1-x2}/3 .dx Then substituting the limits \sqrt{1-x2} and 0 for y I obtained \int (1-x2)2)/3 .dx Limits between 0 and 1 for this final integral