Hi physics people,
This is a past (3rd year university level) exam question, so I hope it's ok that I didn't post this in the homework section even if it's set out like a homework question.
The Question:
Suppose we are observing the collision
Anti-electron-neutrino + electron ---> W-minus...
I think we need to add L because the question requires that we find the displacement of the mass when it catches up with the end of the string, which is L distance away from the other end of the string where the mass is attached. Since we've got the time taken for the string to slacken again...
So I've been working on this problem a bit more... I've been trying to find the total distance adding the distance taken for the string to become slack x(t), when
t = \pi\sqrt{\frac{K}{m}} to the distance traveled by the point B x = Vt_{2} and then adding the length of the string to that giving...
Ok, so our answer for x(t) = Vt - \frac{V}{\omega } sin (wt).
So assuming that at the instant the string looses tension the following is true
Vt-x=0
we can replace Vt with x in the equation for x(t) and after some cancelations figure out that
sin(wt) = 0
So wt = n\pi
Assuming...
Oh of course thank you!
So A is
A = -V\sqrt{\frac{m}{K}}
So the general solution becomes
x(t) = Vt - V\sqrt{\frac{m}{K}} sin (\sqrt{\frac{m}{K}} t)
So after I've figured out the time taken for the string to become slack, I figured that the total distance traveled until the mass...
So I did what you said and used m(dv/dt) = K (Vt-x) to generate the differential equation
m(d^2x/dt^2) + Kx = KVt
The homogeneous solution that I got is
xh(t) = Asin(t(K/m)^0.5) + Bcos(t(K/m)^0.5)
I found the particular solution by guessing the solution for x to be of the form
x = bt^2 +...
Can T=K log(1+au) be rearranged to give us u?
u=1/a * ((10^(T/K)) -1))
If that is substituted back into the equation for force...
F(T)=-(10^T/K) / Ka
Does this get us anywhere?
Classical Mechanics Homework question
Question - A light elastic string AB of natural length L and spring constant K, lies slack on a horisontal plane. A particle of mass m also at rest, is attached to end A of the string. The other end B is pulled along the plane with constant velocity V...