GGR111H5 Final: _Document%2002-May%2c%202019%20523%20PM

A particle of mass m moves under the influence of a central force.The potential energy function is
V(r) = {
K (a2−r2) /2 if r ≤ a
0 if r > a

as illustrated in the figure. The initial position is (x,y) = (a/2,0) and the initial velocity is (vx,vy) = (0, v0).
[Data: m = 1.0 kg; K =17.1 N/m; a = 1.0 m; r0 = 0.5 a; v0 = 2.0m/s.]

(A) Determine the time ta when the mass crosses the boundary r =a.

(B) Determine angular position φa where the particle crosses theboundary r = a.

***HINT GIVEN***

Use the constants of the motion to analyze the orbit.

Write L = mr^2 dφ/dt = the initial value.
What is the initial value?

Write E = (1/2)m (dr/dt)^2 + L^2/(2mr2) +V(r )
= the initial value. What is the
initial value?

Part A. In the energy equation,
change the variable from r to u,
where r =sqrt(u). The result can be
solved by u(t) = c1+c2 cosh(αt).
Determine c1, c2, and α.

Part B.
φ = Integral 0 to t [L/(mr^2)] dt

Suppose there is a little block attached to a cord passing overapulley through a hole in the

Show that the spontaneous emission rate (Equation 9.56 A=[w0^3|p|^2]/[3*pi*eo*hbar*c^3]) for a transition from n,l to n',l' inhydrogen is


9.97 [e^2 w^3 I^2]/[3*pi*eo*hbar*x^3] * { [l+1] /[2l+1] if l' =l+1
'' ' ' '' ' ' '' { l/[2l+1] if l'=l-1


Where I=Integra[r^3 Rnl Rn'l'] dr