Electric Potential Energy — Question 5

Section 24-1 · Physics pages 3800–3801

Question 5 Equal endpoint energies do not guarantee passage over an intermediate barrier Along a frictionless constrained path 0 ≤x ≤L, a charged particle has U(x) = Ub sin2(πx/L) with Ub > 0. It starts at x=0 toward positive x with kinetic energy K0 = .800Ub. x U L Axes and labels specify the model; the sketch is schematic. (a) Find the first turning point and explain why the endpoint x=L is inaccessible. (b) Find the threshold initial kinetic energy for reaching the barrier top, and distinguish the threshold from passage in finite time.
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Question 5 - Solution Equal endpoint energies do not guarantee passage over an intermediate barrier x U L (a) Method and calculation K(x) = K0 −U(x) = Ub[.800 −sin2(πx/L)]. xt = L π sin−1 √ .800 ≈.352416L. (b) Interpretation and further calculation Umax = Ub at x = L/2, K0,threshold = Ub. K0 > Ub ⇒finite-time passage; K0 = Ub ⇒asymptotic approach to the smooth maximum. At threshold the speed vanishes linearly with distance from the top, so the arrival-time integral diverges. Physical check Energy conservation must be checked at every intermediate position, not only at the endpoints.

Question 5

Equal endpoint energies do not guarantee passage over an intermediate barrier

Along a frictionless constrained path 0≤x≤L0\le x\le L, a charged particle has U(x)=Ubsin⁡2(πx/L)U(x)=U_b\sin^2(\pi x/L) with Ub>0U_b>0. It starts at x=0 toward positive x with kinetic energy K0=.800UbK_0=.800U_b.

Diagram shown in the original SVG worksheet.

  1. Find the first turning point and explain why the endpoint x=L is inaccessible.

  2. Find the threshold initial kinetic energy for reaching the barrier top, and distinguish the threshold from passage in finite time.

Question 5 - Solution

Equal endpoint energies do not guarantee passage over an intermediate barrier

Diagram shown in the original SVG worksheet.

(a) Method and calculation

K(x)=K0−U(x)=Ub[.800−sin⁡2(πx/L)].K(x)=K_0-U(x)=U_b[.800-\sin^2(\pi x/L)].xt=Lπsin⁡−1.800≈.352416L.x_t=\frac L\pi\sin^{-1}\sqrt{.800}\approx.352416L.

(b) Interpretation and further calculation

Umax=Ubat x=L/2,K0,threshold=Ub.U_{\max}=U_b\quad\text{at }x=L/2,\qquad K_{0,\mathrm{threshold}}=U_b.K0>Ub⇒finite-time passage;K0=Ub⇒asymptotic approach to the smooth maximum.K_0>U_b\Rightarrow\text{finite-time passage};\quad K_0=U_b\Rightarrow\text{asymptotic approach to the smooth maximum}.At threshold the speed vanishes linearly with distance from the top, so the arrival-time integral diverges.

Physical check

Energy conservation must be checked at every intermediate position, not only at the endpoints.