Reading a Potential Energy Curve — Question 9

Section 8-5 · Physics pages 1108–1109

Question 9 Inferring mass and energy from two curve readings Assume one-dimensional classical motion with K = E −U(x) and Fx = −dU/dx, with no dissipation. At one position a particle has U=2 J and speed 4 m/s; at another it has U=8 J and speed 2 m/s. Both observations belong to the same conservative motion. state 1 U = 2 J, v = 4 m/s state 2 U = 8 J, v = 2 m/s The same conserved total relates the two measured energy gaps. (a) Infer the mass and conserved total energy. (b) Find speed at a point with U=5 J, if that point lies in the accessible region. (c) Would an additional point with U=11 J be accessible at this energy?
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Question 9 - Solution Inferring mass and energy from two curve readings state 1 U = 2 J, v = 4 m/s state 2 U = 8 J, v = 2 m/s (a) Equate the two total energies. Equate the measured total energies: 2 + 1 2m(4)2 = 8 + 1 2m(2)2 ⇒6m = 6. m = 1 kg , E = 2 + 8 = 10 J . (b) Use the newly inferred energy gap. For an accessible point at U = 5 J, K = 10 −5 = 5 J ⇒v = √ 2(5)/1 = √ 10 m/s . The accessibility proviso excludes a disconnected region behind a higher barrier. (c) Reject a negative kinetic energy. At U = 11 J, the required kinetic energy would be K = 10 −11 = −1 J < 0. That point is classically inaccessible at this total energy.

Question 9

Inferring mass and energy from two curve readings

Assume one-dimensional classical motion with K=E−U(x)K=E-U(x) and Fx=−dU/dxF_x=-dU/dx, with no dissipation. At one position a particle has U=2 J and speed 4 m/s; at another it has U=8 J and speed 2 m/s. Both observations belong to the same conservative motion.

Diagram shown in the original SVG worksheet.

  1. Infer the mass and conserved total energy.

  2. Find speed at a point with U=5 J, if that point lies in the accessible region.

  3. Would an additional point with U=11 J be accessible at this energy?

Question 9 - Solution

Inferring mass and energy from two curve readings

Diagram shown in the original SVG worksheet.

(a) Equate the two total energies.

Equate the measured total energies: 2+12m(4)2=8+12m(2)2⇒6m=6.2+\tfrac12m(4)^2=8+\tfrac12m(2)^2\Rightarrow6m=6. m=1kg,E=2+8=10J.\boxed{m=1\ \mathrm{kg}},\qquad\boxed{E=2+8=10\ \mathrm J}.

(b) Use the newly inferred energy gap.

For an accessible point at U=5U=5 J, K=10−5=5J⇒v=2(5)/1=10m/s.K=10-5=5\ \mathrm J\Rightarrow\boxed{v=\sqrt{2(5)/1}=\sqrt{10}\ \mathrm{m/s}}. The accessibility proviso excludes a disconnected region behind a higher barrier.

(c) Reject a negative kinetic energy.

At U=11U=11 J, the required kinetic energy would be K=10−11=−1J<0.K=10-11=-1\ \mathrm J<0. That point is classically inaccessible at this total energy.