Conservation of Mechanical Energy — Question 9

Section 8-4 · Physics pages 1088–1089

Question 9 A conserved scalar leaves velocity direction undetermined A 2 kg particle has constant mechanical energy E=25 J in a conservative planar field. At point P the potential is 9 J. Its x velocity there is 2 m/s. vy vx vx = 2 Two velocity directions can share one speed and the same horizontal component. (a) Find kinetic energy and speed at P. (b) Find the possible y velocity components. (c) Can conservation of mechanical energy alone choose between those possibilities or determine arrival time?
Show solutionHide solution
Question 9 - Solution A conserved scalar leaves velocity direction undetermined vy vx vx = 2 (a) Subtract the potential from the total. At P, the kinetic energy is the energy gap: K = 25 −9 = 16 J , v = √ 2K/m = √ 16 = 4 m/s . (b) Use the component relation. Resolve the speed into components: v2 x + v2 y = v2 ⇒v2 y = 16 −22 = 12, vy = ±2 √ 3 m/s . (c) Retain the missing trajectory information. The scalar constraint 1 2m(v2 x + v2 y) + U = E is unchanged by vy 7→−vy. Initial direction and the force-driven trajectory are needed to select the sign and determine arrival time.

Question 9

A conserved scalar leaves velocity direction undetermined

A 2 kg particle has constant mechanical energy E=25 J in a conservative planar field. At point P the potential is 9 J. Its x velocity there is 2 m/s.

Diagram shown in the original SVG worksheet.

  1. Find kinetic energy and speed at P.

  2. Find the possible y velocity components.

  3. Can conservation of mechanical energy alone choose between those possibilities or determine arrival time?

Question 9 - Solution

A conserved scalar leaves velocity direction undetermined

Diagram shown in the original SVG worksheet.

(a) Subtract the potential from the total.

At P, the kinetic energy is the energy gap: K=25−9=16J,v=2K/m=16=4m/s.\boxed{K=25-9=16\ \mathrm J},\qquad\boxed{v=\sqrt{2K/m}=\sqrt{16}=4\ \mathrm{m/s}}.

(b) Use the component relation.

Resolve the speed into components: vx2+vy2=v2⇒vy2=16−22=12,v_x^2+v_y^2=v^2\Rightarrow v_y^2=16-2^2=12, vy=±23m/s.\boxed{v_y=\pm2\sqrt3\ \mathrm{m/s}}.

(c) Retain the missing trajectory information.

The scalar constraint 12m(vx2+vy2)+U=E\tfrac12m(v_x^2+v_y^2)+U=E is unchanged by vy↦−vyv_y\mapsto-v_y. Initial direction and the force-driven trajectory are needed to select the sign and determine arrival time.