Newton's First Law — Question 5

Section 5-2 · Physics pages 640–641

Question 5 A steadily translating laboratory is also an inertial candidate In inertial laboratory L a free body has x = 8 + 6t m. Frame S has origin X = 2 + 4t m and parallel axes. Use numerical seconds. body v = 6 m/s in L S U = 4 m/s origin Both velocities refer to the same inertial laboratory; axes stay parallel. (a) Find its motion in S. (b) Does it satisfy the first law in both frames? (c) Choose a constant frame velocity that makes the body stationary and explain why this changes no interaction.
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Question 5 - Solution A steadily translating laboratory is also an inertial candidate body v = 6 m/s in L S U = 4 m/s origin (a) Subtract the moving origin. xS = xL −X = 6 + 2t, vS = 2 m/s, aS = 0 xS = 6 + 2t m, so vS = 2 m/s and aS = 0. (b) Compare accelerations. aS = aL −¨X = 0 −0 = 0 Both descriptions have zero acceleration. Constant relative translation preserves the first-law free-particle test. (c) Move with the body. v′ = vL −U = 6 −U = 0 ⇒U = 6 m/s An origin moving at 6 m/s in L makes the body’s velocity zero. Its position may remain a nonzero constant if the initial origins differ. The coordinate choice changes the measured velocity, not the physical forces.

Question 5

A steadily translating laboratory is also an inertial candidate

In inertial laboratory L a free body has x=8+6tx=8+6t m. Frame S has origin X=2+4tX=2+4t m and parallel axes. Use numerical seconds.

Diagram shown in the original SVG worksheet.

  1. Find its motion in S.

  2. Does it satisfy the first law in both frames?

  3. Choose a constant frame velocity that makes the body stationary and explain why this changes no interaction.

Question 5 - Solution

A steadily translating laboratory is also an inertial candidate

Diagram shown in the original SVG worksheet.

(a) Subtract the moving origin.

xS=xL−X=6+2t,vS=2m/s,aS=0x_S=x_L-X=6+2t,\quad \boxed{v_S=2\,\mathrm{m/s},\quad a_S=0} xS=6+2tx_S=6+2t m, so vS=2v_S=2 m/s and aS=0a_S=0.

(b) Compare accelerations.

aS=aL−Ẍ=0−0=0a_S=a_L-\ddot X=0-0=0 Both descriptions have zero acceleration. Constant relative translation preserves the first-law free-particle test.

(c) Move with the body.

v′=vL−U=6−U=0⇒U=6m/sv'=v_L-U=6-U=0\ \Rightarrow\ U=6\,\mathrm{m/s} An origin moving at 6 m/s in L makes the body’s velocity zero. Its position may remain a nonzero constant if the initial origins differ. The coordinate choice changes the measured velocity, not the physical forces.