Newton's Second Law — Question 9

Section 5-5 · Physics pages 708–709

Question 9 Scaling mass and force together preserves acceleration Use ∑⃗F = m⃗a for constant mass in an inertial frame. Run A uses mass 2 kg and net force (4, 6) N. Run B uses mass 6 kg and force (12, 18) N. Both begin with the same position and velocity. 2 kg 6 kg⃗ FA = (4, 6) N⃗ FB = (12, 18) N run A run B The forces are parallel; arrow lengths are schematic. Both runs share the initial state. (a) Compare accelerations. (b) Compare their trajectories while those forces persist. (c) Would doubling only mass preserve the original trajectory?
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Question 9 - Solution Scaling mass and force together preserves acceleration 2 kg 6 kg⃗ FA = (4, 6) N⃗ FB = (12, 18) N run A run B (a) Divide each force by its own mass.⃗ aA = (4, 6)/2 =⃗aB = (12, 18)/6 = (2, 3) m/s2 Both accelerations are (2, 3) m/s2. (b) Use identical evolution and initial data.⃗ rB −⃗rA = ∆⃗r0 + ∆⃗v0t + 1 2∆⃗at2 =⃗0 Their velocity and position histories are identical over the specified interval. Equality of acceleration alone would not suffice if initial data differed. (c) Change one parameter at a time.⃗ a′ =⃗F 2m = 1 2⃗a,⃗ r′ −⃗r = −1 4⃗at2 Doubling mass alone halves acceleration for the same net force. With the same initial data, displacement contributions from acceleration are halved; the original trajectory is generally not preserved.

Question 9

Scaling mass and force together preserves acceleration

Use ∑F→=ma→\sum\vec F=m\vec a for constant mass in an inertial frame. Run A uses mass 2 kg and net force (4,6)(4,6) N. Run B uses mass 6 kg and force (12,18)(12,18) N. Both begin with the same position and velocity.

Diagram shown in the original SVG worksheet.

  1. Compare accelerations.

  2. Compare their trajectories while those forces persist.

  3. Would doubling only mass preserve the original trajectory?

Question 9 - Solution

Scaling mass and force together preserves acceleration

Diagram shown in the original SVG worksheet.

(a) Divide each force by its own mass.

a→A=(4,6)/2=a→B=(12,18)/6=(2,3)m/s2\boxed{\vec a_A=(4,6)/2=\vec a_B=(12,18)/6=(2,3)\,\mathrm{m/s^2}} Both accelerations are (2,3)(2,3) m/s2^2.

(b) Use identical evolution and initial data.

r→B−r→A=Δr→0+Δv→0t+12Δa→t2=0→\vec r_B-\vec r_A=\Delta\vec r_0+\Delta\vec v_0t+\tfrac12\Delta\vec at^2=\vec0 Their velocity and position histories are identical over the specified interval. Equality of acceleration alone would not suffice if initial data differed.

(c) Change one parameter at a time.

a→′=F→2m=12a→,r→′−r→=−14a→t2\vec a'=\frac{\vec F}{2m}=\tfrac12\vec a,\quad\vec r'-\vec r=-\tfrac14\vec at^2 Doubling mass alone halves acceleration for the same net force. With the same initial data, displacement contributions from acceleration are halved; the original trajectory is generally not preserved.