Mass — Question 6

Section 5-4 · Physics pages 682–683

Question 6 One scalar mass must fit every vector component Use the inertial calibration relation⃗Fnet = m⃗a for a constant-mass body in an inertial frame. A measured net force is (6, 8) N and acceleration is (1.5, 2) m/s2. A second report changes only ay to 1 m/s2.⃗ F⃗ a The valid force and acceleration are parallel; drawn lengths are schematic. (a) Find mass from the original report. (b) Test the second report for a single positive scalar mass. (c) Why is using only the ratio of vector magnitudes insufficient for consistency?
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Question 6 - Solution One scalar mass must fit every vector component⃗ F⃗ a (a) Check each nonzero component. mx = 6 1.5 = 4, my = 8 2 = 4 ⇒ m = 4 kg 6/1.5 = 8/2 = 4 kg. The force and acceleration are parallel, and the magnitude ratio 10/2.5 agrees. (b) Compare independent component estimates. m′ x = 4 kg̸ = m′ y = 8/1 = 8 kg The altered report gives 4 kg from x and 8 kg from y. No single mass satisfies both equations. (c) Retain vector direction. Fxay −Fyax = 0 required; 6(1) −8(1.5) = −6̸ = 0 A magnitude ratio can always yield a number when acceleration is nonzero, even for nonparallel vectors. A scalar mass multiplies every component equally; both direction and proportionality must hold.

Question 6

One scalar mass must fit every vector component

Use the inertial calibration relation F→net=ma→\vec F_{\mathrm{net}}=m\vec a for a constant-mass body in an inertial frame. A measured net force is (6,8)(6,8) N and acceleration is (1.5,2)(1.5,2) m/s2^2. A second report changes only aya_y to 1 m/s2^2.

Diagram shown in the original SVG worksheet.

  1. Find mass from the original report.

  2. Test the second report for a single positive scalar mass.

  3. Why is using only the ratio of vector magnitudes insufficient for consistency?

Question 6 - Solution

One scalar mass must fit every vector component

Diagram shown in the original SVG worksheet.

(a) Check each nonzero component.

mx=61.5=4,my=82=4⇒m=4kgm_x=\frac6{1.5}=4,\quad m_y=\frac82=4\ \Rightarrow\ \boxed{m=4\,\mathrm{kg}} 6/1.5=8/2=46/1.5=8/2=4 kg. The force and acceleration are parallel, and the magnitude ratio 10/2.510/2.5 agrees.

(b) Compare independent component estimates.

mx′=4kg≠my′=8/1=8kgm_x'=4\,\mathrm{kg}\ne m_y'=8/1=8\,\mathrm{kg} The altered report gives 4 kg from x and 8 kg from y. No single mass satisfies both equations.

(c) Retain vector direction.

Fxay−Fyax=0required;6(1)−8(1.5)=−6≠0F_xa_y-F_ya_x=0\ \text{required};\quad6(1)-8(1.5)=-6\ne0 A magnitude ratio can always yield a number when acceleration is nonzero, even for nonparallel vectors. A scalar mass multiplies every component equally; both direction and proportionality must hold.