Newton's Second Law for a System of Particles — Question 10

Question 10

When a measured center curve fails a force model

Two point masses, $2.0$ and $3.0\,\mathrm{kg}$, are claimed to form a horizontally isolated system. For $t$ in seconds their measured positions in meters are $x_1=t^2$ and $x_2=4-\tfrac12t^2$. Masses remain constant.

1. Test the isolation claim from the center trajectory.

2. If $x_1$ is correct and $x_2=4+bt^2$, find the value of $b$ required by horizontal isolation.

Question 10 - Solution

When a measured center curve fails a force model

(a) Derivation and calculation

Form the measured center trajectory: $$X_C=\frac{2t^2+3(4-\tfrac12t^2)}5=2.4+0.10t^2.$$ $$a_C=0.20\ \mathrm{m/s^2}\Rightarrow\boxed{F_{{\mathrm{ext}},x}=5(0.20)=1.0\ \mathrm N}.$$ The claimed horizontal isolation is inconsistent with these trajectories.

(b) Result and interpretation

With the proposed corrected second trajectory, $$X_C=\frac{12+(2+3b)t^2}{5}.$$ $$a_C=\frac{2(2+3b)}5=0\Rightarrow\boxed{b=-\frac23\ \mathrm{m/s^2}}.$$ Internal forces cannot account for a nonzero acceleration of an otherwise isolated center.

Physical check

The corrected weighted accelerations cancel: $2(2)+3(-4/3)=0$.