Newton's Second Law for a System of Particles — Question 6

Question 6

A force pulse changes center motion

A $4.0\,\mathrm{kg}$ system initially has $X=0$ and $V=1.0\,\mathrm{m/s}$. Its total external horizontal force is $+8.0\,\mathrm N$ for $0<t<1.0\,\mathrm s$ and $-4.0\,\mathrm N$ for $1.0<t<3.0\,\mathrm s$.

1. Find the center velocity and displacement at $t=3.0\,\mathrm s$.

2. Explain why returning to the initial velocity does not imply returning to the initial position.

Question 6 - Solution

A force pulse changes center motion

(a) Derivation and calculation

Apply each constant force over its own interval: $$a_1=8/4=2,\quad V_1=1+2(1)=3,\quad\Delta X_1=1(1)+\tfrac12(2)(1)^2=2.$$ $$a_2=-4/4=-1,\quad\boxed{V_3=3-1(2)=1.0\ \mathrm{m/s}},\quad \Delta X_2=3(2)-\tfrac12(1)(2)^2=4.$$ $$\boxed{X(3)=2+4=6.0\ \mathrm m}.$$

(b) Result and interpretation

The signed force areas cancel, but the velocity areas do not: $$\Delta V_C=\frac{8(1)-4(2)}4=0,\qquad \Delta X_C=\int_0^3V_C(t)\,dt=6\ \mathrm m>0.$$ Velocity stays positive throughout, so returning to the initial velocity does not restore the initial position.

Physical check

The first interval average velocity is 2 m/s and the second is also 2 m/s.