Newton's Second Law for a System of Particles — Question 3

Question 3

A spring release does not accelerate the center

Two carts of masses $1.0$ and $3.0\,\mathrm{kg}$ interact only through a light spring on a frictionless track. Initially the system center moves right at $2.0\,\mathrm{m/s}$. At an instant the spring pushes the lighter cart left with $12\,\mathrm N$.

1. Find both instantaneous accelerations and the center acceleration.

2. Find the center displacement in the next $0.50\,\mathrm s$, even though the spring force changes.

Question 3 - Solution

A spring release does not accelerate the center

(a) Derivation and calculation

The spring exerts equal and opposite forces: $$\boxed{a_1=-12/1=-12\ \mathrm{m/s^2}},\qquad \boxed{a_2=12/3=4\ \mathrm{m/s^2}}.$$ $$\boxed{a_C=\frac{1(-12)+3(4)}4=0}.$$

(b) Result and interpretation

At every later instant the total horizontal external force remains zero: $$M\dot V_C=0\Rightarrow V_C(t)=2.0\ \mathrm{m/s}.$$ $$\boxed{\Delta X_C=\int_0^{0.50}2.0\,dt=1.0\ \mathrm m}.$$ The changing spring force alters relative motion but not center motion.

Physical check

The two acceleration magnitudes differ inversely with mass, so their weighted sum vanishes.