Newton's Second Law for a System of Particles — Question 4

Question 4

Following an object after it breaks apart

A projectile breaks into fragments in midair at $(x,y)=(0,20)\,\mathrm m$. Just before breakup its velocity is $(6,4)\,\mathrm{m/s}$. Afterward, all fragments remain airborne for $1.0\,\mathrm s$. Air resistance is negligible and $g=9.8\,\mathrm{m/s^2}$.

1. Find the system center position and velocity $1.0\,\mathrm s$ after breakup.

2. Explain why knowing the energy released is unnecessary, and state when the stated trajectory would cease to apply.

Question 4 - Solution

Following an object after it breaks apart

(a) Derivation and calculation

For all airborne fragments together, $$M\vec a_C=(0,-Mg)\Rightarrow\vec a_C=(0,-g).$$ $$\vec R_C(t)=(6t,20+4t-4.9t^2),\qquad\vec V_C(t)=(6,4-9.8t).$$ $$\boxed{\vec R_C(1)=(6,19.1)\ \mathrm m},\qquad \boxed{\vec V_C(1)=(6,-5.8)\ \mathrm{m/s}}.$$

(b) Result and interpretation

The breakup impulses obey $$\sum_i\vec J_{i,\mathrm{internal}}=\vec0,$$ so released internal energy is unnecessary for the center trajectory. Ground contact or appreciable drag adds external force and invalidates the gravity-only equation.

Physical check

The center follows the trajectory of the unbroken projectile while gravity is the only external force.