Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion of the center of mass affected by the explosion? How would it be affected if the pieces experienced significantly more air resistance than the intact shell?

The product of an object's mass and velocity is called linear momentum, translational momentum, or simply momentum. It's a two-dimensional vector quantity with a magnitude and a direction.

The explosion of the shell occurs due to internal forces.The firework shell explodes, breaking in to three large pieces for which air resistance is negligible.These three pieces after explosion will go in different direction in such a manner that net momentum of three pieces becomes zero.

As initially fireworks shell is at rest. All the forces working on the shell are internal forces and the net sum of all internal forces is zero.

\(\begin{array}{l}{P_{initial}} = M \times 0 = 0\\{F_{net}} = \dfrac{{\Delta P}}{{\Delta t}} = 0\\\dfrac{{{P_{final}} - {P_{initial}}}}{{\Delta t}} = 0\\{P_{final}} - {P_{initial}} = 0\\{P_{final}} = {P_{initial}} = 0\end{array}\)

Therefore, initial and final momentum of the system is zero. So, the center of mass of the system will be at rest and the three pieces after explosion will move in such a way that their net momentum becomes zero.If the pieces are of equal mass and moves with the equal velocity then they have to move at an angle of 120 degree from each other so that net momentum of the system become zero.

Hence, there is no external force so the center of mass of the system remains at rest when air resistance is negligible.

Due to the air resistance, there is an external force on the system.Now, net force will not be zero.Hence, momentum of the system will not be conserved.

According to Newton’s second law

Net force =\(mass \times acceleration\)

Thus, in the presence of air resistance centre of mass of the system will accelerate.Now, it will not remain at rest.