91Ó°ÊÓ

If thenet external force on a system is zero, the total momentum of the system $\stackrel{\mathbf{→}}{\mathbf{P}}$is constant or conserved. Each component of total momentum is separately conserved.

If$\underset{\mathbf{}}{\mathbf{∑}}\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{0}$

Then

$\stackrel{\mathbf{→}}{\mathbf{P}}$= constant.

Yes, the kinetic energy of the system of two objects can be zero, if the magnitude of initial momentum of both the objects are the same and the collision is head-on. And initial kinetic energy of the system is not zero.

For example

Consider two objects A and B with initial momentum p and -p

From the conservation of momentum

$$\begin{gathered} P=P_A+P_B \\ P+\left(-P\right)=m_A{\stackrel{→}{v}}_A+m_B{\stackrel{→}{v}}_B \\ 0=m_A{\stackrel{→}{v}}_A+m_B{\stackrel{→}{v}}_B \end{gathered}$$

Since two objects stick together so they will move with the same velocity after collision. Suppose the final velocity is$\stackrel{→}{v}$ .

$$\begin{gathered} m_A\stackrel{→}{v}+m_B\stackrel{→}{v}=0 \\ \left(m_A+m_B\right)\stackrel{→}{v}=0 \end{gathered}$$

Since mass cannot be zero.

Therefore,

$\stackrel{→}{v}=0$

Since final velocity is zero so kinetic energy is also zero.

Hence the kinetic energy of the system of two objects can be zero if the magnitude of initial momentum of both the objects is the same and the collision is head-on.

Since kinetic energy is a scalar quantity

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ K=\frac{1}{2}\frac{m^2{v}^2}{m}=\frac{1}{2}\frac{P^2}{m} \end{gathered}$$

So initial kinetic energy

$$\begin{gathered} K=K_A+K_B \\ =\frac{1}{2}\frac{P^2}{m_A}+\frac{1}{2}\frac{{\left(-P\right)}^2}{m_B} \\ =\frac{1}{2}\frac{P^2}{m_A}+\frac{1}{2}\frac{{\left(P\right)}^2}{m_B}≠0 \end{gathered}$$

Therefore, the initial kinetic energy of the system is not zero.