Consider the following two statements: 1. Linear momentum of a system of particles is zero 2. Kinetic energy of a system of particles is zero Then

The correct option is: 1 does not imply 2 but 2 implies 1

Let us analyze the two statements step-by-step to understand the logical relationship between them.

Statement 1: Linear momentum of a system of particles is zero.
The total linear momentum $P$ of a system of $n$ particles is the vector sum of the individual linear momenta of the particles:
$P=p_1+p_2+...+p_n=\sum _{i=1}^n{m}_i{v}_i$
If the total linear momentum is zero ($P=0$), it means the vector sum of the individual momenta is zero. This does not require each individual velocity $v_i$ to be zero. For example, in a two-particle system, if one particle moves with velocity $v$ and another identical particle moves with velocity $-v$, the total linear momentum is:
$P=mv+m(-v)=0$
However, since the particles are in motion, the total kinetic energy of the system is:
$K=\frac{1}{2}m{v}^2+\frac{1}{2}m{(-v)}^2=m{v}^2\ne 0$
Thus, Statement 1 does not imply Statement 2.

Statement 2: Kinetic energy of a system of particles is zero.
The total kinetic energy $K$ of a system of particles is the sum of the individual kinetic energies:
$K=\sum _{i=1}^n\frac{1}{2}$
Since mass $m_i$ is always positive and the term $v_i^2$ is non-negative, the kinetic energy of each particle is non-negative ($K_i\ge 0$). For the sum of non-negative terms to be zero, each individual term must be zero:
$\frac{1}{2}{m}_i{v}_i^2=0\Rightarrow v_i=0$
for all $i$. Since all particles are at rest, their individual linear momenta are also zero:
$p_i=m_i{v}_i=0$
Consequently, the total linear momentum of the system must be zero:
$P=\sum p_i=0$
Thus, Statement 2 implies Statement 1.

Therefore, Statement 1 does not imply Statement 2, but Statement 2 implies Statement 1.