Question

Consider the following two statements:
(A) Linear momentum of a system of particles is zero.
(B) Kinetic energy of a system of particles is zero.
(a) A implies B and B implies A.
(b) A does not imply B and B does not imply A.
(c) A implies B but B does not imply A.
(d) B implies A but A does not imply B.

Answer 1

(d) B implies A but A does not imply B.

If the linear momentum of a system is zero,
$\Rightarrow m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+...$=0

Thus, for a system of comprising two particles of same masses,
${\overrightarrow{v}}_1=-{\overrightarrow{v}}_2$ ...(1)

The kinetic energy of the system is given by,
$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\overrightarrow{v}}_1^2+\frac{1}{2}m{\overrightarrow{v}}_2^2$

Using equation (1) to solve above equation, we can say:
$\mathrm{K}.\mathrm{E}.\ne 0$
i.e. A does not imply B.

Now,
If the kinetic energy of the system is zero,
$\Rightarrow \frac{1}{2}m{\overrightarrow{v}}_1^2+\frac{1}{2}m{\overrightarrow{v}}_2^2=0$
$v_1=\pm v_2$

On calculating the linear momentum of the system, we get:
$$\begin{gathered} \overrightarrow{P}=m{\overrightarrow{v}}_1+m{\overrightarrow{v}}_2 \\ \mathrm{taking}{v}_1=-v_2,\mathrm{we}\mathrm{can}\mathrm{write}: \\ \overrightarrow{P}=0 \end{gathered}$$

Hence, we can say, B implies A but A does not imply B.