This is an example of Newton's Third Law of Motion applicable in all circumstances

$$\eqalign{ & S = \frac{{{V^2} - {U^2}}}{{2a}}\,{\text{and}}\,a = \frac{F}{m},{\text{so}} \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{{\left( {\frac{{0 - {U^2}}}{{\frac{{2F}}{m}}}} \right)}}{{\left( {\frac{{0 - {U^2}}}{{\frac{{2F}}{{2m}}}}} \right)}} = 1:2 \cr} $$

When the resultant external torque acting on a system is zero, the total angular momentum of a system remains constant. This is the principle of the conservation of angular momentum.
Given, force $$F = \alpha \hat i + 3\hat j + 6\hat k$$    is acting at a point $$r = 2\hat i - 6\hat j - 12\hat k$$
As, angular momentum about origin is conserved.
i.e. $$\tau = {\text{constant}}$$
$$ \Rightarrow {\text{Torque,}}\,\tau = 0 \Rightarrow r \times F = 0$$
\[\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 2&{ - 6}&{ - 12}\\ \alpha &3&6 \end{array}} \right| = 0\]
$$\eqalign{ & \Rightarrow \left( { - 36 + 36} \right)\hat i - \left( {12 + 12\alpha } \right)\hat j + \left( {6 + 6\alpha } \right)\hat k = 0 \cr & \Rightarrow 0\hat i - 12\left( {1 + \alpha } \right)\hat i + 6\left( {1 + \alpha } \right)\hat k = 0 \cr & \Rightarrow 6\left( {1 + \alpha } \right) = 0 \cr & \Rightarrow \alpha = - 1 \cr} $$
So, value of $$\alpha $$ for angular momentum about origin is conserved, $$\alpha = - 1$$

Momentum is conserved before and after collision.
We have, $${p_1} + {p_2} + {p_3} = 0\,\,\left[ {\because p = mv} \right]$$
$$\eqalign{ & \therefore 1 \times 12i + 2 \times 8j + {p_3} = 0 \cr & \Rightarrow 12i + 16j + {p_3} = 0 \cr & \Rightarrow {p_3} = - \left( {12i + 16j} \right) \cr & \therefore {p_3} = \sqrt {{{\left( {12} \right)}^2} + {{\left( {16} \right)}^2}} \cr & = \sqrt {144 + 256} \cr & = 20\,kg{\text{-}}m/s \cr & {\text{Now,}}\,\,{p_3} = {m_3}{v_3} \cr & \Rightarrow {m_3} = \frac{{{p_3}}}{{{v_3}}} = \frac{{20}}{4} = 5\,kg \cr} $$

According to 1^st equation of motion, velocity of particle after $$5\,s$$
$$\eqalign{ & v = u - gt \cr & v = 100 - 10 \times 5 \cr & = 100 - 50 = 50\,m/s\,\,\left( {{\text{upwards}}} \right) \cr} $$
Applying conservation of linear momentum gives
$$Mv = {m_1}{v_1} + {m_2}{v_2}\,.......\left( {{\text{i}}} \right)$$
Taking upward direction positive, the velocity $${v_1}$$ will be negative.
$$\eqalign{ & \therefore {v_1} = - 25\,m/s, \cr & v = 50\,m/s \cr} $$
$$\eqalign{ & {\text{Also,}}\,\,M = 1\,kg,\,{m_1} = 400\,g = 0.4\,kg \cr & {\text{and}}\,\,\,{m_2} = \left( {M - {m_1}} \right) = 1 - 0.4 = 0.6\,kg \cr} $$
Thus, Eq. (i) becomes,
$$\eqalign{ & 1 \times 50 = 0.4 \times \left( { - 25} \right) + 0.6{v_2} \cr & {\text{or}}\,\,50 = - 10 + 0.6{v_2} \cr} $$
$$\eqalign{ & {\text{or}}\,\,0.6{v_2} = 60 \cr & {\text{or}}\,\,{v_2} = \frac{{60}}{{0.6}} \cr & = 100\,m/s \cr} $$
As $${v_2}$$ is positive, therefore the other part will move upwards with a velocity of $$100\,m/s.$$

If we consider the two particles as a system then the external force acting on the system is the gravitational pull $$2\left( {{m_1} + {m_2}} \right)g.$$
$$\eqalign{ & {F_{{\text{ext}}}} = \frac{{\Delta p}}{{\Delta t}} \cr & \therefore \Delta p = {F_{{\text{ext}}}}\Delta t = \left( {{m_1}{{\vec v}_1}' + {m_2}{{\vec v}_2}'} \right) - \left( {{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}} \right) \cr & = \left( {{m_1} + {m_2}} \right)g \times 2{t_0} \cr} $$

If two bodies collide head on with coefficient of restitution
$$e = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\,......\left( {\text{i}} \right)$$
From, the law of conservation of linear momentum
$$\eqalign{ & {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2} \cr & \Rightarrow {v_1} = \left[ {\frac{{{m_1} - e{m_2}}}{{{m_1} + {m_2}}}} \right]{u_1} + \left[ {\frac{{\left( {1 + e} \right){m_2}}}{{{m_1} + {m_2}}}} \right]{u_2} \cr & {\text{Substituting }}{u_1} = 2\,m{s^{ - 1}},{u_2} = 0,{m_1} = m\,{\text{and}}\,{m_2} = 2m,\,e = 0.5 \cr} $$
we get, $${v_1} = \left[ {\frac{{m - m}}{{m + 2m}}} \right] \times 2$$
$$ \Rightarrow {v_1} = 0$$
Similarly, $${v_2} = \left[ {\frac{{\left( {1 + e} \right){m_1}}}{{{m_1} + {m_2}}}} \right]{u_1} + \left[ {\frac{{{m_1} - e{m_1}}}{{{m_1} + {m_2}}}} \right]{u_2}$$
$$\eqalign{ & = \left[ {\frac{{1.5 \times m}}{{3m}}} \right] \times 2 \cr & = 1\,m{s^{ - 1}} \cr} $$

Just after collision
$${v_c} = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}} = \frac{{10 \times 14 + 4 \times 0}}{{10 + 4}} = 10\,m/s;$$
Note : Spring force is an internal force, it cannot change the linear momentum of the (two mass + spring) system. Therefore $${v_c}$$ remains the same.

In completely inelastic collision, all energy is not lost (so, statement -1 is true) and the principle of conservation of momentum holds good for all kinds of collisions (so, statement -2 is true) . Statement -2 explains statement -1 correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body.

When, shell explodes in mid air. The potential energy of the shell changes into kinetic energy of fragments. Hence, kinetic energy increases.