$$\eqalign{ & {\text{Thrust}}\, = \frac{{udM}}{{dt}} = mg \cr & \Rightarrow \frac{{dM}}{{dt}} = \frac{{mg}}{u} = \frac{{600 \times 10}}{{1000}} = 6\,kg\,{s^{ - 1}} \cr} $$

$$\eqalign{ & \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{3}{2}\left( {\frac{1}{2}mv_0^2} \right) \cr & \Rightarrow v_1^2 + v_2^2 = \frac{3}{2}v_0^2\,......\left( {\text{i}} \right) \cr} $$
From momentum conservation
$$m{v_0} = m\left( {{v_1} + {v_2}} \right)\,......\left( {{\text{ii}}} \right)$$
Squarring both sides,
$$\eqalign{ & {\left( {{v_1} + {v_2}} \right)^2} = v_0^2 \cr & \Rightarrow v_1^2 + v_2^2 + 2{v_1}{v_2} = v_0^2 \cr & 2{v_1}{v_2} = - \frac{{v_0^2}}{2} \cr & {\left( {{v_1} - {v_2}} \right)^2} = v_1^2 + v_2^2 - 2{v_1}{v_2} = \frac{3}{2}v_0^2 + \frac{{v_0^2}}{2} \cr} $$
Solving we get relative velocity between the two particles
$${v_1} - {v_2} = \sqrt 2 {v_0}$$

Let $$n$$ be the number of bullets that the man can fire in one second.
$$\therefore $$ change in momentum per second $$ = n \times mv = F$$
[ $$m$$ = mass of bullet, $$v$$ = velocity] ($$\because F$$  is the force)
$$\therefore n = \frac{F}{{mv}} = \frac{{144 \times 1000}}{{40 \times 1200}} = 3$$

Let the velocity and mass of $$4 kg$$  piece be $${v_1}$$ and $${m_1}$$ and that of $$12 kg$$  piece be $${v_2}$$ and $${m_2.}$$

Applying conservation of linear momentum
$$\eqalign{ & {m_2}{v_2} = {m_1}{v_1} \Rightarrow {v_1} = \frac{{12 \times 4}}{4} = 12m{s^{ - 1}} \cr & \therefore K.E{._1} = \frac{1}{2}{m_1}v_1^2 = \frac{1}{2} \times 4 \times 144 = 288J \cr} $$

Apply conservation of momentum with direction.
Let $$u$$ be the velocity and $$\theta $$ the direction of the third piece as shown.

Equating the momentum of the system along $$OA$$  and $$OB$$  to zero, we get
$$\eqalign{ & m \times 30 - 3m \times v\cos \theta = 0\,......\left( {\text{i}} \right) \cr & {\text{and}}\,m \times 30 - 3m \times v\sin \theta = 0\,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii), we get
$$\eqalign{ & 3mv\cos \theta = 3mv\sin \theta \cr & {\text{or}}\,\,\cos \theta = \sin \theta \cr & \therefore \theta = {45^ \circ } \cr & {\text{Thus,}}\,\,\angle AOC = \angle BOC \cr & = {180^ \circ } - {45^ \circ } = {135^ \circ } \cr} $$
Putting the value of $$\theta $$ in Eq. (i), we get
$$\eqalign{ & 30\;m = 3\;mv\cos {45^ \circ } = \frac{{3mv}}{{\sqrt 2 }} \cr & \therefore v = 10\sqrt 2 \,m/s \cr} $$
The third piece will go with a velocity of $$10\sqrt 2 \,m/s$$   in a direction making an angle of $${135^ \circ }$$ with either piece.
Alternative
The square of momentum of third piece is equal to sum of squares of momentum first and second pieces.
$$\eqalign{ & p_3^2 = p_1^2 + p_2^2 \cr & {\text{or}}\,\,{p_3} = \sqrt {p_1^2 + p_2^2} \cr & {\text{or}}\,\,3m{v_3} = \sqrt {{{\left( {m \times 30} \right)}^2} + {{\left( {m \times 30} \right)}^2}} \cr & {\text{or}}\,\,{v_3} = \frac{{30\sqrt 2 }}{3} \cr & = 10\sqrt 2 \,m/s \cr} $$

Activity B to M for particle thrown upwards
$$\eqalign{ & v_1^2 - u_0^2 = 2( - g)\quad \left[ {\frac{{u_0^2{{\sin }^2}\alpha }}{{2g}}} \right] \cr & \therefore v_1^2 = u_0^2\left( {1 - {{\sin }^2}\alpha } \right) = u_0^2{\cos ^2}\alpha \cr & \therefore {v_1} = {u_0}\cos \alpha \,......\left( {\text{i}} \right) \cr} $$

Applying conservation of linear momentum in $$Y$$-direction
$$2mv\sin \theta = m{v_1} = m{u_0}\cos \alpha \,......\left( {{\text{ii}}} \right)\,[{\text{from}}\,\left( {\text{i}} \right)]$$
Applying conservation of linear momentum in $$X$$-direction
$$2mv\cos \theta = m{u_0}\cos \alpha \,.....\left( {{\text{iii}}} \right)$$
on dividing (ii) and (iii) we get
$$\tan \theta = 1\,\therefore \theta = \frac{\pi }{4}$$

For head on elastic collision we have
$$\eqalign{ & {V_1} = \frac{{\left( {{m_1} - {m_2}} \right){u_1}}}{{{m_1} + {m_2}}} + \frac{{2{m_2}{u_2}}}{{{m_1} + {m_2}}} \cr & {\text{Here }}{m_1} = 2kg,{u_1} = x,{u_2} = 0 \cr & {v_1} = \frac{x}{4} \cr & \therefore \frac{x}{4} = \frac{{\left( {2 - {m_2}} \right)x}}{{2 + {m_2}}} \cr & \Rightarrow {m_2} = 1.2kg \cr} $$

Kinetic energy of block $$A$$
$${k_1} = \frac{1}{2}mv_0^2$$
$$\therefore $$ From principle of linear momentum conservation
$$\eqalign{ & m{v_0} = \left( {2m + M} \right){v_f} \Rightarrow {v_f}\frac{{m{v_0}}}{{2m + M}} \cr & K.{E_f} = \frac{1}{6}K.{E_i}\,\,\,\,\left( {{\text{given}}} \right) \cr & \frac{1}{2}\left( {2m + M} \right)v_f^2 = \frac{1}{6} \times \frac{1}{2}mv_0^2 \cr & 6\left( {2m + M} \right)\frac{{{m^2}v_0^2}}{{{{\left( {2m + M} \right)}^2}}} = mv_0^2 \cr & \Rightarrow 6m = 2m + M \cr & \Rightarrow 4m = M \cr & \therefore \frac{M}{m} = \frac{4}{1} \cr} $$

$$\eqalign{ & \overrightarrow p \left( t \right) = A\left[ {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right] \cr & \overrightarrow F = \frac{{d\overrightarrow p }}{{dt}} = Ak\left[ { - \hat i\sin \left( {kt} \right) - \hat j\cos \left( {kt} \right)} \right] \cr & {\text{Here,}}\,\,\overrightarrow F .\overrightarrow P = 0\,\,{\text{But}}\,\,\overrightarrow F .\overrightarrow p = Fp\cos \theta \cr & \therefore \cos \theta = 0 \Rightarrow \theta = {90^ \circ }. \cr} $$

Here, $${p_i} = {m_2}vi + {m_1} \times 0$$
$${p_f} = {m_2}\frac{v}{2}\hat j + {m_1} \times {v_1}$$
Law of conservation of momentum
$$\eqalign{ & {p_i} = {p_f} \cr & {m_2}v\hat i = {m_2}\frac{v}{2}\hat j + {m_1} \times {v_1} \cr & {v_1} = \frac{{{m_2}}}{{{m_1}}}v\hat i + \frac{{{m_2}}}{{{m_1}}}\frac{v}{2}\hat j \cr} $$

From this equation, we can find
$$\tan \theta = \frac{y}{x} = \frac{1}{2},\,\theta = {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$$      to the $$x$$-axis.