Let $$m$$ be the mass of the beads.
By energy conservation
$$\eqalign{ & mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2\,.....\left( {\text{i}} \right) \cr & \frac{{{v_1}}}{{{v_2}}} = \frac{4}{3}\,......\left( {{\text{ii}}} \right) \cr & {v_1} = \frac{{4\sqrt 2 }}{5},{v_2} = \frac{{3\sqrt 2 }}{5} \cr & K.E. = \frac{1}{2} \times 100 \times {10^{ - 3}} \times \frac{{16 \times 2}}{{25}} = 64 \times {10^{ - 3}}\,J \cr & x = 8. \cr} $$

$$\eqalign{ & u = 0;v = u + aT;v = aT \cr & {\text{Instantaneous power}} = F \times v = m.a. \cr & {\text{at}} = m.{a^2}.t \cr & \therefore {\text{Instantaneous power}} = m\frac{{{v^2}}}{{\;{T^2}}}t \cr} $$

$$\eqalign{ & {\text{Using, }}F = ma = m\frac{{dV}}{{dt}} \cr & 6t = 1\,.\,\frac{{dV}}{{dt}}\,\,\,\,\left[ {\because \,\,\,m = 1{\text{ kg given}}} \right] \cr & \int\limits_0^v {dV} = \int {6t} \,dt \cr & V = 6\left[ {\frac{{{t^2}}}{2}} \right]_0^1 = 3\,m{s^{ - 1}}\,\,\,\,\,\left[ {\because \,\,\,t = 1{\text{ sec given}}} \right] \cr} $$
From work-energy theorem,
$$W = \Delta KE = \frac{1}{2}m\left( {{V^2} - {u^2}} \right) = \frac{1}{2} \times 1 \times 9 = 4.5\,J$$

Make two equations, one from conservation of momentum and other from conversation of energy and solve it.
Let the velocity of shell be $$v$$ and that of gun be $$V.$$ Then, according to conservation of linear momentum.
$$4V + 0.2v = 0\,......\left( {\text{i}} \right)$$
Using conservation of energy
$$\frac{1}{2} \times 4 \times {V^2} + \frac{1}{2} \times 0.2 \times {v^2} = 1050\,......\left( {{\text{ii}}} \right)$$
Solving Eqs. (i) and (ii), we get
$$v = 100\,m/s$$
NOTE
Apply problem solving strategy as in question.

$${W_F} + {W_{Sp}} + {W_{{\text{fric}}}} = \Delta k$$

$$\eqalign{ & \Rightarrow F - \frac{1}{2}k{x^2} - \mu {m_1}g\,x = 0\,and\,kx = \mu {m_2}g \cr & \Rightarrow F - \frac{1}{2}\mu {m_2}g - \mu {m_1}g = 0 \Rightarrow F = \mu {m_1}g + \frac{{\mu {m_2}g}}{2} \cr} $$

Momentum of first part $$ = 1 \times 12 = 12\,kg\,m{s^{ - 1}}$$
Momentum of the second part $$ = 2 \times 8 = 16\,kg\,m{s^{ - 1}}$$
∴ Resultant momentum $$ = \sqrt {{{\left( {12} \right)}^2} + {{\left( {16} \right)}^2}} $$
$$ = 20\,kg\,m{s^{ - 1}}$$
The third part should also have the same momentum.
Let the mass of the third part be $$M,$$ then
$$\eqalign{ & 4 \times M = 20 \cr & M = 5\,kg \cr} $$
Alternative

$$\eqalign{ & Mv\cos \theta = 12\,......\left( {\text{i}} \right) \cr & Mv\sin \theta = 16\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing Eqs. (ii) and (i), we get
$$\tan \theta = \frac{{16}}{{12}} = \frac{4}{3}$$
$$M = \frac{{12 \times 5}}{{4 \times 3}} = \frac{{60}}{{12}} = 5\,kg$$

Let $$E$$ be the total energy then
$$\frac{{P.E}}{{K.E}} = \frac{{mgh}}{{E - mgh}} = \frac{2}{3} \Rightarrow E = \frac{5}{2}mgh$$
When velocity is double then initial energy becomes $$4E.$$
$${\text{So,}}\,\,\frac{{mgh}}{{4E - mgh}} = NL = \frac{{mgh}}{{10mgh - mgh}}$$
On solving we get $$\frac{{P.E}}{{K.E}} = \frac{1}{9}.$$

Kinetic energy
$$\eqalign{ & K = \frac{1}{{2m}}\left( {{p^2}} \right) \cr & {\text{or}}\,\,p = \sqrt {2mK} \cr} $$
If kinetic energy of a body is increased by $$300\% ,$$  let its momentum becomes $$p'.$$
New kinetic energy \[K' = K + \frac{{300}}{{100}}K = 4K\,\,\left( {\begin{array}{*{20}{l}} {{\rm{initial }}\,KE = K}\\ {{\rm{final }}\,KE = K'} \end{array}} \right)\]
Therefore, momentum is given by \[p' = \sqrt {2m \times 4K} \,\,\left( {\begin{array}{*{20}{l}} {{\rm{initial }}\,{\rm{momentum}} = p}\\ {{\rm{Final }}\,{\rm{momentum}} = p'} \end{array}} \right)\]
$$ = 2\sqrt {2mK} = 2p$$
Hence, percentage change (increase) in momentum
$$\eqalign{ & \frac{{\Delta p}}{p} \times 100 = \frac{{p' - p}}{p} \times 100 \cr & = \left( {\frac{{p'}}{p} - 1} \right) \times 100 \cr & = \left( {\frac{{2p}}{p} - 1} \right) \times 100 \cr & = 100\% \cr} $$

$$\left| F \right| = \frac{{dU}}{{dx}},$$    Which is greatest in the region $$CD.$$

$$\eqalign{ & {\text{Given : retardation}} \propto {\text{displacement}} \cr & {\text{i}}{\text{.e}}{\text{., }}a = - kx \cr & {\text{But }}a = v\frac{{dv}}{{dx}} \cr & \therefore \frac{{vdv}}{{dx}} = - kx \Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x x \,dx \cr & \left( {v_2^2 - v_1^2} \right) = - k\frac{{{x^2}}}{2} \cr & \Rightarrow \frac{1}{2}m\left( {v_2^2 - v_1^2} \right) = \frac{1}{2}mk\left( {\frac{{ - {x^2}}}{2}} \right) \cr & \therefore {\text{Loss in kinetic energy,}}\,\,\therefore \Delta K \propto {x^2} \cr} $$