By Newton’s second law of motion $$F = n\left( {mv} \right) = nmv$$

For vertical motion of bullet or ball $$u = 0,\,s = 5m,\,t = ?,\,a = 10m/{s^2}$$

$$\eqalign{ & S = ut + \frac{1}{2}a{t^2} \Rightarrow 5 = \frac{1}{2} \times 10 \times {t^2} \cr & \Rightarrow t = 1\sec \cr} $$
For horizontal motion of ball
$${x_{ball}} = {V_{ball}}t \Rightarrow 20 = {V_{ball}} \times 1 = {V_{ball}}$$
For horizontal motion of bullet
$${x_{bullet}} = {V_{bullet}} \times t \Rightarrow 100 = {V_{bullet}} \times 1 = {V_{bullet}}$$
Applying conservation of linear momentum during collision, we get
$$\eqalign{ & mV = m{V_{bullet}} + M{V_{ball}} \cr & 0.01V = 0.01 \times 100 + 0.2 \times 20 \cr & \therefore V = \frac{5}{{0.01}} = 500m/s \cr} $$

$$\eqalign{ & F = \frac{{d\left( {Mv} \right)}}{{dt}} = M\frac{{dv}}{{dt}} + v\frac{{dM}}{{dt}} \cr & \therefore v\,{\text{is}}\,{\text{constant,}} \cr & \therefore F = v\frac{{dM}}{{dt}}\,\,{\text{But}}\,\,\frac{{dM}}{{dt}} = M\,kg/s \cr & \therefore F = vM\,{\text{newton}}{\text{.}} \cr} $$

Thrust force on the rocket $${F_t} = {v_r}\left( { - \frac{{dm}}{{dt}}} \right)\,\,\left( {{\text{upwards}}} \right)$$
Weight of the rocket $$w = mg\,\,\left( {{\text{downwards}}} \right)$$
Net force on the rocket $${F_{{\text{net}}}} = {F_t} - w$$
$$\eqalign{ & \Rightarrow ma = {v_r}\left( {\frac{{ - dm}}{{dt}}} \right) - mg \cr & \Rightarrow \left( {\frac{{ - dm}}{{dt}}} \right) = \frac{{m\left( {g + a} \right)}}{{{v_r}}} \cr} $$
∴ Rate of gas ejected per second $$ = \frac{{5000\left( {10 + 20} \right)}}{{800}}$$
$$\eqalign{ & = \frac{{5000 \times 30}}{{800}} \cr & = 187.5\,kg\,{s^{ - 1}} \cr} $$

Whenever there is change in the mass w.r.t. time, apply $$F = - v\frac{{dm}}{{dt}}$$
Thrust force on the rocket $${F_t} = {v_r}\left( { - \frac{{dm}}{{dt}}} \right)\,\,\left( {{\text{upwards}}} \right)$$
Rate of combustion of fuel $$ - \frac{{dm}}{{dt}} = \frac{{{F_t}}}{{{v_r}}}$$
Given, $${F_t} = 210\,N$$
$$\eqalign{ & {v_r} = 300\,m/s \cr & \therefore - \frac{{dm}}{{dt}} = \frac{{210}}{{300}} = 0.7\,kg/s \cr} $$

Initial momentum of the system
$$\eqalign{ & {p_{net}} = \sqrt {{{\left\{ {\left( {2m} \right)v} \right\}}^2} + {{\left\{ {m\left( {2v} \right)} \right\}}^2}} \cr & {p_{net}} = \sqrt {4{m^2}{v^2} + 4{m^2}{v^2}} \cr & {p_{net}} = \sqrt {8{m^2}{v^2}} \cr & {p_{net}} = 2\sqrt 2 mv \cr} $$
Final momentum of the system $$= 3mV$$
By the law of conservation of momentum
$$\eqalign{ & {p_{net}} = 3m{V_{combined}} \cr & \Rightarrow 2\sqrt 2 mv = 3m{V_{combined}} \cr & \Rightarrow \frac{{2\sqrt 2 v}}{3} = {V_{combined}} \cr} $$
Loss in energy
$$\eqalign{ & \Delta E = \frac{1}{2}{m_1}V_1^2 + \frac{1}{2}{m_2}V_2^2 - \frac{1}{2}\left( {{m_1} + {m_2}} \right)V_{combined}^2 \cr & DE = 3m{v^2} - \frac{4}{3}m{v^2} = \frac{5}{3}m{v^2} = 55.55\% \cr} $$
Percentage loss in energy during the collision $$= 56\%$$

In $$x$$-direction, $$mv = m{v_1}\cos \theta \,......\left( 1 \right)$$
where $${v_1}$$ is the velocity of second mass
In $$y$$-direction, $$\frac{{mv}}{{\sqrt 3 }} = m{v_1}\sin \theta \,......\left( 2 \right)$$

Squaring and adding eqns. (1) and (2)
$$v_1^2 = {v^2} + \frac{{{v^2}}}{{\sqrt 3 }} \Rightarrow {v_1} = \frac{2}{{\sqrt 3 }}v$$

According to the law of conservation of momentum, $${p_i} = {p_f}$$
$$ \Rightarrow \left( {0.01} \right) \times 400 + 0 = 2v + \left( {0.01} \right)v'\,......\left( {\text{i}} \right)$$
Also velocity $$v$$ of the block just after the collision is
$$v = \sqrt {2gh} = \sqrt {2 \times 10 \times 0.1} = \sqrt 2 \,.......\left( {{\text{ii}}} \right)$$
$$ \Rightarrow $$ From Eqs. (i) and (ii), we have
$$v' \approx 120\,m/s$$

Thrust on the satellite,
$$F = \frac{{ - vdM}}{{dt}} = - v\left( {\alpha v} \right) = - \alpha {v^2}$$
Acceleration $$ = \frac{F}{M} = \frac{{ - \alpha {v^2}}}{M}$$

Momentum $$P = mv = m\sqrt {2gh} \,\,\left( {\because {v^2} = {u^2} + 2gh;\,{\text{Here}}\,u = 0} \right)$$
When stone hits the ground momentum
$$P = m\sqrt {2gh} $$
when same stone dropped from $$2h$$ ($$100\% $$  of initial) then momentum $$P' = m\sqrt {2g\left( {2h} \right)} = \sqrt 2 P$$
Which is changed by $$41\% $$  of initial.