$$\eqalign{ & {\text{Power}} = \frac{{{\text{total work done}}}}{{{\text{time}}}} \cr & = \frac{{\frac{1}{2}M{v^2}}}{t} = \frac{1}{2}\left( {m{v^2}} \right)n\left( {\because \frac{M}{t} = mn} \right) \cr & = Kn\left[ {\because KE \cdot K = \frac{1}{2}m{v^2}} \right] \cr} $$

Kinetic energy is given by $$E = \frac{1}{2}m{v^2} = \frac{1}{{2m}}{\left( {mv} \right)^2}$$
but, $$mv =$$  momentum of the particle = $$p$$
$$\eqalign{ & \therefore E = \frac{{{p^2}}}{{2m}} \cr & {\text{or}}\,\,p = \sqrt {2mE} \cr} $$
Therefore, $$\frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}{E_1}}}{{{m_2}{E_2}}}} $$
but it is given that, $${p_1} = {p_2}$$
$$\therefore {m_1}{E_1} = {m_2}{E_2}$$
$$\eqalign{ & {\text{or}}\,\,\frac{{{E_1}}}{{{E_2}}} = \frac{{{m_2}}}{{{m_1}}}\,......\left( {\text{i}} \right) \cr & {\text{Now,}}\,\,{m_1} > {m_2} \cr & {\text{or}}\,\,\frac{{{m_1}}}{{{m_2}}} > 1\,.......\left( {{\text{ii}}} \right) \cr} $$
Thus, from Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{{{E_1}}}{{{E_2}}} < 1 \cr & {\text{or}}\,\,{E_1} < {E_2} \cr} $$

Energy required $$= mgh$$
In both cases, $$h$$ is the same. Hence energy both is the same.

$$\eqalign{ & {W_1} = \frac{1}{2} \times 5 \times {10^3}{\left( {0.05} \right)^2} \cr & \Rightarrow {W_2} = \frac{1}{2} \times 5 \times {10^3}{\left( {0.10} \right)^2} \cr & \therefore \Delta W = \frac{1}{2} \times 5 \times {10^3} \times 0.15 \times 0.05 = 18.75\,J. \cr} $$

Loss in potential energy $$=$$ gain in kinetic energy
$$\eqalign{ & m \times g \times 80 = \frac{1}{2}m{v^2},\,\,10 \times 80 = \frac{1}{2}{v^2} \cr & {v^2} = 1600\,\,or,\,v = 40\,m/s \cr} $$

Change in potential energy
So, $$\Delta U = - \frac{{GMm}}{{R + 2R}} - \left( { - \frac{{GMm}}{R}} \right)$$
where, $${U_{{\text{Final}}}} = \frac{{GMm}}{{R + 2R}}$$
$$\eqalign{ & {U_{{\text{initial}}}} = \frac{{ - GMm}}{R} \cr & \therefore \Delta U = - \frac{{GMm}}{{3R}} + \frac{{GMm}}{R} \cr & = \frac{{2GMm}}{{3R}} = \frac{2}{3}mgR\,\,\left[ {\because g = \frac{{GM}}{{{R^2}}}} \right] \cr} $$

Conserving angular momentum
$$\eqalign{ & {L_i} = {L_f} \cr & \Rightarrow m{v_0}{R_0} = mv'\left( {\frac{{{R_0}}}{2}} \right) \cr & \Rightarrow v' = 2{v_0} \cr} $$
So, final kinetic energy of the particle is
$$\eqalign{ & {K_f} = \frac{1}{2}m{{v'}^2} = \frac{1}{2}m{\left( {2{v_0}} \right)^2} \cr & = 4\frac{1}{2}mv_0^2 = 2mv_0^2 \cr} $$

Initial kinetic energy of the system
$$K.{E_i} = \frac{1}{2}m{u^2} + \frac{1}{2}M{\left( 0 \right)^2} = \frac{1}{2} \times 0.5 \times 2 \times 2 + 0 = 1\,J$$
For collision, applying conservation of linear momentum
$$\eqalign{ & m \times u = \left( {m + M} \right) \times v \cr & \therefore 0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = \frac{2}{3}\,m/s \cr} $$
Final kinetic energy of the system is
$$K.{E_f} = \frac{1}{2}\left( {m + M} \right){v^2} = \frac{1}{2}\left( {0.5 + 1} \right) \times \frac{2}{3} \times \frac{2}{3} = \frac{1}{3}\,J$$
$$\therefore $$ Energy loss during collision $$ = \left( {1 - \frac{1}{3}} \right)J = 0.67\,J$$

$$\eqalign{ & J = m\left( {{v_f} - {v_i}} \right)\,\,{\text{or}}\,\,7.5 = 15\left( {{v_f} - 0} \right) \cr & \Rightarrow {v_f} = 0.5\,m/s \cr & \therefore K = \frac{1}{2}m{v^2} = \frac{1}{2} \times 15 \times {0.5^2} = 1.9\,J \cr} $$

$$U = - \int_0^x {Fdx} = - \int_0^x {kxdx} = - \frac{1}{2}k{x^2}.$$
It is correctly drawn in (A)