Given, $$h = 60\,m,g = 10\,m{s^{ - 2}},$$
Rate of flow of water $$= 15\,kg/s$$
$$\therefore $$ Power of the falling water $$ = 15\,kg{s^{ - 1}} \times 10\,m{s^{ - 2}} \times 60\,m = 900\,watt.$$
Loss in energy due to friction $$ = 9000 \times \frac{{10}}{{100}} = 900\,watt.$$
$$\therefore $$ Power generated by the turbine $$ = \left( {9000 - 900} \right)watt$$
$$\eqalign{ & = 8100\,watt \cr & = 8.1\,kW \cr} $$

The power of body is given by $$ = \vec F.\vec v$$   as the body is moving in circular path, centripetal force and velocity are at $${90^ \circ },$$ or power $$ = 0.$$

Let the radius of the circle be $$r.$$ Then the two distance travelled by the two particles before first collision is $$2\pi r.$$   Therefore $$2v \times t + v \times t = 2\pi r$$     where $$t$$ is the time taken for first collision to occur.
$$\therefore t = \frac{{2\pi r}}{{3v}}$$
$$\therefore $$ Distance travelled by particle with velocity $$v$$ is equal to $$v \times \frac{{2\pi r}}{{3v}} = \frac{{2\pi r}}{3}.$$
Therefore the collision occurs at $$B.$$

As the collision is elastic and the particles have equal masses, the velocities will interchange as shown in the figure. According to the same reasoning as above, the 2^nd collision will take place at $$C$$ and the velocities will again interchange.
With the same reasoning the 3^rd collision will occur at the point $$A.$$ Thus there will be two elastic collisions before the particles again reach at $$A.$$

The centripetal acceleration
$$\eqalign{ & {a_c} = {k^2}r{t^2}\,{\text{or}}\,\frac{{{v^2}}}{r} = {k^2}r{t^2} \cr & \therefore v = krt \cr} $$
So, tangential acceleration, $${a_t} = \frac{{dv}}{{dt}} = kr$$
Work is done by tangential force.
$$\eqalign{ & {\text{Power}} = {F_t} \cdot v \cdot \cos {0^ \circ } = \left( {m{a_t}} \right)\left( {krt} \right) = \left( {mkr} \right)\left( {krt} \right) \cr & = m{k^2}{r^2}t \cr} $$

The degree of elasticity of a collision is determined by a quantity called coefficient of restitution or coefficient of resilience of the collision. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. It is represented by $$e.$$
$$\eqalign{ & e = \frac{{{\text{relative velocity of separation}}\left( {{\text{after collision}}} \right)}}{{{\text{relative velocity of approach}}\left( {{\text{before collision}}} \right)}} \cr & e = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}} \cr} $$
where, $${u_1},{u_2}$$  are the velocities of two bodies before collision and $${v_1},{v_2}$$  are their respective velocities after collision.
For a perfectly elastic collision, relative velocity of separation after collision is equal to relative velocity of approach before collision
$$\therefore e = 1$$

The forces acting on the bead as seen by the observer in the accelerated frame are : (a) $$N ;$$  (b) $$mg ;$$  (c) $$ma$$  (pseudo force).

Let $$\theta $$  is the angle which the tangent at $$P$$  makes with the X- axis. As the bead is in equilibrium with respect to the wire, therefore $$N\sin \theta = ma\,\,{\text{and }}N\cos \theta = mg$$
$$\therefore \tan \theta = \frac{a}{g}.....(i)$$
But $$y = k{x^2}$$
Therefore,
$$\frac{{dy}}{{dx}} = 2kx = \tan \theta .....(ii)$$
From $$(i)\,\& \,(ii)$$
$$2kx = \frac{a}{g}\,\,\,\, \Rightarrow x = \frac{a}{{2kg}}$$

At the highest point
$$\eqalign{ & {v_x} = u\cos \theta \cr & {v_y} = 0 \cr & {K_H} = \frac{1}{2}mv_x^2 \cr & {\text{or}}\,\,{K_H} = \frac{1}{2}m{u^2}{\cos ^2}\theta \,......\left( {\text{i}} \right) \cr} $$
Initial kinetic energy is
$$K = \frac{1}{2}m{u^2}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$\eqalign{ & {K_H} = K{\cos ^2}\theta \cr & = K{\cos ^2}{45^ \circ } \cr & = K \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = \frac{K}{2} \cr} $$

We know that $$\Delta U = - W$$   for conservative forces
$$\eqalign{ & \Delta U = - \int_0^x {Fdx} \;{\text{or }}\Delta U = - \int_0^x {k\,xdx\,\,} \cr & \Rightarrow {U_{\left( x \right)}} - {U_{\left( 0 \right)}} = - \frac{{k{x^2}}}{2} \cr & {\text{Given }}{U_{\left( 0 \right)}} = 0\,\,\,\,\,\,\,\,\,{U_{\left( x \right)}} = - \frac{{k{x^2}}}{2} \cr} $$

$$\eqalign{ & K.E. = \frac{{{p^2}}}{{2m}} \cr & {E_1} = {E_2}\,\,\,\,\,\therefore \frac{{p_1^2}}{{{m_1}}} = \frac{{p_2^2}}{{{m_2}}} \cr & \therefore \frac{{p_1^2}}{{p_2^2}} = \frac{{{m_1}}}{{{m_2}}} \cr & \Rightarrow \frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} = \sqrt {\frac{1}{4}} = \frac{1}{2} \cr} $$

When two bodies of equal masses undergo head on elastic collision in one dimension their velocities are just interchanged.