Radius =R
\(v_{1}^{2}-v_{2}^{2}=4 g R\)
\(\text { Tension } T_{2}=\frac{m v_{2}^{2}}{R_{2}}-m g\)
\(\text {To find minimum speed, let } T_{2}=0\)
\(0 =\frac{m v_{2}^{2}}{R}-m g \)
\(\frac{m v^{2}}{R} =m g \)
\(v_{2}^{2}=R g \ v_{2} =\sqrt{g R} \)
\(\text { sub (2) in the eqn (1) we get }\)
\(v_{1}^{2}-(\sqrt{g R})^{2} =4 g R \)
\(v_{1}^{2}-g R =4 g R \)
\(v_{1}^{2} =4 g R+g R \)
\(=5 g R \)
\(v_{1} =\sqrt{5 g R} \)

For a perfectly elastic collision
\(\left|v_{1}-v_{2}\right|=\left|u_{1}-u_{2}\right| \)
\(v_{e}=\frac{\left|v_{1}-v_{2}\right|}{\left|u_{1}-u_{2}\right|}=1 \)

m = 2 kg, d = 10 m, F_ext = 20 N, \(\mu\)_k = 0.9.
when an object is in motion on he horizontal surface, it experiences two forces.
(a) External force, F_ext = 20 N
(b) Kinetic friction,
f_k = \(\mu\)_k mg = 0.9 \(\times\) (2) \(\times\) 10 = 18N
The work done by the external force W_ext = Fd = 20 x 10 = 200J
The work done by the force of kinetic friction W_k = f_kd = (-18) \(\times\) 10 = -180 J. Here the negative sign implies that the force of kinetic friction is opposite to the direction of displacement.
The total work done on the object W_total = W_ext + W_k = 200 J - 180 J = 20 J.
Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.

K.E=\(\frac{1}{2}\)mv²
=\(\frac{1}{2}\)m(2 as) [From equation of motion II]
∴ K.E.α S.

The impulse on the ball acts perpendicular to the smooth plane.
(i) The component of velocity of ball parallel to the surface. 
(ii) For the component of velocity of ball perpendicular to the surface, apply law of restitution.
The component of velocity parallel to the surface will be changed.

v cos α = u cos 60°
v cos α = 5\(\times\frac{1}{2}=\frac{5}{2}\)
According to law of restitution
v sin α = e.u sin 60° ....(2)
v sin α =\(\frac{1}{2}\times5\times\frac{\sqrt3}{2}=5\frac{\sqrt3}{4}\)
Squaring and adding (1) and (2)
v² (sin² α + cos² α) =\(\left[ \frac { 25 }{ 4 } +\frac { 25\times 3 }{ 16 } \right] \)
v² = \(\left[ \frac { 25 }{ 4 } +\frac { 75 }{ 16 } \right] \)
v² = 10.9
∴ v = 3.3 ms^-1

Characteristics of elastic collision are
1. Total momentum remains conserved
2. Total kinetic energy remains conserved.
3. In elastic collision conservative forces are involved. Hence total kinetic energy is conserved.
4. In elastic collision, mechanical energy is not dissipated.
Characteristics of inelastic collision are
1. Total momentum is conserved.
2. Total kinetic energy is not conserved.
3. Forces involved are non-conservative forces
4. Mechanical energy is dissipated into heat, light, sound etc.

The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from ground to that height h with constant velocity. 

(i) Potential energy of an object at a point P is defined as the amount of work done by an external force in moving the object at constant velocity from the point 0 (initial location) to the point P (final location).
(ii) At initial point O potential energy can be taken as zero.
(iii) Mathematically, potential energy is defined as \(U=\int { { \vec { F } }_{ a } } .\vec { dr } \)
where the limit of integration ranges from initial, location point O to final location point P.

(a) The kinetic energy of the mass is given by \(KE=\frac { { p }^{ 2 } }{ 2m } \)
For the object of mass 2 kg, kinetic energy is KE₁ = \(\frac { ({ 20 })^{ 2 } }{ 2\times 2 } =\frac { 400 }{ 4 } =100 \ J\)
For the object of mass 4 kg, kinetic energy is KE₂ = \(\frac { { (20) }^{ 2 } }{ 2\times 4 } =\frac { 400 }{ 8 } =50 \ J\)
Note that KE₁ \(\neq \) KE₂ i.e., even though both are having the same momentum, the kinetic energy of both masses is not the same. The kinetic energy of the heavier object has lesser kinetic energy than smaller mass. It is because the kinetic energy is inversely proportional to the mass (KE \(\infty \frac { 1 }{ m } \) ) for a given momentum.
(b) As the momentum, p = mv, the two objects will not have same speed.

It implies the following.
(i) If the work done by the force on the body is positive then its kinetic energy increases.
(ii) If the work done by the force on the-body is negative then its kinetic energy decreases.
(iii) If there is no work done by the force on the body then there is no change in its, kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

(i) The energy produced by mechanical means is called mechanical energy.
(ii) It is classified into 2 types : (1) Kinetic energy (2) Potential energy.
(iii) The energy possessed by a body due to its motion is called kinetic energy. The energy possessed by the body by virtue of its position is called potential energy. SI unit of energy: N m (or) joule (J).

(i) When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation
dw=(f cosθ)dr
[F cos θ is the component of the variable force F]
where, F and θ are variables.
(ii) The total work. done for a displacement from initial position r_i to final position r_f is given by the relation,
\(W=\int _{ { r }_{ i } }^{ { r }_{ f } }{ dW } =\int _{ { r }_{ i } }^{ { r }_{ f } }{ F\cos\theta \ dr } \ \) ---- (1)
(iii) A graphical representation of the work done by a variable force is shown in Figure 1. The area under the graph is the work done by the variable force.

Applied force = 10 N
Opposing friction forcej= M_k. N = M_k·mg.
= 0.3\(\times\)3\(\times\)9.8 = 8.82 N.
Net accelerating forceF - f = 10N - 8.82N
=1.18N
Acceleration a =\(\frac{force}{mass}\)=\(\frac{8.82N}{3Kg}=2.94 ms^{-2}\)
Distance covered in 10s (assuming w = 0)
\(s=0+\frac{1}{2}at^{2}=\frac{1}{2}\times2.94\times(10^{2}) = 147 m\)
there force workdone by a applied force,
W = Fs = 10\(\times\)147
W = 1470 J

(i) The potential energy possessed by a spring due to a deforming force which stretches or compresses the spring is termed as elastic potential energy. The work done by the applied force against the restoring force of the spring is stored as the elastic potential energy in the spring.
(ii) Consider a spring-mass system. Let us assume a mass, m lying on a smooth horizontal table as shown in Figure. Here, x = 0 is the equilibrium position. One end of the spring is attached to a rigid wall and the other end to the mass.
               
(iii) As long as the spring remains in equilibrium position, its potential energy is zero. Now an external force \(\overrightarrow { { F }_{ a } } \) is applied so that it is stretched by a distance (x) in the direction of the force.
(iv) There is a restoring force called spring force \(\overrightarrow { { F }_{ s } } \) developed in the spring which tries. to bring the mass back to its original position. This applied force and the spring force are equal in magnitude but opposite in direction i.e \(\overrightarrow { { F }_{ a } } \) = - \(\overrightarrow { { F }_{ s } } \) According Hooke's law, the restoring force developed in the spring is
\(\overrightarrow { { F }_{ s } } \) = -k\(\overrightarrow { x } \)
(v) The negative sign in the above expression implies that the spring force is always opposite to that of displacement \(\overrightarrow { x } \) and k is the force constant. There, fore applied force is \(\overrightarrow { { F }_{ a } } \) = +k\(\overrightarrow { x } \) The positive sign implies. that the applied force is in the direction of displacement \(\overrightarrow { x } \). The spring force is an example of variable force as it depends on the displacement \(\overrightarrow { x } \). Let the spring be stretched to a small distance d\(\overrightarrow { x } \). The work done by the applied force on the spring to stretch it by a displacement \(\overrightarrow { x } \) is stored as elastic potential energy.
 \(U=\int \overrightarrow { { F }_{ a } } .d\overrightarrow { r } =\overset { x }{ \underset { 0 }{ \int } } \left| \overrightarrow { { F }_{ a } } \right| \left| d\overrightarrow { r } \right| \cos { \theta } \)
\(=\overset { x }{ \underset { 0 }{ \int } } { F }_{ a }dx\cos { \theta } \)
(vi) The applied force \(\overrightarrow { { F }_{ a } } \) and the displacement d\(\overrightarrow { { r } } \) (i.e., here dx ) are in the same direction. As, the initial position is taken as the equilibrium position or mean position, x = 0 is the lower limit of integration.
\(U=\overset { x }{ \underset { 0 }{ \int } } Kxdx\)
\(U={ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ x }\)
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }\)
(vii) If the initial position is not zero, and if the mass is changed from position x_i to x_f then the elastic potential energy is
\(U=\frac { 1 }{ 2 } k\left( { x }_{ f }^{ 2 }-{ x }_{ i }^{ 2 } \right) \)
From equations (1) and (2), we observe that the' potential energy of the stretched spring depends on the force constant k and elongation or compression x.

If there is a loss of kinetic energy during a collision, then it is called as an inelastic collision
In the case of inelastic collision,
(i) Total kinetic energy is not conserved.
(ii) Some or all of the forces involved are non-conservative.
(iii) A part of the mechanical energy is transformed into heat, sound, light etc.
Examples for inelastic collision:
(i) Collision between ball and floor
(ii) Collision between two vehicles
Examples for perfectly inelastic collision:
(i) Mud thrown on a wall and sticking to it
(ii) a man jumping into a moving trolley
(iii) a bullet fired into a wooden block and remaining embedded in it.

Work done by a constant force:
(i) When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW= (F cos\(\theta\) ) dr
(ii) The total work done in producing a displacement from initial position r_i to final position r_f is,
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dw\)
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } \left( F\cos { \theta } \right) dr=\left( F\cos { \theta } \right) \)\(\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dr=\left( F\cos { \theta } \right) \left( { r }_{ f }-{ r }_{ i } \right) \)
(iii) The graphical representation of the work done by a constant force is shown in Figure. The area under the graph shows the work done by the constant force.
              
Work done by a variable force:
(i) When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation
dw = (F cos\(\theta\)) dr
[F cos\(\theta\) is the component of the variable force F]
where, F and \(\theta\) are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation, 
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dw=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } F\cos { \theta } dr\)
(ii) A graphical representation of the work done by a variable force is shown in Figure. The area under the graph is the work done by the variable force.