Using law of conservation of momentum,
\(2 m v =\sqrt{m^{2} v^{2}+m^{2} v^{2}} \)
\(=\sqrt{2 m^{2} v^{2}} \)
\(v =\frac{\sqrt{2} m v}{2 m}=\frac{v}{\sqrt{2}} \)
Energy released in explosion = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

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

\(\text { Force }=v \frac{d m}{d t}\)
\(=v \frac{d}{d t} \text { (volume } \times \text { density) }\)
\(=v \frac{d}{d t}(A x p) \)
\(=v A p \frac{d x}{d t} \)
\(=v \times A p \times v \)
\(=A p v^{2} \)
Power = Force x Velocity
\(=A p v^{2} \times v=A p v^{3}\)
\(\therefore \text { Power } \alpha v^{3}\)

(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.

(i) A body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius \(\vec{r}\) of the circular path.
(ii) The motion of the body by taking the free body diagram (FBD) at a position where the position vector \(\vec{r}\) makes an angle e with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton's second law on the mass, In the tangential direction,

mg sinθ = ma_t
mg sinθ = -m \((\frac{dv}{dt})\)
where, a_t = -\((\frac{dv}{dt})\) is tangential retardation
In the radial direction,
T - mg cosθ = m a_r
T - mg cosθ = \(\frac{mv^{2}}{r}\)
where, a_r = \(\frac{v^{2}}{r}\) is the centripetal acceleration.

Total energy is always conserved.
But K.E. is not conserved

P.E of the spring when stretched through a distance x,
\(U={1\over2}{kx}^{2}=10J\)
When x becomes lx, the potential energy will be
\(u'={1\over2}k(2x)^2=4\times{1\over2}{kx}^{2}=4\times10=40J\)
\(\therefore\) Workdone = u' - u = 40 - 10 = 30 J

m = 60 + 20 = 80 kg
h = 20\(\times\)0.2 = 4m
g = 9.8 ms^-2, t = 10s
\(P={W\over t}={mgh \over t}={80\times 9.8 \times 4 \over 10}={3136 \over 10}=313.6W\)

Energy exchange will be maximum if the two colliding bodies are of equal masses.

No, only that much amount of kinetic energy is lost as is necessary for the conservation of momentum.

The work done in striking the two stones against each other gets converted into heat. This appears as a spark.

The energy is dissipated in the form of heat. The heat energy so produced is not available for work.

Work done is more, while moving up a hill.

1 electrical unit = 1 kWh = 1\(\times\)(103 W)\(\times\)(3600 s)
1 electrical unit = 3600\(\times\)103 W s
1 electrical unit = 3.6\(\times\)10⁶ J
1 kWh = 3.6\(\times\)10⁶ J

When bodies have the same mass i.e., m₁ = m₂ and second body (usually called target) is at rest (u₂ = 0), By substituting m₁ = m₂ and u₂ = 0 in equations
\({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\)...(1)
and equations \({ v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\) , we get...(2)
from equations (1) => v₁ = 0
from equations (2) => v₂ = u₁

When the earth revolves around the sun in a perfectly circular orbit the force of the sun on the earth is along the radius while displacement of the earth is along the tangent to the orbit. Now in a circle, radius and tangent are always orthogonal i.e., angle between them at any point on the orbit is 90° and so, work done by the sun on the earth is zero.

\(w=\bar{F}.\bar{a}\)
\(=(-2\hat{i}+15\hat{j}+6\hat{k})-10\hat{j}\)
= 0 + 15\(\times\)10 + 0
= 150 Joule

Initial kinetic energy,
\(k_1={1\over2}\times{50\over 1000}\times200\times 300\ J\)
= 1000 J
Final kinetic energy, \({k}_{f}={10\over100}\times1000\ J=100\ J\)
If v_f is emergent speed of the bullet, then
\({1\over2}\times{50\over1000}\times{v}_{f}^{2}=100\)
or \({v}_{f}^{2}=4000\) (or) v_f = 63.2 ms^-1