For any spring kl = constant
Length of the longer piece
\(=\frac{2 l}{3} \)
\(\therefore k^{1} \times \frac{2 l}{3} =k l \)
\(\therefore k^{1}=\frac{k l \times 3}{2 l}=\frac{3}{2} k \)
\(\therefore k^{1}=\frac{3}{2} k \)

\(\text {Potential energy } P . E=\alpha-\frac{\beta}{2} x^{2}\)
P.E = Work = Fx
\(P=\alpha-\frac{\beta}{2} x^{2}\)
\(\text {Force }=\frac{d p}{d x}=\frac{d}{d x}\left(\alpha-\frac{\beta}{2} x^{2}\right)\)
\(=0-\frac{\beta}{2} \times 2 x =-\beta x \)

\(\text { Force } \overrightarrow{\mathbf{F}}=(2 i+\vec{j}) N\)
\(\text { Displacement } d=(5 \vec{i}+3 \vec{j})-(3 \vec{j}+\vec{k})\)
\(=(5 i-k) m\)
\(\text { Work done } W=F . d\)
\(=(2 \vec{i}+\vec{j})(5 i-k)\)
\(=10-0-0=10 J \)

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.

a) When the weight lifter lifts the mass, force and displacement are in the same direction, which means that the angle between them θ = 0°. Therefore, the work done by the weight lifter,
W_{weight lifter} = F_wh cos θ = F_wh (cos 0°)
= 5000\(\times\)5\(\times\)(1) = 25,000 joule= 25 kJ
(b) When the weight lifter lifts the mass, the gravity acts downwards which means that the force and displacement are in opposite direction. Therefore, the angle between them θ = 180°.
W_gravity = F_gh cos θ = mgh( cos 180°)
= 250\(\times\)10\(\times\)5\(\times\)(-1) = -12,500 joule = -12.5 kJ
(c) The net work done (or total work done) on the object
W_net = W_{weight lifter}+ W_gravity
= 25 kJ -12.5 kJ = +12.5 kJ

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

In Physics, work is said to be done by the force when applied on a body displaces it. To do work, energy is required. But, generally work refers to both physical and mental work. In fact, any activity can be called as work.

F = k_Ax_A = k_Bx_B
\({ x }_{ A }=\frac { F }{ { k }_{ A } } ,{ x }_{ B }=\frac { F }{ { k }_{ B } } \)
The work done on the springs are stored as potential energy in the springs.
\({ U }_{ A }=\frac { 1 }{ 2 } { k }_{ A }{ x }_{ A }^{ 2 };\quad { U }_{ B }=\frac { 1 }{ 2 } { k }_{ B }{ x }_{ B }^{ 2 }\)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ A }{ x }_{ A }^{ 2 } }{ { k }_{ B }{ x }_{ B }^{ 2 } } =\frac { { { k }_{ A }\left( \frac { F }{ { k }_{ A } } \right) }^{ 2 } }{ { { k }_{ B }\left( \frac { F }{ { k }_{ B } } \right) }^{ 2 } } =\frac { \frac { 1 }{ { k }_{ A } } }{ \frac { 1 }{ { k }_{ B } } } \)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ B } }{ { k }_{ A } } \)
k_A > k_B implies that U_B > U_A.Thus, more work is done on B than A.

Various types of potential energy are
(i) The energy possessed by the body due to gravitational force gives rise to gravitational potential energy.
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h.
U = mgh
(ii) The energy due to spring force and other similar forces give rise to elastic potential energy.
a) At the equilibrium position x = 0 potential energy is \(U=\frac{1}{2} k x^{2}\)
b) If the initial position is not zero and if the mass is changed from position x_i to x_f, if then elastic potential energy is \(U=\frac{1}{2} k\left(x_{f}^{2}-x_{i}^{2}\right)\)
(iii) The energy due to electrostatic force on charges gives rise to electrostatic potential energy.
Electrostatic potential energy is the work done to arrange two charges q₁ and q₂ at a separation \(r=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)

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.

Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x-direction) on a frictionless horizontal.
      
 

(i) In order to have collision, we assume that the mass m₁ moves faster than mass m₂ i.e., u₁ > u₂  For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (p_i) = Totai momentum after collision (P_f)
Further,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v_s ...........(1)
or
m₁ (u₁ - v₁) = m₂ (v₂ - u₂) ...............(2)

For elastic collision,
Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f.
       \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }=\frac { 1 }{ 2 } { m }_{ 1 }{ v }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ v }_{ 2 }^{ 2 }\)   ................(3)
After simplifying and rearranging the terms,
        \({ m }_{ 1 }\left( { u }_{ 1 }^{ 2 }-{ v }_{ 1 }^{ 2 } \right) ={ m }_{ 2 }\left( { v }_{ 2 }^{ 2 }-{ u }_{ 2 }^{ 2 } \right) \)
Using the formula a² - b² = (a + b) (a - b), we can rewrite the above equation as
        m₁ (u₁ + v₁) (u₁ - v₁) = m₂ (v₂ + u₂) (v₂ - u₂) .................(4)
Dividing equation (4) by (2) gives,
          \(\frac { { m }_{ 1 }\left( { u }_{ 1 }+{ v }_{ 1 } \right) \left( { u }_{ 1 }-{ v }_{ 1 } \right) }{ { m }_{ 1 }\left( { u }_{ 1 }-{ v }_{ 1 } \right) } =\frac { { m }_{ 2 }\left( { u }_{ 2 }+{ v }_{ 2 } \right) \left( { u }_{ 2 }-{ v }_{ 2 } \right) }{ { m }_{ 2 }\left( { u }_{ 2 }-{ v }_{ 2 } \right) } \)
   u₁ + v₁ = v₂ + u₂
    u₁ - u₂ = v₂ - v₁ .............(5)
Equation (5) can be rewritten as
 (u₁ - u₂) = -(v₁ - v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂+ u₂ - u₁ .........(6)
or
v₂ = u₁ + v₁ - u₂ ............(7)
To find the final velocities v₁ and v₂:
Substituting equation (7) in equation (2) gives the velocity of ml as
m₁ (u₁ - v_I) = m₂ (u₁ + v₁ - u₂ - u₂)
m₁ (u₁ - v₁) = m₂ (u₁ + v₁ - 2u₂)
m₁u₁ - m₁v₁ = m₂u₁ + m₂v₁ - 2m₂u₂
m₁u₁ - m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ - m₂)u₁ + 2m₂u₂ = (m₁ + m₂) v₁
or \({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\) ............(8)
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of m₂ as
\({ v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\) ............... (9)
Case 1:
When bodies has the same mass i.e., m₁ = m₂,
equation (8) \(\Rightarrow { v }_{ 1 }=\left( 0 \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { 2m }_{ 2 } } \right) { u }_{ 2 }\)
v₁ = u₂ ...........................(10)
equation (9)  \(\Rightarrow { v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { 2m }_{ 1 } } \right) { u }_{ 1 }+\left( 0 \right) { u }_{ 2 }\)
v₂ = u₁ ..........................(11)
The equations (10) and (11) show that in one dimensional elastic collision when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2:
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 (8) and (9).
we get,
from equation (8) => v₁ = 0 (..................... 12)
from equation (9) => v₂ = u₁ ( .................. 13)
Equations (12) and (13) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3:
The first body is very much lighter than the second body
\(\left( { m }_{ 1 }<{ m }_{ 2 },\frac { { m }_{ 1 } }{ { m }_{ 2 } } <1 \right) \) then the ratio \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } = 0\) and also if the target is at rest (u₂ = 0)
Dividing numerator and denominator of equation (8) by m₂, we get
\({ v }_{ 1 }=\left( \frac { \frac { { m }_{ 1 } }{ { m }_{ 2 } } -1 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 2 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 1 }=\left( \frac { 0-1 }{ 0+1 } \right) { u }_{ 1 }\)
v₁ = - u₁ ............................(14)
Similarly,
Dividing numerator and denominator of equation (9) by m₂, we get 
\({ v }_{ 2 }=\left( \frac { 2\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 2 }=\left( 0 \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
v₂ = 0  ..............................(15)
The equation (14) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (15) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.
Case 4:
The second body is very much lighter than the first body
\(\left( { m }_{ 2 }<<{ m }_{ 1 },\frac { { m }_{ 2 } }{ { m }_{ 1 } } <<1 \right) \) then the ratio \(\frac { { m }_{ 2 } }{ { m }_{ 1 } } = 0\) and also if the target is at rest (u₂ = 0).
Dividing numerator and denominator of equation 8 by m₁ we get
\({ v }_{ 1 }=\left( \frac { 1-\frac { { m }_{ 2 } }{ { m }_{ 1 } } }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) { u }_{ 1 }+\left( \frac { 2\frac { { m }_{ 2 } }{ { m }_{ 1 } } }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) \left( 0 \right) \)
\({ v }_{ 1 }=\left( \frac { 0-1 }{ 0+1 } \right) { u }_{ 1 }+\left( \frac { 0 }{ 1+0 } \right) \left( 0 \right) \)
v₁  = u₁  .....................(16)
Dividing numerator and denominator of equation (14) by m₁ we get
\({ v }_{ 2 }=\left( \frac { 2 }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) { u }_{ 1 }+\left( \frac { \frac { { m }_{ 2 } }{ { m }_{ 1 } } -1 }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) \left( 0 \right) \)
\({ v }_{ 2 }=\left( \frac { 2 }{ 1+0 } \right) { u }_{ 1 }\)
v₂ = 2u₁   ....................(17)
The equation (16) implies that the first body which is heavier continues to move with the same initial velocity. The equation (17) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.