Initial Kinetic energy = \(\frac { 1 }{ 2 } { mv }^{ 2 }\)
Final Energy = \(\frac { 1 }{ 2 } m{ \left( \frac { v }{ 2 } \right) }^{ 2 }+\frac { 1 }{ 2 } { kx }^{ 2 }\)
By the principle of conservation of energy,
\(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } \frac { { mv }^{ 2 } }{ 4 } +\frac { 1 }{ 2 } { k }x^{ 2 }\)
\(\therefore \quad K=\frac { 3{ mv }^{ 2 } }{ 4{ x }^{ 2 } } \)

Graphs depicting variation of (i) gravitational potential energy (P.E.), (ii) kinetic energy (K.E.),and (iii) the total sum of potential and kinetic energies for a freely falling body are as shown in adjoining Fig. From the graphs, it is clear that:
(a) Gravitational potential energy decreases as the body falls downwards and is zero at the Earth.

(b) Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground.
(c) The sum of kinetic and potential energies remains constant at all points during its free fall.

Work done  by the force = force\( \times \)displacement
or W = F \( \times \) S
But from Newton's 2nd law , we have
Force = mass\( \times \)acceleration
i.e.  F = ma                      ...(ii)
hence , from Eqs (i) and (ii), we get
W = m( \(\frac{d^2s}{dt^2}\) ) S    [ \(\because\) a =  [ \(\frac{d^2s}{dt^2}\)]........(iii)
now, we have
S =  \(\frac{1}{3}\)t²
\(\therefore\)  \(\frac{d^2s}{dt^2}\) =  \(\frac{d}{dt}\)[\(\frac{d}{dt}\)( \(\frac{1}{3}\)t²)]
=  \( \frac{d}{dt}\) \( \times \)( \(\frac{2}{3}\)t )
=  \( \frac{2}{3}\)\(\frac{dt}{dt}\)
=  \(\frac{2}{3}\)
Hence Eq. (iii) becomes
W = \( \frac{2}{3} \)ms
= \(\frac{2}{3}\)\( \times \) m\( \times \)\(\frac{1}{3}\)t²
= \(\frac{2}{9}\)mt²
We have , m = 3 kg , t = 2s
\(\therefore\)      W = \(\frac{2}{9}\)\( \times \)3\( \times \)(2)²
= \(\frac{8}{3}\) J

At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring.
The kinetic energy of the moving car is
\(K=\frac{1}{2} m v^{2}\)
\(=\frac{1}{2} \times 10^{3} \times 5 \times 5\)
K = 1.25 x 10⁴ J
where we have converted 18 km h^–1 to 5 m s^–1 [It is useful to remember that 36 km h^–1 = 10 m s^–1]. At maximum compression x_m , the potential energy V of the spring is equal to the kinetic energy K of the moving car from the principle of conservation of mechanical energy.
\(V=\frac{1}{2} k x_{m}^{2}\)
= 1.25 x 10⁴ J
We obtain
x_m = 2.00 m
We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction.

Given ₁H² = 2.014 u, ₂He³ = 3.0160 u, Mn =1.0087 u
1 u = 1661 \(\times\)10^-27 kg
Energy released =?
mass of reacant, mass_r = 2 \(\times\)2.0141=4.0282 u
mas of product, mass_p = 3.0160 +1.0087 = 4.0247 u
Loss of mass =mass_p - mass_r 
\(\Delta\)M =(4.0282 - 4.0247) u =0.0035 u
= 0.0035 \(\times\)1.66\(\times\) 10^-13 J
Energy released =\(​​\frac { 52.2\times { 10 }^{ -14 } }{ 1.602\times { 10 }^{ -13 } } =3.26\quad Mev\quad \)

Force whose either magnitude or direction or both change.

Coefficient of restitution.

The workdone by a car against gravity is zero because force of gravity is vertical and motion of car is along a straight horizontal road. As angle \(\theta\) between directions of force and displacement is 90⁰ , hence work done is zero.

Here, F = \((-\hat{i}+\hat{2j}+\hat{3k})\) N and s =  \((\hat{4k})\) m
W = F.s = F_z.S_z 
= 3 \(\times\) 4
= 12N - m
= 12 J

Here, \((\frac { 1 }{ 2 } m{ \mu }^{ 2 }){ 30 }^{ o }\)
\(\\ { M }_{ 1 }={ M }_{ 2 }=m,{ \mu }_{ 1 }=v\quad and\quad { \mu }_{ 2 }=-v\)
\(\\ Now,\quad { v }_{ 1 }=\frac { ({ M }_{ 1 }-{ M }_{ 2 }){ \mu }_{ 1 }+{ 2M }_{ 2 }{ \mu }_{ 2 } }{ { M }_{ 1 }+{ M }_{ 2 } } \)
\(= \frac { (m-m)v+2m(-v) }{ m+{ m } } =-v\)
\(and\ { v }_{ 2 }=\frac { ({ M }2-{ M }_{ 1 }){ \mu }_{ 2 }+{ 2M }_{ 1 }{ \mu }_{ 1 } }{ { M }_{ 1 }+{ M }_{ 2 } } \)
\(=\frac { (m-m)v+2m(-v) }{ m+{ m } } =v\)
After collision, the two ball bearings will move with same speed but their directions of motion will be reserved.