Solution:

Kinetic energy is defined as the energy associated with a body under motion.

` W = int F dx = int m (dv)/(dt) dx = int mv dv = 1/2 mv^2`

The work done is transformed as kinetic energy for anybody capable of moving.

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Solution:

Elastic collision. A collision in which there is absolutely no loss of kinetic energy is called elastic collision.

Characteristics :
(i) The linear momentum is conserved.
(ii) Total energy of system is conserved.
(iii) Kinetic energy is conserved.
(iv) Forces involved during elastic collisions must be conservative forces.
Inelastic collision. A collision in which there occurs some loss of kinetic energy is called inelastic collision.

Characteristics :
(i) Linear momentum is conserved.
(ii) Total energy is conserved.
(iii) K.E. is not conserved.
(iv) Some or all forces involved may be non-conservative.
Consider two particles A and B of masses `m_1` and `m_2` moving with initial velocities ` u_1` and `u_2` along x-axis. They collide and move as one entity. Let V be the common velocity when they move as single mass.

`m_1 =` Mass of bullet
`u_1 =` Initial velocity of bullet
`m_2 =` Mass of wood.
`u_ 2 =` Initial velocity of wood

`= 0`
According to conservation of momentum.
`m_1 u_1 + m_2 u_2 = (m_1 + m_2) V`
`u_2 = 0`

` V = (m_1 u_1)/(m_1 + m_2)` ..............(i)

Initial K.E. `k_i = 1/2 m_1 u_1^2`

Final K.E., `K_f = 1/2 (m_1 + m_2 ) V^2`

` K_f/K_i = ( (1//2) (m_1 + m_2 ) V^2)/( 1//2 m_1 u_1^2)`

` = ( (m_1 + m_2)/m_1 ) V^2/u_1^2` .............(ii)

from (i) and (ii)

` K_f/K_i = ( (m_1 + m_2)/m_1 ) ( m_1/(m_1 + m_2) )^2`

` = m_1/(m_1 + m_2)`

When, `K_f < K_i` there is loss in K.E.

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Solution:

Head-on collision. If the bodies move along the same straight line before and after collision it is called head-on collision.
Glancing collision. If the bodies do not move along the same straight line after the collision it is called glancing collision.
Coefficient of restitution. It is the measure of degree of elasticity of a collision and is defined as relative speed of separation

` e = - text( after collision)/text( relative speed of approach before collision)`

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Solution:

Power. The rate of doing work is called power.

` P = Lt_(Delta t -> 0) ( Delta W)/(Delta t) = (d W)/(dt)`

` = d/(dt) (vec F . vec s) = vec F * vec (ds)/(dt)`

` P = vec F * vec v`

` => P = F v`

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Solution:

Watt. Power is said to be one watt if it can work at the rate of `1` joule per second
`1` watt `= 1` joule/sec.
Kilowatt hour `= 1` kWh is work done in `1` hour at a constant rate of `1` kilowatt. (kWh)
`1 \ \ k W h = 3.6 xx 10^6 \ \ J`

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Solution:

Electron Volt. Electron volt is the amount of energy possessed by an electron in falling through a potential difference of `1` volt.
`1 (e V) = 1 e xx 1 V`

`= (1.6 xx 10^(-19) C) xx 1 V`

`= 1.6 xx 10^(-19) \ \ J`

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Solution:

Elastic collisions are those collisions in which the momentum and kinetic energy will be conserved. In inelastic collision only momentum will remain conserved. In inelastic collision, loss in K.E. of moving body may not be `100%` but in complete or perfect inelastic collision, the K.E. of moving body is lost so that the bodies move together after collision.

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Solution:

Let `F_f` be the force of friction in a surface. The work done to carry a mass `m` from a point A to another point B is, `-F_f A B`. In the return path B to A also, the work done is `- F _f A B`, since the `F _f` acts against the motion. The net work done is therefore, `- 2 F_f (A B)`. Since friction is dependent on the nature of the surface it is dependent on path.

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Solution:

One dimensional elastic collision is one in which both momentum and K.E. are conserved and the body moves in the same line of motion even after the collision.
If `m_1 , m_2` are the masses, `u_1 , u_2` are the initial velocities and `v_ 1 , v_2` are the final velocities, then

`m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2` ...........(i)

`1/2 m_1 u_1^2 + 1/2 m_2 u_2^2 = 1/2 m_1 v_1^2 + 1/2 m_2 v_2^2` ............(ii)

(i.e.,) `m_1 ( v_1^2 - u_1^2 ) = m_2 (v_2^2 - u_2^2)` from (ii)

`m_1 (v_1 - u_1) = m_2(v_2 - u_ 2)` from (i)

Dividing both sides

` v_1 + u_1 = v_2 - u_2`

`v_1 = v_2 + u_2 - u_1`

substituting in (i) we have,

`m_1u_1 + m_2 u_2 = m_1 (v_2 + u_2 - u_1 ) + m_2 v_2`

`2 m_1 u_1 + u_2 (m_2 - m_1) = v_2 (m_1 + m_2)`

`:. v_2 = ( u_2 (m_2 - m_2) + 2m_1 u_1)/( m_1 + m_2)`

Similarly

` v_1 = ( u_1 (m_1 - m_2) + 2m_2 u_2)/( m_1 + m_2)`

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Solution:

At. A

P.E. `m g h \ \ K.E. = 0`

Total energy ` = m g h`

At B

Velocity ` v = sqrt (2 gx)`

`K.E. = 1/2 mv^2 = 1/2 m . 2g x = m g x`

P.E. `= mg (h - x)`
Total energy `= m g h`.

At C

Velocity `v = sqrt (2 gh)`

`K.E. = 1/2 m v^2 = 1/2 m . 2 g h = m g h`

`P . E .= 0`
`:.` total energy `= m g h`.
Total mechanical energy is therefore `m g h` at all states as a body is dropped.

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Solution:

Conservation of momentum
`m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2`

Conservation of energy

` 1/2 m_1 v_1 + 1/2 m_2 v_2 = 1/2 m_1 u_1^2 + 1/2 m_2 u_2^2`

Solving the above results, we can get

` v_2 - v_1 = u_1 - u_2`

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