Definition And Proof Based Problems

Q 3254580454

Define kinetic energy. Prove that K.E. associated with a mass 'm' moving with velocity v is `1//2 \ \ mv^2`.

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.

Q 3214080859

Define elastic and inelastic collisions. Write their basic characteristics. A bullet is fired into a block of wood. If it gets totally embedded in it and the system moves together as one entity, then state what happens to the initial kinetic energy and linear momentum of the bullet ?

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.

Q 3214456350

What is the difference between head-on collision and glancing collision ? Define coefficient of restitution.

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)`

Q 3204456358

Define power. Obtain an expression for it in terms of force and velocity.

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`

Q 3244380253

Define the term watt, and kilo-watt hour.

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`

Q 3264380255

Define an electron volt. Express it in terms of joule.

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`

Q 3224580451

Define the terms elastic collision and inelastic collision. What is the difference between inelastic collision and a completely inelastic collision ?

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.

Q 3254356254

Prove that the work done in a frictional surface is non-zero in a closed path.

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.

Q 3214180950

Discuss elastic collision in one dimension. Obtain expression for velocities of two Also prove bodies after such a collision.

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)`

Q 3224180951

Prove that the total mechanical energy remains constant for a ball of mass `m` dropped from a tower of height `h`.

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.

Q 3274180956

Show that in case of one dimensional elastic collision of two bodies, the relative velocity of separation after the collision is equal to the relative velocity of approach before the collision.

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`