**This article is formulated according to the 6th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.**

β΄ π·_{ππ} = πΎ/π = π¬/π

## Instantaneous power

It is the limiting value of the average power as the time interval approaches zero. We have, π = lim _{π‘}_{β0 }πππ£

π· = π πΎ/π π

But ππ = **πΉβ** β π**π β** where π**π β** β diplacment

β΄ π = **πΉβ** . (π**π β**/ππ‘)

π· = **πβ**** β ****πβ**

**Note:** If the Force πΉ acts in the direction of motion then π = 0, cos 0 = 1 β΄ π = πΉπ£ cos π = πΉπ£ cos 0

π· = ππ

## Unit of Power

SI unit of power is watt (W), its dimensions are [ML^{2}T^{-3}] and 1 π€ππ‘π‘ = 1J/1π

## Watt

The power is said to be 1 watt of one-joule work is done/energy is consumed in one second by any agent.

## Practical unit

The practical unit of power is horsepower (hp). 1 horsepower (hp) = 746 W

## Collisions

The term collision refers to the interaction between two bodies due to which the direction and magnitude of the velocity of the colliding bodies change. In a collision, the external force acting on the colliding particles is neglected.

## Types of collisions

## Elastic collisions

In a collision, if both the linear momentum and kinetic energy of the system are conserved then such a collision is called an elastic collision. It means the linear momentum and the kinetic energy of the system before and after the collision is the same.

Ex:

(i) Perfectly elastic collision is a rare event, collisions between atomic particles are nearly elastic.

(ii) A ball dropped from a certain height will rebound to the same height if the collision with the surface is elastic.

## Inelastic (plastic) collision

A collision is said to be inelastic if the linear momentum of the system remains conserved, but its kinetic energy is not conserved. In an inelastic collision, the loss of kinetic energy appears in the form of heat, elastic potential energy sound, and light energy.

**Ex:** Collisions between macroscopic bodies are inelastic A ball dropped from a certain height will not rebound to some height if the collision with the surface is inelastic.

## Perfectly inelastic collision

A collision is said to be perfectly inelastic if the two bodies after collision stick together and move as one body.

**Ex:** When a moving bullet hits the stationary wooden block, then it is embedded into the wooden block, and both move as one body.

## Difference between elastic and inelastic collisions

Elastic collision | Inelastic collision |

Both the momentum and kinetic energy are conserved | Momentum is conserved but the kinetic energy is not conserved |

Force involved in the collision are conservative | Forces involved are non-conservative |

Mechanical energy is not converted into other forms of energy | Mechanical energy is converted into other forms of energy |

## Collisions in one Dimension

If the initial velocities and final velocities of both bodies are along the same straight line, then it is called a one-dimensional collision or Head-on collision.

## Expression for loss in Kinetic energy in a completely inelastic collision

Consider two masses m_{1} and m_{2}. The particle m_{1} is moving with speed u_{1} and m_{2} is at rest. After collision both masses m_{1} and m_{2} stick to each other and move as one body with velocity π£. In this type of collision, the linear momentum is conserved.

π_{1}π’_{1} + π_{2}π’_{2} = (π_{1} + π_{2})π£

π_{1}π’_{1} = (π_{1} + π_{2})π£ (π£_{2} = 0 β π_{2}π’_{2} = 0)

π£ = (π_{1}π’_{1})/(π_{1} + π_{2})

As the final kinetic energy is less than the initial kinetic energy, The change in kinetic energy is,

βπΎ = (1/2)ππ’_{1}^{2} β (1/2)(π_{1} + π_{2})π£^{2}

Substituting for π£ from the above equation

βπΎ = (1/2)ππ’_{1}^{2} β [(1/2)(π_{1} + π_{2})(π_{1}^{2}/(π_{1} + π_{2})^{2}π’_{1}^{2})]

βπΎ = (1/2)π_{1}π’_{1}^{2} β (1/2)(π_{1}^{2}/(π_{1} + π_{2}))π’_{1}^{2}

βπΎ = (1/2)π_{1}π’_{1}^{2}[1 β (π_{1}/(π_{1} + π_{2})]

βπΎ = (1/2)π_{1}π’_{1}^{2}[(π_{1} + π_{2} β π_{1})/(π_{1} + π_{2})]

βπΎ = (1/2)π_{1}π’_{1}^{2}[π_{2}/(π_{1} + π_{2})]

βπ² = (π/π)[(π_{π}π_{π}/(π_{π} + π_{π})]π_{π}^{π}

This loss in kinetic energy appears as the sound and heat energies, Thus total energy is conserved.

## Expression for final velocities of the bodies in an elastic collision

Consider two masses m_{1} and m_{2}. The body m_{1} is moving with speed u_{1} and m_{2} is at rest. After the collision, their velocities are v1 and v2 respectively, and are moving in the same straight line. As the collision is elastic both the linear momentum and kinetic energies are conserved.

ππππππ‘π’π β π_{1}π’_{1} + π_{2}π’_{2} = π_{1}π£_{1} + π_{2}π£_{2}

π_{1}π’_{1} = π_{1}π£_{1} + π_{2}π£_{2} (π_{2}π’_{2} = 0)

π_{1}π’_{1} β π_{1}π£_{1} = π_{2}π£_{2}

π_{1}(π’_{1} β π£_{1}) = π_{2}π£_{2} —-Γ (1)

Kinetic energy β (1/2)π_{1}π’_{1}^{2} = (1/2)π_{1}π£_{1}^{2} + (1/2)π_{1}π£_{2}^{2}

π_{1}π’_{1}^{2} = π_{1}π£_{1}^{2} + π_{2}π£_{2}^{2}

π_{1}π’_{1}^{2} β π_{1}π£_{1}^{2} = π_{2}π£_{2}^{2}

π_{1}(π’_{1}^{2} β π£_{1}^{2}) = π_{2}π£_{2}^{2} —-Γ (2)

Dividing equation (2) by equation (1)

π_{1}(π’_{1}^{2} β π£_{1}^{2})/π_{1}(π’_{1} β π£_{1}) = π^{2}π£_{2}^{2}/π_{2}π£_{2}

π_{1}(π’_{1} + π£_{1})(π’_{1} β π£_{1})/π_{1}(π’_{1} β π£_{1}) = π_{2}π£_{2}^{2}/π_{2}π£_{2}

π’_{1} + π£_{1} = π£_{2}

π£_{2} = π’_{1} + π£_{1} β β β β β> (3)

Substituting equation (3) in equation π_{1}π’_{1} β π_{1}π£_{1} = π_{2}π£_{2}

π_{1}π’_{1} β π_{1}π£_{1} = π_{2}(π’_{1} + π£_{1})

π_{1}π’_{1} β π_{1}π£_{1} = π_{2}π’_{1} + π_{2}π£_{1}

π_{1}π’_{1} β π_{2}π’_{1} = π_{1}π£_{1} + π_{2}π£_{1}

π’_{1}(π_{1} β π_{2}) = π£_{1}(π_{1} + π_{2})

π_{π} = [(π_{π} β π_{π})/(π_{π} + π_{π})]π_{π} β β β β β> (4)

Substituting (4) in (3)

π£_{2} = π’_{1} + ((π_{1} β π_{2})/(π_{1} + π_{2}))π’_{1}

π£_{2} = π’_{1}[1 + ((π_{1} β π_{2})/(π_{1} + π_{2}))]

π£_{2} = π’_{1}[(π_{1} + π_{2} + π_{1} β π_{2})/(π_{1} + π_{2})]

π£_{2} = π’_{1}[2π_{1}/(π_{1} + π_{2})]

π_{π} = (ππ_{π}/(π_{π} + π_{π}))π_{π} β β β β β> (5)

Equation (4) and (5) gives the expression for the final velocities of the masses m_{1 }and m_{2} respectively. π. π.,

π£_{1} = [(π_{1} β π_{2})/(π_{1} + π_{2})]π’_{1} and π£_{2} = [2π_{1}/(π_{1} + π_{2})]π’_{1}

**Special cases: Case (1): **If two masses one equal. π. π.,

π_{1} = π_{2} = π

π£_{1} = [(π β π)/(π + π)]π’_{1}

π£_{1} = 0.

The final velocity of mass m_{1} will become zero. It means after collision mass m_{1} comes to rest and pushes off the second mass m_{2}.

π£_{2} = [2π/(π + π)]π’_{1} = (2π/2π)π’_{1}

π£_{2} = π’_{1}

The final velocity of mass m_{2} = Initial velocity of mass m_{1}.

It means the second mass m_{2} moves with a velocity of m_{1}

**Case (2):** If one the mass dominates. If m_{2} >> m_{1}

π£_{2} = [(π_{1} β π_{2})/(π_{1} + π_{2})]π’_{1} ππ π_{2} > π_{1} then π_{1} β π_{2} = βπ£π

π£_{1} = βπ’_{1}

The final velocity of mass m_{1} gets reversed which means it comes back with the same velocity.

π£_{2} = [2π_{1}/(π_{1 }+ π_{2})]π’_{1}

π£_{2} = 2π_{1}π’_{1}/π_{2} β 0

(πππaπ’π π π_{2} β« π_{1}, 2π_{1}π’_{1}/π_{2} β 0)

π£_{2} = 0

Heavy mass m_{2} practically remains at rest.

**Case (3):** If π_{1} β« π_{2}

π£_{1} = [(π_{1} β π_{2})/(π_{1} + π_{2})]π’_{1}

π£_{1} = (π_{1}/π_{1})π’_{1}

(π_{2} πππ ππ πππππππ‘ππ)

π£_{1} = π’_{1}

The mass m_{1 }moves with the same velocity.

π£_{2} = [2π_{1}/(π_{1} + π_{2})]π’_{1} = (2π_{1}/π_{1})π’_{1}

π£_{2} = 2π’_{1}

The lighter mass m_{2} moves with twice the velocity of the heavy body.

## Collisions in two Dimensions

Consider two masses m_{1} and m_{2}. The body m_{1} is moving with speed u_{1} and m_{2} is at rest. The mass m_{1} collides with the stationary mass m_{2} and this is shown in the figure. After the collision, the masses m_{1} and m_{2 }fly off in different directions. The body m_{1} moves with velocity v_{1} making an angle π_{1}, called deflecting angle, with π₯ β ππ₯ππ and mass m_{2} move with a velocity v_{2} making an angle π_{2}, called the angle of recoil, with π₯ β ππ₯ππ . In all types of collision, momentum is conserved. Therefore, the conservation of momentum along π₯ β ππ₯ππ and π¦ β ππ₯ππ gives,

(i) π_{1}π’_{1} = π_{1}π£_{1} cos π_{1} + π_{2}π£_{2} cos π_{2} along π₯ β ππ₯ππ

(ii) 0 = π_{1}π£_{2} sin π_{1} β π_{2}π£_{2} sin π_{2} along π¦ β ππ₯ππ

π_{1}π£_{2 }sin π_{1} = π_{2}π£_{2} sin π_{2}

If the collision is perfectly elastic, then kinetic energy is also conserved.

(iii) (1/2)π_{1}π’_{1}^{2} = (1/2)π_{1}π£_{1}^{2} + (1/2)π_{2}π£_{2}^{2}

Assume that π_{1}, π_{2}, and π’_{1} are known. Now the motion after collision involves four unknowns i.e., π£_{1}, π£_{2}, π_{1}, and π_{2}. To evaluate these four quantities, we need a fourth equation. However, we have only three equations. Therefore, the fourth equation is to be developed, but this process of developing the fourth equation is quite complicated. To overcome this problem the easiest way of developing the fourth equation is the measure the angle of defection π_{1} and the angle of recoil π_{2} experimentally.

## Problems

1. Show that **π΄** = 2πΜ+ πΜ+ 2πΜ and **π΅** = 2πΜβ 2πΜβ πΜ are perpendicular to each other.

**π΄** β **π΅** = π΄_{π₯}π΅_{π₯} + π΄_{π¦}π΅_{π¦} + π΄_{π§}π΅_{π§}

**π΄** β **π΅** = (2)(2) + (1)(β2) + (2)(β1) = 4 β 2 β 2 = 0

**π΄** . **π΅** = 0, So, vectors are perpendicular to each other.

2. Find the angle between two vectors **π΄** = πΜ+ 2πΜ+ πΜ and **π΅** = 2πΜβ πΜ+ πΜ.

**π΄** β **π΅** = (1)(2) + (2)(β1) + (1)(1) = 2 β 2 + 1 = 1

|**π΄**| = β[(1)^{2} + (2)^{2} + (1)^{2}] = β(1 + 4 + 1) = β6

|**π΅**| = β[(2)^{2 }+ (β1) ^{2} + (1)^{2}] = β(4 + 1 + 1) = β6

π = πππ ^{β1} [**π΄** . **π΅**/(|**π΄**||**π΅**|)] = πππ ^{β1}[1/(β6 Γ β6)]

π = πππ ^{β1}(1/6) = πππ ^{β1} 0.1667

π = 80Β°24β

3. Find the angle between force **πΉ** = (3πΜ+ 4πΜβ 5πΜ) unit and displacement **π** = (5πΜ+ 4πΜ+ 3πΜ unit.

**πΉ** β **π** = (3)(5) + (4)(4) + (β5)(3) = 15 + 16 β 15 = 16 units

|**πΉ**| = β(3^{2} + 4^{2} + (β5)^{2}) = β(9 + 16 + 25) = β50 unit

|**π**| = β(5^{2} + 4^{2} + 3^{2}) = β(25 + 16 + 9) = β50 units

cos π = **πΉ** . **π** /|**πΉ**||**π**|

cos π = 16 /(β50 Γ β50) = 16/50

cos π = 0.32

π = πππ ^{β1}(0.32) β 71Β°

4. A body constrained to move along the z-axis of a coordinating system is subjected to a constant force πΉ given by πΉ = βπΜ+ 2πΜ+ 3πΜ π, where π,ΜπΜ and πΜ are unit vectors along x,y and z – axis of the system respectively. What is the work done by this force in moving the body a distance of 4m along the z-axis.

πΉ = βπΜ+ 2πΜ+ 3πΜ π

π = πΉ β π = πΉ_{π₯}ππ₯ + πΉ_{π¦}ππ¦ + πΉ_{π§}ππ§

π = 4ππ

π = (βπΜ+ 2πΜ+ 3πΜ) β (0πΜ+ 0πΜ+ 4πΜ)

π = (β1)(0) + (2)(0) + (3)(4) = 12 π½

5. In a ballistics demonstration a police officer fires a bullet of mass 50g with speed 200ms^{-1} on soft plywood of thickness 2 cm. The bullet emerges with only 10% of its initial kinetic energy what is the emergent speed of the bullet?

π = 50 Γ 10^{β3}ππ

Initial kinetic energy is π_{π}_{ }= (1/2)ππ£_{π}^{2}

π£_{π} = 200ππ ^{β1}

π_{π} = (1/2) Γ 50 Γ 100 Γ 200 Γ 10^{β3}

π_{π} = 1000000 Γ 10^{β3} = 1000π½

Final kinetic energy is 10% of initial kinetic energy.

π_{π} = 10% π_{π} = (10/100) Γ 1000 = 100π½

π£_{π}^{2} = 2π_{π}/π

π£_{π}^{2} = (2 Γ 100)/(50 Γ 10^{3}) = 4 Γ 10^{3}

π£_{π} = β(4 Γ 10^{3}) = 63.2ππ ^{β1}

6. An electron and proton are detected in a cosmic ray experiment, the first with kinetic energy 10πππ, and the second with 100πππ, which is faster, the electron or the proton? Obtain the ratio of their speeds?

π_{π} = 9.11 Γ 10^{β31}ππ

π_{π} = 1.67 Γ 10^{β27}ππ

π_{π} = 10 Γ 10^{3} Γ 1.6 Γ 10^{β19}π½

πΎ_{π} = 100 Γ 10^{3} Γ 1.6 Γ 10^{β19}π½

Velocity of electron is,

π£_{π} = β (2 Γ πΎ_{π}/π_{π})

π£_{π} = [(2 Γ 10 Γ 10^{3} Γ 1.6 Γ 10^{β19})/(9.11 Γ 10^{β31})]^{1/2}

π£_{π} = 5.9267 Γ 10^{7}ππ ^{β1}

Velocity of the proton,

π£_{π} = β(2πΎ_{π}/π_{π})

π£_{π} = [(2 Γ 100 Γ 10^{3} Γ 1.6 Γ 10^{β19})/(1.67 Γ 10^{β27}]^{1/2}

π£_{π} = 4.3774 Γ 10^{6}ππ ^{β1}

Comparing v_{e} and v_{p} we get electron is moving faster,

π£_{π}/π£_{π} = (5.9267 Γ 10^{7})/(4.3774 Γ 10^{6}) = 13.54

7. A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height it attains its maximum (terminal) speed and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire Journey if its speed on reaching the ground is 10ms^{-1}.

π = 2ππ = 2 Γ 10^{-3}π

β_{1} = 250 π

β_{2} = 250 π

β = β1 + β2 = 500π

Mass of the drop is mass = volume Γ density

Volume = (4/3)ππ^{3} = (4/3) Γ 3.14 Γ (2 Γ 10^{-3})^{3} = 3.3493 Γ 10^{β8}ππ/π^{3}

Mass = 3.3493 Γ 10^{β8} Γ 10^{3} = 3.3493 Γ 10^{β5}ππ

π_{π}_{1} =? π_{π}_{1} = ππβ = 3.3493 Γ 10^{β5} Γ 9.8 Γ 250 = 0.082 π½

π_{π}_{2} =? π_{π} =?

Since the distance travelled in the second half is same as first halt therefore π_{g}_{2} = 0.082π½

Total work done on it is, π = π_{π}_{1} + π_{π}_{2} + π_{π}

From work energy theorem, π = βπΎ

π_{π}_{1} + π_{π}_{2 }+ π_{π} = (1/2)ππ£_{π}^{2} β (1/2)ππ£_{π}^{2}

0.082 + 0.08 + W_{r} = [(1/2) Γ 3.3493 Γ 10^{5} Γ 10^{2}) β (0) [β΅ π£_{π} = 0]

0.164 + W_{π} = 1.6746 Γ 10^{β3} π_{π} = 1.6746 Γ 10^{β3} β 0.164 π_{π} = β0.1623 π½

8. A body of mass 0.5 Kg travels in a straight line with velocity π£ = ππ₯^{3/2}, where π = 5π^{β1/2}π ^{β1}. What is the work done by the net force during its displacement from π₯ = 0 to π₯ = 2π.

Initial velocity at π₯ = 0. π = 0.5ππ. π£_{π} = ππ₯^{3/2} = 5 Γ (0)^{3/2} = 0

Final velocity at π₯ = 2, π£_{π} = ππ₯^{3/2} = 5 Γ (2)^{3/2} = 5 Γ 2.8284 = 14.1421ππ ^{β1}

From work energy theorem, π = βπΎ = (1/2)ππ£_{π}^{2} β (1/2)ππ£_{i}^{2} = ((1/2) Γ 0.5 Γ 14.1421^{2}) = (0) β΅ π£_{π}_{ }= 0

π = 49.999 π½ β 50π½

9. The potential energy of a certain spring when stretched through a distance π₯ is 10J. What is the amount of work done on the same spring to stretch it through an additional distance?

We have (1/2)ππ₯^{2} = 10π½

When π₯ becomes 2π₯, then, New P.E = (1/2)πΎ(2π₯)^{2} = (1/2)πΎ(4π₯^{2}) = (1/2)πΎ π₯^{2} Γ 4 = 10 Γ 4 = 40π½

Work done = 40 π½ β 10π½ = 30π½

10. A 100W bulb is on for 10 hours – calculate the energy consumed in kWh. To find energy in kWh, power must be in kW and time in hours.

π = 100π€ = 100 Γππ

1000 = 100 Γ 10^{β3}ππ β΄ π = πΈ/π‘

t = 10 h

πΈ = ππ‘ = 100 Γ 10^{β3} Γ 10 = 1000 Γ 10^{β3}

πΈ = 1000/10^{3} = 1 ππ€β

11. A pump on the ground floor of a building can pump up water to fill a tank of volume 30m^{3} in 15 min. If the tank is 40m above the ground and the efficiency of the pump is 30% how much electric power is consumed by the pump? (When water is pumped up against the gravitational force, work has to be done, and this work done is stored as potential energy.)

Volume π£ = 30π^{3}

Density of water π = 1000 ππ/π^{3} β΄ Mass of the water, m = volume Γ density

π = 30 Γ 1000 = 3 Γ 10_{4}ππ

β΄ Work done = ππβ = 3 Γ 10^{4} Γ 9.8 Γ 40

Work done = 1176 Γ 10^{4} π½

To do the above work actual power needed is π_{ππ} = π/π‘ = (1176Γ10^{4})/(15Γ60) = 13.0667 Γ 10^{3} watt

But the efficiency of the motor is only 30% therefore the power required is

π = (πππ‘π’ππ πππ€ππ/π π’ππππππ πππ€ππ) Γ 100

Supplied power = (πππ‘π’ππ πππ€ππ/π) Γ 100

Supplied power = (13 Γ 0.667 Γ 10^{3}/30) Γ 100

Supplied power = 43555.5556 π

12. A family uses 8kw of power. Direct solar energy is incident on the horizontal surface at an average rate of 200W/m^{2}. It 20% of this energy can be converted to useful electrical energy how large an area is needed to supply 8 kW?

Power used by the family = 8ππ = 8000 π

But it is only 20% of the total solar power supplied.

β΄ 20% of solar power = 8000N

Solar power supplied = (8000/20) Γ 100 = 40000π

Now, Area of the surface receives power of 200W = 1π^{2}

Area of the surface receives 1π power = 1π^{2}/200

Area of the surface receives 40000π power is = (1/200) Γ 40000 = 200π^{2}

β΄ Area needed = 200π^{2}

Area needed is nearly equal to 14π Γ 14π roof.

13. The blades of a wind – mill sweep out a circle of area βAβ (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time βtβ ? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the winds energy into electrical energy and that A = 30m^{2},v= 36 km/h and the density of the air is 1.2kgm^{-3} what is the electrical power produced?

(a) Volume of air passing per see = Area Γ velocity = Av

Mass of the wind passing per sec = volume Γ density = π΄π£ Γ π = π΄π£π

Mass of the wind passing in time π‘ = π΄π£ππ‘

(b) Kinetic energy of wind (Air) = (1/2)ππ£^{2}

Kinetic energy of wind (Air) = (1/2)(π΄π£ππ‘)π£^{2}

Kinetic energy of wind (Air) = (1/2)π΄ππ£^{3}π‘

(c) The wind mill converts only 25% of the wind energy

β΄ Energy available is = 25% of KE

Energy available = (25/100) Γ (1/2)π΄ππ£^{3}π‘ = (1/8)π΄ππ£^{3}π‘

β΄ πππ€ππ πππππ’πππ = πΈπeπππ¦/π‘πππ = (1/8)π΄ππ£^{3}π‘/π‘ πππ€ππ πππππ’πππ = π΄ππ£^{3}/8 = (30 Γ 1.2 Γ (10)^{3})/8

πππ€ππ πππππ’πππ = 4500π = 4.5 ππ

14. A neutron having mass of 1.67 Γ 10^{β27}kg and moving at 10^{8}ππ ^{β1} collides with deuteron at rest and stick to it. Calculate the speed of the combination given mass of the deuteron is 3.34 Γ 10^{β27}kg.

Initial momentum = π_{π}π’_{π} + π_{π}π’_{π} = 1.67 Γ 10^{β27} Γ 10^{8} + 0 = 1.67 Γ 10^{β19}

Final momentum = (π_{π} + π_{π})π£ = (1.67 Γ 10^{β27} + 3.34 Γ 10^{β27})π£

As the momentum is conserved, Initial momentum = final momentum

1.67 Γ 10^{β19} = 5.01 Γ 10^{β27} Γ π£

π£ = (1.67 Γ 10^{19})/(5.01 Γ 10^{β27}) = 3.3 Γ 10^{β6}ππ ^{β1}

## Important Questions for Exams.

**One mark.**

1. State work-energy theorem.

2. When work is said to be done? or What is meant by βWork done by a forceβ? or Define work done by the force.

3. Write an expression for potential energy of a compressed spring.

4. Mention the expression for work done by a force in vector form.

5. What type of energy decreases when a body is falling freely?

6. What does the area under βforce-displacementβ curve represents?

7. What is potential energy?

8. What is meant by collision?

9. When the work done by a force does is zero?

10. Define power.

**Two marks.**

1. Distinguish between elastic and inelastic collision.

2. Give one example for elastic and inelastic collision.

3. What are elastic and inelastic collision?

**Three marks.**

1. What is the value of (a) πΜ. π Μ (b) πΜ.πΜ (c) πΜ. πΜ

2. What is the value of (a) πΜ.π Μ (b) πΜ. πΜ (c) πΜ.πΜ

3. Give an example for (a) positive work (b) Negative work (c) zero work.

4. What is potential energy? Derive an expression for it.

5. Define kinetic energy. Give its unit and dimensional formula.

6. Obtain an expression for gravitational potential energy.

7. Define power. Show that the power is equal to the dot product of force and velocity.

8. Prove Work-Energy theorem for the motion of a particle under a constant acceleration. or Show that πΎ_{π} β πΎ_{π} = π, Where πΎ_{π} and πΎ_{π} represent final and initial kinetic energies respectively and π represents work. or Prove the work-energy theorem in case of a rectilinear motion under constant acceleration. or Show that work done by the system is equal to change in energy.

9. Distinguish between Conservative force and Non-conservative force.

**Five marks.**

1. State the principle of conservation of energy and illustrate it in case of freely falling body.

2. State the law of conservation of mechanical energy. Show that the mechanical energy of a body dropped from a height is conserved. or State the law of conservation of mechanical energy. Show that the mechanical energy of a body falling freely under gravity is conserved. or State the law of conservation of mechanical energy and show that the mechanical energy is conserved in the case of a freely falling body. or State and prove the law of conservation of mechanical energy for a body falling under gravity. or Verify the law of conservation of mechanical energy in case of freely falling body.

3. State and Prove Work-Energy theorem for a constant force.