WORK, ENERGY AND POWER

E f −−−−− E i = W nc where W nc is the total work done by the non-conservative forces over the path. Note that Our electricity bills carry the energy consumption in units of kWh. Note that kWh is a unit of energy and not of power. Example .

An elevator can carry a maximum load of kg (elevator + passengers) is moving up with a constant speed of m s – . The frictional force opposing the motion is N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. Answer The downward force on the elevator is F = m g + F f = ( × ) + = 22000 N The motor must supply enough power to balance this force.

Hence, P = F. v = 22000 × = 44000 W = hp ⊳ . COLLISIONS In physics we study motion (change in position). At the same time, we try to discover physical quantities, which do not change in a physical process.

The laws of momentum and energy conservation are typical examples. In this section we shall apply these laws to a commonly encountered phenomena, namely collisions. Several games such as billiards, marbles or carrom involve collisions.We shall study the collision of two masses in an idealised form. Consider two masses m and m .

The particle m is moving with speed v 1i , the subscript ‘ i ’ implying initial. We can cosider m to be at rest. No loss of generality is involved in making such a selection. In this situation the mass m collides with the stationary mass m and this is depicted in Fig.

Collision of mass m , with a stationary mass m . The masses m and m fly-off in different directions. We shall see that there are relationships, which connect the masses, the velocities and the angles. u unlike the conservative force, W nc depends on the particular path i to f.

. POWER Often it is interesting to know not only the work done on an object, but also the rate at which this work is done. We say a person is physically fit if he not only climbs four floors of a building but climbs them fast. Power is defined as the time rate at which work is done or energy is transferred.

The average power of a force is defined as the ratio of the work, W , to the total time t taken P t av = The instantaneous power is defined as the limiting value of the average power as time interval approaches zero, P t ( . ) The work dW done by a force F for a displacement d r is d W = F. d r . The instantaneous power can also be expressed as P t = F.

r = F.v ( . ) where v is the instantaneous velocity when the force is F . Power, like work and energy, is a scalar quantity. Its dimensions are [ML T – ].

In the SI, its unit is called a watt (W). The watt is J s – . The unit of power is named after James Watt, one of the innovators of the steam engine in the eighteenth century. There is another unit of power, namely the horse-power (hp) hp = W This unit is still used to describe the output of automobiles, motorbikes, etc.

We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A watt bulb which is on for hours uses kilowatt hour (kWh) of energy. (watt) × (hour) = watt hour = kilowatt hour (kWh) = (W) × (s) = . × J .

. Elastic and Inelastic Collisions In all collisions the total linear momentum is conserved; the initial momentum of the system is equal to the final momentum of the system. One can argue this as follows. When two objects collide, the mutual impulsive forces acting over the collision time ∆ t cause a change in their respective momenta : ∆ p = F ∆ t ∆ p = F ∆ t where F is the force exerted on the first particle by the second particle.

F is likewise the force exerted on the second particle by the first particle. Now from Newton’s third law, F = − F . This implies ∆ p + ∆ p = The above conclusion is true even though the forces vary in a complex fashion during the collision time ∆ t . Since the third law is true at every instant, the total impulse on the first object is equal and opposite to that on the second.

On the other hand, the total kinetic energy of the system is not necessarily conserved. The impact and deformation during collision may generate heat and sound. Part of the initial kinetic energy is transformed into other forms of energy. A useful way to visualise the deformation during collision is in terms of a ‘compressed spring’.

If the ‘spring’ connecting the two masses regains its original shape without loss in energy, then the initial kinetic energy is equal to the final kinetic energy but the kinetic energy during the collision time ∆ t is not constant. Such a collision is called an elastic collision . On the other hand the deformation may not be relieved and the two bodies could move together after the collision. A collision in which the two particles move together after the collision is called a completely inelastic collision .

The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision . . . Collisions in One Dimension Consider first a completely inelastic collision in one dimension.

Then, in Fig. . , θ = θ = m v i = ( m +m ) v f (momentum conservation) ( . ) The loss in kinetic energy on collision is 1i ∆ [using Eq.

( . )]       m m which is a positive quantity as expected. Consider next an elastic collision . Using the above nomenclature with θ = θ = , the momentum and kinetic energy conservation equations are m v i = m v f + m v f ( .

) it follows that, or, )( Hence, ∴ ( . ) Substituting this in Eq. ( . ), we obtain ( .

) and ( . ) Thus, the ‘unknowns’ { v f , v 2f } are obtained in terms of the ‘knowns’ { m , m , v i }. Special cases of our analysis are interesting. Case I : If the two masses are equal v f = v f = v i The first mass comes to rest and pushes off the second mass with its initial speed on collision.

Case II : If one mass dominates, e.g. m > > m v f ~ − v i v f ~ The heavier mass is undisturbed while the lighter mass reverses its velocity. Example . Slowing down of neutrons : In a nuclear reactor a neutron of high speed (typically m s – ) must be slowed to m s – so that it can have a high probability of interacting with isotope U and causing it to fission.

Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D O) or graphite, is called a moderator. Answer The initial kinetic energy of the neutron is while its final kinetic energy from Eq. ( .

)       The fractional kinetic energy lost is       while the fractional kinetic energy gained by the moderating nuclei K 2f / K 1i is f = − f (elastic collision) m m One can also verify this result by substituting from Eq. ( . ). For deuterium m = m and we obtain f = / while f = / .

Almost % of the neutron’s energy is transferred to deuterium. For carbon f = . % and f = . %.

In practice, however, this number is smaller since head-on collisions are rare. If the initial velocities and final velocities of both the bodies are along the same straight line, then it is called a one-dimensional collision, or head-on collision. In the case of small spherical bodies, this is possible if the direction of travel of body passes through the centre of body which is at rest. In general, the collision is two- dimensional, where the initial velocities and the final velocities lie in a plane.

. . Collisions in Two Dimensions Fig. .

also depicts the collision of a moving mass m with the stationary mass m . Linear momentum is conserved in such a collision. Since momentum is a vector this implies three equations for the three directions { x, y, z }. Consider the plane determined by the final velocity directions of m and m and choose it to be the x-y plane.

The conservation of the z -component of the linear momentum implies that the entire collision is in the x-y plane. The x - and y- component equations are m v i = m v f cos θ + m v f cos θ ( . ) = m v f sin θ − m v f sin θ ( . ) One knows { m , m , v i } in most situations.

There are thus four unknowns { v f , v f , θ and θ }, and only two equations. If θ = θ = , we regain Eq. ( . ) for one dimensional collision.

If, further the collision is elastic, ( . ) We obtain an additional equation. That still leaves us one equation short. At least one of the four unknowns, say θ , must be made known for the problem to be solvable.

For example, θ can be determined by moving a detector in an angular fashion from the x to the y axis. Given { m , m , v 1i , θ } we can determine { v 1f , v 2f , θ } from Eqs. ( . )-( .

). Example . Consider the collision depicted in Fig. .

to be between two billiard balls with equal masses m = m . The first ball is called the cue while the second ball is called the target. The billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle θ = °. Assume that the collision is elastic and that friction and rotational motion are not important.

Obtain θ . Answer From momentum conservation, since the masses are equal 2f 1f 1i or