**What is the difinition of momentum ?**

A motortruck requires a larger force to set in motion when it is heavily loaded than when it is empty. Similarly, far more powerful brakes are needed to stop a heavy goods vehicle than a light car moving with the same speed. The heavier vehicle is said to posses a greater quantity of motion or

*momentum*than the lighter one.**The momentum of a body is defined as the product of its mass and its velocity.**

The unit of momentum in the SI system is, therefore, 1 kilogramme multiplied by 1 metre per second ( unit sympol, kg m / s ).

When two bodies, a heavy one and a light one, are acted upon by the same force for the same time, the lighter body builds up a higher velocity than the heavy one. But the momentum they gain is the same in both cases. This important connection between force and momentum was recognized by Sir Isaac Newton and expressed by him in a

**second law of motion.****Newton**

**'s second law of motion**

*The rate of change in momentum of a body is proportional to the applied force and takes place in the direction in which the force acts.*Suppose a force

*F*acts on a body of mass*m*for a time*t*, and causes its velocity to change from*u*to*v*.The momentum changes uniformly from

*mu*to*mv*in time*t*, therefore, the rate of change of momentum =*mv – mu**/ t*.By Newton ’s second law, the rate of change of momentum is proportional to the applied force and hence,

*F*is proportional to

*mv – mu / t*

factorizing

*F*is proportional to*m ( v – u ) /t**But ( v – u ) /t*= change in velocity / time = acceleration

*= a*

Therefore

*F*is proportional to*ma**Thus*

*, F = constant * ma.*

It is this equation which enables us to define an

*absolute unit of force*. If*m*= 1 kg and*a*= 1 m / s^{2}, the value of the unit of force is chosen so as to make*F*= 1 when the constant = 1 .The SI unit of force is called the Newton ( N ), and is the force which produce an acceleration of 1 m / s

^{2 }when it acts on a mass of a kg.**Thus,**

*F = ma.***Weight of a body :**

**Galileo showed that all bodies, whatever their masses, have the same acceleration when they fall freely under the action of gravity.**

**Thus, F = mg, where g is the acceleration due to gravity ( 9.81 m / s**

^{2 }).**Weight of a body in an elevator:**

**Everyone is acquainted with the feeling of extra personal weight when the elevator accelerates from rest to attain its steady upward velocity, together with the feeling of lightness when the elevator decelerates to rest. Similar feelings are experienced in a descending elevator.**

**1.***Elevator at rest***The case is no different from that when the body is placed on any other support fixed relative to the earth. It therefore presses down on the floor with its rest weight of**

*mg*in newtons.

*2.**Elevator moving up or down with uniform velocity***From**

*, no force is required to keep a body moving with uniform velocity in a straight line. The air inside the elevator moves with the body, so the question of air resistance does not concern us.*Newton ’s first law of motion**Gravity pulls the body down with a force of mg in N . The body presses on the elevator floor with the same force. By**

*– we’ll discuss it later - the floor exerts an equal upward reaction on the body. Consequently the resultant force on the body is zero which, as we have seen, is a requirement for uniform velocity in any direction.*Newton ’s third law**Thus, if the elevator is moving up or down with uniform velocity, the weight of the body is mg, the same when at rest on earth.**

*3.**Elevator accelerating upwards***Arising out of Newton’s second law of motion we saw that when a resultant force,**

*F*, acts on a body of mass*m*, it will move with an acceleration of a which is given by the equation,

*F = ma***If the elevator moves upwards with acceleration**

*a*its floor must push upwards on the body inside to give that the same acceleration also.**By**

*Newton*

*’s third law*the body will therefore exert an equal and opposite reaction of*ma*downwards on the floor. The force of gravity, however, has not ceased to act and is still causing the body to press down on the floor with a force of*mg*. The resultant force which the body exerts on the floor is, therefore,

*mg + ma*=*m ( g + a )***Hence, new weight of body in newtons =**

*m ( g + a )*

*4.**Elevator moving downwards with acceleration a, less than g***In this case, new weight of body =**

*m ( g - a )*

*5.***Elevator falling freely ( the rope broke !)****Its weight has now become,**

**Mg – mg = 0 newtons**

**That is a**

*weightless*body ! .