Principle of linear momentum/ NBE question/physics/class11/

         

                              Principle of linear momentum  

    Principle of linear momentumstate that “Total linear momentum before collision is equal to total linear momentum after collision if there is no external force acts to them.

Let m1 and m2 be the two objects A and B, their initial velocity become u1 and u2 respectively after certain interval. They collide with each other and moves with final velocity v1 and v2 respectively. Force on A is FA and force on B is FB . 

  Momentum of A before collision =m1u1 

  Momentum of B before collision = m2u2

  Momentum of A after collision =m1 v1

  Momentum of B after collision =m2v2 

Also,

  Change in momentum on A=m1v1-m1u1

  Change in momentum on B=m2v2-m2u2

Rate of change of momentum on A=(m1v1-m1u1)/t

Rate of change of momentum on B=(m2v2-m2u2)/t

According to  Newton’s second law of motion , 

  FA= Rate of change of momentum on A=(m1v1-m1u1)/t

 FB=Rate of change of momentum on B=(m2v2-m2u2)/t

According to Newton’s third law of motion ,

      FA=-FB

 or,(m1v1-m1u1)/t=-[(m2v2-m2u2)/t]

 or, (m1v1-m1u1)=-m2v2+m2u2

 or, m1v1+m2v2=m2u2+m1u1

Therefore, m1u1+m2u2=m1v1+m2v2 

This is the mathematical expression for the principle of linear momentum. 

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#Angle of friction:The angle between resultant value of normal reaction and friction force to the normal reaction is called angle of friction. It is denoted by alpha.

# Angle of repose: The angle made by the inclined surface with the horizontal for which the body kept over the inclined surface just begins to slide is called angle of repose.

For angle of friction 

       Consider ‘m’ be mass of body, ‘Fs’ be friction force, ‘ma’ be force applied and ‘R’ be normal reaction.
From figure and definition of angle of friction           

     Tanα=PQ/OP
or, Tanα=ON/OP
or, Tanα=Fs/R ......................(i)

Again,
         Coefficient of friction (ɰ)=Fs/R..............................(ii)
      From equation(i) and (ii)
          Tanα=ɰ..................................(A)
Hence,the coefficient of limiting friction is equal to the tangent of angle of friction.
 
 
For angle of repose
       
     Consider ‘m’be mass of object with weight mg.mg resolved into two component mg cosፀ and mg sinፀ.mg cosፀ balance the normal reaction ‘R’ and mg sinፀ balance the friction force ‘Fs’ where ፀ is the angle of repose.
We know,
  Normal reaction (R)=mg cosፀ.............(1)
  Frictional force (Fs)=mg sinፀ...............(2).                  

 Dividing equation (2 )by (1)
     Fs/R=mg sinፀ/mg cosፀ
or,Fs/R=Tanፀ
Also, Fs/R=ɰ [ ɰ is cofficient of friction ]
 or,Tanፀ=ɰ............................(B)
From equation (A) and(B)
     Tanα= Tanፀ
Therefore, α=ፀ
Hence, angle of friction is equal to angle of repose



















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