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Conservation of Linear Momentum and Mechanical Energy

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					Conservation of Linear Momentum
    and Mechanical Energy
                    In absence of
                 external forces the
                 linear momentum of
                    the system is
                      conserved.
        Conservation Of Linear Momentum
 If no external force acts on a system of masses,
    then its linear momentum remains constant.


  Consider a system of masses m1, m2, m3, .......
moving with velocities v1, v2, v3,... respectively. The
       net linear momentum of the system is


     p  m1v1  m2v2  m3v3  ...



                                                          2
If Fext = O, then In terms of X & Y components


         px  initial    px final

         py  initial    py final
    Conservation Of Linear Momentum

 As linear momentum depends on frame of reference,
   observers in different frames would find different
    values of linear momentum of a given system but
      each would agree that his own value of linear
   momentum does not change with time provided the
              system is isolated and closed,

    i.e., law of conservation of linear momentum is
       independent of the frame of reference though
     linear momentum depends on frame of reference.




                                                      4
 Conservation of linear momentum is equivalent to
            Newton’s III law of motion.

   This law is universal, i.e., it applies to both
   macroscopic as well as microscopic systems. It
    holds good even in atomic and nuclear physics
           where classical mechanics fails.




                                                      5
Further it is more generally applicable than the
   law of ‘conservation of mechanical energy’
    because ‘internal forces’ are often non-
  conservative and so mechanical energy is not
    conserved but momentum is (if Fext = 0).
 Principal applications of conservation of linear
    momentum are in the field of collisions.




                                                    6
     Problem 6 .

 An object of mass m is projected with a velocity u at
angle q with horizontal. At the top of the trajectory it
 explode into two pieces in the ratio of mass 1 : 2. If
  heavier piece starts falling in vertical direction find
     Velocity of lighter piece just after explosion.
      y             2m      1
                    3       3m
           u



           q           M            A
      O
                                              N
Problem 6 .




y             2m   1
              3    3m
     u



     q         M        A
 O
                            N
   Problem 38.
  Two friends A and B (each weighing 40 kg) are
sitting on a frictionless platform some distance d
apart. A rolls a ball of mass 4 kg on the platform
 towards B which B catches. Then B rolls the ball
   toward A and A catches it. The ball keeps on
moving back and forth between A and B. The ball
    has a fixed speed of 5 m/s on the platform.
  (a) Find the speed of A after he rolls the ball for
                      the first time.
   (b) Find the speed of A after he catches the ball
                     for the first time.
  (c) Find the speed of A and B after the ball has
           made 5 round trips and is held by A.
  (d) How many times can A roll the ball ?

                                                        9
A          B
    ball
A          B
    ball
A          B
    ball
A          B
    ball
Solution.
         After every throw ‘A’ or ‘B’ gain the
         momentum 20 kgm/s in the direction
            opposite to the motion of ball.
                        Ist round
               A                                         B
pA= 20kg m/s
                          p = 20 kg m/s
                          ball

               A                                 B
pA= 20kg m/s                                         20kg m/s
                   20 kg m/s




    =
  pA,1
                                  20kg m/s       40kg m/s
  40kg m/s




                                                             14
                         Second round
                A                                     B
pA = 60kg m/s                                             40kg m/s
                                           20kg m/s
                       20 kg m/s



pA =60kg m/s                               20kg m/s       80kg m/s
                      20kg m/s
  pA,2 =
  80kg m/s


   At the end of each round both the man receive a
                momentum 40 kg m/s.
                (a)              pA = 20 = 40 vA
                                        1
                                 vA      m/ s
                                        2
                                                                 15
            (b)      pA,1 = 40 = (40+4) vA,1

                    40 10
           vA,1         m/ s
                    44 11

(c) At the end of 5 round trip the linear momentum of
                    (ball + ‘A) system
            pA,5  40  5  200 kg m / s

            200  (40  4) vA,5              50
                                    vA,5       m/ s
                                             11




                                                        16
(d) ‘A’ can catch the ball untill the velocity of ‘A’ is
    less than 5 m/s. After 5 round when ‘A’ through
    the ball i.e., 6th time the velocity of ‘A’ become
    greater than 5 m/s hence A can through the ball
                       only 6 times.
(e) As no external force is acting on ‘A’, ‘B’ and ‘ball’
     system hence the position of centre of mass of
          system should remain constant always.
    Let the centre of mass is at a distance x cm from
                 the initial position of A.

             (40  40  4) x cm  40  0  4  0  40 d

                      10
              xcm       m     From initial position of A.
                      21




                                                             18
        CONSIDER THE SITUATION
     A smooth wedge of mass M is placed on
horizontal surface and a block of mass m is kept
 at the highest position of wedge. Initially, the
                 system is held.
                 m


        h
             M
                                      v


                                                    19
        CONSIDER THE SITUATION
     A smooth wedge of mass M is placed on
horizontal surface and a block of mass m is kept
 at the highest position of wedge. Initially, the
                 system is held.
              m


        h
             M



                                                    20
If the wedge is held and block is released
               from rests.
                 m


           h
                M
                                    v

  The block starts sliding on the curved
           path of the wedge.


                                             21
   If we take the wedge and block as
                system
Can we apply conservation of linear momentum
              for this system?
  Can we apply conservation of mechanical
         energy for this system ?.
             m


        h
             M
                                   v
 We can not apply conservation of linear
 momentum as we have external forces
acting on the system in horizontal as well
           as vertical direction.
External force in vertical direction is gravity
    as well as normal reaction of ground,
              m


         h
              M
                                      v
       We can use conservation of
    mechanical energy for this system.
    We have external force in horizontal
    direction but it is not doing any work.
 While applying conservation of mechanical
energy earth is the part of the system hence
 no work done by external force in vertical
                  direction.
 Calculate the velocity of the block
     when it reaches at ground.
Apply C.O.M.E.     K + U = 0
1          
  mv 2  0    0  mgh   0    v  2 gh
2          

                m


         h
                M
                                               v
                m



            h
                    M



  What will happen if we remove stopping
force on the wedge and system is released.
h
    M
h
    M
h
    M
V
    h
        M
            v
When system is released both the block
    and wedge will starts moving.

  The wedge starts moving backward in
horizontal direction while the block starts
         sliding over the wedge.
                        m

                h
                    M


                    m           Can we apply C.O.L.M in
            h                   horizontal direction for
                M                 the system (block +
                                       wedge) ?
V                               Can we apply C.O.M.E for
    h               m                this system ?.
        M                   v
    We can apply C.O.L.M. in horizontal
  direction as there are not any external
      forces in horizontal direction.

We can also apply C.O.M.E. for the system as no
 external forces are doing work on the system
  What is the velocity of the block and
  wedge, when the block reaches at the
          base of the wedge ?
We can apply C.O.L.M in horizontal direction
   As no force is acting on the system in
horizontal direction linear momentum during
      entire process will be conserved
                                                  m
Initial linear momentum
                                          h
          Linitial = 0                        M

Final linear momentum
 Lfinal = mv - MV                     V
                                          h           m   v
As     Linitial = Lfinal                      M

     0 = mv - MV
           MV
      v                  ...( i )
             m
                                                 m
       Now applying C.O.M.E
         for this system.                h
                                             M
       K + U = 0
1       1         
  MV 2  mv 2  0    0  mgh   0
2       2         
                         2
                                    V
    1     1  MV 
      MV  m 
        2
                   mgh                 h           m   v
    2     2  m                             M
                2mgh
         V  2

                  M2 
              M  m 
                     
 The velocity of the block and wedge,
when the block reaches at the base of
              the wedge

The velocity of wedge       V
                                  2m 2 gh
                                 M (m  M )

                                 2 Mgh
The velocity of the block    v
                                (m  M )
   Now an another identical wedge is
     placed in front of wedge (I).
The block will slide over wedge I and reaches
         at its base with velocity v.
Then the block slide over smooth horizontal
   surface and interact with wedge II.
         m

    h
        M                  M
    m


h
    M   M
h
    M   M
h
    M   M
h
    M   M
V
    h
        M   v
                M
V
    h
        M   v   M
h
        v
    M       M
    II   V
M   M
    v   II

M       M
v
    II

M
                     m
             h
                 M             M


                     m
         h
                 M                 M

     V
         h               m v
                 M                     M


Vh
                               mv
         M                             M

                                                   v
                                           m
h
     M                                         M
                                   2 Mgh
 The velocity of the block v               …(i)
                                  (m  M )
Now considering the ‘m’ and wedge ‘2’ as system.
Using conservation of linear
        momentum                             m
  mv  (m  M )V ' …(ii)                             h’ V’
                                m v
                                                 M
   Conservation of
  mechanical energy
    1       1
      mv 2  (M  m)V '2  mg h ' …(iii)
    2       2
 From equations(i) (ii) and (iii) we get
                                             2
                                     M 
                           h'  h       
                                   M  m
  Illustration 88.
A smooth wedge of mass M rests on a smooth
  horizontal surface. A block of mass m is
   projected from its lowermost point with
                 velocity v0 .

 What is the maximum height
   reached by the block?   m   v0
                                       M




                                             51
m v0
       M




           52
m
    M




        53
m


    M




        54
m




    M




        55
   At the instant the block breaks contact with
  the wedge, they have common x-component of
  velocity. In addition, the block has a vertical
              component of velocity.
 The trajectory of
  the block will be
     parabola.
                         m
 Due to this vertical
 component the block
rises upwards till the
vertical component of
  velocity vanishes.         M



                                                    56
m



    M




        57
m




M




    58
m




M




    59
m




    M




        60
m




    M




        61
m



    M




        62
  From momentum conservation along x-axis,
 m0  (m  M )V …(i)
       m0
V               …(ii)
      (m  M )


             m                      m



                 M                      M




                                             63
From energy conservation between initial and final
               positions of block,
           1       1
             m2  (m  M )V 2  mgh …(iii)
                0
          2        2
           1       1  m2  2
             m2            0  mgh
           2    0
                   2 m  M 

                 M 
                     2
            h       0
               2g  m  M 
                         



                                                     64
    Illustration
Calculate the velocity of the wedge when the ball
      reaches at the bottom of the groove.




                   r       O

                       m
                               r


                           M
r       O

    m
            r


        M
Two identical buggies each of mass ‘M’ moves one after
                                                         
due to inertia (without friction) with the some velocity V0
. A man of mass m rides the rear buggy. At a certain
  moment, the man jumps into the front buggy with
          
 velocity u relative to this buggy . Knowing that the
          mass of each buggy is equal to M.
    Find the velocity with which the buggies will move
                       after that .
                V0      m
                                  V0
         M                                   M
Solution.

   Considering the rear buggy and man ‘m’ as system.
                                     
            (M  m)V0  MVr  m (Vr  u)
                                
                       mu
            Vr  V0 
                      ( M  m)

       Vr
                         (Vr + u)




        Rear buggy
Considering front buggy and mass as system
                                          
               M0 V0  m (u  Vr )  (m  M )Vf
                                 
                                       
                              mu                  
          MV0  m  u  V0               (m  M )Vf
                            (M  m) 
                                       
                         
                                            
                         u(M  m)  mu               
        (M  m)V0  m                         (M  m)Vf
                             (M  m)       
                                                     
    m                                   Mmu
                              Vf  V0 
                          Vf             (M  m)2
           M
  Illustration 85.


Two identical buggies 1 and 2 with one man in each
  move along parallel rails. When the buggies are
opposite to each other, the men jump in a direction
perpendicular to the direction of motion of buggies,
          so as to exchange their places.

                       Buggy 1

                                      v1


                       Buggy 2

            v2

                      Top view
                        (a)

                                                       70
  Illustration 85.

 As a consequence buggy 1 stop and buggy 2 keeps
moving in the same direction with its final velocity v.
Find the initial velocities v1 and v2 of buggies. Mass
of each buggy (without man) equals M, mass of each
man is m; ignore frictional effects anywhere and the
buggies are constrained to move along the rails only.
                        Buggy 1

                                        v1


                        Buggy 2

             v2

                       Top view
                         (a)

                                                          71
  Solution.

    In the problems involving mass ejection and mass
 addition, the ejected as well as added mass both are
part of a system. When the men jump they push on the
  floor of the buggy (action); the same force acts on
                them as well (reaction).

                       Buggy 1

                                     v1


                       Buggy 2

              v2

                      Top view
                        (a)

                                                    72
 This action- reaction pair is internal force for the
system of man and buggy. Similarly, when men jump
 in, there is an action-reaction force between them
                    and the buggy.


                      Buggy 1

                                      v1


                      Buggy 2

            v2

                     Top view
                       (a)

                                                        73
Let the buggy “1” move to the right and buggy “2” to the
 left. When the men jump. Man in “1” brings momentum
         mv1 directed towards right in the buggy “2”.
  Similarly the man in “2” brings momentum mv2 in “1”
                 directed towards left.
                       Buggy 1

                                      v1


                       Buggy 2

             v2

                       Top view
                         (a)
                  For system 1:
      Pi  (M  m)1  m1  m2  M 1  m2
      Pf  0
                                       System 1
From law of conservation
     of momentum,
       Pi  Pf                                       v1

 M 1  m2  0   …(i)
                                         Moving in


                                       v1   v2




                           Going out


                                                      75
                   For system 2:
  Pi  (M  m)2  m1  m2                                   Going out


     Pf  (M  m)
                                                    v1   v2


 From law of conservation                      Moving in

      of momentum,                     v2


         Pi  Pf
                                                    System 2
(M  m)2  m1  m2  (M  m) (b)        …(ii)




                                                                      76
On solving equations (i) and (ii) simultaneously, we
                       obtain


         m                         M
  1                        2 
       (M  m )                   (M  m )
 Problem 19.

A man of mass M having a bag of
mass m slips from the roof of a
  tall building of height H and
                                        
    starts falling vertically.
  When at a height h from the
  ground, he notices that the
                                   H
                                          h
ground below him is pretty hard,
    but there is a pond at a
 horizontal distance x from the               x
            line of fall.

                                   hard ground    pond


                                                     78
Problem 19.

  In order to save himself he
  throws the bag horizontally
(with respect to himself) in the
                                        
direction opposite to the pond.
    Calculate the minimum          H
horizontal velocity imparted to           h
the bag so that the man lands
         in the water                         x

                                   hard ground    pond


                                                     79
Solution.
 Considering (man + bag) as a
 system; No external forces
   in horizontal direction is
        acting on man.
                                     
  To avoid the man falling
   on hard floor the man
                                H
                                       h
  should be displaced by a
 horizontal displacement x
   during which he fall a                  x
          height h.

                                hard ground    pond


                                                  80
   Solution.
  The time taken by man to fall a
              height h
T = time to fall H – time to fall (H – h)   
           2H      2(H  h)
      t      
            g         g
                                        H
 Hence horizontal velocity of man             h
            required                              …(i)
            Vman 
                   x                              x
                    t

                                       hard ground       pond


                                                            81
  Solution.

  As no external force in horizontal
direction hence, the linear momentum
    of ‘man + bag’ system should be
                                         
     conserved (i.e. equal to zero).
                       
     Mman .V man  Mbag V bag  0   H
                                           h
           x
       M    mv b  0
           t
              M x                              x
      v bag   .
              m t
                    Mx g            hard ground
      v bag                                       pond
                m[ 2H  2(H  2)]


                                                      82
    Illustration 89.
      A wedge of mass m1 with its upper surface
hemispherical in shape, as shown in figure (a), rests on
      a smooth horizontal surface near the wall.
 A small block of mass m2 slides
     without friction on the
  hemispherical surface of the          m2
             wedge.                            r
  What is the maximum velocity
     attained by the wedge?

                                                   m1

                                             (a)




                                                           83
   Solution.

    As long as the block moves from A to B, the
reaction on the wedge presses it to the wall. When
   the block reaches the lowermost positions, its
         velocity from energy conservation is   2gr


                              A    N N’
      m2                                    C
                          N          2gr        N’
                 r

                                    B

                     m1
                                   (b)
           (a)

                                                     84
  When the block moves along the right half of the
wedge, during its upward journey as well as downward
  journey the reaction of the block on the wedge is
        towards right as shown in figure (b).
 Therefore during the entire motion of the block from B
to C and C to B, the wedge is accelerated towards right.
                               A    N N’
       m2                                    C
                           N          2gr        N’
                  r

                                     B

                      m1
                                    (b)
            (a)

                                                       85
  Thus maximum velocity is attained by the
 wedge at the instant when the block passes
     point B during the return journey.
From conservation of momentum, at the instant when
       the wedge is separated from the wall,
      Pi  Pf          m2 2gr  m11  m2 2    …(i)
                                   A     N N’
       m2                                           C
                            N              2gr          N’
                  r

                                          B

                       m1
                                         (b)
            (a)

                                                             86
 From M.E. energy conservation,     E i  Ef
                m11 m2 2
                   2     2
         m2gr                 …(ii)
                 2     2
On solving equations (i) and (ii) simultaneously, we
                 obtain two solutions
               1  0,    2  2gr

          1 
                 2m2                 m2  m1
  and                    2gr ,  2          2gr
               m1  m2               m1  m2
The first solution corresponds to the instant
when the block reaches for the first time at
 point B. At this instant the   block moves
 with velocity v2 and the wedge is at rest.


                                                       87
The second solution corresponds to the instant
 when the    block has the maximum velocity

                     2m2 2gr
        (1 )max   
                      m1  m2
   Problem 27.
  A gun is mounted on a railroad car. The mass of the car,
the gun, the shells and the operator is 50 m where m is the
 mass of one shell. If the muzzle velocity of the shells is
  200 m/s, what is the recoil speed of the car after the
              second shot ? Neglect friction.




                       gun   vs,g =200 m/s
            v1




                                                         89
  Solution.

 Taking Gun and car as a system no external force are
 acting on the system in horizontal direction at any time.
After Ist shot     49 mv1  m .200
                           200
                      v1      m/ s
                            49




                       gun    vs,g =200 m/s
              v1
After IInd shot
             200                       200 
   49 m          48mv 2  m  200      
              49                        49 


                           1   1 
                v 2  200       
                           49 48 



                    gun    vs,g =200 m/s
       v1




                                                91

				
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