# 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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