DC Pandey Solutions: Centre of Mass, Conservation of Linear Momentum- 2 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


Introductory Exercise 8.8
1. v v
0
sin cos ( ) a a b = +
v
0
cos sin ( ) a a b = + (Impact being elastic)
\ tan cot( ) a a b = +
or cot cot( )
p
a a b
2
- = +   
or          
p
a a b
2
- = +
or b
p
a = -
2
2
2. Speed after n impacts
Speed before first impact = = u gh ( ) 2
Speed before one impact = eu
Speed after 2 impacts = = e eu e u ( )
2
 
........ ......... ........ ........ ........ ......... 
Speed after n impacts = e u
n
Height ( ) H upto which the ball rebounds
after nth rebound
            H
e u
g
n
=
( )
2
2
              =
e u
g
n 2 2
2
              = × e h
n 2
3. mv MV + = + 0 0
Þ      V
m
M
v =
e=
Velocityof separation
Velocityofapproach
          =
-
-
V
v
0
0
          =
V
v
          =
m
M
4.
As the balls are of same size, the centres
of the balls P Q , and R will be at vertices
of an equilateral triangle when ball C just
strikes balls A and B symmetrically and
as such the balls A and B will follow the
path as shown below
Applying law of conservation of
momentum
mu= Resultant momentum of A and B
balls along the axis of X.
or mu mv = ° 2 30 cos
or u v = 2
3
2
   
or u v = 3
As the ball C will strike ball A (and as well 
as ball B) with velocity ucos 30°
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 169
a
v cos (a + b) a
v cos a
0
v
0
a
v sin a
0
v sin (a + b)
0
b
v
m M m
V
M
Before collision After collision
A B B A
Rest Rest
X
C
R
P
Q
B
A
u
Rest
Rest
ÐPRX = 30°
ÐQRX = 30°
v
30°
30°
C
A
v
B
X Angle O
A
Rest
Page 2


Introductory Exercise 8.8
1. v v
0
sin cos ( ) a a b = +
v
0
cos sin ( ) a a b = + (Impact being elastic)
\ tan cot( ) a a b = +
or cot cot( )
p
a a b
2
- = +   
or          
p
a a b
2
- = +
or b
p
a = -
2
2
2. Speed after n impacts
Speed before first impact = = u gh ( ) 2
Speed before one impact = eu
Speed after 2 impacts = = e eu e u ( )
2
 
........ ......... ........ ........ ........ ......... 
Speed after n impacts = e u
n
Height ( ) H upto which the ball rebounds
after nth rebound
            H
e u
g
n
=
( )
2
2
              =
e u
g
n 2 2
2
              = × e h
n 2
3. mv MV + = + 0 0
Þ      V
m
M
v =
e=
Velocityof separation
Velocityofapproach
          =
-
-
V
v
0
0
          =
V
v
          =
m
M
4.
As the balls are of same size, the centres
of the balls P Q , and R will be at vertices
of an equilateral triangle when ball C just
strikes balls A and B symmetrically and
as such the balls A and B will follow the
path as shown below
Applying law of conservation of
momentum
mu= Resultant momentum of A and B
balls along the axis of X.
or mu mv = ° 2 30 cos
or u v = 2
3
2
   
or u v = 3
As the ball C will strike ball A (and as well 
as ball B) with velocity ucos 30°
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 169
a
v cos (a + b) a
v cos a
0
v
0
a
v sin a
0
v sin (a + b)
0
b
v
m M m
V
M
Before collision After collision
A B B A
Rest Rest
X
C
R
P
Q
B
A
u
Rest
Rest
ÐPRX = 30°
ÐQRX = 30°
v
30°
30°
C
A
v
B
X Angle O
A
Rest
Velocity of approach of ball C towards ball A
            = ° - ucos30 0
            = ° ucos30
            =u 3
Velocity of separation of ball A away from
ball C = v
\            e
v
u
=
3
2
  =
v
v ( ) 3
3
2
               =
2
3
5. x-component of velocity
before impact = 2 i
^
\ after impact = - e2 i
^
            = - ´
1
2
2i
^
            = - i
^
y-component of velocity
before impact = 2 j
^
\ after impact = 2 j
^
\ Velocity after impact = - + i j
^ ^
2
6. At A, u u
| |
cos = q
    u u
^
= sinq
\ At B
u u
| |
cos = q      
u eu
^
= sinq     
\    
T
T
u
g
eu
g
e
1
2
2
2
1
= =
sin
sin
q
q
UsingT
u
g
=
é
ë
ê
ù
û
ú
^
2
R
R
u u
eu u
g
1
2
2
2
2
=
( sin )( cos )
( sin ) cos
q q
q q
   Using R
u u
g
=
é
ë
ê
ù
û
ú
^
2
| |
      =
1
e
H
H
u
g
eu
g
1
2
2
2
2
2
=
( sin )
( sin )
q
q
                    Using H
u
g
=
é
ë
ê
ù
û
ú
^
2
2
   =
1
2
e
AIEEE Corner
Subjective Questions (Level 1)
1.
  x
CM
=
× + × + × + ×
+ + +
1 0 2 1 3 1 4 0
1 2 3 4
      =
5
10
 m
    y
CM
=
× + × + × + ×
+ + +
1 0 2 0 3 1 4 1
1 2 3 4
      =
7
10
 m
170 | Mechanics-1
45°
2 j
2 i
45°
45°
X
v = 2i + 2j
^
^ ^
^
q
A
u sin q
H
1
e u sin q
B
T
1
T
2
H
2
R
2
R
1
u cos q
v
u sin q
B 2kg
C
3kg
D 4kg
A 1kg
P
(CM)
1 m
1 m 1 m
1 m
x
y
Page 3


Introductory Exercise 8.8
1. v v
0
sin cos ( ) a a b = +
v
0
cos sin ( ) a a b = + (Impact being elastic)
\ tan cot( ) a a b = +
or cot cot( )
p
a a b
2
- = +   
or          
p
a a b
2
- = +
or b
p
a = -
2
2
2. Speed after n impacts
Speed before first impact = = u gh ( ) 2
Speed before one impact = eu
Speed after 2 impacts = = e eu e u ( )
2
 
........ ......... ........ ........ ........ ......... 
Speed after n impacts = e u
n
Height ( ) H upto which the ball rebounds
after nth rebound
            H
e u
g
n
=
( )
2
2
              =
e u
g
n 2 2
2
              = × e h
n 2
3. mv MV + = + 0 0
Þ      V
m
M
v =
e=
Velocityof separation
Velocityofapproach
          =
-
-
V
v
0
0
          =
V
v
          =
m
M
4.
As the balls are of same size, the centres
of the balls P Q , and R will be at vertices
of an equilateral triangle when ball C just
strikes balls A and B symmetrically and
as such the balls A and B will follow the
path as shown below
Applying law of conservation of
momentum
mu= Resultant momentum of A and B
balls along the axis of X.
or mu mv = ° 2 30 cos
or u v = 2
3
2
   
or u v = 3
As the ball C will strike ball A (and as well 
as ball B) with velocity ucos 30°
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 169
a
v cos (a + b) a
v cos a
0
v
0
a
v sin a
0
v sin (a + b)
0
b
v
m M m
V
M
Before collision After collision
A B B A
Rest Rest
X
C
R
P
Q
B
A
u
Rest
Rest
ÐPRX = 30°
ÐQRX = 30°
v
30°
30°
C
A
v
B
X Angle O
A
Rest
Velocity of approach of ball C towards ball A
            = ° - ucos30 0
            = ° ucos30
            =u 3
Velocity of separation of ball A away from
ball C = v
\            e
v
u
=
3
2
  =
v
v ( ) 3
3
2
               =
2
3
5. x-component of velocity
before impact = 2 i
^
\ after impact = - e2 i
^
            = - ´
1
2
2i
^
            = - i
^
y-component of velocity
before impact = 2 j
^
\ after impact = 2 j
^
\ Velocity after impact = - + i j
^ ^
2
6. At A, u u
| |
cos = q
    u u
^
= sinq
\ At B
u u
| |
cos = q      
u eu
^
= sinq     
\    
T
T
u
g
eu
g
e
1
2
2
2
1
= =
sin
sin
q
q
UsingT
u
g
=
é
ë
ê
ù
û
ú
^
2
R
R
u u
eu u
g
1
2
2
2
2
=
( sin )( cos )
( sin ) cos
q q
q q
   Using R
u u
g
=
é
ë
ê
ù
û
ú
^
2
| |
      =
1
e
H
H
u
g
eu
g
1
2
2
2
2
2
=
( sin )
( sin )
q
q
                    Using H
u
g
=
é
ë
ê
ù
û
ú
^
2
2
   =
1
2
e
AIEEE Corner
Subjective Questions (Level 1)
1.
  x
CM
=
× + × + × + ×
+ + +
1 0 2 1 3 1 4 0
1 2 3 4
      =
5
10
 m
    y
CM
=
× + × + × + ×
+ + +
1 0 2 0 3 1 4 1
1 2 3 4
      =
7
10
 m
170 | Mechanics-1
45°
2 j
2 i
45°
45°
X
v = 2i + 2j
^
^ ^
^
q
A
u sin q
H
1
e u sin q
B
T
1
T
2
H
2
R
2
R
1
u cos q
v
u sin q
B 2kg
C
3kg
D 4kg
A 1kg
P
(CM)
1 m
1 m 1 m
1 m
x
y
   AP x y
2 2 2
= +
CM CM
 =
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
5
10
7
10
2 2
      =0.74m
2
2. x
A x A x
A A
CM
=
+
+
1 1 2 2
1 2
=
× + ×
+
a
a
a
a
a
2
2
2
2
0
4
4
p
p
( )
            =
+
p
p 4
a
3. Let A = area of rectangle
  x
A
A a
A
A
CM
=
× + -
æ
è
ç
ö
ø
÷
+ -
æ
è
ç
ö
ø
÷
0
4 4
4
      = - × ×
A
A
a
4
4
3 4
      = -
a
12
  y
A
A b
A
A
CM
=
× + -
æ
è
ç
ö
ø
÷
+ -
æ
è
ç
ö
ø
÷
0
4 4
4
     = -
b
12
Centre of mass - -
æ
è
ç
ö
ø
÷
a b
12 12
,
4.   x
V a b
V a
CM
=
× + -
æ
è
ç
ö
ø
÷
-
0
4
3
4
3
3
3
p
p
       =
-
-
4
3
4
3
4
3
3
3 3
p
p p
a b
R a
       = -
-
æ
è
ç
ö
ø
÷
a
R a
b
3
3 3
By symmetry y
CM
= 0
5. In Fig. 1,  C is CM.
\ m a m l a
1 2
= - ( ) …(i)
In Fig. 2, C¢ is CM.
m a b l m l a l b
1 1 2 2
( ) ( ) + - = - + - …(ii)
Substituting Eq. (i) in Eq. (ii),
m b l m l b
1 1 2 2
( ) ( ) - = -
or m b m l m l m b
1 1 1 2 2 2
- = -
or ( ) m m b m l m l
1 2 1 1 2 2
+ = +
 
or         b
m l m l
m m
=
+
+
1 1 2 2
1 2
6. x
x dm
dm
CM
=
ò
ò
 
where, dm = mass of element of length dx.
          =
ò
ò
x dx A
dx A
r
r
          =
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
ò
ò
x
x
l
dx
x
l
a
l
x
l
0
0
2
2
0
2
2
0
r
r
(Q r r =
0
2
2
x
l
)
          =
ò
ò
x dx
x dx
l
l
3
0
2
0
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 171
A =
2
p
a
4
2
2
A = a
1
a
2
a a
a
O
b
a
O
X
Y
(       )
a
4
,
b
4
l
a
C m
1
m
2
l
2
b
l
1
m
1
m
2
C'
Fig. 1
Fig. 2
dx
x O
Page 4


Introductory Exercise 8.8
1. v v
0
sin cos ( ) a a b = +
v
0
cos sin ( ) a a b = + (Impact being elastic)
\ tan cot( ) a a b = +
or cot cot( )
p
a a b
2
- = +   
or          
p
a a b
2
- = +
or b
p
a = -
2
2
2. Speed after n impacts
Speed before first impact = = u gh ( ) 2
Speed before one impact = eu
Speed after 2 impacts = = e eu e u ( )
2
 
........ ......... ........ ........ ........ ......... 
Speed after n impacts = e u
n
Height ( ) H upto which the ball rebounds
after nth rebound
            H
e u
g
n
=
( )
2
2
              =
e u
g
n 2 2
2
              = × e h
n 2
3. mv MV + = + 0 0
Þ      V
m
M
v =
e=
Velocityof separation
Velocityofapproach
          =
-
-
V
v
0
0
          =
V
v
          =
m
M
4.
As the balls are of same size, the centres
of the balls P Q , and R will be at vertices
of an equilateral triangle when ball C just
strikes balls A and B symmetrically and
as such the balls A and B will follow the
path as shown below
Applying law of conservation of
momentum
mu= Resultant momentum of A and B
balls along the axis of X.
or mu mv = ° 2 30 cos
or u v = 2
3
2
   
or u v = 3
As the ball C will strike ball A (and as well 
as ball B) with velocity ucos 30°
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 169
a
v cos (a + b) a
v cos a
0
v
0
a
v sin a
0
v sin (a + b)
0
b
v
m M m
V
M
Before collision After collision
A B B A
Rest Rest
X
C
R
P
Q
B
A
u
Rest
Rest
ÐPRX = 30°
ÐQRX = 30°
v
30°
30°
C
A
v
B
X Angle O
A
Rest
Velocity of approach of ball C towards ball A
            = ° - ucos30 0
            = ° ucos30
            =u 3
Velocity of separation of ball A away from
ball C = v
\            e
v
u
=
3
2
  =
v
v ( ) 3
3
2
               =
2
3
5. x-component of velocity
before impact = 2 i
^
\ after impact = - e2 i
^
            = - ´
1
2
2i
^
            = - i
^
y-component of velocity
before impact = 2 j
^
\ after impact = 2 j
^
\ Velocity after impact = - + i j
^ ^
2
6. At A, u u
| |
cos = q
    u u
^
= sinq
\ At B
u u
| |
cos = q      
u eu
^
= sinq     
\    
T
T
u
g
eu
g
e
1
2
2
2
1
= =
sin
sin
q
q
UsingT
u
g
=
é
ë
ê
ù
û
ú
^
2
R
R
u u
eu u
g
1
2
2
2
2
=
( sin )( cos )
( sin ) cos
q q
q q
   Using R
u u
g
=
é
ë
ê
ù
û
ú
^
2
| |
      =
1
e
H
H
u
g
eu
g
1
2
2
2
2
2
=
( sin )
( sin )
q
q
                    Using H
u
g
=
é
ë
ê
ù
û
ú
^
2
2
   =
1
2
e
AIEEE Corner
Subjective Questions (Level 1)
1.
  x
CM
=
× + × + × + ×
+ + +
1 0 2 1 3 1 4 0
1 2 3 4
      =
5
10
 m
    y
CM
=
× + × + × + ×
+ + +
1 0 2 0 3 1 4 1
1 2 3 4
      =
7
10
 m
170 | Mechanics-1
45°
2 j
2 i
45°
45°
X
v = 2i + 2j
^
^ ^
^
q
A
u sin q
H
1
e u sin q
B
T
1
T
2
H
2
R
2
R
1
u cos q
v
u sin q
B 2kg
C
3kg
D 4kg
A 1kg
P
(CM)
1 m
1 m 1 m
1 m
x
y
   AP x y
2 2 2
= +
CM CM
 =
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
5
10
7
10
2 2
      =0.74m
2
2. x
A x A x
A A
CM
=
+
+
1 1 2 2
1 2
=
× + ×
+
a
a
a
a
a
2
2
2
2
0
4
4
p
p
( )
            =
+
p
p 4
a
3. Let A = area of rectangle
  x
A
A a
A
A
CM
=
× + -
æ
è
ç
ö
ø
÷
+ -
æ
è
ç
ö
ø
÷
0
4 4
4
      = - × ×
A
A
a
4
4
3 4
      = -
a
12
  y
A
A b
A
A
CM
=
× + -
æ
è
ç
ö
ø
÷
+ -
æ
è
ç
ö
ø
÷
0
4 4
4
     = -
b
12
Centre of mass - -
æ
è
ç
ö
ø
÷
a b
12 12
,
4.   x
V a b
V a
CM
=
× + -
æ
è
ç
ö
ø
÷
-
0
4
3
4
3
3
3
p
p
       =
-
-
4
3
4
3
4
3
3
3 3
p
p p
a b
R a
       = -
-
æ
è
ç
ö
ø
÷
a
R a
b
3
3 3
By symmetry y
CM
= 0
5. In Fig. 1,  C is CM.
\ m a m l a
1 2
= - ( ) …(i)
In Fig. 2, C¢ is CM.
m a b l m l a l b
1 1 2 2
( ) ( ) + - = - + - …(ii)
Substituting Eq. (i) in Eq. (ii),
m b l m l b
1 1 2 2
( ) ( ) - = -
or m b m l m l m b
1 1 1 2 2 2
- = -
or ( ) m m b m l m l
1 2 1 1 2 2
+ = +
 
or         b
m l m l
m m
=
+
+
1 1 2 2
1 2
6. x
x dm
dm
CM
=
ò
ò
 
where, dm = mass of element of length dx.
          =
ò
ò
x dx A
dx A
r
r
          =
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
ò
ò
x
x
l
dx
x
l
a
l
x
l
0
0
2
2
0
2
2
0
r
r
(Q r r =
0
2
2
x
l
)
          =
ò
ò
x dx
x dx
l
l
3
0
2
0
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 171
A =
2
p
a
4
2
2
A = a
1
a
2
a a
a
O
b
a
O
X
Y
(       )
a
4
,
b
4
l
a
C m
1
m
2
l
2
b
l
1
m
1
m
2
C'
Fig. 1
Fig. 2
dx
x O
          = ×
é
ë
ê
ù
û
ú
3
4
3
4
0
x
x
l
          =
3
4
l
7. x
CM
=
´ + ´
+
1 10 2 12
1 2
        =
34
3
 m
     v
CM
=
´ - + ´ +
+
1 6 2 4
1 2
( ) ( )
  = +
-
2
3
1
ms
x¢
CM
 (new position of CM) 
        = +
æ
è
ç
ö
ø
÷
-
34
3
2
2
3
1
m s ms
        =
38
3
 m = 12.67 m
8. v
CM
=
´ + + ´ -
+
1 2 2 2 1
1 2
( ) ( )
= 0 ms
-2
\ Displacement of CM in 1 s = 0 m.
9. Acceleration ( ) a
®
 = -
-
10
2
j
^
ms
       v u a
® ® ®
= + t
         =
+ +
-
( )( ) ( )(
$ $
)
$
1 0 2 10 10
3
10
i j
j
                 = -
æ
è
ç
ö
ø
÷
20
3
10
3
$ $
i j m/s
New position vector ( ) r
1
®
 of particle A
   ( ) r s u a
1 0
2
1
2
® ® ® ®
= + + t t
    = + + + - ×
®
( ) ( )
^ ^ ^
10 20 0
1
2
10 1
2
i j j
      = + 10 15 i j
^ ^
m
New position vector ( ) r
2
®
 of particle B
r i j i j j
2
2
20 40 10 10 1
1
2
10 1
®
= + + + × + - × ( ) ( ) ( )
^ ^ ^ ^ ^
  = + 30 45 i j
^ ^
New position of CM
     R
r r ®
® ®
=
+
+
1 2
1 2
1 2
( ) ( )
       =
+ + + 10 15 60 90
3
i j i j
^ ^ ^ ^
       =
+ 70 105
3
i j
^ ^
 m
10.
(a)  3
0 0 1 12
0 1
=
× + × ´
+
m
m .
Þ m = 0.3 kg
(b) Momentum of system = Momentum of CM
           = + ´
-
( . )
^
m 01 6
1
kg ms j
           =
-
2.4 kgms j
^ 1
(c) v
v ®
® ®
=
+
+
CM
.3) 0.10)
0.3 .1
m
( ( 0 0
0
   v v
m
® ®
=
4
4
CM
       = ´
-
4
3
6
1
j
^
ms
       =
-
8
1
j
^
m s
11.   A t
m
· =0s
  B t
m
· =
2
100 ( ) ms
Position of 1st particle ( ) A at t = 300 ms
s
1
3 2
1
2
10 300 10 = ´ ´ ´
-
( )
     =0.45 m
Position of 2nd particle ( ) B at t = 300 ms
   (B is at the position of A at t = 100 ms)
    s
2
3 2
1
2
100 200 10 = ´ ´ ´
-
( )
      =0.20 m 
\ Position of CM =
´ + ´
+
2 0.2 0.45
2
m m
m m
= 28.3 cm
172 | Mechanics-1
1 kg
–1
6ms
+ ive
2 kg
10 m 10 m
–1
4ms
m
2 kg 1 kg
–1
2 ms
–1
1 ms
+ ive
12 m
O
CM
–1
6 j ms
 
m kg
3 m
0.10 kg
^
Page 5


Introductory Exercise 8.8
1. v v
0
sin cos ( ) a a b = +
v
0
cos sin ( ) a a b = + (Impact being elastic)
\ tan cot( ) a a b = +
or cot cot( )
p
a a b
2
- = +   
or          
p
a a b
2
- = +
or b
p
a = -
2
2
2. Speed after n impacts
Speed before first impact = = u gh ( ) 2
Speed before one impact = eu
Speed after 2 impacts = = e eu e u ( )
2
 
........ ......... ........ ........ ........ ......... 
Speed after n impacts = e u
n
Height ( ) H upto which the ball rebounds
after nth rebound
            H
e u
g
n
=
( )
2
2
              =
e u
g
n 2 2
2
              = × e h
n 2
3. mv MV + = + 0 0
Þ      V
m
M
v =
e=
Velocityof separation
Velocityofapproach
          =
-
-
V
v
0
0
          =
V
v
          =
m
M
4.
As the balls are of same size, the centres
of the balls P Q , and R will be at vertices
of an equilateral triangle when ball C just
strikes balls A and B symmetrically and
as such the balls A and B will follow the
path as shown below
Applying law of conservation of
momentum
mu= Resultant momentum of A and B
balls along the axis of X.
or mu mv = ° 2 30 cos
or u v = 2
3
2
   
or u v = 3
As the ball C will strike ball A (and as well 
as ball B) with velocity ucos 30°
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 169
a
v cos (a + b) a
v cos a
0
v
0
a
v sin a
0
v sin (a + b)
0
b
v
m M m
V
M
Before collision After collision
A B B A
Rest Rest
X
C
R
P
Q
B
A
u
Rest
Rest
ÐPRX = 30°
ÐQRX = 30°
v
30°
30°
C
A
v
B
X Angle O
A
Rest
Velocity of approach of ball C towards ball A
            = ° - ucos30 0
            = ° ucos30
            =u 3
Velocity of separation of ball A away from
ball C = v
\            e
v
u
=
3
2
  =
v
v ( ) 3
3
2
               =
2
3
5. x-component of velocity
before impact = 2 i
^
\ after impact = - e2 i
^
            = - ´
1
2
2i
^
            = - i
^
y-component of velocity
before impact = 2 j
^
\ after impact = 2 j
^
\ Velocity after impact = - + i j
^ ^
2
6. At A, u u
| |
cos = q
    u u
^
= sinq
\ At B
u u
| |
cos = q      
u eu
^
= sinq     
\    
T
T
u
g
eu
g
e
1
2
2
2
1
= =
sin
sin
q
q
UsingT
u
g
=
é
ë
ê
ù
û
ú
^
2
R
R
u u
eu u
g
1
2
2
2
2
=
( sin )( cos )
( sin ) cos
q q
q q
   Using R
u u
g
=
é
ë
ê
ù
û
ú
^
2
| |
      =
1
e
H
H
u
g
eu
g
1
2
2
2
2
2
=
( sin )
( sin )
q
q
                    Using H
u
g
=
é
ë
ê
ù
û
ú
^
2
2
   =
1
2
e
AIEEE Corner
Subjective Questions (Level 1)
1.
  x
CM
=
× + × + × + ×
+ + +
1 0 2 1 3 1 4 0
1 2 3 4
      =
5
10
 m
    y
CM
=
× + × + × + ×
+ + +
1 0 2 0 3 1 4 1
1 2 3 4
      =
7
10
 m
170 | Mechanics-1
45°
2 j
2 i
45°
45°
X
v = 2i + 2j
^
^ ^
^
q
A
u sin q
H
1
e u sin q
B
T
1
T
2
H
2
R
2
R
1
u cos q
v
u sin q
B 2kg
C
3kg
D 4kg
A 1kg
P
(CM)
1 m
1 m 1 m
1 m
x
y
   AP x y
2 2 2
= +
CM CM
 =
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
5
10
7
10
2 2
      =0.74m
2
2. x
A x A x
A A
CM
=
+
+
1 1 2 2
1 2
=
× + ×
+
a
a
a
a
a
2
2
2
2
0
4
4
p
p
( )
            =
+
p
p 4
a
3. Let A = area of rectangle
  x
A
A a
A
A
CM
=
× + -
æ
è
ç
ö
ø
÷
+ -
æ
è
ç
ö
ø
÷
0
4 4
4
      = - × ×
A
A
a
4
4
3 4
      = -
a
12
  y
A
A b
A
A
CM
=
× + -
æ
è
ç
ö
ø
÷
+ -
æ
è
ç
ö
ø
÷
0
4 4
4
     = -
b
12
Centre of mass - -
æ
è
ç
ö
ø
÷
a b
12 12
,
4.   x
V a b
V a
CM
=
× + -
æ
è
ç
ö
ø
÷
-
0
4
3
4
3
3
3
p
p
       =
-
-
4
3
4
3
4
3
3
3 3
p
p p
a b
R a
       = -
-
æ
è
ç
ö
ø
÷
a
R a
b
3
3 3
By symmetry y
CM
= 0
5. In Fig. 1,  C is CM.
\ m a m l a
1 2
= - ( ) …(i)
In Fig. 2, C¢ is CM.
m a b l m l a l b
1 1 2 2
( ) ( ) + - = - + - …(ii)
Substituting Eq. (i) in Eq. (ii),
m b l m l b
1 1 2 2
( ) ( ) - = -
or m b m l m l m b
1 1 1 2 2 2
- = -
or ( ) m m b m l m l
1 2 1 1 2 2
+ = +
 
or         b
m l m l
m m
=
+
+
1 1 2 2
1 2
6. x
x dm
dm
CM
=
ò
ò
 
where, dm = mass of element of length dx.
          =
ò
ò
x dx A
dx A
r
r
          =
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
ò
ò
x
x
l
dx
x
l
a
l
x
l
0
0
2
2
0
2
2
0
r
r
(Q r r =
0
2
2
x
l
)
          =
ò
ò
x dx
x dx
l
l
3
0
2
0
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 171
A =
2
p
a
4
2
2
A = a
1
a
2
a a
a
O
b
a
O
X
Y
(       )
a
4
,
b
4
l
a
C m
1
m
2
l
2
b
l
1
m
1
m
2
C'
Fig. 1
Fig. 2
dx
x O
          = ×
é
ë
ê
ù
û
ú
3
4
3
4
0
x
x
l
          =
3
4
l
7. x
CM
=
´ + ´
+
1 10 2 12
1 2
        =
34
3
 m
     v
CM
=
´ - + ´ +
+
1 6 2 4
1 2
( ) ( )
  = +
-
2
3
1
ms
x¢
CM
 (new position of CM) 
        = +
æ
è
ç
ö
ø
÷
-
34
3
2
2
3
1
m s ms
        =
38
3
 m = 12.67 m
8. v
CM
=
´ + + ´ -
+
1 2 2 2 1
1 2
( ) ( )
= 0 ms
-2
\ Displacement of CM in 1 s = 0 m.
9. Acceleration ( ) a
®
 = -
-
10
2
j
^
ms
       v u a
® ® ®
= + t
         =
+ +
-
( )( ) ( )(
$ $
)
$
1 0 2 10 10
3
10
i j
j
                 = -
æ
è
ç
ö
ø
÷
20
3
10
3
$ $
i j m/s
New position vector ( ) r
1
®
 of particle A
   ( ) r s u a
1 0
2
1
2
® ® ® ®
= + + t t
    = + + + - ×
®
( ) ( )
^ ^ ^
10 20 0
1
2
10 1
2
i j j
      = + 10 15 i j
^ ^
m
New position vector ( ) r
2
®
 of particle B
r i j i j j
2
2
20 40 10 10 1
1
2
10 1
®
= + + + × + - × ( ) ( ) ( )
^ ^ ^ ^ ^
  = + 30 45 i j
^ ^
New position of CM
     R
r r ®
® ®
=
+
+
1 2
1 2
1 2
( ) ( )
       =
+ + + 10 15 60 90
3
i j i j
^ ^ ^ ^
       =
+ 70 105
3
i j
^ ^
 m
10.
(a)  3
0 0 1 12
0 1
=
× + × ´
+
m
m .
Þ m = 0.3 kg
(b) Momentum of system = Momentum of CM
           = + ´
-
( . )
^
m 01 6
1
kg ms j
           =
-
2.4 kgms j
^ 1
(c) v
v ®
® ®
=
+
+
CM
.3) 0.10)
0.3 .1
m
( ( 0 0
0
   v v
m
® ®
=
4
4
CM
       = ´
-
4
3
6
1
j
^
ms
       =
-
8
1
j
^
m s
11.   A t
m
· =0s
  B t
m
· =
2
100 ( ) ms
Position of 1st particle ( ) A at t = 300 ms
s
1
3 2
1
2
10 300 10 = ´ ´ ´
-
( )
     =0.45 m
Position of 2nd particle ( ) B at t = 300 ms
   (B is at the position of A at t = 100 ms)
    s
2
3 2
1
2
100 200 10 = ´ ´ ´
-
( )
      =0.20 m 
\ Position of CM =
´ + ´
+
2 0.2 0.45
2
m m
m m
= 28.3 cm
172 | Mechanics-1
1 kg
–1
6ms
+ ive
2 kg
10 m 10 m
–1
4ms
m
2 kg 1 kg
–1
2 ms
–1
1 ms
+ ive
12 m
O
CM
–1
6 j ms
 
m kg
3 m
0.10 kg
^
Velocity of 1st particle ( ) A  at t = 300 ms
   v
1
3
10 300 10 = ´ ´
-
             =3 ms
-1
Velocity of 2nd particle at t = 300 ms
   v
2
3 1
10 200 10 = ´ ´
- -
ms
           =
-
2
1
ms
\       v
m m
m m
CM
=
´ + ´
+
2 2 3
2
            =
7
3
 =
-
2.33ms
1
12. 24
0 80
=
× + ×
+
m m
m m
A B
A B
or 24 80 ( ) m
A
+ = ´ 0.6 0.6
or       m
A
=1.4 kg
\ Total mass of system = + 1.4 kg 0.6 kg
            =2.0kg
          v t
CM
6.0 =
2
j
^
\                a t
CM
ms =
-
12
2
j
^
 
Net force acting on system (at t = 3 s)
= Total mass of system ´ (a
COM
 at t = 3 s)
= ´
-
2.0 kg ms 36
2
j
^
= 72 N j
^
13.
            x
xdm
dm
CM
=
ò
ò
             =
ò
ò
x Axdx
Axdx
L
o
L
0
( )
  
             =
ò
ò
x dx
xdx
L
L
2
0
0
             =
L
L
3
2
3
2
/
/
 
=
2
3
L   
14. Let x= displacement of wedge (30 kg)
towards right.
\ Displacement of block A towards right
(along x-axis) when it arrives at the
bottom of the wedge
= - QR x
= - 0.5 x
Now, as net force on the system (wedge +
block) along x-axis is zero, the position of
CM of the system, along x-axis, will not
change
\ 5 30 ( ) 0.5 - = x x
i.e., x =
0.5
7
 m
= 71.4 mm
15. As no external force acts on the system,
the velocity of CM will be zero.
i.e.,
m v m v
m m
A A B B
A B
+
+
= 0
i.e.,     
v
v
m
m
A
B
B
A
= - = -
2
1
(a) \ Ratio of speeds = 2
(b) 
p
p
m v
m v
m
m
m
m
A
B
A A
B B
A
B
B
A
= = -
æ
è
ç
ö
ø
÷ = - 1
(c) 
K
K
p m
p m
p
p
m
m
A
B
A A
B B
A
B
B
A
= = ´ =
2
2
2
2
2
2
2
1
/
/
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 173
dx
x
O
L
30° B (50 kg)
A
x R
X
Q
5 kg
1
A
A B
A B
F
A
F
B