Chapter 9 : Centre of Mass, Linear Momentum, Collision - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

JEE : Chapter 9 : Centre of Mass, Linear Momentum, Collision - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


9.1
SOLUTIONS TO CONCEPTS 
CHAPTER 9
1. m
1 
= 1kg, m
2 
= 2kg, m
3
= 3kg,
x
1
= 0, x
2
= 1, x
3
=1/2
y
1
= 0, y
2
= 0, y
3
= 2 / 3
The position of centre of mass is
C.M = 
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ? ? ? ?
? ?
? ? ? ? ?
3 2 1
)) 2 / 3 ( 3 ( ) 0 2 ( ) 0 1 (
,
3 2 1
) 2 / 1 3 ( ) 1 2 ( ) 0 1 (
= 
?
?
?
?
?
?
?
?
12
3 3
,
12
7
from the point B. 
2. Let ? be the origin of the system
In the above figure
m
1
= 1gm, x
1
= – (0.96×10
–10
)sin 52° y
1
= 0
m
2
= 1gm, x
2
= – (0.96×10
–10
)sin 52° y
2
= 0
x
3
= 0 y
3
= (0.96 × 10
–10
) cos 52°
The position of centre of mass
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ?
? ? ? ? ? ? ?
? ?
18
y 16 0 0
,
16 1 1
0 16 52 sin ) 10 96 . 0 ( 52 sin ) 10 96 . 0 (
3
10 10
=  ? ? ? ?
o 10
52 cos 10 96 . 0 9 / 8 , 0
?
?
3. Let ‘O’ (0,0) be the origin of the system.
Each brick is mass ‘M’ & length ‘L’. 
Each brick is displaced w.r.t. one in contact by ‘L/10’
?The X coordinate of the centre of mass
m 7
2
L
m
10
L
2
L
m
10
L
10
L 3
2
L
m
10
L 3
2
L
m
10
L 2
2
L
m
10
L
2
L
m
2
L
m
X
cm
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
= 
7
2
L
10
L
2
L
5
L
2
L
10
L 3
2
L
5
L
2
L
10
L
2
L
2
L
? ? ? ? ? ? ? ? ? ? ?
= 
7
5
L 2
10
L 5
2
L 7
? ?
= 
7 10
L 4 L 5 L 35
?
? ?
= 
70
L 44
= L
35
11
4. Let the centre of the bigger disc be the origin.
2R = Radius of bigger disc
R = Radius of smaller disc
m
1
= ?R
2
× T × ?
m
2
= ?(2R)
2
I T × ?
where T = Thickness of the two discs
? = Density of the two discs
? The position of the centre of mass 
(0, 0)
C B
1m
(1, 1)
?
?
?
?
?
?
?
?
2
3
,
2
1
1m
A
1m
X
Y
0.96×10
–10
m
m 2 H
(0, 0)
m 1
Q
104°
52°
52°
H
m 3
0.96×10
–10
m
X O
L
L/10
m 1
(0, 0)
O
m 2
R
(R, 0)
Page 2


9.1
SOLUTIONS TO CONCEPTS 
CHAPTER 9
1. m
1 
= 1kg, m
2 
= 2kg, m
3
= 3kg,
x
1
= 0, x
2
= 1, x
3
=1/2
y
1
= 0, y
2
= 0, y
3
= 2 / 3
The position of centre of mass is
C.M = 
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ? ? ? ?
? ?
? ? ? ? ?
3 2 1
)) 2 / 3 ( 3 ( ) 0 2 ( ) 0 1 (
,
3 2 1
) 2 / 1 3 ( ) 1 2 ( ) 0 1 (
= 
?
?
?
?
?
?
?
?
12
3 3
,
12
7
from the point B. 
2. Let ? be the origin of the system
In the above figure
m
1
= 1gm, x
1
= – (0.96×10
–10
)sin 52° y
1
= 0
m
2
= 1gm, x
2
= – (0.96×10
–10
)sin 52° y
2
= 0
x
3
= 0 y
3
= (0.96 × 10
–10
) cos 52°
The position of centre of mass
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ?
? ? ? ? ? ? ?
? ?
18
y 16 0 0
,
16 1 1
0 16 52 sin ) 10 96 . 0 ( 52 sin ) 10 96 . 0 (
3
10 10
=  ? ? ? ?
o 10
52 cos 10 96 . 0 9 / 8 , 0
?
?
3. Let ‘O’ (0,0) be the origin of the system.
Each brick is mass ‘M’ & length ‘L’. 
Each brick is displaced w.r.t. one in contact by ‘L/10’
?The X coordinate of the centre of mass
m 7
2
L
m
10
L
2
L
m
10
L
10
L 3
2
L
m
10
L 3
2
L
m
10
L 2
2
L
m
10
L
2
L
m
2
L
m
X
cm
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
= 
7
2
L
10
L
2
L
5
L
2
L
10
L 3
2
L
5
L
2
L
10
L
2
L
2
L
? ? ? ? ? ? ? ? ? ? ?
= 
7
5
L 2
10
L 5
2
L 7
? ?
= 
7 10
L 4 L 5 L 35
?
? ?
= 
70
L 44
= L
35
11
4. Let the centre of the bigger disc be the origin.
2R = Radius of bigger disc
R = Radius of smaller disc
m
1
= ?R
2
× T × ?
m
2
= ?(2R)
2
I T × ?
where T = Thickness of the two discs
? = Density of the two discs
? The position of the centre of mass 
(0, 0)
C B
1m
(1, 1)
?
?
?
?
?
?
?
?
2
3
,
2
1
1m
A
1m
X
Y
0.96×10
–10
m
m 2 H
(0, 0)
m 1
Q
104°
52°
52°
H
m 3
0.96×10
–10
m
X O
L
L/10
m 1
(0, 0)
O
m 2
R
(R, 0)
Chapter 9
9.2
?
?
?
?
?
?
?
?
?
?
?
?
2 1
2 2 1 1
2 1
2 2 1 1
m m
y m y m
,
m m
x m x m
x
1
= R y
1
= 0
x
2
= 0 y
2
= 0
?
?
?
?
?
?
?
?
?
? ? ? ? ?
? ? ?
2 1
2 2
2
m m
0
,
T ) R 2 ( T R
0 R T R
???
?
?
?
?
?
?
?
?
? ?
? ?
0 ,
T R 5
R T R
2
2
??? ?
?
?
?
?
?
0 ,
5
R
?
At R/5 from the centre of bigger disc towards the centre of smaller disc.
5. Let ‘0’ be the origin of the system.
R = radius of the smaller disc
2R = radius of the bigger disc
The smaller disc is cut out from the bigger disc
As from the figure
m
1
= ?R
2
T ? x
1
= R y
1
= 0
m
2
= ?(2R)
2
T ? x
2
= 0 y
2
= 0
The position of C.M. = 
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
? ? ? ?
2 1
2 2
2
m m
0 0
,
R T ) R 2 ( T R
0 R T R
= 
?
?
?
?
?
?
?
?
? ?
? ? ?
0 ,
T R 3
R T R
2
2
= ?
?
?
?
?
?
? 0 ,
3
R
C.M. is at R/3 from the centre of bigger disc away from centre of the hole. ?
6. Let  m be the mass per unit area.
? Mass of the square plate = M
1
= d
2
m
Mass of the circular disc = M
2
= m
4
d
2
?
Let the centre of the circular disc be the origin of the system.
? Position of centre of mass
=
?
?
?
?
?
?
?
?
?
?
? ?
? ? ?
2 1
2 2
2 2
M M
0 0
,
m ) 4 / d ( m d
0 m ) 4 / d ( md d
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
0 ,
4
1 m d
m d
2
3
= ?
?
?
?
?
?
? ?
0 ,
4
d 4
The new centre of mass is ?
?
?
?
?
?
? ? 4
d 4
right of the centre of circular disc.
7. m
1
= 1kg.
1
v
?
= –1.5 cos 37 i
ˆ
– 1.55 sin 37 j
ˆ
= – 1.2 i
ˆ
– 0.9 j
ˆ
m
2
= 1.2kg.
2
v
?
= 0.4 j
ˆ
m
3
= 1.5kg
3
v
?
= – 0.8 i
ˆ
+ 0.6 j
ˆ
m
4
= 0.5kg
4
v
?
= 3 i
ˆ
m
5
= 1kg
5
v
?
= 1.6 i
ˆ
– 1.2 j
ˆ
So, 
c
v
?
= 
5 4 3 2 1
5 5 4 4 3 3 2 2 1 1
m m m m m
v m v m v m v m v m
? ? ? ?
? ? ? ?
? ? ? ? ?
= 
2 . 5
) j
ˆ
2 . 1 i
ˆ
6 . 1 ( 1 ) i
ˆ
3 ( 5 . 0 ) j
ˆ
6 . 0 i
ˆ
8 . 0 ( 5 . 1 ) j
ˆ
4 . 0 ( 2 . 1 ) j
ˆ
9 . 0 i
ˆ
2 . 1 ( 1 ? ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
2 . 1 i
ˆ
6 . 1 i
ˆ
5 . 1 j
ˆ
90 . i
ˆ
2 . 1 j
ˆ
8 . 4 j
ˆ
9 . 0 i
ˆ
2 . 1 ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
72 . 0
2 . 5
i
ˆ
7 . 0
?
m 1
(0, 0)
O
m 2
R
(R, 0)
(x 1, y 1)
M 1 d
M 1
d/2 d/2
d/2 O (0, 0)
(d, 0)
(x 2, y 2)
37°
1.5m/s
1kg 0.4m/s
1.2kg
37°
1m/s
1.5kg
37°
2m/s
1kg
05kg
3m/s
Page 3


9.1
SOLUTIONS TO CONCEPTS 
CHAPTER 9
1. m
1 
= 1kg, m
2 
= 2kg, m
3
= 3kg,
x
1
= 0, x
2
= 1, x
3
=1/2
y
1
= 0, y
2
= 0, y
3
= 2 / 3
The position of centre of mass is
C.M = 
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ? ? ? ?
? ?
? ? ? ? ?
3 2 1
)) 2 / 3 ( 3 ( ) 0 2 ( ) 0 1 (
,
3 2 1
) 2 / 1 3 ( ) 1 2 ( ) 0 1 (
= 
?
?
?
?
?
?
?
?
12
3 3
,
12
7
from the point B. 
2. Let ? be the origin of the system
In the above figure
m
1
= 1gm, x
1
= – (0.96×10
–10
)sin 52° y
1
= 0
m
2
= 1gm, x
2
= – (0.96×10
–10
)sin 52° y
2
= 0
x
3
= 0 y
3
= (0.96 × 10
–10
) cos 52°
The position of centre of mass
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ?
? ? ? ? ? ? ?
? ?
18
y 16 0 0
,
16 1 1
0 16 52 sin ) 10 96 . 0 ( 52 sin ) 10 96 . 0 (
3
10 10
=  ? ? ? ?
o 10
52 cos 10 96 . 0 9 / 8 , 0
?
?
3. Let ‘O’ (0,0) be the origin of the system.
Each brick is mass ‘M’ & length ‘L’. 
Each brick is displaced w.r.t. one in contact by ‘L/10’
?The X coordinate of the centre of mass
m 7
2
L
m
10
L
2
L
m
10
L
10
L 3
2
L
m
10
L 3
2
L
m
10
L 2
2
L
m
10
L
2
L
m
2
L
m
X
cm
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
= 
7
2
L
10
L
2
L
5
L
2
L
10
L 3
2
L
5
L
2
L
10
L
2
L
2
L
? ? ? ? ? ? ? ? ? ? ?
= 
7
5
L 2
10
L 5
2
L 7
? ?
= 
7 10
L 4 L 5 L 35
?
? ?
= 
70
L 44
= L
35
11
4. Let the centre of the bigger disc be the origin.
2R = Radius of bigger disc
R = Radius of smaller disc
m
1
= ?R
2
× T × ?
m
2
= ?(2R)
2
I T × ?
where T = Thickness of the two discs
? = Density of the two discs
? The position of the centre of mass 
(0, 0)
C B
1m
(1, 1)
?
?
?
?
?
?
?
?
2
3
,
2
1
1m
A
1m
X
Y
0.96×10
–10
m
m 2 H
(0, 0)
m 1
Q
104°
52°
52°
H
m 3
0.96×10
–10
m
X O
L
L/10
m 1
(0, 0)
O
m 2
R
(R, 0)
Chapter 9
9.2
?
?
?
?
?
?
?
?
?
?
?
?
2 1
2 2 1 1
2 1
2 2 1 1
m m
y m y m
,
m m
x m x m
x
1
= R y
1
= 0
x
2
= 0 y
2
= 0
?
?
?
?
?
?
?
?
?
? ? ? ? ?
? ? ?
2 1
2 2
2
m m
0
,
T ) R 2 ( T R
0 R T R
???
?
?
?
?
?
?
?
?
? ?
? ?
0 ,
T R 5
R T R
2
2
??? ?
?
?
?
?
?
0 ,
5
R
?
At R/5 from the centre of bigger disc towards the centre of smaller disc.
5. Let ‘0’ be the origin of the system.
R = radius of the smaller disc
2R = radius of the bigger disc
The smaller disc is cut out from the bigger disc
As from the figure
m
1
= ?R
2
T ? x
1
= R y
1
= 0
m
2
= ?(2R)
2
T ? x
2
= 0 y
2
= 0
The position of C.M. = 
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
? ? ? ?
2 1
2 2
2
m m
0 0
,
R T ) R 2 ( T R
0 R T R
= 
?
?
?
?
?
?
?
?
? ?
? ? ?
0 ,
T R 3
R T R
2
2
= ?
?
?
?
?
?
? 0 ,
3
R
C.M. is at R/3 from the centre of bigger disc away from centre of the hole. ?
6. Let  m be the mass per unit area.
? Mass of the square plate = M
1
= d
2
m
Mass of the circular disc = M
2
= m
4
d
2
?
Let the centre of the circular disc be the origin of the system.
? Position of centre of mass
=
?
?
?
?
?
?
?
?
?
?
? ?
? ? ?
2 1
2 2
2 2
M M
0 0
,
m ) 4 / d ( m d
0 m ) 4 / d ( md d
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
0 ,
4
1 m d
m d
2
3
= ?
?
?
?
?
?
? ?
0 ,
4
d 4
The new centre of mass is ?
?
?
?
?
?
? ? 4
d 4
right of the centre of circular disc.
7. m
1
= 1kg.
1
v
?
= –1.5 cos 37 i
ˆ
– 1.55 sin 37 j
ˆ
= – 1.2 i
ˆ
– 0.9 j
ˆ
m
2
= 1.2kg.
2
v
?
= 0.4 j
ˆ
m
3
= 1.5kg
3
v
?
= – 0.8 i
ˆ
+ 0.6 j
ˆ
m
4
= 0.5kg
4
v
?
= 3 i
ˆ
m
5
= 1kg
5
v
?
= 1.6 i
ˆ
– 1.2 j
ˆ
So, 
c
v
?
= 
5 4 3 2 1
5 5 4 4 3 3 2 2 1 1
m m m m m
v m v m v m v m v m
? ? ? ?
? ? ? ?
? ? ? ? ?
= 
2 . 5
) j
ˆ
2 . 1 i
ˆ
6 . 1 ( 1 ) i
ˆ
3 ( 5 . 0 ) j
ˆ
6 . 0 i
ˆ
8 . 0 ( 5 . 1 ) j
ˆ
4 . 0 ( 2 . 1 ) j
ˆ
9 . 0 i
ˆ
2 . 1 ( 1 ? ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
2 . 1 i
ˆ
6 . 1 i
ˆ
5 . 1 j
ˆ
90 . i
ˆ
2 . 1 j
ˆ
8 . 4 j
ˆ
9 . 0 i
ˆ
2 . 1 ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
72 . 0
2 . 5
i
ˆ
7 . 0
?
m 1
(0, 0)
O
m 2
R
(R, 0)
(x 1, y 1)
M 1 d
M 1
d/2 d/2
d/2 O (0, 0)
(d, 0)
(x 2, y 2)
37°
1.5m/s
1kg 0.4m/s
1.2kg
37°
1m/s
1.5kg
37°
2m/s
1kg
05kg
3m/s
Chapter 9
9.3
8. Two masses m
1
& m
2
are placed on the X-axis
m
1
= 10 kg, m
2
= 20kg.
The first mass is displaced by a distance of 2 cm
?
2 1
2 2 1 1
cm
m m
x m x m
X
?
?
? = 
30
x 20 2 10
2
? ?
? 0 = 
30
x 20 20
2
?
? 20 + 20x
2
= 0 
? 20 = – 20x
2
? x
2
= –1.
? The 2
nd
mass should be displaced by a distance 1cm towards left so as to kept the position of centre 
of mass unchanged.
9. Two masses m
1
& m
2
are kept in a vertical line 
m
1
= 10kg, m
2
= 30kg
The first block is raised through a height of 7 cm.
The centre of mass is raised  by 1 cm.
? 1 = 
2 1
2 2 1 1
m m
y m y m
?
?
= 
40
y 30 7 10
2
? ?
? 1 = 
40
y 30 70
2
?
? 70 +30y
2
= 40 ? 30y
2
= – 30 ? y
2
= –1.
The 30 kg body should be displaced 1cm downward inorder to raise the centre of mass through 1 cm.
10. As the hall is gravity free, after the ice melts, it would tend to acquire a 
spherical shape. But, there is no external force acting on the system. So, the 
centre of mass of the system would not move.
11. The centre of mass of the blate will be on the symmetrical axis.
?
2
R
2
R
3
R 4
2
R
3
R 4
2
R
y
2
1
2
2
1
2
1 2
2
2
cm
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
3
2
1
2
2
1
3
2
) R R ( 2 /
R ) 3 / 2 ( R ) 3 / 2 (
? ?
?
= 
) R R )( R R (
) R R R R )( R R (
3
4
1 2 1 2
2 1
2
1
2
2 1 2
? ?
? ? ?
?
= 
2 1
2 1
2
1
2
2
R R
) R R R R (
3
4
?
? ?
?
above the centre.
12. m
1
= 60kg, m
2
= 40kg , m
3
= 50kg,
Let A be the origin of the system.
Initially Mr. Verma & Mr. Mathur are at extreme position of the boat. 
? The centre of mass will be at a distance 
= 
150
4 50 2 40 0 60 ? ? ? ? ?
= 
150
280
= 1.87m from ‘A’
When they come to the mid point of the boat the CM lies at 2m from ‘A’. 
? The shift in CM = 2 – 1.87 = 0.13m towards right.
But as there is no external force in longitudinal direction their CM would not shift. 
So, the boat moves 0.13m or 13 cm towards right.
13. Let the bob fall at A,. The mass of bob = m.
The mass of cart = M.
Initially their centre of mass will be at 
m M
0 M L m
?
? ? ?
= L
m M
m
?
?
?
?
?
?
?
Distance from P
When, the bob falls in the slot the CM is at a distance ‘O’ from P.
m M
L
R 1
R 2
60kg
A
20kg 40kg
B
M
m
A
P
Page 4


9.1
SOLUTIONS TO CONCEPTS 
CHAPTER 9
1. m
1 
= 1kg, m
2 
= 2kg, m
3
= 3kg,
x
1
= 0, x
2
= 1, x
3
=1/2
y
1
= 0, y
2
= 0, y
3
= 2 / 3
The position of centre of mass is
C.M = 
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ? ? ? ?
? ?
? ? ? ? ?
3 2 1
)) 2 / 3 ( 3 ( ) 0 2 ( ) 0 1 (
,
3 2 1
) 2 / 1 3 ( ) 1 2 ( ) 0 1 (
= 
?
?
?
?
?
?
?
?
12
3 3
,
12
7
from the point B. 
2. Let ? be the origin of the system
In the above figure
m
1
= 1gm, x
1
= – (0.96×10
–10
)sin 52° y
1
= 0
m
2
= 1gm, x
2
= – (0.96×10
–10
)sin 52° y
2
= 0
x
3
= 0 y
3
= (0.96 × 10
–10
) cos 52°
The position of centre of mass
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ?
? ? ? ? ? ? ?
? ?
18
y 16 0 0
,
16 1 1
0 16 52 sin ) 10 96 . 0 ( 52 sin ) 10 96 . 0 (
3
10 10
=  ? ? ? ?
o 10
52 cos 10 96 . 0 9 / 8 , 0
?
?
3. Let ‘O’ (0,0) be the origin of the system.
Each brick is mass ‘M’ & length ‘L’. 
Each brick is displaced w.r.t. one in contact by ‘L/10’
?The X coordinate of the centre of mass
m 7
2
L
m
10
L
2
L
m
10
L
10
L 3
2
L
m
10
L 3
2
L
m
10
L 2
2
L
m
10
L
2
L
m
2
L
m
X
cm
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
= 
7
2
L
10
L
2
L
5
L
2
L
10
L 3
2
L
5
L
2
L
10
L
2
L
2
L
? ? ? ? ? ? ? ? ? ? ?
= 
7
5
L 2
10
L 5
2
L 7
? ?
= 
7 10
L 4 L 5 L 35
?
? ?
= 
70
L 44
= L
35
11
4. Let the centre of the bigger disc be the origin.
2R = Radius of bigger disc
R = Radius of smaller disc
m
1
= ?R
2
× T × ?
m
2
= ?(2R)
2
I T × ?
where T = Thickness of the two discs
? = Density of the two discs
? The position of the centre of mass 
(0, 0)
C B
1m
(1, 1)
?
?
?
?
?
?
?
?
2
3
,
2
1
1m
A
1m
X
Y
0.96×10
–10
m
m 2 H
(0, 0)
m 1
Q
104°
52°
52°
H
m 3
0.96×10
–10
m
X O
L
L/10
m 1
(0, 0)
O
m 2
R
(R, 0)
Chapter 9
9.2
?
?
?
?
?
?
?
?
?
?
?
?
2 1
2 2 1 1
2 1
2 2 1 1
m m
y m y m
,
m m
x m x m
x
1
= R y
1
= 0
x
2
= 0 y
2
= 0
?
?
?
?
?
?
?
?
?
? ? ? ? ?
? ? ?
2 1
2 2
2
m m
0
,
T ) R 2 ( T R
0 R T R
???
?
?
?
?
?
?
?
?
? ?
? ?
0 ,
T R 5
R T R
2
2
??? ?
?
?
?
?
?
0 ,
5
R
?
At R/5 from the centre of bigger disc towards the centre of smaller disc.
5. Let ‘0’ be the origin of the system.
R = radius of the smaller disc
2R = radius of the bigger disc
The smaller disc is cut out from the bigger disc
As from the figure
m
1
= ?R
2
T ? x
1
= R y
1
= 0
m
2
= ?(2R)
2
T ? x
2
= 0 y
2
= 0
The position of C.M. = 
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
? ? ? ?
2 1
2 2
2
m m
0 0
,
R T ) R 2 ( T R
0 R T R
= 
?
?
?
?
?
?
?
?
? ?
? ? ?
0 ,
T R 3
R T R
2
2
= ?
?
?
?
?
?
? 0 ,
3
R
C.M. is at R/3 from the centre of bigger disc away from centre of the hole. ?
6. Let  m be the mass per unit area.
? Mass of the square plate = M
1
= d
2
m
Mass of the circular disc = M
2
= m
4
d
2
?
Let the centre of the circular disc be the origin of the system.
? Position of centre of mass
=
?
?
?
?
?
?
?
?
?
?
? ?
? ? ?
2 1
2 2
2 2
M M
0 0
,
m ) 4 / d ( m d
0 m ) 4 / d ( md d
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
0 ,
4
1 m d
m d
2
3
= ?
?
?
?
?
?
? ?
0 ,
4
d 4
The new centre of mass is ?
?
?
?
?
?
? ? 4
d 4
right of the centre of circular disc.
7. m
1
= 1kg.
1
v
?
= –1.5 cos 37 i
ˆ
– 1.55 sin 37 j
ˆ
= – 1.2 i
ˆ
– 0.9 j
ˆ
m
2
= 1.2kg.
2
v
?
= 0.4 j
ˆ
m
3
= 1.5kg
3
v
?
= – 0.8 i
ˆ
+ 0.6 j
ˆ
m
4
= 0.5kg
4
v
?
= 3 i
ˆ
m
5
= 1kg
5
v
?
= 1.6 i
ˆ
– 1.2 j
ˆ
So, 
c
v
?
= 
5 4 3 2 1
5 5 4 4 3 3 2 2 1 1
m m m m m
v m v m v m v m v m
? ? ? ?
? ? ? ?
? ? ? ? ?
= 
2 . 5
) j
ˆ
2 . 1 i
ˆ
6 . 1 ( 1 ) i
ˆ
3 ( 5 . 0 ) j
ˆ
6 . 0 i
ˆ
8 . 0 ( 5 . 1 ) j
ˆ
4 . 0 ( 2 . 1 ) j
ˆ
9 . 0 i
ˆ
2 . 1 ( 1 ? ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
2 . 1 i
ˆ
6 . 1 i
ˆ
5 . 1 j
ˆ
90 . i
ˆ
2 . 1 j
ˆ
8 . 4 j
ˆ
9 . 0 i
ˆ
2 . 1 ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
72 . 0
2 . 5
i
ˆ
7 . 0
?
m 1
(0, 0)
O
m 2
R
(R, 0)
(x 1, y 1)
M 1 d
M 1
d/2 d/2
d/2 O (0, 0)
(d, 0)
(x 2, y 2)
37°
1.5m/s
1kg 0.4m/s
1.2kg
37°
1m/s
1.5kg
37°
2m/s
1kg
05kg
3m/s
Chapter 9
9.3
8. Two masses m
1
& m
2
are placed on the X-axis
m
1
= 10 kg, m
2
= 20kg.
The first mass is displaced by a distance of 2 cm
?
2 1
2 2 1 1
cm
m m
x m x m
X
?
?
? = 
30
x 20 2 10
2
? ?
? 0 = 
30
x 20 20
2
?
? 20 + 20x
2
= 0 
? 20 = – 20x
2
? x
2
= –1.
? The 2
nd
mass should be displaced by a distance 1cm towards left so as to kept the position of centre 
of mass unchanged.
9. Two masses m
1
& m
2
are kept in a vertical line 
m
1
= 10kg, m
2
= 30kg
The first block is raised through a height of 7 cm.
The centre of mass is raised  by 1 cm.
? 1 = 
2 1
2 2 1 1
m m
y m y m
?
?
= 
40
y 30 7 10
2
? ?
? 1 = 
40
y 30 70
2
?
? 70 +30y
2
= 40 ? 30y
2
= – 30 ? y
2
= –1.
The 30 kg body should be displaced 1cm downward inorder to raise the centre of mass through 1 cm.
10. As the hall is gravity free, after the ice melts, it would tend to acquire a 
spherical shape. But, there is no external force acting on the system. So, the 
centre of mass of the system would not move.
11. The centre of mass of the blate will be on the symmetrical axis.
?
2
R
2
R
3
R 4
2
R
3
R 4
2
R
y
2
1
2
2
1
2
1 2
2
2
cm
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
3
2
1
2
2
1
3
2
) R R ( 2 /
R ) 3 / 2 ( R ) 3 / 2 (
? ?
?
= 
) R R )( R R (
) R R R R )( R R (
3
4
1 2 1 2
2 1
2
1
2
2 1 2
? ?
? ? ?
?
= 
2 1
2 1
2
1
2
2
R R
) R R R R (
3
4
?
? ?
?
above the centre.
12. m
1
= 60kg, m
2
= 40kg , m
3
= 50kg,
Let A be the origin of the system.
Initially Mr. Verma & Mr. Mathur are at extreme position of the boat. 
? The centre of mass will be at a distance 
= 
150
4 50 2 40 0 60 ? ? ? ? ?
= 
150
280
= 1.87m from ‘A’
When they come to the mid point of the boat the CM lies at 2m from ‘A’. 
? The shift in CM = 2 – 1.87 = 0.13m towards right.
But as there is no external force in longitudinal direction their CM would not shift. 
So, the boat moves 0.13m or 13 cm towards right.
13. Let the bob fall at A,. The mass of bob = m.
The mass of cart = M.
Initially their centre of mass will be at 
m M
0 M L m
?
? ? ?
= L
m M
m
?
?
?
?
?
?
?
Distance from P
When, the bob falls in the slot the CM is at a distance ‘O’ from P.
m M
L
R 1
R 2
60kg
A
20kg 40kg
B
M
m
A
P
Chapter 9
9.4
Shift in CM = 0 –
m M
mL
?
=  –
m M
mL
?
towards left 
= 
m M
mL
?
towards right.
But there is no external force in horizontal direction.
So the cart displaces a distance 
m M
mL
?
towards right.
14. Initially the monkey & balloon are at rest.
So the CM is at ‘P’
When the monkey descends through a distance ‘L’
The CM will shift
t
o
= 
m M
0 M L m
?
? ? ?
= 
m M
mL
?
  from P
So, the balloon descends through a distance 
m M
mL
?
15. Let the mass of the to particles be m
1
& m
2
respectively
m
1
= 1kg, m
2
= 4kg
?According to question
½ m
1
v
1
2
= ½ m
2
v
2
2
?
2
1
2
2
2
1
v
v
m
m
? ?
2
1
1
2
m
m
v
v
? ?
1
2
2
1
m
m
v
v
?
Now, 
2
1
1
2
2
1
2 2
1 1
m
m
m
m
m
m
v m
v m
? ? ? = 
4
1
= 1/2
?
2 2
1 1
v m
v m
= 1 : 2
16. As uranium 238 nucleus emits a ?-particle with a speed of 1.4 × 10
7
m/sec. Let v
2
be the speed of the 
residual nucleus thorium 234.
? m
1
v
1
= m
2
v
2
? 4 × 1.4 × 10
7
= 234 × v
2
? v
2
= 
234
10 4 . 1 4
7
? ?
= 2.4 × 10
5
m/sec.
17. m
1
v
1
= m
2
v
2
? 50 × 1.8 = 6 × 10
24
× v
2
? v
2
= 
24
10 6
8 . 1 50
?
?
= 1.5 × 10
–23
m/sec
so, the earth will recoil at  a speed of  1.5 × 10
–23
m/sec.
18. Mass of proton = 1.67 × 10
–27
Let ‘V
p
’ be the velocity of proton
Given momentum of electron = 1.4 × 10
–26
kg m/sec
Given momentum of antineutrino = 6.4 × 10
–27
kg m/sec
a) The electron & the antineutrino are ejected in the same direction. As the total momentum is 
conserved the proton should be ejected in the opposite direction.
1.67 × 10
–27
× V
p
= 1.4 × 10
–26
+ 6.4 × 10
–27
= 20.4 × 10
–27
? V
p
= (20.4 /1.67) = 12.2 m/sec in the opposite direction.
b) The electron & antineutrino are ejected ?
r
to each other.
Total momentum of electron and antineutrino,
=
27 2 2
10 ) 4 . 6 ( ) 14 (
?
? ? kg m/s = 15.4 × 10
–27
kg m/s
Since, 1.67 × 10
–27
V
p
= 15.4 × 10
–27
kg m/s
So V
p
= 9.2 m/s
p
e a
p
a
e
L
M
mg
Page 5


9.1
SOLUTIONS TO CONCEPTS 
CHAPTER 9
1. m
1 
= 1kg, m
2 
= 2kg, m
3
= 3kg,
x
1
= 0, x
2
= 1, x
3
=1/2
y
1
= 0, y
2
= 0, y
3
= 2 / 3
The position of centre of mass is
C.M = 
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ? ? ? ?
? ?
? ? ? ? ?
3 2 1
)) 2 / 3 ( 3 ( ) 0 2 ( ) 0 1 (
,
3 2 1
) 2 / 1 3 ( ) 1 2 ( ) 0 1 (
= 
?
?
?
?
?
?
?
?
12
3 3
,
12
7
from the point B. 
2. Let ? be the origin of the system
In the above figure
m
1
= 1gm, x
1
= – (0.96×10
–10
)sin 52° y
1
= 0
m
2
= 1gm, x
2
= – (0.96×10
–10
)sin 52° y
2
= 0
x
3
= 0 y
3
= (0.96 × 10
–10
) cos 52°
The position of centre of mass
?
?
?
?
?
?
?
?
? ?
? ?
? ?
? ?
3 2 1
3 3 2 2 1 1
3 2 1
3 3 2 2 1 1
m m m
y m y m y m
,
m m m
x m x m x m
= 
?
?
?
?
?
?
?
?
? ?
? ?
? ? ? ? ? ? ?
? ?
18
y 16 0 0
,
16 1 1
0 16 52 sin ) 10 96 . 0 ( 52 sin ) 10 96 . 0 (
3
10 10
=  ? ? ? ?
o 10
52 cos 10 96 . 0 9 / 8 , 0
?
?
3. Let ‘O’ (0,0) be the origin of the system.
Each brick is mass ‘M’ & length ‘L’. 
Each brick is displaced w.r.t. one in contact by ‘L/10’
?The X coordinate of the centre of mass
m 7
2
L
m
10
L
2
L
m
10
L
10
L 3
2
L
m
10
L 3
2
L
m
10
L 2
2
L
m
10
L
2
L
m
2
L
m
X
cm
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
= 
7
2
L
10
L
2
L
5
L
2
L
10
L 3
2
L
5
L
2
L
10
L
2
L
2
L
? ? ? ? ? ? ? ? ? ? ?
= 
7
5
L 2
10
L 5
2
L 7
? ?
= 
7 10
L 4 L 5 L 35
?
? ?
= 
70
L 44
= L
35
11
4. Let the centre of the bigger disc be the origin.
2R = Radius of bigger disc
R = Radius of smaller disc
m
1
= ?R
2
× T × ?
m
2
= ?(2R)
2
I T × ?
where T = Thickness of the two discs
? = Density of the two discs
? The position of the centre of mass 
(0, 0)
C B
1m
(1, 1)
?
?
?
?
?
?
?
?
2
3
,
2
1
1m
A
1m
X
Y
0.96×10
–10
m
m 2 H
(0, 0)
m 1
Q
104°
52°
52°
H
m 3
0.96×10
–10
m
X O
L
L/10
m 1
(0, 0)
O
m 2
R
(R, 0)
Chapter 9
9.2
?
?
?
?
?
?
?
?
?
?
?
?
2 1
2 2 1 1
2 1
2 2 1 1
m m
y m y m
,
m m
x m x m
x
1
= R y
1
= 0
x
2
= 0 y
2
= 0
?
?
?
?
?
?
?
?
?
? ? ? ? ?
? ? ?
2 1
2 2
2
m m
0
,
T ) R 2 ( T R
0 R T R
???
?
?
?
?
?
?
?
?
? ?
? ?
0 ,
T R 5
R T R
2
2
??? ?
?
?
?
?
?
0 ,
5
R
?
At R/5 from the centre of bigger disc towards the centre of smaller disc.
5. Let ‘0’ be the origin of the system.
R = radius of the smaller disc
2R = radius of the bigger disc
The smaller disc is cut out from the bigger disc
As from the figure
m
1
= ?R
2
T ? x
1
= R y
1
= 0
m
2
= ?(2R)
2
T ? x
2
= 0 y
2
= 0
The position of C.M. = 
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
? ? ? ?
2 1
2 2
2
m m
0 0
,
R T ) R 2 ( T R
0 R T R
= 
?
?
?
?
?
?
?
?
? ?
? ? ?
0 ,
T R 3
R T R
2
2
= ?
?
?
?
?
?
? 0 ,
3
R
C.M. is at R/3 from the centre of bigger disc away from centre of the hole. ?
6. Let  m be the mass per unit area.
? Mass of the square plate = M
1
= d
2
m
Mass of the circular disc = M
2
= m
4
d
2
?
Let the centre of the circular disc be the origin of the system.
? Position of centre of mass
=
?
?
?
?
?
?
?
?
?
?
? ?
? ? ?
2 1
2 2
2 2
M M
0 0
,
m ) 4 / d ( m d
0 m ) 4 / d ( md d
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
0 ,
4
1 m d
m d
2
3
= ?
?
?
?
?
?
? ?
0 ,
4
d 4
The new centre of mass is ?
?
?
?
?
?
? ? 4
d 4
right of the centre of circular disc.
7. m
1
= 1kg.
1
v
?
= –1.5 cos 37 i
ˆ
– 1.55 sin 37 j
ˆ
= – 1.2 i
ˆ
– 0.9 j
ˆ
m
2
= 1.2kg.
2
v
?
= 0.4 j
ˆ
m
3
= 1.5kg
3
v
?
= – 0.8 i
ˆ
+ 0.6 j
ˆ
m
4
= 0.5kg
4
v
?
= 3 i
ˆ
m
5
= 1kg
5
v
?
= 1.6 i
ˆ
– 1.2 j
ˆ
So, 
c
v
?
= 
5 4 3 2 1
5 5 4 4 3 3 2 2 1 1
m m m m m
v m v m v m v m v m
? ? ? ?
? ? ? ?
? ? ? ? ?
= 
2 . 5
) j
ˆ
2 . 1 i
ˆ
6 . 1 ( 1 ) i
ˆ
3 ( 5 . 0 ) j
ˆ
6 . 0 i
ˆ
8 . 0 ( 5 . 1 ) j
ˆ
4 . 0 ( 2 . 1 ) j
ˆ
9 . 0 i
ˆ
2 . 1 ( 1 ? ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
2 . 1 i
ˆ
6 . 1 i
ˆ
5 . 1 j
ˆ
90 . i
ˆ
2 . 1 j
ˆ
8 . 4 j
ˆ
9 . 0 i
ˆ
2 . 1 ? ? ? ? ? ? ? ?
= 
2 . 5
j
ˆ
72 . 0
2 . 5
i
ˆ
7 . 0
?
m 1
(0, 0)
O
m 2
R
(R, 0)
(x 1, y 1)
M 1 d
M 1
d/2 d/2
d/2 O (0, 0)
(d, 0)
(x 2, y 2)
37°
1.5m/s
1kg 0.4m/s
1.2kg
37°
1m/s
1.5kg
37°
2m/s
1kg
05kg
3m/s
Chapter 9
9.3
8. Two masses m
1
& m
2
are placed on the X-axis
m
1
= 10 kg, m
2
= 20kg.
The first mass is displaced by a distance of 2 cm
?
2 1
2 2 1 1
cm
m m
x m x m
X
?
?
? = 
30
x 20 2 10
2
? ?
? 0 = 
30
x 20 20
2
?
? 20 + 20x
2
= 0 
? 20 = – 20x
2
? x
2
= –1.
? The 2
nd
mass should be displaced by a distance 1cm towards left so as to kept the position of centre 
of mass unchanged.
9. Two masses m
1
& m
2
are kept in a vertical line 
m
1
= 10kg, m
2
= 30kg
The first block is raised through a height of 7 cm.
The centre of mass is raised  by 1 cm.
? 1 = 
2 1
2 2 1 1
m m
y m y m
?
?
= 
40
y 30 7 10
2
? ?
? 1 = 
40
y 30 70
2
?
? 70 +30y
2
= 40 ? 30y
2
= – 30 ? y
2
= –1.
The 30 kg body should be displaced 1cm downward inorder to raise the centre of mass through 1 cm.
10. As the hall is gravity free, after the ice melts, it would tend to acquire a 
spherical shape. But, there is no external force acting on the system. So, the 
centre of mass of the system would not move.
11. The centre of mass of the blate will be on the symmetrical axis.
?
2
R
2
R
3
R 4
2
R
3
R 4
2
R
y
2
1
2
2
1
2
1 2
2
2
cm
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
3
2
1
2
2
1
3
2
) R R ( 2 /
R ) 3 / 2 ( R ) 3 / 2 (
? ?
?
= 
) R R )( R R (
) R R R R )( R R (
3
4
1 2 1 2
2 1
2
1
2
2 1 2
? ?
? ? ?
?
= 
2 1
2 1
2
1
2
2
R R
) R R R R (
3
4
?
? ?
?
above the centre.
12. m
1
= 60kg, m
2
= 40kg , m
3
= 50kg,
Let A be the origin of the system.
Initially Mr. Verma & Mr. Mathur are at extreme position of the boat. 
? The centre of mass will be at a distance 
= 
150
4 50 2 40 0 60 ? ? ? ? ?
= 
150
280
= 1.87m from ‘A’
When they come to the mid point of the boat the CM lies at 2m from ‘A’. 
? The shift in CM = 2 – 1.87 = 0.13m towards right.
But as there is no external force in longitudinal direction their CM would not shift. 
So, the boat moves 0.13m or 13 cm towards right.
13. Let the bob fall at A,. The mass of bob = m.
The mass of cart = M.
Initially their centre of mass will be at 
m M
0 M L m
?
? ? ?
= L
m M
m
?
?
?
?
?
?
?
Distance from P
When, the bob falls in the slot the CM is at a distance ‘O’ from P.
m M
L
R 1
R 2
60kg
A
20kg 40kg
B
M
m
A
P
Chapter 9
9.4
Shift in CM = 0 –
m M
mL
?
=  –
m M
mL
?
towards left 
= 
m M
mL
?
towards right.
But there is no external force in horizontal direction.
So the cart displaces a distance 
m M
mL
?
towards right.
14. Initially the monkey & balloon are at rest.
So the CM is at ‘P’
When the monkey descends through a distance ‘L’
The CM will shift
t
o
= 
m M
0 M L m
?
? ? ?
= 
m M
mL
?
  from P
So, the balloon descends through a distance 
m M
mL
?
15. Let the mass of the to particles be m
1
& m
2
respectively
m
1
= 1kg, m
2
= 4kg
?According to question
½ m
1
v
1
2
= ½ m
2
v
2
2
?
2
1
2
2
2
1
v
v
m
m
? ?
2
1
1
2
m
m
v
v
? ?
1
2
2
1
m
m
v
v
?
Now, 
2
1
1
2
2
1
2 2
1 1
m
m
m
m
m
m
v m
v m
? ? ? = 
4
1
= 1/2
?
2 2
1 1
v m
v m
= 1 : 2
16. As uranium 238 nucleus emits a ?-particle with a speed of 1.4 × 10
7
m/sec. Let v
2
be the speed of the 
residual nucleus thorium 234.
? m
1
v
1
= m
2
v
2
? 4 × 1.4 × 10
7
= 234 × v
2
? v
2
= 
234
10 4 . 1 4
7
? ?
= 2.4 × 10
5
m/sec.
17. m
1
v
1
= m
2
v
2
? 50 × 1.8 = 6 × 10
24
× v
2
? v
2
= 
24
10 6
8 . 1 50
?
?
= 1.5 × 10
–23
m/sec
so, the earth will recoil at  a speed of  1.5 × 10
–23
m/sec.
18. Mass of proton = 1.67 × 10
–27
Let ‘V
p
’ be the velocity of proton
Given momentum of electron = 1.4 × 10
–26
kg m/sec
Given momentum of antineutrino = 6.4 × 10
–27
kg m/sec
a) The electron & the antineutrino are ejected in the same direction. As the total momentum is 
conserved the proton should be ejected in the opposite direction.
1.67 × 10
–27
× V
p
= 1.4 × 10
–26
+ 6.4 × 10
–27
= 20.4 × 10
–27
? V
p
= (20.4 /1.67) = 12.2 m/sec in the opposite direction.
b) The electron & antineutrino are ejected ?
r
to each other.
Total momentum of electron and antineutrino,
=
27 2 2
10 ) 4 . 6 ( ) 14 (
?
? ? kg m/s = 15.4 × 10
–27
kg m/s
Since, 1.67 × 10
–27
V
p
= 15.4 × 10
–27
kg m/s
So V
p
= 9.2 m/s
p
e a
p
a
e
L
M
mg
Chapter 9
9.5
19. Mass of man = M, Initial velocity = 0
Mass of bad = m
Let the throws the bag towards left with a velocity v towards left. So, 
there is no external force in the horizontal direction.
The momentum will be conserved. Let he goes right with a velocity 
mv = MV ? V = 
M
mv
? v = 
m
MV
..(i)
Let the total time he will take to reach ground = g / H 2 = t
1
Let the total time he will take to reach the height h = t
2
= g / ) h H ( 2 ?
Then the time of his flying = t
1
– t
2
=  g / H 2 –  g / ) h H ( 2 ? = ? ? h H H g / 2 ? ?
Within this time he reaches the ground in the pond covering a horizontal distance x 
? x  = V × t ? V = x / t
? v = 
t
x
m
M
= 
? ? h H H 2
g
m
M
? ?
?
As there is no external force in horizontal direction, the x-coordinate of CM will remain at that position.
? 0 = 
m M
x m ) x ( M
1
?
? ? ?
? x
1
= x
m
M
?
? The bag will reach the bottom at a distance (M/m) x towards left of the line it falls.
20. Mass = 50g = 0.05kg
v = 2 cos 45° i
ˆ
– 2 sin 45° j
ˆ
v
1
= – 2 cos 45° i
ˆ
– 2 sin 45° j
ˆ
a) change in momentum = m v
?
– m
1
v
?
= 0.05 (2 cos 45° i
ˆ
– 2 sin 45° j
ˆ
) – 0.05 (– 2 cos 45° i
ˆ
– 2 sin 45° j
ˆ
)
= 0.1 cos 45° i
ˆ
– 0.1 sin 45° j
ˆ
+0.1 cos 45° i
ˆ
+ 0.1 sin 45° j
ˆ
= 0.2 cos 45° i
ˆ
? magnitude = 
2
2
2 . 0
?
?
?
?
?
?
?
?
= 
2
2 . 0
= 0.14 kg m/s
c) The change in magnitude of the momentum of the ball 
–
i
P
?
–
f
P
?
= 2 × 0.5 – 2 × 0.5 = 0.
21.
incidence
P
?
= (h/ ?) cos ? i
ˆ
– (h/ ?) sin ? j
ˆ
P
Reflected
= – (h/ ?) cos ? i
ˆ
– (h/ ?) sin ? j
ˆ
The change in momentum will be only in the x-axis direction. i.e.
P ? = (h/ ?) cos ? – ((h/ ?) cos ?) = (2h/ ?) cos ? ?
22. As the block is exploded only due to its internal energy. So net 
external force during this process is 0. So the centre mass will not change.
Let the body while exploded was at the origin of the co-ordinate system.
If the two bodies of equal mass is moving at a speed of 10m/s in + x & +y axis 
direction respectively,
o 2 2
90 cos 10 . 210 10 10 ? ? = 10 2 m/s 45° w.r.t. + x axis
If the centre mass is at rest, then the third mass which have equal mass with other 
two, will move in the opposite direction (i.e. 135° w.r.t. + x- axis) of the resultant at the same velocity.
23. Since the spaceship is removed from any material object & totally isolated from surrounding, the 
missions by astronauts couldn’t slip away from the spaceship. So the total mass of the spaceship 
remain unchanged and also its velocity.  
v
45°
45°
P 1 – h/ ? ?cos ? ?
? ?
x
y 1
x 1
P R – h/ ??cos ??
P R – h/ ? ??
P 1 – h/ ? ?sin ??= P R
x
y
Hard ground
h
h
pound
Read More
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