Class 11-science H C VERMA Solutions Physics Chapter 9 - Centre of Mass, Linear Momentum, Collision
Centre of Mass, Linear Momentum, Collision Exercise 159
Solution 1
Take A as origin (0,0)
then C= (1,0)
if we take AC as
x-axis.
Similarly, if we take AB as x-axis,
Solution 2
O
Ao
Take 
as origin.
then 
=0
Distance from O-atom= 
=
m
m
Solution 3
take O as (0,0)
brick A and E is from 
brick B and D is from 
brick C is from 
,F are from (0 to L)
For and F
bricks, C.O.M=
For B and D,
C.O.M=
(as they are displaced by 
)
Similarly,
For A and E
C.O.M=
For C, C.O.M=
of all bricks
Solution 4
C.O.M of 2R disc =(2R,0)
C.O.M of R disc=(R,0)
C.O.M of system=
}
{
}
C.O.M of system=
=
Distance of C.O.M (system) from C.O.M of 2R disc
=
Centre of Mass, Linear Momentum, Collision Exercise 160
Solution 5
C.O.M of 2R disc =(2R,0)
C.O.M of R disc=(R,0)
C.O.M of shaded area=
=
From C.O.M of 2R disc, it is 
distance away.
Solution 6
Area density of square=
(Mass/unit area)
Mass/unit area of circle=
C.O.M of square = (
)
C.O.M of circle = (
)
C.O.M of the whole system= 
=
=
From C.O.M of disc it is
=
distance away.
Solution 7
Velocity coordinates of 1Kg
For another 1Kg
For 1.2Kg
For 1.5Kg
For 0.5Kg
COM= (0.13487, -0.13915)
Solution 8
Initially, let
10Kg be 
cm
away
20Kg be 
cm
away
Let 20Kg be moved by distance of 2cm.
cm
20Kg should move 1cm towards left.
Solution 9
cm
should be moved 1cm
downward.
Solution 10
As ice melts, COM would no shift, as there is no external force (not even gravity).
Solution 11
C.O.M of shaded region=
{
=density
of material}
{l=thickness of disks}
C.O.M of shaded area
=
=
=
=
above center
Solution 12
V-Verma
M-Mathur
B-Boat
m
When they come to center
m
Shift in 
m
So, the boat would shift 0.13m so that COM of system remains same.
Solution 14
Let balloon be at origin.
Both monkey and balloon are at origin.
Shift=
=
So, the balloon descends by distance.
Solution 15
Solution 16
m/s
Solution 17
Conservation of lin. Momentum (C.O.L.M)
M-Man
E-Earth
m/s
Solution 18
n
Kg-m/s
Kg-m/s
(a)
If 
and 
are along same direction.
Resultant
velocity, 
in same direction.
C.O.L.M
m/s
(b) If 
, 
or 
m/s
Centre of Mass, Linear Momentum, Collision Exercise 161
Solution 19
When he throws a bag to left, momentum is conserved.
velocity
of bag
speed
of man
Let pond start from (
)
Time to reach height h=
Time to reach ground, 
Time taken by man=
=
COM of the system will be at (0,0)
after he reaches ground (let bag reach from origin) (
)
From law of kinematics
Solution 20
(a) change in momentum
kg-m/s
(b) change in momentum magnitude
Solution 21
Solution 22
resultant velocity of 2 particles moving along x and y-axis
m/s
C.O.L.M
So,
V is along 
below x-axis.
Solution 23
In a closed spaceship, there is no external force (not even gravity). So the spaceship will move with a constant speed of 15m/s.
Solution 24
Volume of 1 hailstorm
Mass=
kg
Mass of 2000 hailstorms=
kg
average force on 1
roof
N
Average force on 10m×10m (100
)roof
N
Solution 25
For falling down,
Time taken 
For bouncing up (back)
Solution 26
C.OL.M
velocity
with which man in car approaches.
Solution 27
After 1 bullet,
C.O.L.M
After 2nd bullet C.O.L.M
m/s
(after 2nd shot)
Solution 28
After left person jumps,
u is left, 
is right
After right person jumps
u is left, 
is right
(as 
is left)
Net velocity=
=
towards left
Solution 29
Use C.O.L.M
Centre of Mass, Linear Momentum, Collision Exercise 162
Solution 30
C.O.L.M (conservation of linear momentum)
s-school boy, b-bugghi
m/s
Solution 31
Use C.O.L.M
m/s
Solution 32
Use C.O.L.M
m/s
=1200J
Solution 33
Use C.O.L.M, 
Solution 34
Use C.O.L.M,
Solution 40
K.E. of nucleus=
Decrease in internal energy
Solution 41
We know that for the spring,
Use law of kinematics,
m
x=10cm
Solution 42
FBD of block after collision
Use C.O.L.M
m/s
Use law of kinematics,
Solution 43
e=co-efficient of restitution
After collision with ground,
is the equation of trajectory.
{y=0 at second projectile after it falls down from 1st projectile}
From starting point, it falls after distance.
Solution 44
After falling down, ball will have a projectile motion.
Equation of trajectory
Solution 46
At equilibrium 
(before 120g falls) 
N/m
m/s
Use C.O.L.M,
m/s
If spring is stretched extra by 
According to conservation of energy,
cm
Centre of Mass, Linear Momentum, Collision Exercise 163
Solution 35
Use C.O.L.M
v=initial velocity
final
velocities
-(1)
Given
-(2)
Solve (1) and (2)
Solution 36
(a) Use C.O.L.M
For maximizing, take derivative.
m/s
J
(b) 
v
Solution 37
Use C.O.L.M
J
When 
m/s
when 
or 
if 
if 
as 
m/s
Solution 39
Solution 45
and 
=10m/s
m/s
Angle of reflection
Angle of projection
l=18.5m
Solution 47
Use C.O.L.M
From C.O.E.L (conservation of energy law)
m/s
Solution 48
Use C.O.L.M
From C.O.E.L,
m/s
m/s
Solution 49
Let 
and 
be travelled by 
and 
{integrate}
Use C.O.E.L,
Similarly,
Solution 50
(a) 
(b) Use C.O.E.L
Solution 51
If 
and is travelled by 
under F
Work done
Use C.O.E.L 
Use C.O.L.M 
Solution 52
Net force of 
Net force of 
Use C.O.E.L,
Solution 53
ft/s
(velocity of pillow w.r.t man)
Use C.O.L.M,
{ ft/s}
ft/s
Time taken for going down=
sec
Total time taken (up+ down) =
sec
Solution 54
Use C.O.L.M,
When B reaches man,
Use C.O.E.L,
Use C.O.L.M of whole system
Solve (1) and (2) to get
Use 
Block's initial velocity
Solution 55
Use C.O.L.M,
m/s
(
=velocity
at position B)
Use C.O.E.L
Angle of projection=
Distance in horizontal direction
m
Vertical direction distance=
Total distance=
=0.22m
Solution 56
(a) For initial velocities,
Use C.O.E.L
Use C.O.L.M
-(1)
Since collision is elastic, e=1
-(2)
Solve (1) and (2),
Similarly, 
(b) Use C.O.E.L
For mass m, use C.O.E.L,
2m will rise up to, 
m will rise up to, 
Solution 57
Consider 
Mass
per unit length of chain
After chain's release,
Weight of chain due to x length
{M'=Mass
of x length}
Total force=
Solution 58
(a) Use C.O.L.M
-(1)
Use C.O.E.L,
-(2)
Solving (1) and (2)
If 
Use kinematics law,
m
(b) If it is inelastic,
C.O.L.M
m/s
m
Solution 59
Using work energy principle
V=velocity of 2kg near collision
m/s
Use C.O.L.M, {
=final
velocities}
- (1)
As collision is elastic, e=1
- (2)
Solving (1) and (2)
m/s
m/s
Use C.O.E.L for 1st block,
cm
Use C.O.E.L for 2nd block,
cm
Distance between them=4+1=5cm
Solution 60
Along x-direction 
=resultant
Vertical distance travelled by m=
Use law of kinematics, 
{velocity
of M block}
Centre of Mass, Linear Momentum, Collision Exercise 164
Solution 61
(a) Use C.O.L.M
(b) Use C.O.E.L
(c) 
{resultant
velocity}
Use C.O.E.L
Total height=
=
(d) time for flight
Total time=
Solution 62
b-block
e-earth
Use C.O.L.M
Use C.O.E.L
h<<R
Solution 63
If m is not colliding head-on, along x-axis, use C.O.L.M
-(1)
Along y-axis, 
-(2)
Use C.O.E.L
Squaring (1) and (2) and add them,
Solution 64
{after
collision}
Centre of Mass, Linear Momentum, Collision Exercise 169
Solution 13
Take
as origin
Cart should be displaced 
cm for
COM of system to be same.
